THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES SOUTHERN BRANCH' UNIVERSITY OF CALIFORNIA LIBRARY, , CALIF. LIBRARY OF PRACTICAL ELECTRICITY VOLUME I JM Book Co. Jn& PUBLISHERS OP BOOKS F O P-_/ Electrical Ubrld v Engineering News-Record Power v Engineering and Mining Joumal-Riess Qiemical and Metallurgical Engineering Electric Railway Journal v Coal Age American Machinist * Ingenieria Internacional Electrical Merchandising v BusTransportation Journal of Electricity and Western Industry Industrial Engineer PRACTICAL MATHEMATICS FOR HOME STUDY BEING THE ESSENTIALS OF ARITHMETIC, GEOMETRY, ALGEBRA AND TRIGONOMETRY BY CLAUDE IRWIN PALMER ASSOCIATE PROFESSOR OF MATHEMATICS ARMOUR INSTITUTE OF TECHNOLOGY FIRST EDITION TWELFTH IMPRESSION McGRAW-HILL BOOK COMPANY, INC. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 & 8 BOUVERIE ST., E. C. 4 r* n 4 > *** r* 5 u & i ;> COPYRIGHT, 1919, BY THE McGRAW-HiLL BOOK COMPANY, INC. PRINTED IN THE UNITED STATES OF AMERICA TMK M A I- *, K J'KKHH VOMK ! A College Library PREFACE During the past fifteen years the author has taught classes in practical mathematics in the evening school at the Armour Institute of Technology, Chicago. These classes have been H composed of men engaged in practical pursuits of various ^ kinds. The needs of these men have been carefully studied; :v - and, so far as possible, those mathematical subjects of interest to them have been taken up. The matter presented to the classes has necessarily been of an intensely practical nature. ^ This has been worked over and arranged in a form that was ^ thought most suitable for class use; and was printed in Pal- Ui mer's Practical Mathematics, four volumes, in 1912 and ap- v peared in a revised edition in 1918. The four volume edition has been used by thousands of men for home study. It is to meet the needs especially of such men that this one volume edition has been made. The vj subject matter includes all that is in the four volumes; and to * this has been added a few new topics together with many solu- 3 tions of exercises, and suggestions that make the text more I suitable for home study. It is hoped that it will find a place ^ in the library of the man who applies elementary mathematics. v 5 and who wishes occasionally to brush up his mathematics. Usually when the practical man appreciates the fact for himself that mathematics is a powerful tool that he must be able to use in performing his work, he finds that even the arithmetic that he learned at school has left him. A student of this kind is discouraged if required to pursue the study of mathematics in the ordinary text-books. This work has been written for the adult. The endeavor has been to make the student feel that he is in actual touch with real things. The intention has been to lay as broad a foundation as is consistent with the scope of the work. The nearly 3000 drill exercises and problems are, in most vi PREFACE cases, new. Many of them are adapted from engineering and trade journals, from handbooks of various kinds, and from treatises on the steel square and other mechanical devices; other problems are from the author's experience; and a largo number of the specially practical problems were proposed by members of the classes pursuing the course during its growth. Much information on various matters to which mathe- matics is applied, is incidentally given in the problems. Many devices and methods used by the practical man are given. Care has been taken to make these true to practice; but, in so wide a range of matter, there are undoubtedly errors. It is thought that the answers to the exercises are given to a reason- able degree of accuracy. It is hoped that the volume, as a whole, will not be found unmathematical. The main features of Part I are the concise treatment of various subjects in arithmetic and their applications, checks of processes, degree of accuracy possible in solutions, and contracted processes. In Part II, the endeavor has been to state definitions so as to give a clear idea of the term or object defined, and yet not to be too technical. Wherever possible, the attempt is made to discuss a fact or principle of geometry in such a way that its reasonableness will be apparent. While the subjects are treated in the mathematical order, many applications are given under separate headings. Such are brickwork, lumber, the steel square, screw threads, circular mils, belt pulleys, and gear wheels. In Part III, the intention is to give sufficient drill in algebra for one who wishes to make direct applications to practical problems. Much attention is given to formulas and their transformations. The equation is applied to many practical problems. Graphical methods are considered, and many articles on special subjects are given. In Part IV, the intention is to give sufficient work in logarithms to secure a fair degree of skill in computations. In trigonometry, those parts are emphasized that may be applied directly to practical problems; while the portions chiefly necessary as an aid in the study of more advanced mathematical subjects, are either treated very slightly or PREFACE vii omitted. Many applications are given. The tables are given to four decimal places. The author wishes to acknowledge his great indebtedness to the more than 1000 men who made up his classes during the growth of this work, and to the hundreds of men from various parts of the country who have offered kindly criti- cisms and suggestions; for, without their help and sympathy, the present results would have been impossible. Because of the remarkable success of the previous editions, it is with the greatest pleasure that this special edition is sub- mitted to our practical men. C. I. PALMER. CHICAGO, June, 1919. A WORD WITH THE STUDENT One of the lessons of the Great War and the strenuous efforts necessary to carry it on, has been to bring forcibly to our minds the great usefulness of mathematics. The war activities have exhibited the extensive mathematical needs of those who aim to render the most efficient service under the most trying circumstances. The young men of the country realize the need for a working knowledge of mathematics, and see clearly that the need so emphasized by the war con- ditions is being carried over into the days of peace and into the period of great industrial activity that is sure to follow. This volume being entitled Practical Mathematics does not mean that all exercises are such as would be called prac- tical. It means that, in the main, the exercises, outside of those intended for pure drill, are such as may arise in some practical field of work. The endeavor has been to utilize the material afforded by the shops and the laboratories as well as in the trades and in engineering. The practical man realizes that, for him, mathematics is a chest of tools together with many more or less complicated pieces of machinery that he may use to accomplish his purpose. To apply mathematics, then, he must be able to run its machinery not only accurately but speedily. To do this a great deal of work must be done in the arithmetical processes themselves. The student must drill himself on the funda- mental operations addition, subtraction, multiplication, and division both in whole numbers and in fractions, until the processes become to a large degree mechanical. That is, he should be able to do these operations with but little expenditure of mental energy. This drill is best gained by doing many exercises especially set for this purpose. Each student who is studying alone, that is, without being in a class, must of necessity be his own judge as to how much drill he needs. For most people such drill is tedious and x PRACTICAL MATHEMATICS uninteresting, and it requires a strong will to force oneself to do the proper amount of such work. That these ideas are not new is evident from the following quotation in quaint old English, taken from an arithmetic printed more than two hundred years ago: "Therefore, Courteous Reader, if thou intendest to be a Proficient in the Mathematicks, begin cheerfully, proceed gradually, and with Resolution, and the end will crown thy Endeavours with Success; and be not so slothfully Studious, as at every Diffi- culty thou meetest withal to cry, Ne plus ultra, for Pains and Diligence will overcome the greatest Difficulty: To conclude, That thou may'st so read as to understand, and so understand, as to become a Proficient, is the hearty cfesire of him who wisheth thy Welfare, and the Progress of Arts. From my School at St. George's Church in Southwork, October 27, 1684." CONTENTS PAGE PREFACE , v A WORD WITH THE STUDENT ix PART I ARITHMETIC CHAPTER I PRELIMINARY WORK AND REVIEW ART. PAGE 1. Language of mathematics 1 2. How to attack a problem 3 3. Definitions 4 4. Rules for finding divisor of numbers 7 5. Relative importance of signs of operation 2 6. Cancellation 9 7. Applying Rules 2 CHAPTER II COMMON FRACTIONS 9. Definitions 12 10. Mixed number 12 11. Proper and improper fractions 12 12. Comparison of fractions 13 13. Remarks 13 14. Principles 14 15. Reduction of a whole or a mixed number to an improper frac- tion 14 16. Reduction of an improper fraction to a whole or mixed number . 14 17. Reduction of fractions to lowest terms 15 18. Reduction of several fractions to fractions having the same denominator 15 19. Least common denominator 16 20. Addition of fractions 17 21. Subtraction of fractions 19 22. Resultants 21 xi xii CONTENTS ART. PACK 23. Multiplication of fraction and integer 21 24. Multiplication of a fraction by a fraction 22 25. Multiplication of mixed numbers and integers 23 26. Division of a fraction by an integer 26 27. Division by a fraction 27 28. Special methods in division 27 29. Pitch and lead of screw threads 31 30. The micrometer 32 32. Screw gearing 33 CHAPTER III DECIMAL FRACTIONS 32. Definition 38 33. Reading numbers 39 34. Reduction of a common fraction to a decimal fraction .... 40 35. Decimal fraction to common fraction 40 36. Addition of decimals 41 37. Subtraction of decimals 41 38. Multiplication of decimals 42 39. Division of decimals 42 40. Accuracy of results 43 41. Proportions of machines screw heads. A. S. M. E. standard. 49 CHAPTER IV SHORT METHODS AND CHECKS 42. Contracted methods and approximate results 52 43. Other methods 53 44. Checking 55 CHAPTER V WEIGHTS AND MEASURES 45. English system 58 46. The Metric system 62 47. Measure of length. The Meter 63 48. Legal units 63 49. Measure of surface 64 50. Measures of volume. Cubic and capacity measures 64 51. Measures of mass 64 52. Tables and terms used 65 53. Equivalents 67 54. Simplicity of the metric system 68 55. Relations of the units 69 56. Changing from English to metric or from metric to English systems 69 CONTENTS xiii CHAPTER VI PKHCENTAGK AND APPLICATIONS ART. P AK 57. Per cents as fractions 73 58. Cases 73 59. Rules and formulas 74 60. Solutions 75 61. Applications 76 62. Averages and per cent of error 77 63. List prices and discounts 78 64. Interest . 83 CHAPTER VII RATIO AND PROPORTION 65. Ratio 85 66. Proportion 86 67. Measuring heights 89 68. The lever 91 69. Hydraulic machines 93 CHAPTER VIII DENSITY AND SPECIFIC GRAVITY 70. Density 94 71. Specific gravity 94 72. Standards 95 73. Use 94 CHAPTER IX POWER AND ROOTS 74. Powers 98 75. Exponent of a power 98 76. Squares, cubes, involution 98 77. Roots 99 78. Radical sign and index of root 99 79. Square root 99 80. Process for the square root of a perfect square 100 81. Square root of a number containing a decimal 102 82. Roots not exact 102 83. Root of a common fraction 103 84. Short methods 103 85. Rule for square root 105 86. Cube root 105 87. Similar figures 106 xiv CONTENTS PART II GEOMETRY CHAPTER X PLANE SURFACES. LINES AND ANGLES ART . PACK 89. Definitions 109 90. Angles Ill 91. Polygons 112 92. Concerning triangles 113 93. Concerning quadrilaterals 114 94. The rectangle 115 95. The parallelogram ... .115 96. Formulas .116 97. The triangle 116 98. Area of a triangle when the three sides only are given . . . .117 99. Area of trapezoid 118 100. Measuring lumber 121 101. Estimations 122 102. Shingles 122 CHAPTER XI TRIANGLES 103. A right triangle 126 104. Similar triangles 130 105. Tapers 131 106. Turning 132 107. The steel square 134 108. Rafters and roofs 135 109. Uses of the square 136 110. Isosceles and equilateral triangles 139 111. The isosceles triangle 139 112. The equilateral triangle 139 113. The regular hexagon 141 114. Screwthreads 143 115. Sharp V-thread 144 116. United States standard thread 144 CHAPTER XII CIRCLES 118. Definitions 147 119. Properties of the circle 149 CONTENTS xv ART. PAGE 120. The segment 150 121. Relations between the diameter, radius, and circumference . . 151 122. Area of the circle 151 123. Area of a ring 153 124. Area of a sector 153 125. Area of a segment 154 126. The ellipse 155 127. Regular polygons and circles 166 128. Rules 169 129. Feed 170 130. Cutting speeds 170 131. Belt pulleys and gear wheels 171 132. The circular mil 175 133. The square mil 176 CHAPTER XIII GRAPHICAL METHODS 134. Units 177 135. Circular measure, radian 177 136. The protractor 178 137. To measure an angle with protractor 179 138. To lay off an angle with a pi otr actor 179 139. Angles and chords 180 140. To find a chord length from the table 180 141. To lay off an angle 180 142. To measure an angle . 181 143. Drawing to scale 181 144. To construct a triangle having given two sides and the angle between these sides 181 145. To construct a triangle when given two angles and the side between these angles 182 146. To construct a triangle when the three sides are given. . . . 182 147. Areas found by the use of squared paper 183 148. Other methods for approximating areas 184 149. To divide a line of any length into a given number of equal parts 186 150. To cut off the corners of a square so as to form a regular octagon 187 151. To divide a given circle into any number of equal parts by con- centric circles 187 152. To inscribe regular polygons 188 153. To draw the arc of a segment when the chord and the height of the segment are given 189 154. To find the radius of a circle when only a part of the circumfer- ence is known 189 155. How to cut a strikeboard to a circular arc 190 156. The vernier 191 157. Micrometer with vernier . .192 xvi CONTENTS CHAPTER XIV PltlBMS ART. PACK 158. Definitions 194 159. Surfaces 195 160. Volumes 196 161. Terms 200 162. Estimating cost of stonework 200 163. Brick 201 164. Estimating number, and cost of brickwork 201 CHAPTER XV CYLINDERS 165. Definitions 203 166. Area and volume 203 167. The hollow cylinder 204 CHAPTER XVI PYRAMIDS, CONES, AND FRUSTUMS 168. Pyramid 214 169. Cone 214 170. Frustum 215 171. Areas 216 172. Volumes . . 216 CHAPTER XVII THE SPHERE 173. Definitions 220 174. Area 220 175. Volume 221 176. Zone and segment of sphere 221 CHAPTER XVIII VARIOUS OTHER SOLIDS 177. Anchor ring 226 178. Prtsmatoids . . 227 CONTENTS xvii PART III ALGEBRA CHAPTER XIX NOTATION AND DEFINITIONS ART. PAGE 179. General remarks 231 180. Definite numbers .231 181. General numbers 231 182. Signs 232 183. Algebraic expression 233 184. Coefficient 233 185. Power, exponent 233 186. Terms 234 187. Remarks 234 CHAPTER XX FORMULAS AND TRANSLATIONS 188. Subject matter 237 189. The slide rules 237 190. Evaluation of algebraic expressions 238 CHAPTER XXI POSITIVE AND NEGATIVE NUMBERS 191. Meaning of negative numbers 244 192. Need of negative number 244 193. Representation of negative and positive numbers 245 194. Definitions 245 195. Remarks on numbers 246 CHAPTER XXII ADDITION AND SUBTRACTION 196. Definitions 248 197. Addition of algebraic numbers 248 198. Principles 249 199. Subtraction of algebraic numbers 249 200. Addition and subtraction of literal algebraic expressions . . . 250 201. Polynomials 251 202. Test or proof of results 251 203. Terms with unlike coefficients 253 204. Signs of grouping 254 205. Insertion of signs of grouping 255 xviii CONTENTS CHAPTER XXIir EQUATIONS ART. PAGE 206. Definitions 257 207. The equation 258 208. Solution of equations 268 209. Axioms 259 210. Testing the equation 260 211. The equation in solving problems 261 CHAPTER XXIV MULTIPLICATION 212. Fundamental ideas 265 213. Rules 266 214. Concrete illustration 266 215. Continued products 267 216. Law of exponents 268 217. To multiply a monomial by a monomial 268 218. To multiply a polynomial by monomial 269 219. To multiply a polynomial by a polynomial 269 220' Test 270 221. Representation of products 272 222. Approximate products 273 CHAPTER XXV DIVISION, SPECIAL PRODUCTS, AND FACTORS 223. Division 275 224. Division of one monomial by another 275 225. Test . . . 276 226. Division of a polynomial by a monomial 277 227. Factors of a polynomial when one factor is a monomial . . . 277 228. Squares and square roots of monomials 278 229. The square of a binomial . 279 230. Factors of a trinomial square 280 231. The product of the sum of two numbers by the difference of the same two numbers 280 232. Factors of the difference of two squares 281 233. The product of two binomials having one common term . . . 282 234. To factor a trinomial into two binomials with one common term 283 235. Other forms . . 283 CONTENTS xix CHAPTER XXVI EQUATIONS 236. Solution . 285 237. Equations solved by aid of factoring 287 238. Formulas 290 CHAPTER XXVII FRACTIONS 240. Reduction of a fraction to its lowest terms 293 241. Reduction of fractions to common denominators 294 242. Lowest common multiple . 295 243. Fractions having a L. C. D 295 244. Addition and subtraction of fractions 297 245. Multiplication of fractions 298 246. Division of fractions. 299 CHAPTER XXVIII EQUATIONS AND FORMULAS 247. Subject matter 303 248. Order of procedure 303 249. Clearing of fractions 303 250. Thermometers 311 251. Horse-power 312 252. Relation of resistance, electromotive force, and current. . . 316 253. Resistance of conductors 317 CHAPTER XXIX EQUATIONS WITH MORE THAN ONE UNKNOWN 255. Indeterminate equations 319 256. Simultaneous equations 319 257. Solution of independent equations 320 258. Elimination by adding or subtracting 320 259. Elimination by substitution 321 260. Elimination by comparison 322 261. Suggestions 322 XX CONTENTS CHAPTER XXX EXPONENTS, POWKKS, AND ROOTS ART. PACE 262. General statement 327 263. Laws of exponents 327 264. Zero exponent 329 265. Negative exponent 329 266. Fractional exponent 330 267. Exponents used in writing numbers 331 CHAPTER XXXI QUADRATIC EQUATIONS 268. Definitions ... 332 269. Solution 332 270. Solution by factoring . . 334 272. Completing the square 334 273. Solution by completing the square 335 274. Solution of the affected quadratic equation by the formula . . 337 CHAPTER XXXII VARIATION 275. General statement 340 276. Constants and variables 340 277. Direct variation 340 278. Mathematical statement 341 279. Inverse variation 342 280. Mathematical statement 342 281. Joint variation 343 282. Transverse strength of wooden beams 346 283. The constant 347 284. Factor of safety 348 CHAPTER XXXIII GRAPHICS 285. The graph 351 286. Definitions and terms used 352 287. Plotting points 354 288. Graph of an equation 359 289. Simultaneous equations 360 290. The graph of an equation of any degree 361 291. Simpson's Rule 363 292. The average ordinate rule 364 293. Area in a closed curve 364 294. The steam indicator diagram 365 CONTENTS xxi PART IV LOGARITHMS AND TRIGONOMETRY CHAPTER XXXIV LOGARITHMS ART. PAGE 295. Uses 369 296. Exponents 369 297. Definitions and history 370 298. Notation 370 299. Illustrative computations by means of exponents 372 300. Logarithms of any number 373 301. Logarithms to the base 10 373 302. Rules for determining the characteristic 374 303. The mantissa 375 304. Tables 375 305. To find the mantissa of a number 376 306. Rules for finding the mantissa 377 307. Finding the logarithm of a number 377 308. To find the number corresponding to a logarithm 379 309. Rules for finding the number corresponding to a given loga- rithm 380 310. To find the product of two or more factors by the use of loga- rithms 381 311. To find the quotient of two numbers by logarithms 381 312. To find the power of a number by logarithms 383 313. To find the root of a number by logarithms 383 314. Computations made by logarithms only approximate .... 383 315. Natural logarithms 384 CHAPTER XXXV INTRODUCTION, ANGLES 316. Introductory 396 317. Angles 396 318. Location of angles, quadrants 398 319. Measurement of angles 398 320. Relations between angle, arc, and radius 400 321. Railroad curves 400 CHAPTER XXXVI TRIGONOMETRIC FUNCTIONS 322. Sine, cosine, and tangent of an acute angle 404 323. Ratios for an angle 404 324. General form for ratios 405 325. Acute angle in a right triangle 406 xxh CONTENTS ART. P A c 326. Relation between the functions of an angle and the functions of its complement 407 327. Trigonometric functions by construction and measurement . . 408 328. Use of functions in constructing angles 409 329. Values of functions by computation 410 330. Angles in other quadrants 410 331. Angles of 90, 180, 270, and 411 332. Table of functions 412 CHAPTER XXXVII TABLES AND THEIR USES 333. Nature of trigonometric functions 414 334. Table of functions 414 335. To find the function of an angle from the table 414 336. To find the angle corresponding to a function 415 337. Evaluation of formulas 417 CHAPTER XXXVIII RIGHT TRIANGLES 339. Solving 421 340. The right triangle 422 341. Directions for solving 422 342. Case I. Given A and b, A and a, B and a, or B and 6 ... 423 343. Directions for solution of triangles 424 344. Case II. Given A and c or B and c 425 345. Case III. Given c and a or c and 6 425 346. Remark on inverse functions 426 347. Case IV. Given o and b 426 348. Orthogonal projection 427 349. Vectors 427 350. Definitions 429 35 J. Widening of pavements on curves 438 352. Spirals 439 CHAPTER XXXIX RELATIONS BETWEEN RATIOS, AND PLOTTING 353. Relations between the ratios of an angle and the ratios of its complement 442 354. Relations between the ratios of an angle and the ratios of its supplement 442 355. Relations between ratios of an angle and 90 + 443 356. Relations between the ratios of an angle and the ratios of its negative 444 CONTENTS xxiii ART. PAGE 357. Relations between the ratios of any angle 445 358. Plotting the sine curve 447 359. Curves for cosine, tangent, cotangent, secant, and cosecant. . 449 3CO. Projections of a point having circular motion 449 361. Sine curves of different frequency . . . 452 362. Variation in the amplitude of sine curves 453 CHAPTER XL TRIGONOMETRIC RATIOS OF MORE THAN ONE ANGLE 364. Functions of the sum or difference of two angles 455 365. Functions of twice an angle and half an angle 456 366. Formulas for changing products to sums or differences, and sums and differences to products 457 CHAPTER XLI SOLUTION OP OBLIQUE TRIANGLES 367. Cases 459 368. The law of sines 459 369. The law of cosines 460 370. Directions for solving 461 371. Case I, a side and two angles given 461 372. Case II, two sides and an angle opposite one of them given. . 462 373. Case III, two sides and the angle between them given .... 464 374. Case IV, three sides given 464 375. Resultant of forces . . . . , 466 376. Computation of a resultant 467 TABLES I. SUMMARY OF FORMULAS 470 II. USEFUL NUMBERS 474 III. DECIMAL AND FRACTIONAL PARTS OF AN INCH ..'...- .475 IV. ENGLISH INCHES INTO MILLIMETERS 476 V. U. S. STANDARD AND SHARP V-THREADS 477 VI. CHORDS OF ANGLES IN CIRCLES OF RADIUS UNITY .... 478 VII. STANDARD GAGES FOR WIRE AND SHEET METALS .... 479 VIII. SPECIFIC GRAVITIES AND WEIGHTS OF SUBSTANCES .... 480 IX. STRENGTH OF MATERIALS 481 X. FOUR-PLACE TABLE OF LOGARITHMS 482 XI. TABLE OF NATURAL AND LOGARITHMIC SINES, COSINES, TAN- GENTS, AND COTANGENTS OF ANGLES DIFFERING BY TEN MI.NUTES 484 INDEX , . 489 PRACTICAL MATHEMATICS CHAPTER I PRELIMINARY WORK AND REVIEW 1. Language of mathematics. Mathematics has a lan- guage of its own, with certain signs and symbols peculiar to it. It is as necessary to become familiar with these signs and symbols and their uses, in order to understand the lan- guage of mathematics, as it is for the shorthand writer to become familiar with the symbols used in his work. Failure to fix them in mind, and to learn the definitions and technical terms keeps many students from mastering the mathematical subjects they take up. Some of the best known symbols of mathematics are the Arabic numerals, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, the signs of addition, -)-, subtraction, , multiplication, X, division, -*-, and equality, =, and the letters of the alphabet. Other symbols will be explained as used. In the study of mathematics much time should be devoted: (1) to the expressing of verbally stated facts in mathematical language, that is, in the signs and symbols of mathematics; (2) to the translating of mathematical expressions into common language. The signs and symbols of mathematics are used for con- venience. They have gradually come into use by general agreement. In some cases the symbols are abbreviations of words, but often have no such relation to the thing they stand for. We cannot tell why they stand for what they do any more than we can tell why the words for cat and dog stand for (he different animals they do. They mean what they do by common agreement or by definition. 1 2 PRACTICAL MATHEMATICS 2. How to attack a problem. A problem in mathematics should not be attacked as a puzzle. No guesswork has any place in its consideration. The statement of the problem should be clear and so leave but one solution possible. This is the business of the author or the one who states the problem. The following points concern the student: (1) The problem should be read and analyzed so carefully that all cjnditions are well fixed in mind. If the problem cannot be understood there is no use in trying to solve it. Of course, if the answer is given, a series of guess-operations may obtain it, but the work is worse than useless. (2) In the solution there should be no unnecessary work. Shorten the processes whenever possible. (3) Always apply some proof or check to the work if possible. A wrong answer is valueless. Accuracy is of the highest im- portance, and to no one more than to the practical man. If a check can be applied there is no need of an answer being given to the problem. In this text the answers follow most of the exercises. They are given for the convenience of the student in checking his work, and great care must be taken not to misuse them. An answer should never assist in determining how to solve a problem. It is best, then, not to look at the answer till the problem is solved. 3. Definitions. In order to be exact in ideas and statements, it is necessary to give certain definitions. It would seem, however, that for the practical man, technical terms should be omitted so far as possible. It is usually sufficient to make the term understood, though the definition may not be a good one technically. In mathematics more than in almost any other subject, each word used has a definite and fixed meaning. The following definitions are inserted here to help to recall to mind some of the terms used : (1) An integer, or an integral number, is a whole number. (2) A factor, or a divisor, of a whole number is any whole number that will exactly divide it. (3) An even number is a number that is exactly divisible by 2. Thus, 4, 8. and 20 arc even numbers. PRELIMINARY WORK AND REVIEW 3 (4) An odd number is an integer that is not exactly divisible by 2. Thus, 5, 11, and 47 are odd numbers. (5) A prime number is a number that has no factors except itself and 1. Thus, 1, 2, 7, 11, and 17 are prime numbers. (6) A composite number is a number that has other factors than itself and 1. Thus, 6, 22, 49, and 100 are composite numbers. (7) A common factor, or divisor, of two or more numbers is a factor that will exactly divide each of them. If this factor is the largest factor possible it is called the greatest common divisor: abbreviated to G. C. D. Thus, 4 is a common divisor of 16 and 24 but 8 is the G. C. D. of 16 and 24. (8) A multiple of a number is a number that is divisible by the given number. If the same number is exactly divisible by two or more numbers it is a common multiple of them. The least such number is called the least common multiple : abbreviated to L. C. M. Thus, 36 and 72 are common multiples of 12, 9, and 4, but 36 is the L. C. M. 4. Rules for finding divisor of numbers. It is often convenient to be able to tell without performing the division, whether or not a given number is divisible by another. The following rules will assist in this. Their proofs are simple but are not given here. (1) A number is divisible by 2 if its right-hand figure is or one divisible by 2. (2) A number is divisible by 3 if the sum of its digits is divisible by 3. Thus, 73245 is divisible by 3 since 7+3+2+4+5 =21 is divisible by 3. (3) A number is divisible by 4 if the number represented by its two last digits on the right is divisible by 4, or if it ends in two zeros. Thus, 87656 is divisible by 4 since 56 is divisible by 4. 4 PRACTICAL MATHEMATICS (4) A number is divisible by 5 if the last figure on the right is or 5. (5) An even number the sum of whose digits is divisible by 3 is divisible by 6. (6) No convenient rule can be given for 7; the best thing to do is to test by trial. (7) A number is divisible by 8 if the number represented by the last three digits on the right is divisible by 8. Thus, 987672 is divisible by 8 since 672 is divisible by 8. (8) A number is divisible by 9 if the sum of its digits is divisible by 9. (9) A number is divisible by 11 if the difference between the sum of the odd digits and the even digits, counting from the right, is divisible by 11. Thus, 47679291 is divisible by 11 since (9+9+6+4) -(1+2+ 7+7) =11 is divisible by 11. Note. This rule is of little value since the division can be tried about as easily as the rule can be applied. The following facts are of some value : (10) A factor of a number is a factor of any of its multiples. (11) A common factor of any two numbers is a factor of the sum or the difference of any two multiples of the numbers. 5. Relative importance of signs of operation. (1) In a series of operations denoted by the signs of addition, +, subtraction, , multiplication, X, and division, -5-, the multi- plications must be performed first, the divisions next, and lastly the additions and subtractions. (2) If several additions, or several multiplications, occur together they may be performed in any order. (3) If several subtractions, or several divisions, occur together they must be performed in the order in which they come from left to right. The rules as stated here are in agreement with the best usage in algebra and in formulas used in practical work. Examples. (1) 12+3-2+9+7-3 = 26, by performing the operations in the order in which they occur. (2) 120-^3X5X2-^2 = 2, by first performing the multi- PRELIMINARY WORK AND REVIEW 5 plications and then the divisions in the order in which they occur. (3) 12H-3+8X2-6-7-2+7X2X3-9 = 12-r3+ 16- 6 -=-2+42- 9 = 4+16-3+42-9 = 50, by first performing the multi- plications, then the divisions, and then the additions and subtractions. EXERCISES 1 In the exercises 1 to 10, first perform the multiplications, then the divisions, and finally the additions and subtractions, each in the order in which they occur. 1. 14 + 16-3 + 10-4-6 = ? Ans. 27. 2. 16-7-8+4X2X3-16X2-7-4 = ? Ans. 18. 3. 15-2X3-15-=-5+4 = ? Ans. 10. 4. 60-25-=-5 + 15-100-r-4x5 = ? Ans. 65. 6. 17X3+27 -=-3 -40X2 -=-5 = ? Ans. 44. 6. 56-7+525-7-5X7X3 + 15-7X8 = ? Ans. 13. 7. 864-=-12-124-=-31+54-=-27 = ? Ans. 70. 8. 4X27 -=-9X4+9X2-3X6-7-9 = ? Ans. 19. 9. 4963-7-7 + 144-7-72-14X9 = ? Ans. 585. 10. 13X9X62+444-4-17X22 = ? Ans. 6891. 11. Time yourself in doing the following ten multiplications: (1)347X371. (6)3249X987. (2) 547X682. (7) 4444X888. (3) 433X925. (8) 8764X2233. (4) 986X478. (9) 9898X4257. (5) 3587X729. (10) 9999X8888. 12. Check your work in the above ten multiplications by doing the multiplying, using the first number in each case as the multiplier. 13. Time yourself in doing the following divisions. Check your work by finding the product of the divisor and quotient, and comparing it with the dividend. (1) 395,883 -=-9. (5) 4,518.976 -=-784. (2) 64,362 -=-17. (6) 783,783^-4147. (3) 306,192 -=-48. (7) 1,312,748 -=-437. (4) 87,168 -=-384. (8) 4,495,491 -=-499. 14. Do the following multiplications and check the work by dividing the product by the multiplier and comparing the result with the multi- plicand : (1) 843X329. (3) 4493X345. (2) 4327X987. (4) 8397X9327. 6 PRACTICAL MATHEMATICS 16. Do the following divisions and check the work by finding the product of the divisor and quotient, then adding the remainder and comparing the result with the dividend: (1) 43,9624-97. (4) 9,372,4684-375. (2) 842,6374-233. (5) 4,343,7644-983. (3) 467,2344-487. (6) 3,784,3284-2345. 16. The cost of constructing 275 miles of railway was $4,195,400 What was the cost per mile? Ans. $15,256. 17. The circumference of a drive wheel of a locomotive is 22 ft. How many revolutions will it make in going 44 miles if there are 5280 ft. in one mile? Ans. 10,560. 18. If a power plant consumes 277 tons of coal at $3.10 per ton each day, what is the cost of the coal to run the plant one year of 365 days? Check the work. 19. A hog weighing 78 Ib. requires 400 Ib. of grain for each 100 Ib. gain in weight. If the price of 56 Ib. of grain increases in price from 42 cents to $1.54 what should be the increase in price of hogs per 100 Ib.? Ans. $8. 20. When 56 Ib. of corn cost 56 cents, hogs sold at $5.60 per 100 Ib. live weight; and when corn cost $1.82 for 56 Ib., they sold at $15.10 per 100 Ib. How much more or less does the farmer make per hundred if it takes 400 Ib. of corn for each 100 Ib. increase in weight of hogs? Ans. In second case 50 cents more. 21. How many tons of silage, 2000 Ib. per ton, should be stored for 20 cows, the intention being to feed each cow 40 Ib. a day for 5 months, then 30 Ib. a day for 2 months, then 20 Ib. a day for 2 months? Con- sider 30 days to a month. Ans. 90. 22. Give three divisors of 192. Give three multiples of 72. Give a common divisor of 144 and 192. Give the greatest common divisor of these numbers. Give three common multiples of 15, 8 and 20. Give the least common multiple of these numbers. Is there a greatest common multiple? 23. Find the prime factors of each of the following: 1188; 148,225; 89,964; 36,992,000. Solution. 2)1188 The work is best carried out by selecting the small- 2)594 est prime factors, leaving the larger till later. The 3)297 prime factors of the number are all of the divisors used. 3"\99 The prime factors of 1188 are 2, 2, 3, 3, 3, 11. .> jo3 njTi i 24. Tell which of the following numbers are exactly divisible by each of 2, 3, 4, 5, 6, 7, and 9: 324; 7644; 3,645,111; 4550: 3645: 49.875: 23,147,355. PRELIMINARY WORK AND REVIEW 7 6. Cancellation. Often in solving problems, a fractional form like the following is obtained: 64X25X8X12X17 48X15X32X17X24' If we do all the multiplications above the line and below the line, and then perform the division which the line indicates, we shall obtain the result. It is often easy to avoid much of this work, however, by applying a principle of fractions. The process which is explained below is called cancellation. ? 5 63 ? (1) It is seen that 17 is found both above and below the line; we draw a line through each of these. These numbers are then said to be cancelled. (2) Now notice that the numbers 64 and 32 are divisible by 32. Cancel 64 and 32 and place 2 above 64 which is the number of times 32 is contained in 64. (3) Next, divide 48 and 8 by 8 and cancel them, writing the quotient 6 below 48. (4) Divide 25 and 15 by 5 and cancel them, writing the quotient 5 above 25 and 3 below 15. (5) In a similar manner 12 and 24 are cancelled; also 2 and 2. In this manner we have replaced the given form by the 5 5 simpler one = j^ This is the answer. D Xo ID It should be noted that when no factor remains either above or below the line after the cancellation is finished, we retain one of the unit factors which we neglected to write when cancelling. _i. A "9 Ans - Remark. It should be remembered that this method of simplifying cannot be used when there are additions or sub- tractions indicated in the problem. In such a case the operations above the line must be per- 8 PRACTICAL MATHEMATICS formed first, then those below, and lastly the result above must be divided by the result below. 4+200-6X2 192 These processes may be restated in the following: RULE. (1) Any factor above may be divided into any factor below the line. (2) Any factor below may be divided into any factor above the line. (3) Any factor common to factors one above and one belou may be divided into each. (4) The answer is obtained by dividing the product of the numbers remaining above the line by the product of the numbers remaining below the line. If no number remains above or below, use 1. EXERCISES 2 Use cancellation to find the results in the following: 5X8X3X16 20X56X12 *' 8X15X4 3> 21X10X18 _ 57X119X16 77X100X18X14 3 ' 17XI2X19~' fc 25X11X49X16" 18X_100XJ3ja2 16X12X7X11 26X25X9X3 ' *' 24X22X7X18* 7 90X89X88X87 Ans. 8 1200X515X70X100 Ans. 1X2X3X4 ' 2555,190. 5X35X103 240,000. 180X132X140X75 Ans. 10 750X4500X5760 . 15X70X44X36 ' 150. 2400X750X50* 11 Ifj!. XI 728X999 Ans. ia 1320X432X660 96 X 270 X 33 ' 290| |. 4400 X 297 X 288' ,,'^45X4 ,, ^. IB '6+3X7-4X2 + 167 Ans. 16 256X6 + 125X3-14X76 Ans. 94-7X9+3X6 4. 17X27+32X40-1618 ' 7. Analyze the following and shorten the computation as much as (lOMsiblc by cancellation. 17. If 18 men can do a piece of work in 14 days, how many men will do the work in 21 days? Analysis. If 18 men can do a piece of work in 14 Operation. days, one man can do the work in 18X14 days. It 62 will take an many men to do the work in 21 days as 21 is contained times in 18 X 14. Hence it takes 12 men to do the work in 21 days. PRELIMINARY WORK AND REVIEW 9 18. A man worked 16 days for 30 bushels of potatoes worth 88 cents a bushel. What did he earn per day? Ans. $1.65. 19. How many days at $1.50 must 24 men work to pay for 360 bushels of wheat worth $1.20 a bushel? Ans. 12. 20. How many acres of potatoes yielding 150 bushels to the acre and worth 25 cents a bushel will amount to as much as 65 acres of wheat yielding 18 bushels to the acre and worth $1.05 per bushel? Ans. 32||. 21. If 8 men, in 15 days of 10 hours each, can throw 1000 cu. yd. of earth into wheelbarrows, how many men will be required to throw 2000 cu. yd. of earth into wheelbarrows in 20 days of 8 hours each? Ans. 15. Suggestion. Analyze the problem and state in the following form for cancellation: 8X15X10X2000 20X8X1000 22. A gardener sells 75 crates of berries, 24 boxes in a crate, at 8 cents a box, and receives in return 12 rolls of matting, 40 yards in a roll. Find the price of the matting per yard. Ans. 30 cents. 23. A merchant bought 15 car-loads of apples of 212 barrels each, 3 bushels in each barrel, at 45 cents per bushel. He paid for them in cloth at 25 cents a yard. How many bales of 500 yd. each did he give? Ans. 34 and 172 yd. over. 24. How many bushels of potatoes at 55 cents a bushel must be given in exchange for 44 sacks of corn, each containing 2 bushels, at 30 cents a bushel? Ans. 48. 7. Applying rules. The practical man often has to apply a rule in solving a problem. This rule may; be given to him by a fellow workman, or it may be taken from a handbook. The rule may be one, the reasonableness of which is apparent, but often it is not. Many rules are the results of experience, others of experiment, and still others are mere "rules of thumb," that is, they merely state a combination of numbers which gives the result desired. In the following problems, read the rule carefully before applying it. RULE. To find the number of revolutions of a driven pulley in a given time, multiply the diameter of the driving pulley by its number of revo- lutions in the given time and divide by the diameter of the driven pulley. 26. A pulley 48 in. in diameter and making 65 revolutions per minute (R. P. M.) is driving a pulley 26 in. in diameter. Find its number of R. P. M. Ans. 120. 26. Find the R. P. M. of a pulley 8 in. in diameter driven by a 28-in. pulley, making 36 R. P. M. Ans. 126. 10 PRACTICAL MATHEMATICS 27. Find the R. P. M. of a pulley 44 in. in diameter driven by a 32-in. pulley, making 60 II. P. M. Ans. 48. RULE. To determine the width of belt required to transmit a given horse-power at a given speed of the belt: For single leather or 4-ply rubber belts, multiply the number of horse-power to be transmitted by 33,000 and divide the product by the product of the speed of the belt, in feet per minute, multiplied by 60. The quotient will be the width of the bolt in inches. 28. What is the required width of belt to transmit 100 horse-power with a belt speed of 3500 ft. per minute? 110 29. Find the width of a single leather belt to transmit 75 horse-power, with a belt speed of 3000 ft. per minute. Ans. 13} in. 30. For heavy double leather or 6-ply rubber belts, use 100 instead of 60 in the rule. Find the width of such a belt to transmit 135 horse- power with a belt speed of 3600 ft. per minute. Ans, 12| in. RULE. To determine the horse-power a belt of given width will transmit when running at a given speed : For single leather or 4-ply rubbei belts, multiply width of belt in inches by 60 and the product by speed ol belt in feet per minute and divide the product by 33,000. The quotient will be the number of horse-power that the belt will transmit with safety. 31. How many horse-power will a 10-in. single leather belt transmit. if running at 4000 ft. per minute? Ans. 72^. 32. How many horse-power will a 36-in. heavy double leather belt transmit, running at 4500 ft. per minute? (Use 100 instead of 60 in the rule.) Ans. 491 nearly. The first letter of a word is often used in mathematics instead of th( word itself. When two or more such letters are written together with nc sign between them it is understood that multiplication is indicated. If H stands for horse-power, P for effective pressure in pounds of steam per square inch, L for length of piston stroke in feet, A for area of piston in square inches, and N for number of strokes per minute, then the rule for finding the horse-power of a steam engine may be stated n the following abbreviated form: PLAJf. " 33000 33. Find // if P = 55 Ib. per square inch, L -2 ft., A - 195 sq. in., and AT =80. 55X2X195X80 _ Solutwn. H = - --- 52. PRELIMINARY WORK AND REVIEW 11 34. Find H if P = 70, L = 2, A = 1 65, and 2V = 90. Ans. 63. 35. Find H if P =85, A =95, L =2, and TV = 190. Ans. 93 nearly. 36. A railroad uses 2,240,000 ties each year. If 350 trees grow on one acre and three ties are cut from a locust tree that is 30 years old, how many acres of locust trees must be planted each year to supply the ties? Ans. 2133|. 37. If 9 men can cut 28 cords of wood in 4 days of 6 hours each, how many cords can 15 men cut in 16 days of 9 hours each? Ans. 280. 38. A marble slab 20 feet long, 5 feet wide, and 4 inches thick weighs 850 pounds. What is the weight of another slab of the same marble 16 feet long, 4 feet wide, and 2 inches thick? Ans. 272 pounds. 39. If 24 men in 18 days of 8 hours each can dig a ditch 95 rods long, 12 feet wide, and 9 feet deep, how many men in 24 days of 12 hours each will be required to dig a ditch 380 rods long, 9 feet wide, and 6 feet deep? Ans. 24. CHAPTER II COMMON FRACTIONS DEFINITIONS AND GENERAL PROPERTIES 8. The number 6 when divided by 3 gives a quotient of 2. This may be written f = 2. If now we attempt to divide 6 by 7, we are unable to find the quotient as above. The divi- sion may be written $. This is called a fraction. \ means that a unit is divided into 7 equal parts. The fraction 4 indicates that 6 of the 7 equal parts are taken. 9. Definitions. A fraction is an indicated division, which in a simple form expresses one or more of the equal parts into which a unit is divided. The divisor or the number below the line in the fraction is called the denominator of the fraction. The denominator tells into how many parts the unit is divided. The dividend, or the number above the line in the fraction, is called the numerator of the fraction. The numerator tells how many of the parts, into which the unit is divided, are taken. The numerator and the denominator are called the terms of the fraction. The value of a fraction is the number that it represents. 10. Mixed number. Just as we have whole numbers and fractional numbers, so we have numbers made up of whole numbers and fractions. Thus, we may have 2 which is read 2 and I and means 2 + j. Definition. A mixed number is one composed of a whole number and a fraction. 11. Proper and improper fractions. If the fraction shows fewer parts taken than the unit is divided into, its value is evidently less than 1. If the fraction shows as many or more parts taken than the unit is divided into, the fraction is evi- dently equal to or greater than 1. 12 COMMON FRACTIONS 13 Thus, 2 shows fewer parts taken than the unit is divided into, and is less than 1 ; f shows as many parts taken as the unit is divided into and is equal to 1 ; and \ shows more parts taken than the unit is divided into, and is greater than 1. Then f is a proper fraction, while f and \ are improper fractions. Definitions. A proper fraction is one in which the numera- tor is less than the denominator. An improper fraction is one in which the numerator is equal to or greater than the denominator. It should be noted that an indicated division is often called a fraction, even though the division can be performed exactly, that is, without a remainder. Thus, Jg 2 -, ff, 44 are fractions. 12. Comparison of fractions. If two fractions have equal numerators and equal denominators they are evidently equal in value. If two fractions have equal denominators, the one that has the larger numerator is the greater in value. Explain why. Thus, of f and f , f is the larger. If two fractions have equal numerators, the one that has the larger denominator is the smaller in value. Explain why. Thus, of | and |, $ is the smaller. If two fractions have both numerators and denominators unequal, their values cannot be compared so easily. Thus, the values of f and f can be more easily compared when the fractions are changed to fractions that have the same denominator. See Art. 18. 13. In order to get the right viewpoint, it is well for the student to note that before he took up fractions he had learned to add, subtract, multiply, and divide whole numbers; here he has new numbers, fractions, to deal with. It is now necessary to learn how to perform the fundamental operations on fractions. They must be combined not only with other fractions but with whole numbers. The main thing in this chapter is to do these fundamental operations. But to do these in all cases it is necessary to be able to change the frac- tional numbers in various ways, that is, to reduce to lower or higher terms, change fractions to common denominators, 14 PRACTICAL MATHEMATICS mixed numbers to improper fractions, and improper fractions to mixed numbers. The student studying alone must determine for himself how many exercises he needs to do in order that he may secure the necessary accuracy and speed. 14. Principles. Since a fraction is an indicated division the following principles may be stated for fractions: (1) Multiplying or dividing both numerator and denominator by the same number does not change the value of the fraction. (2) Multiplying the numerator or dividing the denominator by a number multiplies the fraction by that number. (3) Dividing the numerator or multiplying the denominator by a number divides the fraction by that number. 15. Reduction of a whole or a mixed number to an im- proper fraction. Example. Reduce 5 to 6ths. Since 1=|, 5 = 5XJ=V. Ans. By principle (2). Example 2. Reduce 7$ to 5ths. Since 1=1 7 = 7X| = The three dots, .*., as used above form a symbol meaning hence or therefore. RULE. To reduce a whole number to a fraction of a given denominator, first change 1 to a fraction of the given denomi- nator and then multiply the numerator by the given whole number. With a mixed number, reduce the whole number to a fraction and then add to the numerator of this fraction the numerator of the fractional part of the mixed number. 16. Reduction of an improper fraction to a whole or mixed number. Example 1. Reduce -" 4 - to a whole number. J^ = 32-=-4 = 8. Ans, Example 2. Reduce -V" to a mixed number. -Y- = 47-i-9 = 5. Ans. RULE. To reduce an improper fraction to a whole or mixed number, perform the indicated division. The quotient is the number of units. If there is no remainder, it reduces to a whole number. If there is a remainder, it reduces to a mixed number of which the quotient is the whole number part and the remainder the numerator of the fractional part. COMMON FRACTIONS 15 EXERCISES 3 1. Reduce the following numbers to sixths : 7, 1 1, 40, 17, 19. To thirds To tenths. 2. Reduce the following mixed numbers to improper fractions: 2^, 7?, 9i, l>4, 17$, 18, 22H, 46 j 3. Reduce the following improper fractions to whole or mixed numbers : 17 J9 2" 32 49 60 71 97 47 t_58 493 9U96 .1.928 9999 376S4 V> Vi -SO -J-> -?-> -r> T> T8J IS. Tl I 17 ) ?fi5 > a7~j 3J/J -J4T-- 17. Reduction of fractions to lowest terms. Definition. A fraction is in its lowest terms when the numerator and denominator are prime to each other, that is, when there is no integer that will divide both of them. Example. Reduce -j 7 ^ to its lowest terms. 75 _ 1 5 5. TT>% Tl T Since dividing both numerator and denominator by the same number does not change the value of the fraction, both terms may be divided by 5. Thus |f is obtained. Both terms of this fraction are divided by 3, and 4 is obtained. Since 5 and 7 are prime to each other, the fraction is in its lowest terms. Both terms could have been divided by 15 and the reduction made in one step. RULE. To reduce a fraction to its lowest terms, divide both terms successively by their common factors, or divide by the greatest common divisor of the terms. EXERCISES 4 Reduce the following fractions to their lowest terms: 15,3,14,9. 9 7 , 25, SI. 24. 13 12 51 27 35' 3S' 28' ft 3. H'Ji-ii-n- 4. AVri&'l*- 5. HI- Ans. |- 6. m- Ans. I- 7. fli- Ans. t- 8. US- Ans. f 9. T V 5 V Ans. f- 10. ffi- Ans. H- 11. fil?- Ans.lt- 12. r 9 |g- Arw. H- 13. Jg|3- Ans. $J. 14. AUf. Ans. f 15. if|8- Ans. 1 T V 16. mi' -^ns. rfflfc- 17. T ViWo- ^ns. flJ. 18. Um- Ans. HI*- 19. Reduce the following per cents to fractions in their lowest terms : (The sign % takes the place of the denominator 100). 5%, 10%, 40%, 25%, 35%, 42%, 45%, 30%, 28%, 75%, 80%, 95%, 98%, 14%. 18. Reduction of several fractions to fractions having the same denominator. Definition. Fractions that have the 16 PRACTICAL MATHEMATICS same denominator are called similar fractions or fractions with a common denominator. Example 1. Reduce $ and $ to fractions which have for a denominator. The fraction may be changed to 6ths by multiplying both its terms by a number which will make the denominator 6. This will not change the value of the fraction. This multiplier is obtained by dividing 6 by 2 which gives 3. Likewise = 2X3 1X2 3X2 Example 2. Reduce I, f, and | to 72ds. Both terms of ^ are multiplied by 72 -f- 9 = 8, both terms of f are multiplied by 72-7-8 = 9, both terms of are multiplied by 72-7-6=12. RULE. To reduce several fractions to fractions having a common denominator, multiply both terms of each fraction by a number found by dividing the common denominator by the denominator of that fraction. 19. Least common denominator. In the preceding the common denominator, 72, was given. Usually the de- nominator is not given but we are asked to reduce the given fractions to fractions having a least common denominator. When this is the case we find the least common multiple of the denominators of the given fractions, and this is the least common denominator (L. C. D.) for all the fractions. Example. Reduce $, -fc, and ^f to fractions with a L. C. D. The L. C. M. of 9, 12, and 24 is 72. If we divide 72 by each of the given denominators we get the numbers to be used as multipliers. Remark. Usually the fractions dealt with have such denominators that their L. C. D. can be seen by inspection. The student should endeavor to determine it in this way wherever possible. If it cannot be seen by inspection, a good way to find it is as follows: COMMON FRACTIONS 17 RULE. Divide the given denominators by a prime number that will divide two or more of them, then divide the remaining numbers and the quotients by a prime number that will divide two or more of them. Continue this as long as possible. The L. C. D. is the continued product of all the divisors and the quotients or numbers left. Example. Find the L. C. D. of ^, ^ io> an d if- Process. 5)30, 45, 135, 25 3)6, 9, 27, 5 3)2, 3, 9, 5 2, 1, 3, 5 L. C. D. = 5X3X3X2X3X5-1350. Ans. EXERCISES 6 Change as indicated. 1. i, |, and I to 12ths. 2. | and ', t-r 42ds. 3. f , f, and f to 24ths. 4. |, f , and f to 63ds. 5. 2, I, and f to 42ds. 6. 2, f, f, and 5 to SOths. 7. f, f, |, and 2 V to lOOths. 8. f , g, i, and | to 120ths. 9. if, A, |, and f to 208ths. Ans. if, Hf , HI, Hf 10. |, T 5 r, H, and & to 396ths. ylns. HI, HS, Ml, sVr Change the following to fractions having a L. C. D. 11. -| and |- Ans. T 5 , / 12. | and T \- Ans. f f, 4^- 13. fV and if ^^ s - io, IS- 14. T 9 3 and T \- ^ns. Ml, Hf 15. | and f r Ans. f|f, // s - 16. f| and if- Ans. Iff, f |f - 17. f, Y, A, and ? V 18. ^ 5 , 1, ii and f- 19. 2A, 4P , 7/ r 20. Change the following to lOOths and then write as per cents: , 1318341 3 7 9J 3 "i 9111317J91 37 917 5j 4) t Sj o? BJ 1C> 10} TOj 10> 20> "srOj 50> 20) 0) ao> 50> 20, "25, ^Et 25, V5i tt, A,H,f8- ADDITION OF FRACTIONS 20. Example 1. Add ^ ^ and H- Just as 7 apples + 5 apples + 11 apples = 23 apples, so 7 twelfths + 5 twelfths +11 twelfths = 23 twelfths. The work may be arranged as follows : Tz + T 5 2 + 1 i = f i = ITS Aras. Example 2. Find the sum of -^ -^5, -^ i Here the fractions must first be reduced to fractions having a L. C. D. The L. C. M. of 12, 15, and 30 is 60. 18 I'KACTICAL MATHEMATICS W-W- An *' Example 3. Find the sum of 3J, 5$, 2^, 7J- The whole numbers and the fractions may he added sepa- rately, and then these sums united. The work may t>e written as here. A more convenient way of 3* = writing the mixed numbers for 5f = adding, is to write them under 2^ = each other, and add, similar to 7J = the method of adding whole numbers. Ans. RULE. To add fractions that have a L. C. D., add the numera- tors of the fractions and place the sum over the L. C. D. If this gives an improper fraction, it should be reduced to a whole or mixed number. If the fractions do not have a L.C.D., first reduce them to fractions wiih a L. C. D. To add mixed num- bers, add the whole numbers and fractions separately and then unite the sums. EXERCISES 6 Add the following and express the sum in the simplest form. i. i+i+j+i- 2. t +I+J+Y. 3. &+&+H+H- 4. l+i+A- 6. 7 + ?+4i+7J- 6. 9J+3J+6. 7. 41J+40J+3. 8. 9*+7i+8j- 9- A+A + I + I- Ans. 1JH- 10. f + V+l+V Ans. Si- 11. iji+fl+ifi. Ans. 60. 12. V + V+A- Ans. 5W- 13. 214J+517/J + 145&- Ans. 876f| 14. 3! + 17i + 28A+3,V Ans. 53A- 15. S + S + l + i + i+^+A- Ans. 3|- 16. 2|+7|+HA+14!+17H- Ans. 54\. 17. 871A+614J|+81f Ans. 1067|j- 18. 145J+36 + H + 194 + H- Ans. 376Hf- 19. 126i + 35 + 15J+5SJ+9rV Ana. 245f- 20. 16| + 14J + 17i + 19j+27A- Ans. 95j- 21. 16^ + 191+24^+29/0+14. Ans. 1035- 22. A merchant sold to different customers 5 1 yards of cloth, 7J yards. 15J yards, 9} yards, and 3| yards. Find the total number of yards sold. Ans. 41JJ- COMMON FRACTIONS 19 23. A farmer has 10 acres in one field, 8| acres in another, and 30j acres in a third. How many acres in the three fields. Ans. 495- 24. In five days a steamer sails the following distances: 384f miles, 372 1 miles, 356 5 miles, 392 J miles, and 345 1 miles. How far did it sail in the five days? Ans. 1852J miles. SUBTRACTION OF FRACTIONS 21. Example 1. Subtract T 4 r from i 9 i- Since like numbers can be subtracted we can subtract 4 elevenths fiom 9 elevenths and have the remainder 5 elevenths. This may be written -i 9 r~~T 4 r = 'iV Ans. Example 2. Subtract T 7 T from f- Here the fractions must first be reduced to fractions having the same denominator. It may be written 2 7 .22 21 JL A Ans. 10JJ- 33. 7f+6J-2f+?+2ft- Ans. 14\\- 34. 3i+41 + l?-(?+2 2 T )- Ans. 6^- 35. 7|+2!-3f+(lf+lJ)- Ans. 9if- Do as many of the following as you can without a pencil. 36. A boy had $J and spent $i; how much money did he have le^t? 37. A man bought 2 J tons of coal and had 1} tons delivered ; how much was left to be delivered? 38. A man had 5f acres of land and sold 3$ acres; how many acres did he have left? 39. A man weighed 159} Ib. on Monday and 154$ Ib. on the following Saturday; how many pounds did he lose? 40. A man had $7$ and paid a debt of $3}; how much did he have left? 41. A man sold J of his farm at one time and } of it at another; what part of his farm did he sell? What part did ho have left? 42. A coal dealer had 10 tons of coal. He sold 3i tons to one customer, 2} tons to another and the remainder to a third customer. How much did he sell to the third customer? COMMON FRACTIONS 21 43. A tank full of water has two pipes opening from it, one will empty 5 of the water in the tank in one hour and the other of it; what part will both pipes empty in one hour? What part remains in the tank? 44. Find the distance around the figure with di- g, mensions as given. __ v ,, 45. One coal wagon drew 6i 7 o and 8| tons of coal \ \*" on two successive days; another wagon drew 7-fg and $*! \ / 9f tons on the same days. How much more did the \ /^~ latter draw than the former? \______-v 22. Resultants. The combined effect of FIQ * 1 several forces is called the resultant. Thus, a pull of 100 Ib. toward the east and at the same time a pull of 75 Ib. toward the west gives a resultant pull 25 Ib. toward the east. The resultant of pulls of 150 Ib. toward the east, 85 Ib. toward the west, and 75 Ib. toward the west is a pull of 10 Ib. toward the west. 46. Using E for east, W for west, N for north, and S for south; find the resultants of the following: (1) 48 Ib. W, 92 f Ib. E, 76| Ib. E, and 9H Ib. W. (2) 125& Ib. N, 751 Ib. N, 47f Ib. S, and 156^ Ib. S. Ans. (1) 29i Ib. E; (2) 3 Ib. S. MULTIPLICATION OF FRACTIONS 23. Multiplication of fraction and integer. -Example 1. Multiply | by 4. To multiply f by 4 is to find a fraction that is 4 times as large as f . By Art. 14, multiplying the numerator of a frac- tion multiplies the value of the fraction. 3X4 12 .'. |X4 = = =2g- Ans. Example 2. Multiply 8 by f Since in finding the product of two numbers either may be used for the multiplier without changing the product, Example 3. Multiply T 3 ^ by 7. Here we may use the principle that dividing the denomina- 22 PRACTICAL MATHEMATICS tor multiplies the value of the fraction, or the operation may be thought of as one in cancellation. --- An*. Or T \X7 = ~p = $ = U. Am. Here the 7 and 14 are cancelled. RULE. To multiply a fraction by an integer or an integer by a fraction, multiply the numerator or divide the denominator of the fraction by the integer. Remark. When a whole number is multiplied by a whole number the product is larger than the multiplicand; but whenever the multiplier is a proper fraction the product is smaller than the multiplicand. Here we cannot think of multiplication as a shortened addition. We often write | of 6 for |X6. The meaning is the same in each case. 24. Multiplication of a fraction by a fraction. Example 1. Multiply f by 4' f by f is the same as 4 of $, but 4 of f is 5 times j of f and | of f has a value | as large as f . By Art. 14, the value of a fraction is divided when the denomi- nator is multiplied. 2X5 And 4 of f = 5 times ^ = ^- These steps may be combined as follows: 4-0 An Example 2. Multiply }| by f _l . -- Ans. Cancellation should be used when it will shorten the work. Example 3. Multiply f by ^ by $? 2 COMMON FRACTIONS 23 RULE. To multiply a fraction by a fraction, multiply the numerators together for the numerator of the product, and the denominators together for the denominator of the product. Cancel when convenient. Remark. A form like f of f of f is often called a com- pound fraction. 25. Multiplication of mixed numbers and integers. Example 1. Multiply 7| by 6. 5 5 Example 2. Multiply 8| by 3| ?- Ans. 8| by 3|- 5 O p O RULE. To multiply two numbers, one or both of which are mixed numbers, reduce the mixed numbers to improper frac- tions and multiply as with fractions. Remark. The work may often be simplified by using the following methods : Example 3. Multiply 47 by 16f Process. 47 Explanation. Multiply 47 by 4 and divide 16| by 5, which is the same as multiplying 47 by |; 5)188 this gives 37f Then multiply 47 by 16 in the 37 ordinary way for multiplying whole numbers. Add these three partial products and the entire __ product is 789| 789f Example 4. Multiply 25 f by 6^- Process. Explanation, f X i = & ', 25 X \ = 8-| ; | X 6 = 2| ; 25f 25X6=150. Al The entire product equals the sum of these QI partial products. 2| 150 160U If several fractions and mixed numbers are to be multiplied together it is usually best to reduce all to fractions for then the work may be shortened by cancellation. 24 PRACTICAL MATHEMATICS Example 5. Find the product of J X3X9X4 1 2 , 23 ' " __ *"* __ | A ij i 11 EXERCISES 8 Find the product of each of the following; without pencil when possible. 1. 1X4. 10. 25XA- 19. 2. JX4. 11. 5XA- 20. lof 20. 3. $X2. 12. 15 XH 21. I of 30. 4. $X5. 13. 27Xli 22. f of 63. 6. AX8. U. 45X2J- 23. A of 120. 6. AX5- 16. 55X2J- 24. f of 99. 7. 7X|- 16. iof I 25. A of 22. 8. 8X|- 17. SXJ- 26. 6iX8. 9. 9Xf 18. AX A- 27. 4AX6. 28. What is 3 times 4 bushels? 3 times 4-fifths? 3 times }? 3 X t *i ? 29. What is $ of 9 quarts? $ of 9-tenths? $ of A? J Xrj? 30. A can is 3 full of milk. If I of this is drawn off, what part of the whole can is drawn off? What part remains in the can? 31. It took a boy living in the country 50 minutes to walk to school. He could drive with a horse in f of this time. How long did it take him to drive? 32. On one field a farmer harvested 230 bushels of wheat and on a second f as much. How many bushels were harvested on the second field? 33. One-third of the water in a tank will flow from a certain opening in 1 hour. If the tank holds 60 barrels how many barrels will flow out in 2 hours? 34. If two pipes open from a tank, one of which can empty J of the tank in 1 hour, and the other J of it in 1 hour, what part of the tank will both empty in 1 hour? If the tank holds 60 barrels, how many barrels will flow out in 2 hours if both pipes are open? Ans. Ai 54. 35. The circumference of a circle is about 3f times the diameter. Find the circumference of a circle if the diameter is 7 ft., 21 ft., 6 ft. 36. The diagonal of a square is very nearly 1 , 5 j the length of one side. Find the diagonal when one side is 12 in., 6 in., 84 ft. 37. If you have a vacation of 100 days, and spend i in the country, A camping, and the remainder in the city, how many days do you spend in each place? 38. A boy has $2.50. He spends I of it for a fishing rod, A of it for a reel, and the remainder for a line. How much did he spend for each? COMMON FRACTIONS 25 39. A 7 sXl6. Ans. IfJ. 40. 7 V,X96. Ans. 6 T 8 3 . 41. fiiX96. Ans. 62. 42. 28 T 5 2 X14. Ans. 397. 43. 816X17. Ans. 13883. 44. 956* X29. Ans. 277471. 45. 12^X62. Ans. 781i 46. 12fX3f. Ans. 47||. 47. 13iX9f. Ans. 131|. 48. 23|Xl8f. Ans. 441|. 49. 14|X10|. Ans. 153ff. 50. 212fX7i Ans. 1595. 51. Multiply 1| by 1, 2J by 2J, 3| by 3J, 10J by 10J. 52. Can you make a rule for finding the product of two factors that are the same and end in 5? See Art. 43 (5). The product of two factors that are exactly alike is called the square of one of them. Thus, the square of 4 J is 4| X4| =20j- Find the square of each of the following by your rule: 7%, 9J, 11 5, 16J, m, 20|, 100|. 53. fXfXf=what? Ans. T V 54. fXfXlfXl=what? Ans. f. 55. AX2jX7f X 2 9 gX3|=what? Ans. 7f- 56. If hogs are worth 9f cents a pound, what is a hog weighing 325 lb. worth? Ans. $31.68f. 57. In Chicago in 1912, carpenters received 65 cents per hour. How much was this for an 8-hour day? For one week of 5^ days? Ans. $5.20; $28.60. 58. In the same city, a bricklayer received 72 1 cents an hour. How much could a man earn in a year if he worked 225 days of 8 hours each ? Ans. $1305. 59. In 1908, a stonecutter in New York received 56/ cents an hour. If a man worked 50 weeks a year and 5j days per week, 8^ hours a day, how much would he earn a year? Solution. 56/o X8^ X5| X50 = HJP X 1 /- XV- X =$1312.50f . Ans. 60. A man earns 62 J cents an hour and his two sons each 22 f cents an hour. How much do the three earn per week of 5J days of 8? hours each? Ans. $50.49. 61. A gang of men mix and place an average of 43J| cu. yd. of con- crete per hour. How many cubic yards do they place in a day of 8J hours? Ans. 381. 62. An alloy, used for bearings in machinery, is if copper, ^ tin, and ^V zinc. How many pounds of each in 346 lb. of the alloy? .4ns. 286^; 47|J; llfi 63. An alloy, called "anti-friction metal," is ylou copper, \\% tin, and j 3 o antimony. Find the weight of each metal in a mass of the alloy weighing 1250 lb. Ans. 46 lb.; 1110 lb.; 93 f lb. 64. Find the cost of 27f sq. ft. of plate glass at 66| cents per square foot. Ans. $18.40. 65. A pumping engine in Chicago pumps on an average of 17, 361 \ gallons per minute, how many gallons is this in 24 hours? Ans. 25,000,000. 66. An ice-plant has an output of 45 tons daily. What is the value of this output for a year of 320 days at $3g per ton? Ans. $51,840. 26 67. Nickel ateel will stand a pull of 90,000 Ib. per square inch of cross section. What pull will a bar of l|f sq. in. cross section stand? ATM. 174,375 Ib. 68. The average yearly fire loss in the United States from 1897 to 1906 was $2i 7 (j per capita. If the population averaged 75,000,000, what was the average loss per year? Ana. $202,600,000. 69. In the European countries for the same period as in the previous exercise, the average fire loss per capita was $J. What would have been saved in the United States if the fire loss had been the same as in the European countries? Ans. $177,500,000. 70. The circumference of a circle is very nearly f?f times the diameter. What is the circumference of a circle that is 24/0 in. in diameter? Ans. 77ii| in. nearly. 71. If the diagonal of a square is very nearly 1^ the length of one side, find the diagonal in feet of a square \\ miles on a side. A mile is 5280 ft. Ans. 50CO ft. 72. Remembering that 6 per cent means rfof, find the value of the following : (a) 6% of $7.25. (/) 45% of 325 acres. (6) 9% of $820. (0) 75% of $3276. (c) 12% of $75.20. (h) 95% of 396 miles. (d) 20% of 476 bushels (i)' 82% of 7684 bushels. (e) 30% of 9227 bushels. 0') 65% of 4762. Ans. (o) $0.43 J; (e) 2768 & bu.; (g) $2457; (j) 3095^. DIVISION OF FRACTIONS 26. Division of a fraction by an integer. Example 1. Divide 4 by 4. (1) Since to divide by 4 is to find one of the 4 equal parts and to get \ of a number is to find one of the 4 equal parts, we have f-s-4 = i off =& Ans. (2) Or, using the principle that multiplying the denominator of a fraction divides the value of the fraction, we have (3) In division of fractions, we can often divide the numera- tor and thus divide the fraction. Example 2. Divide Yi 4 by 25. 925 925^-25 37 TT -IT = TT =3 *' This may be written W^-25 = WX-sVHi =3^- Ans. COMMON FRACTIONS 27 RULE. To divide a fraction by an integer, divide the numerator, or multiply the denominator, of the fraction by the integer; or multiply the fraction by 1 over the integer. 27. Division by a fraction. Example 1. Divide 6 by f (1) If we reduce 6 to thirds, we may divide the numerators, since then the numbers will both be thirds, and so be like numbers. 6 -5-f= -^-5-1 = 18 thirds ^-2 thirds =9. Ans. (2) Since there are 3 times ^ in 1, and | as many times | there are f times 3, or f times f in 1. Now -f is f in- verted. Hence we can find how many times the fraction | is contained in 1 by inverting the fraction, f will be contained 6 times as many times in 6 as it is contained in 1. .-. 6-hf = 6Xf = 9. Ans. Definition. The reciprocal of a number is 1 divided by that number. Thus, f is the reciprocal of f. f is the recipro- cal of f or 4. Example 2. Divide c by |. 49 _i_ 1 4 49V39 21 O 1 65 39" ~ (55 A 1~ "17 ^1 IT Example 3. Divide 4| by 3|. First reducing each to improper fractions, we have 4*-3-3$=- = Y-XA = -H = ltt. Ans. RULE. To divide a whole number or a fraction by a fraction, invert the divisor and multiply by the dividend. If either or both dividend and divisor are mixed numbers, first change to improper fractions. 28. Special methods in division. The work of division may often be simplified by one of the following methods: Example 1. Divide 56 f by 5. Process. Explanation. 5)56f 56-5-5 = 11 with a remainder of 1. 1H Ans. ]f-=-5 = f-=-5 = ^- Example 2. Divide 75 by 3| Process. Explanation. Since multiplying both dividend and divisor by the same 3 1) 75 number does not change the quotient, 11)225 we can multiply both by the denomi- 20/1 Ans. nator in the divisor, then divide as before. 28 PRACTICAL MATHEMATICS Example 3. Divide 125 by 2f Process. Explanation. The same as in the 2f)125 preceding, multiply both by the denomi- lll**P_?J_ nator in the divisor. Then divide as in 45|f Am. example 6. EXERCISES 9 Divide the following, using the pencil only when necessary: 1. f-5- 4. 6. l jV -5-H. 11. * j-j. 2. \\- -3. 7. _W"X12. 12. f- L A- 3. Y- -3. 8. 5-^-i' 13. l- 5-A' 4. -fi - -4. 9. 17+f- 14. 1*3 -5-1. 6. ?1- -7. 10. 16-5-f. 15. Y - 16. 32 J- 4. 21. 961-5-8. 17. 326 f-5- 2. 22. 122|-3. 18. 764? -4 23. 27|^-5. 19. 211-5-2. 24. 86f -=-3. 20. 28! -r3. 25. 192|-i-5. 26. If the denominator of a fraction is multiplied by 3, how is the unit of the fraction changed? How changed if multiplied by 6? By 5? Illustrate with the fraction . 27. If \ in. on a map represents 1 mile, how many miles are represented by 6 in. on the map? 28. In the drawing of a house, \ in. in the picture represents 1 ft. in the actual house. Find the dimensions of the rooms that measure as follows: 2J in. by 2| in., 1} by 1}, 1& by 1^, ft by ft. In the following, x is used for the number that is to be found : 29. -{-6 = x. 32. x-s-J = 7. 35. j -s-4 = - x 30. 3-5-z = 6. 33. Jp-ns = ?. 36. y-r-9 = . \s X 31. |*.r = 5. 34. V-l = g- 37. ?-hx = ?- 38. If a man can do a piece of work in 3 hours, what part of it can he do in 1 hour? In 2 hours? 39. If a man can do i of a piece of work in 1 hour, in how many hours can he do all the work? of the work? 40. If a man can do f of a piece of work in 2 hours, in how many hours can he do all the work? f of the work? J of the work? A of the work? 41. If a boy can run of a mile in 6 minutes, how many minutes will it take him to run a mile? J a mile? f of a mile? COMMON FRACTIONS 29 42 If John can do ^ of a piece of work in 1 hour and Henry can do of it in 1 hour, what part of the work can they both do in 1 hour? How many hours will it take them to do the whole work if they work together? Ans. f ; 1. 43. One boy can hoe a patch of potatoes in 3 hours and another boy can hoe the same patch in 4 hours. In how many hours can they hoe the potatoes if they work together? Solution. The first boy can hoe | of the patch in one hour, and the second boy J of the patch in one hour. Together they can hoe + } = T V of the patch in one hour. They can hoe the entire patch in l-=-i 7 j = lXV= = lf or If hours. 44. A water tank that holds 60 barrels has two pipes opening from it. One of these can empty the tank in 4 hours when running alone, and the other pipe can empty the tank in 12 hours when running alone. If both of the pipes are running at the same time, how long will it take them to empty the tank? Ans. 3 hours. 45. If one pipe can empty a tank in 4 hours, and another pipe can empty it in 12 hours, in how many hours will both pipes empty the tank when running together? Ans. 3. 46. If a boy earns SJ a day, in how many days will he earn $9. Ans. 12. 47. If a man earns $| in 1 hour, in how many hours will he earn $15? Ans. 10. 48. 49. 60. 51. 100 -4| 52. 53. 31| -I 54. In the following five exercises, reduce all to fractions, take reciprocals of each divisor and cancel. 62. 2^X31 X r 4 r X2 r ^-20f. Process. 2 Xa+20f-xxxx-. Ans. Ans. ^V. 55. 3}i-^7|f. A -MO 295 /ins. 5^7. Ans. |f. 56. T&r-TriT. Ans. 2f4f. Ans. ! T vV 57. lOH-5-lrf*. Ans. 10. Ans. 20 1$. 58. 51-^3- Ans. 24|. Ans. T 9 5 . 59. 7^3MJ. Ans. Iff. Ans. 18. 60. 104 T 3 T H-8&. Ans. 11U. Ana 1253 /I/IS. l^j^. 61. 1151-5-20$. Ana. 5f. 63. SiXQg-T-Sif-s-Qf-s-S&^-riy. Ans. 64. if X29f Xl3f-^15f-r-2|. Ans. 65. &X 13* X*|X9i -5-1! *-(!& XfX26i)-H. Ans. 2ff. 66. 8f XHX&XHXHX71X 11-5-6$ -Hf *12i-5-. Ans. 1*. 67. Find the value of J * (f +f ) -}. Ans. H- Parentheses indicate that the inclosed operations must be performed first. For example, in the above, -f is added to % and then f is divided by the sum. 68. Find the value of 2J-7S+5j-562-. .4ns. 614. 69. Simplify ()*, (b) -?, (c) - , (d) -\l Ans. l{), , H, 11*. 30 PRACTICAL MATHEMATICS 70. From 75 f, take 12ft. Ann. 62J?. 71. Multiply 21 -5- i by } of (i+l). Ant. 2^. 72. Find the value of - 4rw. 2|J. Suggestion. First, multiply 3? by 8 J, second, 4f by 2^j, then divide the first product by the second. 73. 133-2ft-6A+3-l 1 5 a +8J-t?-10H? Ans. 4?*. 74. (si-^-f) X2 75. Evaluate - _1.*.K 4' Solution. The word evaluate means that the indicated operations should be performed and the value of the expression found. At first decide what operations must be performed first, what second, and so on. Then do these operations as simply as possible. 4 76. Evaluate ^r^rSj^- ^rw. 5 . 77. Add f Xy to JX(4J-2J). D 78. Subtract of f from of |. vlrw. /,. 79. f of 20 is ^ of what number? Ans. 111. 1 21 34 4 80. Find the simplest expression for 5T~q+2 ~4" '^ nj * g ^' Perform the operations indicated in the following five exercises: t ,, 32 (41 +7j) +81 ' 3I-2J 83 " 86. In the following four exercises, the letters stand for values as follows: 0,000 Ib. Use 50% as the strength of the joints and a factor of safety of 6. Ans. | in. PRF 120 X 25 X 6 Solution, t = T X % 60000 X = I 130. Find the thickness of the boiler plate for a 72-in. boiler to carry a pressure of 90 Ib. with a tensile strength of 60,000 Ib. Use 50% as the joint strength and a factor of safety of 6. Ans. f in. nearly. 131. The following formula is used in finding the diameter of a steam- oiler 2tT X % D = PF where D = diameter of boiler in inches, t = thickness of plate in inches, T= tensile strength of boiler plate in pounds, P = steam pressure in pounds per square inch, F factor of safety, % = percentage of strength in joints. Find the diameter for a steamboiler having a f-in. plate, allowing 50% for strength of joints and a factor of safety of 6, with a tensile strength of 60,000 Ib., and 125 Ib. pressure per square inch. Ans. 50 in. FIG. 7. 132. Find the number of lines in a paper of 38 pages, two columns to the page, each 10 J in. long, and 15 lines in 2 in. How many words if they average 11 words to the line? How long would it take to read such a paper at the rate of 170 words a minute? 133. The distance F across the flats in a bolt head or nut, either a square or a hexagon (Fig. 7), is equal to 1-| times the diameter of the bolt plus | in. As a formula this is Test the widths across the flats in the following table taken from a manufacturer's catalog: Diameter of bolt D I 5 16 f 7 16 f 3 i ii 1 * A 8 If 2 Width across flats F i 2 3~2 i & 1*' n i& 1 i4 1 16 29 T6 2f Ql *5 5 2 CHAPTER III DECIMAL FRACTIONS 32. Definition. Fractions that have 10, 100, 1000, etc. for denominators are decimal fractions. Thus, -iVo, iWo'o, T^TJ 1888 are decimal fractions. In writing a decimal fraction it is convenient to omit the denominator, and indicate what it is by placing a point (.), called a decimal point, in the numerator so that there shall be as many figures to the right of this point as there are ciphers in the denominator. Thus, T Vo is written 0.53; fWA = 0.3756; T JJ = 0.076; J888 = 4.326. In such numbers as 0.53 and 0.3765, the cipher is printed at the left of the decimal point for clearness; but it is not necessary and is often omitted. It is to be noted that when there are fewer figures in the numerator than there are ciphers in the denominator, ciphers are added on the left of the figures to make the required number. From the meaning of the decimal fraction, it is seen that the misplacing of the decimal point changes the meaning greatly. For each place it is moved to the right, the value of the decimal fraction is multiplied by 10; and for each place it is moved to the left, the value is divided by 10. Thus, 2.75 becomes 27.5 when the point is moved one place to the right, and 0.275 when the point is moved one place to the left. In the first case 2.75 is multiplied by 10, and in the second case it is divided by 10. It is well to recall the fact that when we have a number such as 3333, where the same figure is used throughout, the values expressed by the threes vary greatly. For every place a three is moved toward the left, its value is increased ten times; and as we pass from left toward the right, each three has one-tenth the value of the one to the left of it. Since the above relations hold when we pass to the right of 38 DECIMAL FRACTIONS 39 the place representing units, we have the following relative values of the places: [ -^ 02 -C 1- -c H X 00 ^ -t c c rr -a 3 A ~ Thousands Hundreds Tens Units Decimal poi: Tenths Hundredths Thousandth; Ten-thousan Hundred-thc Million ths Ten-million t Hundred-mil 0000.00000000 33. Reading numbers. The whole number 23,676 is read twenty-three thousand six hundred seventy-six. It should be noticed that no word "and" is used in reading a whole number. A decimal is read like a whole number except that the name of the right-hand place is added. For example, the number 0.7657 is read, seven thousand six hundred fifty-seven ten-thousandths. When a whole number and a decimal fraction are written together the word "and" is used between the two parts in reading. Thus, 73.2658 is read, seventy-three and two thousand six hundred fifty-eight ten-thousandths. Where one person is reading numbers for another to write, it is not customary to proceed in the above manner. Thus, the number 23.6785 may be read twenty-three, point, sixty- seven, eighty-five. Or we may read it, two, three, point, six, seven, eight, five. EXERCISES 10 Write the following in figures : 1. Three hundred fifty-six ten-thousandths. 2. Two hundred fifty-six and twenty-three thousandths. 3. One hundred fifty-five millionths. 4. Four hundred fifty-six thousandths. 5. Four hundred and fifty-six thousandths. 6. Three hundred twenty-five and twenty-five ten-thousandths. 40 PRACTICAL MATHEMATICS Read the following : 7. 23.402. 11. 1200.3604. 8. 2003.203. 12. 10,101.2301. 9. 0.4256. 13. 5867.0067. 10. 4200 0056. 14. 10,0000001. 34. Reduction of a common fraction to a decimal fraction.- A decimal fraction differs from a common fraction only in having 1 with a certain number of ciphers annexed for the denominator. The common fraction can then be changed to a decimal by reducing it to a fraction having 1 with ciphers for a denominator. It is evident from the method of reducing a common fraction to one with a different denominator, that a common fraction can be changed to a decimal only when its denominator is con- tained an exact number of times in 10,100,1000, or 10000, etc. Thus, I =iV or 0.4, and, T 9 8 =-f^o or 0.5625, but ? cannot be expressed exactly as a decimal because 7 is not exactly contained in 10, 100, or JOOO, etc. To reduce a common fraction to a decimal proceed as follows: RULE. Annex ciphers to the numerator and divide by the denominator. Place the decimal point so as to make as many decimal places in the result as there were ciphers annexed. Thus, |=0.875, Process. 8)7.000 0.875 and $=0.2857+ Process. 7)2.0000 0.2857+ The sign, +, placed after the number indicates that there are still other figures if the division is carried further. A common fraction in its lowest terms will reduce to an exact decimal only when its denominator contains no other prime factors than 2 and 5. Thus, 8 8 reduces to an exact decimal for 64 is made up of 2X2X2X 2X2x2, while j 7 z cannot be reduced to an exact decimal for its de- nominator contains the factor 3. 35. Decimal fraction to common fraction. To change a decimal fraction to a common fraction proceed as follows: COMMON FRACTIONS 41 RULE. Replace the decimal point by a denominator having 1 and as many ciphers as there are decimal places in the original fraction. (See Art. 32.) Thus, 2. 375 =f?$, which may be written as a mixed number, 9371 O 3 ''ItiTS'tf ** EXERCISES 11 Reduce the following to decimals : 1. ff- 2. . 3. if. 4. l\\. 5. 21- 6 %. 6. 62|gfg. Reduce the following to common fractions or mixed numbers in their lowest terms: 7. 0.440. 8. 0.98. 9. 0.03125. 10. 0.00096. 11. 14.06225. 12. 42.030125. 13. Reduce the following decimals of an inch to common fractions in their lowest terms: 0.375; 0.359375; 0.28125; 0.171875; 0.078125. 14. Express the following in their simplest common fractional form: 3.04f; 0.00; 0.28f; 0.714?; 0.484f; 0.87i Ans. 304 ? A- 7 ,? a Suggestion. 3.04 = = - 15. Change the following per cents to their simplest common fractional forms: 87J%; 133*%; f%; 185$%; 1.85f%; 2.21H%. Ans. 1; |; T ^; If 5 ?V is phosphorus. How much of each is there in 369.523 pounds of phosphor bronze? Ans. 34 1. 8088 -; 25.8666 + ; 1.8476+ Ib. 66. A certain paper weighs 68 Ib. per ream of 500 sheets. If tin- paper costs 5J cents per pound, how much will 2fg reams cost? Ans. $8.41. 67. Manganese bronze contains the following: copper 0.89, tin 0.1, manganese 0.01. How much of each metal is there in a propeller weigh- ing 2378J Ib.? Ans. 2117.16 + ; 237.88 + ; 23.79- Ib. 58. Add and _*> divide the result by 7Jf, and change the quotient OITT t to a decimal. An*. 0.125. 69. From i X2J subtract the product of 0.075 and Ij, divide tho re- mainder by 12, and change the result to a decimal. Ans. 0.0375. cn Q . ... (3.2+0. 004 -1.11DX0.25 *' Simpllfy (4 +0.2) -17.907 61. Simplify (1+H -0.024) -=-(15* -1.209). Ans. 0.214 +. 62. Simplify (iM> XOOOO^ An*. 0.00051 28+. U.U7O ... (3.71-1. 908) X7.03 63. Simplify ~ Ans. 6.405+. j'ss ... (201 +2.25 X0.004) -=- (1.0337 -31.09 X0.03) 64. Simplify - _____ Ans. 424,573.5-. 66. How many lengths each 0.0275 of a foot are contained in 27.2375 ft.? Ans. 990.45+. 66. The expenditures of the British Naval Service were as follows: for the year 1906-7, 31,472,087; for the year 1907-8, 31,251,156; express these sums to the nearest dollar if 1 =$4.8665. Ans. $153,158,911; $152,083,751. 67. A wood dealer charged $33.62 for a pile of wood containing 7J cords. What error did the dealer make if wood was worth $4.25 a cord? Ans. $0.68. 68. If a circle is 3.1416 times as far around as through it, find the number of feet around a cart wheel 3.75 ft. through. Find the distance around the earth if the diameter is 7918 miles. Ans. 11.781; 24,875+ miles. 69. Find the value of: 27,750 shingles at $4.25 per thousand. Ans. $117.94. 47,256 ft. of lumber at $45 per thousand. Ans. $2126.52. 126,450 bricks at $7.75 per thousand. Ans. $979.99. 45,350 ft. of gas at $0.85 per thousand. Ans. $38.55. 70. What is the cost of carbon-steel rails to lay 6 miles of street car track, if the rails weigh 129 Ib. per yard, and cost $28 per ton. (1 mile = 1760yd.) Ans. $38,142.72. 71. It has been determined by experiment that each square foot of steam radiation will give off to the surrounding air about 3 heat units DECIMAL FRACTIONS 47 per hour per degree difference between the air in the room and the steam radiator. If the temperature of the radiator is 212 and that of the room 70, how many heat units will be given off per hour on 24,000 sq. ft. of radiating surface? How many pounds of coal will it take to make this steam if 1 Ib. of coal contains 10,000 heat units? Ans. 10,224,000; 1022.4. 72. Nickel steel will stand a pull of about 90,000 Ib. per square inch in cross section. What pull will a bar 0.786 in. wide and 0.237 in. thick stand? Ans. 16,765+ Ib. Suggestion. The area of the cross section is found by multiplying 0.786 by 0.237. 73. The composition of white metal as used in the Navy Department is as follows: tin 7.6 parts, copper 2.3 parts, zinc 83.3 parts, antimony 3.8 parts, and lead 30 parts. Find the number of pounds of each in 635 Ib. of the white metal. Ans. tin 38; copper 11.5; zinc 416.5; antimony 19; lead 150. Suggestion. Adding 7.6+2.3 +83.3 +3.8+30 = 127 = whole number of parts. 635 -j-127 =5 = number of pounds for each part. Multiply 5 Ib. by number of parts of each. 74. In 1909 the number of horses in the United States was 20,640,000. They were valued at $1,974,052,000. Find the average price per head. Ans. $95.64+. 76. A reamer that is 6 in. long is 1.2755 in. in diameter at the small end, and 1.4375 in. at the larger end. Find the taper per foot. (The taper per foot means the decrease in diameter per foot of length.) Ans. 0.324 in. 76. A creditor receives $0.76 on each dollar due him. If he loses $326.40, how much was due him? What would he have received if $7642 had been due him? Ans. $1360; $5807.92. 77. It cost, for labor and materials, $38,692.38 to construct 7500 ft. of car track. What was the average cost per foot? What would be the cost for 100 miles at the same rate? (1 mile = 5280 ft.) Ans. $5.159-; $2,723,943.55. 78. During 1908 the Daly- West mine marketed 12,760 tons of crude ore; containing 2,683,830 Ib. of lead; 343,376 Ib. of copper; 454,149 oz. of silver; and 441.86 oz. of gold. Find its value if lead is worth 3.925 cents per pound; copper 13.208 cents per pound; silver 52.864 cents per ounce; and gold $18.842 per ounce. Ans. $399,100.28. 79. One cubic foot of water weighs 62.5 Ib. ; find the volume of 1 Ib. of water. Of 23 Ib. Ans. 0.016 cu. ft.; 0.368 cu. ft. 80. One cubic foot of ice weighs 57.5 Ib. ; find the volume of 1 Ib. of ice. Of 49.3 Ib. Ans. 0.0174- cu. ft.; 0.857+ cu. ft. 81. How many times as heavy as ice is water? How many times as heavy as water is ice? Ans. 1.087 ; 0.92. 82. If the fire under a steam boiler requires 3 Ib. of coal per horse- power per hour, find the cost of coal at $3.75 a ton to run a 160 horse- power boiler for 30 days of 10 hours each. Ans. $270. 48 PRACTICAL MATHEMATICS 83. A column of water 2.302 ft. high gives a pressure at the base of 1 Ib. per square inch. Find the height of a column of water to give a pressure of 256.3 Ib. per square inch. Find the pressure per square inch of a column of water 1 ft. high. A column 237.4 ft. high. Ans. 590.0+ ft.; 0.4344+ Ib.; 103.1+ Ib. 84. The Auditorium Building in Chicago has a cubic content of 9,128,744 cu. ft., and cost 36 cents per cubic foot. Find the total cost. Ans. $3,286,347.84. 86. Using U. S. standard, the gage and thickness for sheet steel is as follows: No. 00, 0.34375 in.; No. 2, 0.265625 in.; No. 4, 0.234375 in.; No. 7, 0.1875 in.; No. 13, 0.09375 in.; No. 28, 0.015625 in. Find the approximate thickness of each in a common fraction of an inch having 8, 16, 32, or 64 for a denominator. Ans. J$, JJ, f}, -fa, -fj, ^j. 86. If it costs $106.50 per day to run a gang of men and a rock crusher giving a daily output of 200 cu. yd. of crushed rock, find the cost per cubic yard. Ans. $0.5325. 87. In building a certain canal lock 2HO cu. yd. of concrete were used. It cost $1.77 a cubic yard for mixing and placing the concrete. The material for the concrete was as follows: 3010 barrels of cement, at $3.02, 1377 cu. yd. of broken stone at $1.37, 393 cu. yd. of screened pebbles at $0.90, 459 cu. yd. of gravel at $0.67, 500 cu. yd. of sand at $1.78. Find the cost of the concrete work. Ans. $16,315.72. 88. In building a concrete viaduct containing 2111 cu. yd., the total cost was 1908 barrels of cement at $1.60; 1105 cu. yd. of sand at $1.95; 1468 cu. yd. of stone at $1.48; lumber for forms $1140; tools, hardware, etc., $527.75; water $63.00; labor $7262. Find the average cost per cubic yard of concrete. Ans. $7.756 + 89. Number 8 (B. & S.) gage sheet steel is 0.1285 in. thick and weighs 5.22 Ib. per square foot. (1) Find the thickness of a pile of 56 such sheets. (2) Find the nearest whole number of sheets to make a pile 1 ft. thick. (3) Find the weight of this number of sheets if each sheet has 4 sq. ft. Ans. 7.196 in.; 93; 1941.8+ Ib. 90. Number 25 (B. & S.) gage sheet copper is 0.0179 in. thick and weighs 0.811 Ib. per square foot. Answer the same questions as in exercise 91. Ans. 1.0024 in.; 670; 2173.5- Ib. 91. An iron chain made of l|-in. round iron has a breaking strain of 88,301 Ib. If the chain weighs 17.5 Ib. per foot, how long would the chain have to be to break of its own weight if suspended from one end? Ans. 5046- ft. 92. Answer the same question as in exercise 93 for a. chain made of -fi-in. round iron, the chain weighing 0.904 Ib. per foot and breaking under a strain of 4794 Ib. Ans. 5303+ ft. 93. How much must be paid for 1600 ft. of 9teel bar weighing 1.87 11>. I>er foot and costing $4.65 per hundred pounds? Am. $139.13. DECIMAL FRACTIONS 49 94. If a steel tape expands 0.00016 in. for every inch when heated, how much will a tape 100 ft. long expand? 95. A round piece of work being turned in a lathe is 1.4275 in. in diameter. What is the diameter after a cut -fa in. deep is taken in the work? Ans. 1.39625 in. 96. The inside diameter of a steam cylinder before boring was 26 in. The diameter after boring was 26.3125 in. How deep a cut was taken in boring? In boring, 20 turns were made in a minute. How long would it take to bore a distance of 28 in. if the feed was 0.0625 in.? Ans. -2 in.; 22.4 minutes. 97. Under a load of 325 Ib. a wire 112 in. long and 0.09074 in. in diameter lengthened 0.265 in. What was the stretch per foot to four decimal places? Ans. 0.0284 in. 98. In mixing a quantity of concrete using 1 part Portland cement, 2 parts of sand, 3 parts of gravel, and 5 parts of broken stone, it was found that 1 barrel of cement averaged 1.18 cu. yd. of concrete. If a barrel of cement contains 3f cu. ft., how many cubic feet of material were put into one cubic yard of concrete? Explain how this could be. Ans. 36.55 cu. ft. FIG. 9. 41. Proportions of machine screw heads. A. S. M. E. standard. In the following four problems are given the four standard heads. The proportions are based on and include the diameter of the screw, diameter and thickness of the head, width and depth of the slot, radius for round and fillister heads, and included angle of flat-headed screw. ( 1 ) Oval fillister head machine screws. A = diameter of body. jB = 1.64A 0.009= diameter of head and radius of oval. C =0.66A - 0.002 =height of side. D=0.173A +0.015= width of slot. E = %F = depth of slot. F =0.1348+ C = height of head. f>o I'RA CTICAL MA Til EM A TICS 99. Given the values of A find those of B, C, D, E, and F. A B C D E F 0.216 0.3452 0.1406 0.052 0.093 0.1868 0.398 0.6437 0.2607 0.084 173 0.3469 0.450 0.729 0.295 0.093 0.196 0.3927 Fie;. 10. FIG. 11. Suggestion. The values of B, C, D, E, and F are to be found from the given value of A. 5 = 1.64X0.216-0.009 = 0.3452. C = 0.66 X0.216 -0.002 =0.1406. (2) Flat fillister head machine screws. A diameter of body. B = 1.64A -0.009 =diameter of head. C =0.66A -0.002 =height of head. >=0.173A +0.015 =width of slot. E = \C= depth of slot. 100. Given the values of A find those of B, C, D, and E. A B C D E 0.112 0.1747 0.0719 0.034 0.036 0.177 0.2813 0.1148 0.046 0.057 0.320 0.5158 0.2092 0.070 0.105 (3) Flat head machine screws. A = diameter of body. B =2 A -0.008 = diameter of head. A -0.008 C= i 7 39~ =depth of head. D = 0.173A +0.015 =width of slot. # = jC = depth of slot. 101. Given the values of A find those of B, C, D, and E. A B C D E 0.086 0.164 0.045 0.030 0.015 0.242 0.476 0.135 0.057 0.045 0.372 0.736 0.209 0.079 0.070 DECIMAL FRACTIONS 51 (4) Round head machine screws. A = diameter of body. B = 1.85 A -0.005 = diameter of head. C = 0.7A = height of head. D=0.173A +0.015= width of slot. = depth of slot. 102. Given the values of A find those of B, C, D, and E. A B C D E 0.073 0.130 0.051 0.028 0.035 0.164 0.298 0.115 0.043 0.067 398 0.731 0.279 0.084 0.149 103. The formula for determining the number of threads per inch on machine screws is v = 6-5 A +0.02 where N is the number of threads per inch, and A the diameter of the. screw. Compute the number of threads per inch for machine screws of the following diameters: 0.242, 0.398, 0.450, 0.563, 0.750. In each casR give the answer to the nearest whole number. Ans. 25; 16; 14; 11; 8. CHAPTER IV SHORT METHODS AND CHECKS 42. Contracted methods and approximate results. As a rule the practical man does not need a large number of decimal places. The results of all measurements are at best only an approximation of the truth. The accuracy depends upon the instruments, the method used, and upon the thing measured. All that is necessary is to be sure that the magni- tude of the error is small compared with the quantity meas- ured. It is clear that in a dimension of several feet, a fraction of an inch would probably not make much difference ; but if the dimension was small, such an error could not be allowed. The man in practical work uses instruments which are of such accuracy as to secure results suitable to his purpose. If he requires measurements accurate to 0.001 in., it is not neces- sary for him in a computation to carry his work to 0.00001 in. A good rule to go by is not to calculate to more than one more decimal place than measurements are made. Thus, if a measurement of 3.265 in. is made, and it is to be multiplied by 3.1416, it is not necessary to multiply in the usual way, as then there would be seven decimal places, while the measurement was accurate to only three places. If the multiplication is performed by multiplying first by the left-hand figure of the multiplier, and then passing toward the right, we have the following forms: Form in full. Contracted form. 3.265 3.265 3.1416 3.1416 9795 9795 3265 3265 13060 1306 3265 32 19590 19 10.2573240 10.2572 52 SHORT METHODS AND CHECKS 53 It will be noticed that one more decimal place is retained than the desired number. In a similar manner, division can be contracted. Suppose it is required to divide 0.04267 by 3.278, and secure an answer correct to four significant figures. The division in the full and contracted forms is as follows: 0.042670000, 3.278 0.042670 3.278 3278 0.013017 3278 0.013017 9890 9890 9834_ _9834 ~5600 56 3278 32 23220 24 22946 22 274 2 Hence the result correct to four significant figures is 0.01302- . It is easy to follow the method in obtaining the above, but it is hardly worth spending time upon unless one is to do much computing of this kind. EXERCISES 13 Solve the following by contracted forms: 1. 3.14159X3.14159 correct to four decimal places. Ans. 9.8696. 2. 9,376,245-^3724 correct to the unit's place. Ans. 2518. 3. 1 00 H-3. 14159 correct to 0.01. Ans. 31.83. 4. 87,659,734-^5467 correct to five significant figures. Ans. 16,034. 5. 45.8636X26.4356 correct to five significant figures. Ans. 1212.4. 6. 6.234X0.05473 correct to four significant figures. Ans. 0.3412. 7. 4.326X0.003457 correct to five significant figures. Ans. 0.014955. 43. Other methods. -Numerous short methods in multi- plication and division can be given. A few of the most useful ones are given here. If benefit is to be derived from them, they must be very carefully fixed in mind, and used whenever occasion arises. 54 PRACTICAL MATHEMATICS (1) To multiply a number by 5, 50, 500, etc., multiply by 10, 100, 1000, etc., and divide by 2. Why will this give the result? Example. 7856 X 50 = 785600 -5- 2 = 392800. A ns. Multiply tho following without using the pencil: 76X50 432X50 5.5X5 96X5 768X500 4.35X50 88X500 47X50 79.2X5000 (2) To multiply by 25, 250, etc., multiply by 100, 1000, etc., and divide by 4. Why will this give the result? Example. 32 X 250 = 32000 -r 4 = 8000. Ans. Multiply the following without using the pencil : 256X25 8956X25 728X250 74.92X250 492X2500 942.3X2500 (3) To multiply a number by 125, multiply by 1000 and divide by 8. Why will this give the result? Example. 848 X 1 25 = 848000 -f- 8 = 106000. Ans. Multiply the following: 920X125 4.76X125 72.88X125 55.5X125 (4) To multiply a number by 33$, 16f , 12$, 8$, or &\, multiply by 100 and divide by 3, 6, 8, 12, or 16. Example. 84 X 8$ = 8400 -5-1 2 = 700. Ans. Multiply the following: 48X33* 42.6X16? 32*X16 41|X8* 96X12* 3.97X8* 33JX33* 19|X6J 72X6J 4.76X33* 98.76X16? 27|X12J This rule can be used easily in multiplying a number by 37*, 62$, 87*, 83}, and other fractional parts of 100 or 1000. Multiply the following: 24X62* 35X333* 421x62* 32X87* 476*X625 71$ X37* 36X83* 672X62* 47X333* 64X37* 272X87* 36* X83J (5) To multiply a number ending in \, as 2$, 4$, 11$, by itself, multiply the whole number by the whole number plus 1 and add J to the product. SHORT METHODS AND CHECKS 55 Examples. 8|X8 = 8X9 + | = 72f. Ans. ll^Xlii = 11X12+| = 132|. Ans. The reason may be shown as follows: But 3X^+1X3=1X3 and |X| = i Hence 3^X3^ Multiply the following: 40JX40J 59|X59| Putting in the decimal form, we have 81x8^ = 8.5X8.5 = 72.25. Now removing the decimal point, we have 85X85 = 7225. Multiply the following : 7.5X7.5 135X135 505X505 12.5X12.5 95X95 615X615 11.5X11.5 155X155 925X925 (6) Divisions. By using the inverse operations to those given in the preceding rules, we may divide by 33|, 16f, 12|, 125, 250, 8i etc. Make the rules for divisions. 84 -M2| = 84 -^100X8 = 6. 72. 32-=- 125 =32-M 000X8 = 0.256. 450 + 61i = 450 -MOO X 16 = 72. 23 ^- 250 = 23 -M 000X4 =0.0092. The multiplications in such problems can usually be per- formed without using the pencil. Divide the following : 800-M2* 492 -H 16f 720 -5-8-J- 37.6 -f- 250 923 4-331 783 -M2i 7.62 -=-12^ 436^-3| 7.29 -=-125 927^-333^ 43.9-^250 8927 -^166f 44. Checking. No check can be made that is absolutely certain to detect an error, but there are many very useful devices for checking the accuracy of the work. (1) Addition. A simple way to check addition is to re-add, taking the figures in some other order. Add first up and then down, is very satisfactory. 56 PRACTICAL MATHEMATICS (2) Subtraction. An error in a subtraction will generally be detected by adding the remainder to the subtrahend. If this gives the minuend the work is correct. Example. Minuend 37249 Subtrahend 18496 Remainder 18753 37249 = subtrahend-^- remainder. (3) Multiplication. A good way to check multiplication is to interchange the multiplicand and multiplier and multiply again. A very convenient and quick method is to proceed a.s follows : (a) Add the digits in the multiplicand. If this sum has more than one digit, add these. Continue till a number of one digit is found. (6) Add the digits of the multiplier as directed in (a). (c) Multiply together the numbers obtained in (a) and (6), and add digits till a number of one digit is found. (d) Add digits of product as directed in (a). (e) Compare results of (c) and (d). If they are the same the work checks. Check. Example. Summing digits as directed 34768 Multiplicand in (a^ 492 Multiplier for multiplicand gives 1, for 69536 multiplier gives 6, 6X1=6. 312912 Sum of digits from product 139072 gives 6. Since this is the same 17105856 Product as obtained before, the work is checked. (4) Division. Division can be checked by multiplying the divisor by the quotient and then adding the remainder. The result should be the dividend. A quicker way to check is to add the digits as directed for checking multiplication: (a) the dividend; (6) the divisor; (c) the quotient; (d) the remainder. Multiply the results in (6) and (c), add the result in (d), and then add the digits in this result which should give the same as the result of (a) if the work is correct. SHORT METHODS AND CHECKS 57 Check. Example. Dividend 4923567476 Divisor (a) Sum of digits, dividend 9, 476 1 10343 Quotient (6) sum of digits, divisor 8, 1635 (c) sum of digits, quotient 2, (d) sum of digits, remainder 2. TO !? 8X2+2 = 18. 1727 Sum of digits of 18 = 9, which ]428 is the same as the sum in "299 Remainder (a) and so checks the work. The preceding rules apply as well to decimals as to whole numbers, but do not check the position of the decimal point. EXERCISES 14 First multiply then divide the following and check as directed in the preceding article. 1. 435678 by 4537. 6. 456.78 by 45.32. 2. 980765 by 789. 7. 1230.8 by 3.876. 3. 60385 by 4327. 8. 32418 by 8.098. 4. 342153 by 7651. 9. 4.6543 by 1.0876. 6. 45.654 by 345. 10. 32 654 by 7.547. CHAPTER V WEIGHTS AND MEASURES 45. English system. The English system of weights and measures is the one in common use in the United States. The most used tables and equivalents of this system follow. The problems which are given later are inserted as material for review of work which it is supposed the student has done previously. Suggestions on solutions are given for some of the exercises that follow but no general methods are given as to how to solve such exercises. (1) Measures of time. 60 seconds (sec.) =1 minute (min.) 60 minutes =1 hour (hr.) 24 hours =1 day (da.) 365 days = 1 common year (yr.) 366 days = 1 leap year. (2) Measures of length. 12 inches (in. or ") = 1 foot (ft. or ') 3 feet = 1 yard (yd.) 5$ yards = 1 rod (rd.) 320 rods = 1 mile (mi.) 5280 feet = 1 mile. 1760 yards = 1 mile. (3) Measures of area. 144 square inches (sq. in. or in. 2 ) = 1 square foot (sq. ft. or ft. z ) 9 square feet = 1 square yard (sq. yd. or yd. 1 ) 30J square yards = 1 square rod (sq. rd. or rd. s ) 160 square rods =1 acre (A.) 640 acres =1 square mile (sq. mi.'* (4) Measures of volume. 1728 cubic inches (cu. in. or in. 3 ) = 1 cubic foot (cu. ft. or ft. 1 ) 27 cubic feet = 1 cubic yard (cu. yd. or yd. 1 ) 128 cubic feet = 1 cord (cd.) (5) Liquid measures. 2 pints (pt.) = 1 quart (qt.) 4 quarts =1 gallon (gal.) 31 J gallons = 1 barrel (bbl.) 231 cubic inches = 1 gallon. 58 WEIGHTS AND MEASURES 59 (6) Dry measures. 2 pints (pt.) =1 quart (qt.) 8 quarts =1 peck (pk.) 4 pecks =1 bushel (bu.) 2150.42 cubic inches = 1 bushel. It should be carefully noted that dry and liquid measures are very different. For instance, 4 quarts in liquid measure contain 231 cu. in., while in dry measure they contain 268.8 cu. in. nearly. (7) Measures of weight (Avoirdupois). 7000 grains (gr.) = 1 pound (Ib.) 16 ounces (oz.) =1 pound. 100 pounds =1 hundred weight (cwt.) 2000 pounds =1 ton (T.) 2240 pounds = 1 long ton. In practice it is customary to consider 1 cu. ft. of water as 62.5 Ib. or 1000 oz. (See Table II.) EXERCISES 15 1. Reduce 27 yd. 2 ft. 11 in. to inches. Ans. 1007 in. 2. Reduce 18 hr. 20 min. 35 sec. to seconds. Ans. 66,035 sec. 3. Reduce 4 T. 7 cwt. 35 Ib. 9 oz. to ounces. Ans. 139,769 oz. 4. Reduce 8 bu. 3 pk. 7 qt. 1 pt. to pints. Ans. 575 pt. 6. Reduce 8 A. 25 sq. rd. 4 sq. yd. to square yards. Ans. 39,480i sq. yd. 6. Reduce 5937 sq. in. to higher denominations. Ans. 4 yd.'-* 5 ft. 2 33 in. 2 Suggestion. First divide 5937 by 144, the number of square inches in a square foot. The quotient is the number of square feet and the re- mainder is square inches. Then divide by the number of square feet in one square yard. 7. Multiply 12 cu. yd. 15 cu. ft. 1115 cu. in. by 6. Ans. 75yd. 3 12ft. 3 1506 in. 3 8. How many iron rails each 30 ft. long will be required to lay a rail- road track 26 miles long? Ans. 9152. 9. Find the value of a field 180 rods long and 94 rods wide, at $18.00 per acre. Ans. $1913.625. 10. Reduce 17 pints to the decimal of a gallon. Ans. 2.125 gal. 11. How many steps does a man take in walking 2 mi. 76 rd. if he goes 2 ft. 8} in. each step? Ans. 4362.1 - Suggestion. Divide the total number of inches by the number of inches in one step. 12. Find the weight of 1 gal. of water. Ans. 133.68+ oz. =8 Ib. 5.68 oz. 13. How many sacks each containing 2 bu. 1 pk. can be filled from a bin containing 245 bu.? Ans. 109 sacks nearly. 60 PRACTICAL MATHEMATICS 14. A large steamship will hold 75 bargo loads of wheat at 8500 bu. to the bargo. A freight car 40 ft. long will carry 950 bu. Find the length of a train carrying enough wheat to load the steamship, allowing 2 ft. between cars. 16. What decimal part of a foot is iV in. ? j in. ? What decimal part of a yard is each? Ana. 0.0052 + ; 0.03125; 0.00173 + ; 0.010417 -. 16. Reduce the following to decimal parts of a foot: (a) 1 in., (6) 2 in., (c) 3* in., (d) 71 in. Ans. (a) 0.0833+, (6) 0.16G7-, (c) 0.2917-, (d) 0.61458+. 17. Reduce each in exercise 16 to a decimal part of a yard. Ans. (a) 0.02778 -, (6) 0.0556-, (c) 0.0972+, (d) 0.20486+. 18. Reduce the following to decimal parts of a pound avoirdupois: (a) | oz., (b) li oz., (c) 3 oz., (d) 7J oz., (e) 13 oz., (/) 4J oz. Ans. (a) 0.046875, (6) 0.09375, (c) 0.1875, (d) 0.46875, (e) 0.8125, (/) 0.28125. 19. Reduce 3.36 in. to a decimal fraction of a rod. Ans. 0.01697- rd. 20. Reduce a pressure of 22.5 Ib. per square foot to ounces per squarr inch. Ans. 2.5 oz. 22 5x16 Suggestion. The cancellation is ~\TA, =2.5. 21. What is the cost per hour for lighting a room with 68 burners each consuming 2J cu. in. per second, the price of gas being 85 cents per thousand cubic feet? Ans. 27.1 cents. 22. A clock that gains 1 min. in 10 hr. is correct at Monday noon. What is the correct time when the clock registers noon on the following Monday? Ans. 43 min. 14 sec. past 11 A. M. 23. How many feet per second are equivalent to 30 miles per hour? Ans. 44 ft. 24. If sound travels at the rate of 1125 ft. per second, in what time would the report of a gun be heard when fired at a distance of 1.276 miles? Ans. 5.989- sec. 26. A train travels 316 miles in 10 hr. 34 min.; what distance will it travel in 27 hr. 17 min. at the same rate? Ans. 815.9+ mi. 26. A tank holding 7 bbl. has 2 pipes opening from it; one empties out 2 qt. in 5 sec., and the other 17 gal. per minute. How long will it take to empty the tank if both pipes are open? Ans. 9.587 min. 27. If it takes 4 qt. of oats for one feeding for a horse; how many bushels of oats will it take to feed 5 horses one year, giving them 2 feedings per day? Ans. 456} bu. 28. A carload of potatoes has a total weight of 55,600 Ib. The car alone weighs 15,675 Ib. How many bushels of potatoes in the carload if potatoes weigh 60 Ib. per bushel? Ans. 665.41 bu. 29. A tank that holds 25.6 bbl. will hold how many bushels? Ans. 86.62+. 30. A bin that holds 13 bu. will hold how many gallons? Ana. 121.02- gal. WEIGHTS AND MEASURES 61 31. How long would it take a cannon ball going at the rate of 1950 ft. per second to reach the sun, if the sun is distant 93,000,000 miles? Ans. 8 yr. nearly. 32. Supposing the distance travelled by the earth about the sun to be 596,440,000 miles per year, what is the average hourly distance travelled, taking the year to be 365J days? Find the average distance per second. Ans. 19 miles per second, nearly. 33. Find the area in acres of a farm which is represented on paper as a rectangle 3f in. by 10 in. on a scale of A in. to the rod. Ans. 63 acres. 34. If 4 oz. of the white of egg is used in cleansing 50 gal. of wine, how many eggs will it take for 17 bbl. of wine? One egg contains 1.1 oz. of white. Ans. 39 eggs. 35. The total cost of making a cement walk 300 ft. long, 5 ft. wide, and 6 in. thick, where the cement was hand mixed, was as follows: Foreman, 8 hours; laborers, 120 hours at a cost of $53.20; cement, $86.00; gravel, $34.08. Find the total cost per square yard and per square foot. 36. A farmer drew a load of potatoes to market for which he received 76 cents a bushel. If the wagon and load weighed 3710 Ib. and the empty wagon weighed 1150 Ib., find what he received for the potatoes. 60 Ib. of potatoes make 1 bu. 37. How many pounds of charcoal does it take to make 3 tons of gun- powder, if the powder is rS sulphur, f saltpeter, and the rest charcoal? Ans. 900. 38. How many barrels of flour, 196 Ib. each, does it take to run a bakery one week of 7 days if the output is 6000 loaves a day, and there are 9J oz. of flour in each loaf? Ans. 127 bbl. 45| Ib. 39. (a) Find the number of cubic feet in a barrel to the nearest 0.001. (6) Find the number of cubic feet in a bushel to the nearest 0.00001. Ans. 4.211 cu. ft.; 1.24446 cu. ft. 40. A new copper cent weighs 48 grains. How many pounds will $50 in these weigh? Ans. 34 7 Ib. 41. One of the largest diamonds in the world weighs 3025f carats. How many pounds avoirdupois is this, correct to the nearest 0.0001? A carat is 3.168 grains. Ans. 1.3694 Ib. 42. If railroad ties are placed 18 in. apart from center to center, how many miles will 54,320 ties reach? Ans. 1511 mi. 43. How many rails, each 30 ft. in length, are used in laying two railroad tracks from New York to Chicago a distance of 870 mi.? Find the weight of these rails at 90 Ib. per yard. Ans. 612,480 rails; 275,616 tons. 44. Supposing the distance from the earth to the sun to be 91,713,000 miles, and that the sun's light reaches the earth in 8 min. 18 sec., what is the velocity of light per second? Ans. 184, Io3 miles nearly. 45. The pressure of the atmosphere is 14.7 Ib. per square inch. Find the pressure in pounds per square foot. Ans. 2116.8 Ib. 62 PRACTICAL MATHEMATICS 46. A column of water how high will give a pressure of 1 Ib. per square inch? Ana. 2.3 ft. nearly. 47. If the ends of an iron beam, bearing 5 tons at its middle, rest upon stone piers, required the necessary bearing surface of each pier if the stone will support 200 Ib. per square inch of surface. Ans. 25 in. 1 By bearing surface is meant the area of the stone in contact with the beam. 48. One voussoir (or block) of an arch ring presses its neighbor with a force of 50 tons. If the joint has a surface of 5 sq. ft., find the pressure per square inch. Ans. 138.9 Ib. per square inch. 49. Work is done when resistance is overcome. Ic is measured by the product of the force times the distance over which the force acts. As a formula this is w=fXs, where w is the work, / the force, and s the distance. If the force is in pounds and the distance in feet then the work is in foot-pounds. A steam crane lifts a block of granite weighing 2 tons 80 ft. Find the work done in foot-pounds. Ans. 320,000 ft. Ib. 60. How many foot-pounds of work is necessary to pump 100 bbl. of water to a height of 120 ft.? Use 1 bbl. =4.211 cu. ft. Ans. 3,158,203 ft. Ib. 61. How many foot-pounds of work is done in lifting an elevator weighing 3 tons to the top of a building 220 ft. high? If the elevator is raised through this height in 2 minutes, how many foot-pounds of work is done per second? If an engine of one horse-power can do 550 foot- pounds of work per second, an engine of what horse-power will be necessary to lift the elevator to the top in 2 minutes? 62. A man weighing 165 pounds ascends a stairs to a height of 60 ft. in 20 seconds. How many horse-power does he exert? THE METRIC SYSTEM 46. From a study of weights and measures in the United States, it is seen that a legal standard, the troy pound, has been established for the use of the mint; but that beyond that, our weights and measures in ordinary use rest on custom only with indirect legislative recognition. It is seen that the metric weights and measures are made legal by direct legislative permission, and that standards of both systems have been equally furnished by the Government to the several states; that the customary system has been adopted by the Treasury Department for use in the custom-houses, but that the same department has by formal order adopted the metric standards as "fundamental standards" from which measures of the customary system shall be derived. WEIGHTS AND MEASURES 63 Commercial intercourse between nations makes it advisable, if not necessary, to have a uniform system of weights and measures. Such relations cause those countries not already using the metric system to make more and more use of that system. For instance, large orders for locomotives placed in the United States by foreign governments or by corporations in those countries, have made it a matter of good business to carry out the manufacturing in metric units of measure. The European war, beginning in 1914, made it necessary for large manufacturing plants in this country to make considerable use of the metric system. It is stated on good authority that the first six months following the entrance of the United States into the great war advanced the use of the metric system in this country more than had the previous ten years. A few manufacturing companies have for several years quoted sizes of reamers, drills, and other tools in the metric system as well as in the ordinary system. At present this custom is becoming increasingly more general. 47. Measure of length. The meter. The length of the meter was at first determined as one ten-millionth part of the distance from the equator to the north-pole. It was afterward found that there had been a slight error in this determination. At present the meter is the length at 0C. of a certain bar, made of 90 per cent platinum and 10 per cent iridium, called the International Meter, and kept at the International Bureau of Weights and Measures, near Paris. The two copies of the meter which the United States has are made of the same material. One of these is used as the working standard, and the other is kept for comparison. To insure still greater accuracy, these are compared at regular intervals with the International Meter. 48. Legal units. As has been stated the Treasury Depart- ment has determined that the meter shall be the " funda- mental standard" of length. By the act of July, 1866, Congress fixed the relation, 1 meter = 39.37 in. This is the only legal relation between the two systems, and is used in the Office of Standards of Weights and Measures in this country in deriving the inch, foot, yard, etc., from the meter. Determined in this way the customary units are gal. In the Philippine Islands, Porto Rico, and Guam the metric 64 PRACTICAL MATHEMATICS system is in general use, and is the sole legalized system for these islands. 1 49. Measure of surface. There is no fundamental standard of surfaces or areas as there is of the measures of length. But as the measures of areas are based upon the units of length, and as these are standards, the measures of areas may be so considered. 60. Measures of volume. Cubic and capacity measures. In the United States the fundamental standards of volume are: (1) the cubes of the linear units based on the International Meter; (2) the liter, which is the volume of the mass of one kilogram of pure water at its greatest density; (3) the gallon, which is 231 cu. in.; (4) the bushel, which is 2150.42 cu. in. The liter here used is almost exactly 1 cubic decimeter, and the inch is derived from the meter according to the relation, 1 meter = 39.37 in. 61. Measures of mass. The fundamental standard of mass (weight) in the United States is the International Kilo- gram, a cylinder of 90 per cent platinum and 10 per cent iridium, preserved at the International Bureau of Weights and Measures, near Paris. As in the case of the meter, one of the two copies of the kilogram possessed by the United States is used as a working standard, and the other is kept under seal and used only to compare with the working standard from time to time. To insure still greater accuracy, these are compared at regular intervals with the International Kilogram. By act of Congress of July 28, 1866, the pound is derived from the kilogram. The relation established at that time was 1 kilogram = 2.2046 avoirdupois pounds. This relation has since been made more nearly accurate and is 1 kilogram = 15,432.35639 grains, which would change the first relation to 1 kilogram = 2.20462234 Ib. avoirdupois, or 1 Ib. avoirdupois = 453.5924277 grams. This value is the one used by the National Bureau of Standards in Washington. It is thus seen that the avoirdupois pounds, ounces, etc., in common use 1 See Introduction to "Laws Concerning Weights and Measures of the United States," compiled by Louis A. Fisher and Henry D. Huhbard of the Bureau of Standards, Washington, D. C. WEIGHTS AND MEASURES 65 are derived from the kilogram, and so are fixed and definite derived units. The established relation between the troy pound and the avoirdupois pound is 1 troy pound = fM$ avoirdupois pounds. When made, the standard kilogram was supposed to be the exact mass of one cubic decimeter or 1 liter of pure water at the temperature of its greatest density. It has been found that this is not exactly true, but the difference is very slight, the kilogram being about 25 parts in 1,000,000 too heavy. This difference is so very small that it could hardly affect any ordinary problem. 52. Tables and terms used. In the customary system of weights and measures we have about 150 different terms and 50 different numbers, ranging all the way from 2 to 1728. These numbers bear no relation to one another. In the metric system we have only 14 different terms and but a single number, and that is the number 10. In the metric system the different terms used are the following : meter the unit of length, liter the unit of volume, are the unit of area, gram the unit of weight, myria which denotes 10,000, kilo which denotes 1000, hecto which denotes 100, deka which denotes 10, deci which denotes 0.1, centi which denotes 0.01, milli which denotes 0.001. Terms which are sometimes used are: millier which denotes 1,000,000, quintal which denotes 100,000, stere which is 1 cubic meter. To these might be added mikron and mikrogram. If the foregoing terms are carefully fixed in mind the tables are easily formed. t)6 PRACTICAL MATHEMATICS (1) Measures of length. 10 millimeters (mm.) = 1 centimeter (cm.) 10 centimeters 10 decimeters 10 meters 10 dekametei-s 10 hectometers = 1 decimeter (dm.) = 1 meter (.m.) = 1 dekameter (Dm.) = 1 hectometer (.Hm.) = 1 kilometer (Km.) = 0.01 meter = 0.1 meter > 10 meters 100 meters = 1000 meters 10 kilometers = 1 myriameter (Mm.) =10,000 meters (2) Measures of surface. 100 square millimeters (mm. 2 ) =1 sq. centimeter (cm. 1 ) 100 square centimeters =1 sq. decimeter (dm. 2 ) 100 square decimeters =1 sq. meter (m. 2 ) = l centare (ca.) 100 square meters =1 sq. dekameter (Dm. 2 ) = 1 are (a.) 100 square dekameters = 1 sq. hectometer (Hm.*) = 1 hektare (Ha.) 100 square hectometers =1 sq. kilometer (Km.*) (3) Measures for land. 100 centares (ca.) =1 are (a.) 100 ares = 1 hectare (Ha.) (4) Measures of volume. 1000 cubic millimeters (mm. 3 ) = 1 cu. centimeter (cm. 3 or cc.) 1000 cubic centimeters =1 cu. decimeter (dm. 1 ) = 1 liter (1.) 1000 cubic decimeters = 1 cu. meter (m. 3 ) = 1 kiloliter (Kl.) (5) Measures of capacity. 10 milliliters (ml.) = 1 centiliter (cl.) 10 centiliters 10 deciliters 10 liters 10 dekaliters 10 hectoliters = 1 deciliter (dl.) = 1 liter (!.) = ! dm.* = 1 dekaliter (Dl.) = 1 hectoliter (HI.) = 1 kiloliter (Kl.) = lm. (6) Measures of weight. 10 milligrams (mg.) = 1 centigram (eg.) 10 centigrams 10 decigrams 10 grams 10 dekagrams 10 hectograms 10 kilograms 10 myriagrams 10 quintals = 1 decigram (dg.) = 1 gram (g.) = 1 dekagram (Dg.) = 1 hectogram (Hg.) = 1 kilogram or kilo (Kg.) = 1 myriagram (Mg.) = 1 quintal (Q.) = 1 millier, tonneau, or metric ton (T.) To these may be added the following used in scientific work : 1 mikron (M) =0.000001 meter. 1 mikrogram (7) =0.000001 gram. WEIGHTS AND MEASURES 67 Note. A chart showing very clearly the relations of the different measures can be secured by addressing the Bureau of Standards, Washington, D. C. 1 i i 'i 11 ' ' ' ' 2 I 1 ' 3 1 1 1 Inches Centimeters Illl i u 2 Illl 345 illl 6 M 7 llllll Illll v > ill Illl 9 FIG. 14. 63. Equivalents. To be remembered. 1 gallon =231 cubic inches (established by law). 1 bushel =2150.42 cubic inches (established by law). 1 meter =39.37 inches (established by law). 1 gram = 15.432 grains. 1 pound avoirdupois =7000 grains. 1 inch =2.54 centimeters (approximately). 1 kilogram =2.2 pounds (approximately). For Reference. Lengths 1 inch 1 foot 1 kilometer 1 mile 1 sq. in. 1 sq. ft. 1 sq. yd. 1 cm. 2 1 m. 2 1 are 1 acre Areas = 2.54001 cm. = 30.4801 cm. = 3280.83 ft, =0.62137 mi. = 1.60935 Km. = 6.45163 cm. 2 = 0.0929034m. 2 = 0.836131 m. 2 = 0.155 sq. in. = 10.76387- sq. ft. =1.19599- sq. yd. = 119. 5985 sq. yd. = 40.4687 ares. Volumes, capacities 1 cu. in. 1 cu. ft. 1 pt, (liquid) 1 pt. (dry) 1 qt. (liquid) 1 qt. (dry) 1 cm. 3 1 liter 1 liter 1 liter = 16.38716 cc. = 28.317 liters or dm. 3 = 473.179 cc. =0.473179 liters or dm. 3 = 550.614 cc. =0.550614 liters or dm. 3 = 946.358 cc. =0.946358 liters or dm. 3 = 1101.228 cc. =1.101228 liters or dm. 3 = 0.0610234 cu. in. = 61.0234 cu. in. = 2.11336 pt. (liquid) = 1.81616 pt. (dry). = 1.05668 qt. (liquid) =0.90808 qt. (dry). f>8 PRACTICAL MATHEMATICS Weights (maw) 1 grain -0.0647989 gram. 1 ounce (avoirdupois) =28.3495 grams. 1 pound (avoirdupois) =453.5924277 grams - 0.45359 + Kg. 1 ton (short) =907.185 kilograms. 1 gram = 15.43235639 grains. 1 kilogram =2.20462 pounds (avoirdupois). 1 metric ton =2204.62 pounds (avoirdupois). 64. Simplicity of the metric system. The men who devised the metric system endeavored to invent a system of weights and measures that would be as simple as possible; and they undoubtedly succeeded in making a system that is simpler than any other in use. Many look upon the system as difficult because they con- sider the difficulties of changing from the English to the metric system, or from the metric to the English, as difficulties of the metric system. In reality this is not the case, as all such difficulties would disappear if the metric system were in univer- sal use. The simplicity of the metric system lies in the two facts: first, it is decimal, and therefore fits our decimal notation; second, its units for lengths, surfaces, solids, and weights are all dependent on one unit, the meter. Ability to handle the metric system easily, depends, in great part, on understanding thoroughly the terms used. It is of first importance then to learn well these terms and their meanings. For instance, the word decimeter should mean, at once, one-tenth of a meter. Because of the decimal relations between the different terms used, the changing from one unit to another is a very simple matter. In reducing to higher denominations, we divide by 10, 100, 1000, etc., by removing the decimal point to the left. Thus, to change 3768 cm. to meters, we divide by 100 by removing the decimal point two places to the left, and have 3768 cm. =37.68 m. In a similar manner, 72,468 g. =72.468 Kg., and 8643 1-86.43 HI. It should be noticed that we never write 4 Km. 7 Hrn. 3 Dm. 5 m. but write it 4735 m. The former way of writing WEIGHTS AND MEASURES 69 it would be similar to writing $7.265 in the form 7 dollars 2 dimes 6 cents 5 mills. In reducing to lower denominations, the multiplication is performed by moving the decimal point to the right. Thus, 25 m. =250 dm. =25,000 mm. and 16 Kg. =16,000 g. 55. Relations of the units. It cannot be impressed upon the mind of the student too strongly that he should understand clearly the relations between the units of different kinds of measure. He must know that a liter is a cubic decimeter, that a kilogram is the weight of a liter of pure water, that an are is a square dekameter, and so on. He should notice that in the surface measures, when using square meters, dekameters, etc., the scale is 100; while in using cubic meters, dekameters, etc., for volumes, the scale is 1000. Thus, 2m. 2 = 200 dm. 2 = 20,000 cm. 2 and 3m. 3 =3000 dm. 3 =3,000,000 cc. =3,000,000,000 mm. 3 56. Changing from English to metric or from metric to English systems. The changing from one system to another is simply a matter of multiplication or division. (1) Thus, to express 17 m. in inches, 1m. =39.37 in. 17 m. = 39.37 in. X 17 = 669.29 in. (2) Also, to express 2468 Ib. in kilograms, 2.2 Ib. (approx.) = 1 Kg. 2468 Ib. = 2468 -5- 2.2 = 1 121 .8 Kg. Or using the equivalent 1 Ib. = 0.45359 Kg., 2468 Ib. = 0.45359 Kg. X 2468 - 1 1 19.46 Kg. The disagreement in the results is on account of 2.2 Ib. being a rough approximation. The United States Bureau of Standards has compiled numerous tables of equivalents for use in the custom houses. By the use of these tables, a conversion from one system to another, is made by simply referring to the proper table and reading the result. For further information the following pamphlets can be obtained gratis from the Bureau of Standards, Washington, D. C.: History of Standard Weights and Measures of the 70 PRACTICAL MATHEMATICS United States, Table of Equivalents, and the International Metric System of Weights and Measures. EXERCISES 16 1. Express the following, first, in meters, and second, in millimeters: 456 cm., 1763 Dm., 27 Km. Ans. 4.56 m., 17,630 m., 27,000 m., 4560 mm., 17,630,000 mm., 27,000,000 mm. 2. Express the following in m. 2 : 75 cm. 1 , 125 mm. 1 , 0.025 Dm.*, 0.0029 Km. s Ans. 0.0075 m., 0.000125 m. 1 , 2.5 m. 1 , 2900 m. 3. Expres the following in terms of m. 3 : 1756 1., 467 KL, 4937 dl., 0.1067 Dl., 735,432 dm.', 764 Dm. Ans. 1.756 m., 467 m. 1 , 0.4937 m. 3 , 10.67 m. 3 , 735.432 m. 3 , 764,000 m. 4. Reduce 750 1. to liquid quarts; 326 1. to dry quarts; 75 m. to inches; 576 cm. to feet; 27 m. 3 to bushels; 9276 mm. 3 to gallons; 12 Dm. 3 to barrels. Ans. 792.51 qt., 296.03408 qt., 2952.75 in., 18.8976 ft, 766.19 bu., 0.00245 gal., 100,636.19 bbl. Solution. From Art. 47 find 1 1. =1.05668 qt. (liquid). .'. 7501. =1.05668 qt.X 750 = 792. 51 qt. Ans. In the fifth part, 27 m. 3 to bushels, the change is not so easy from the equivalents given. 27 m. 3 =27,000 dm. 3 or liters. 11. =0.90808 qt. (dry). /. 27,000 1. =0.90808 qt. X27,000 = 24,518. 16 qt. Divide this by 32 because 1 bu. =32 qt. .'. 27 m. 3 = 766.19 bu. Ans. 6. Reduce 456 in. to meters; 43.5 ft. to centimeters; 327 gal. to liters; 92.87 qt. (dry) to liters; 756 bu. to cubic meters; Ans. 11.58 m., 1325.88 cm., 1237.84 1., 102.27 1., 26.64 m. 3 6. No. 16 gage sheet steel is -? 6 in. thick and weighs 40 oz. per square foot. Find thickness in millimeters (four decimal places), and the weight per square meter in kilograms (two decimal places). Ans. 1.5875 mm., 12.21 -Kg. per m. 2 Solution. To find the weight in Kg. per m. 2 , first find the weight of a square meter in ounces and then change to pounds and to kilograms. 1 m. 2 = 10.76387ft. 2 .'. 1 m. 2 weighs 40 oz. X 10. 76387 =430.5548 oz. =26.9097 Ib. 1 Ib. =0.45359 Kg. .-.26.9097 Ib. =0.45359 Kg. X26.9097 = 12.20597 Kg. Ans. 7. No. 24 gage sheet steel is 0.635 mm. thick and weighs 4.882 Kg. per m. 2 Find thickness in decimal of inch (three decimal places), and weight per ft. 2 in ounces. Ans. 0.025 in., 16 oz. per ft. 2 8. Find the difference between 313 in. and 10 cm. Ans. 3U in. larger by 0.0005 in. 9. Feb. 12, 1912, Oscar Mathieson, of Norway, set a new world's record in ice skating. He made 1500 m. in 2 min. 20 sec. This is a mile in what time? WEIGHTS AND MEASURES 71 Solution. 2 min. 20 sec. = 140 sec. 1500X39.37,^ 1500 m. = j ft. \2t 1500X39.37 n = number of feet m 1 sec. 1^ X 14U 5280X12X140 1500X39.37 150.2 sec. =2 min. 30.2 sec. Ans. = 150.2 = number of sec. to go 1 mi. 10. The same day in Chicago, Harry Kaad won the mile race in 3 min. 23! sec. This is 1500 m. in what time? Ans. 3 min. 9.6 sec. 11. In describing the making of reenforced concrete the necessary pressure is given as 25 kilos per square centimeter. How many pounds is this per square inch? Ans. 355.58+. 12. Find in kilograms the weight of air in a room 10.5 by 8.3 by 4 meters, air being 0.001276 times as heavy as water. Ans. 444.8136 Kg. 13. Find the weight in kilograms of the mercury in a tube of 1 cm. 2 cross section and 760 mm. long, mercury being 13.596 times as heavy as water. Ans. 1.0333- Kg. 14. If a map is made on a scale of 1 to 60,000, how many kilometers do 79 mm. on the map represent? Ans. 4.74. 15. If a person in breathing uses 0.25 m. 3 of air a minute, how long will it take 6 persons to use the air in a room 6 m. long, 3.5 m. high, and 5.3 m. wide? Ans. 74.2 min. 16. A block of stone weighs 7643 Kg. A cubic decimeter of the stone weighs 2.7 Kg. Find the volume of the block in cubic meters. Ans. 2.83074 m. 3 17. Find the area in hectares of a triangular field whose base is 70 m. and the altitude 60 m. Ans. 0.21 Ha. 18. How many liters of water are contained in a reservoir 10 m. X6 m. X4 m.? What is the weight of the water in kilograms? Ana. 240,000 1., 240,000 Kg. 19. Find the capacity in liters of a rectangular tank 2 m. X9 dm. X8 dm. Ans. 1440 1. 20. What is the length of a centigram of wire 255 mm. of which weighs 0.172g.? Ans. 14.83- mm. 21. A liter of mercury weighs 13.596 Kg.; how many mm. 3 of mercury weigh 1 g.? Ans. 73.551. 22. A man's height is 174 cm. What is his height in feet and inches? Ans. 5 ft. 8.5+ in. 23. Express the following readings in centimeters: 29.9 in., 30.0 in., 30.1 in., 30.2 in. Ans. 75.946 cm., 76.200 cm., 76.454 cm., 76.708 cm. 24. Express the following in inches *o the nearest 0.01: 71.119 cm., 73.659 cm., 74.929 cm. Ans. 28.00 in., 29.00 in., 29.50 in. 72 PRACTICAL MATHEMATICS 26. Cast copper being 8.8 times as heavy aa an equal volume of water. what is the weight of 5 cm. J ? An*. 44 g. 26. A velocity of 32.2 ft. per second is how many centimeters per second? An*. 981.5 -. 27. A rate of 1 mile in 2 min. 6 sec. is how many kilometers per minute? How many meters per second? Ans. 0.7664, 12.77+. 28. A rate of 30 miles per hour is at the rate of one kilometer in how many minutes? An*. 1.243 . 29. A pressure of 14.7 Ib. per square inch is how many grams per cm. 1 ? Solution. 7= = number of Ib. per sq. cm. O.451DO 14.7X453.5924 .-,.._ - = 1033.5+ = number of g. per sq. cm. D. 451 Go CHAPTER VI PERCENTAGE AND APPLICATIONS 57. Per cents as fractions. To one who thoroughly understands fractions, percentage offers no new difficulties. The words per cent mean by the hundred. The symbol % means per cent. Thus, 10% means 10 per cent or iVV or 0.10. In a similar way: 5% =0.05 =.} a 50% =0.50 =\ 10% =0.1 = 1 I 60% =0.60 -$ 12J% =0.12* =| 62i% = 0.62^ =f 16|% =0.16 = 75% =0.75 =f 20% =0.20 = !,- 80% =0.80 25% =0.25 =1 83^% =0.83^ = 1} 33^% =0.33^ =J 87i% =0.87| =| 37i% =0.37* =| 90% =0.90 =A 40% =0.40 =| To change a fraction as | to an equivalent form in per cent, reduce it to a fraction having- 100 for a denominator. Thus, 1 = T Y in percentage. rate product corresponds to percentage Case I corresponds to: multiplicand and multiplier given to find the product. Product = multiplicand X multiplier. Case II corresponds to: multiplicand and product given to find the multiplier. Multiplier = product -f- multiplicand. Case III corresponds to: multiplier and product given to find the multiplicand. Multiplicand = product -f- multiplier. 69. Rules and formulas. In the language of percentage these become: Case I. Percentage = base X rate. This may be written as a formula if b stands for base, p for percentage, and r for rate. The formula is p = b X r. Case II. Rate = percentage -=- base. The formula is r = p -f- 6. Case III. Base = percentage -r- rate. The formula is 6 = p -* r. PERCENTAGE AND APPLICATIONS 75 60. Solutions. Problem (1) is solved thus: By fractions. 37|% of 720 = f of 720 = 270. Ans. By formula. Using the formula p= bX r, gives the same result, for then p = 720 X 0.37 = 270. Ans. Problem (2). By fractions. 45 is what per cent of 450 means 45 is how many hundredths of 450, that is, some number of hundredths of 450 is 45. Then 45 is ^ = yV = iVir = 10% of 450. By formula, r = p -4- 6 gives r = 45 -f- 450 = 0.1 = 10%. Problem (3). By fractions. 85 is 62^% of what number is the same as 85 is f of what number. It is now a simple problem in fractions and may be reasoned thus : If 85 is f of some number then 17 is f of that number, and 136 is f of that number. Hence the number = 136. Ans. By formula, b = p -f- r gives p = 85 -H 0.62^ = 136. Ans. EXERCISES 17 Solve the following exercises without a pencil if possible. 1. What is J of 24? 0.25 of 24? 25% of 24? 2. What is f of 36? 0.75 of 36? 75% of 36? 3. What is $ of 45? 0.33$ of 45? 33$% of 45? 4. What is f of 48? 0.66f of 48? 66f% of 48? 6. What is l of 60? 0.20 of 60? 20% of 60? 6. What is f of 70? 0.80 of 70? 80% of 70? 7. What is \ of 72? 0.12* of 72? 12$ % of 72? 8. What is f of 48? 0.625 of 48? 62.5% of 48? 9. What is ^ of 64? 0.06i of 64? 6i% of 64? 10. What is 25% of 16? of 48? of 90? of 240? 11. What is 33$% of 75? of 42? of 96? of 720? 12. What is 4% of 25? of 75? of 300? of 1000? 13. What is 7% of 20? of 14? of 55? of 300? 14. 5 is what per cent of 10? of 20? of 40? 15. 8 is what per cent of 16? of 40? of 80? 16. 30 is what per cent of 90? of 240? of 360? 17. What % is 8 of 150? 7$ of 12? Ans. 5$%; 62$%. 18. What % is f of 12$? 27i of 600? Ans. 4*%; 4|%. 19. 20% of what number is 3? 7? 14? 17? 20. 33$% of what number is 7? 8? 14? 90? 21. 62$% of what number is 5? 20? f? H? 22. 37$% of a number is 72; find the number. Any. 192. 23. 20% off of what number leaves 48? Ans. 60. Suggestion. 20% off a number leaves 80% of the number. 76 PRACTICAL MATHEMATICS 24. 30% off of what number leaves 60? Ant. 71$. 25. 33 J% off of what number leaves 12? Ant. 18. 26. 68 is 15% less than what number? Ans. 80. 27. 49 is 30% less than what number? Ans. 70. 2Q. 18 is 80% more than what number? Ans. 10. 29. 80 is 33 j% more than what number? Ans. 60. 30. 98 is 40% more than what number? Ans. 70. 31. 87 J is 37i% less than what number? Ans. 140. 32. If oranges that cost 25 cents a dozen are sold at 3 for 10 cents what part of the cost is gained? What per cent? 33. Pencils are bought at 15 cents a dozen and sold foi 2 cents each. What part of the cost is gained? What per cent? Ant. 60%. 34. Bought a horse for $75 and sold it for $100; find the gain per cent. Ans. 33$%. 35. Find the gain per cent in each case if a horse was bought at the following prices: $50, $40, $25, $10, $5, and $1 ; and sold for $100. Find the gain per cent if the horse was given to the seller. 36. I bought a bicycle for $100 and after using it one year sold it for $55. Find the per cent of discount. Ans. 45%. 37. A gas bill was 25% higher last month than this. If it is $6.40 this month, what was it last month? 38. A horse was bought for $100 and sold for $90. What was the loss per cent? 39. A man sold a suit of clothes gaining i the cost. What part of the cost was the selling price? What was the gain per cent? What per cent of the selling price was the cost? 40. A quantity of wool was bought for $360, and f of it was then sold for the cost of the whole. \Vhat per cent would have been gained if tho entire amount had been sold at the same rate? Ans. 33J%. 41. A man spent 16% of his salary for board and room. If he spent $6.50 a week for board and room, what was his yearly salary? (52 weeks per year.) Ans. $2028. 61. Applications. Example 1. The population of a certain city in 1900 was 52,600, and in 1905 was 61,805. Find the gain in population in the 5 years. What was the gain for each 100 of the population during the 5 years? State this increase as a per cent of the population in 1900. What was the average per cent of increase per year? Discussion. 61,805 - 52,600 = 9205 = gain in 5 years. 9205 -^526 =17. 5 = gain per 100 of the population. Since, asking for the per cent of gain is the same as asking for the number of increase for each 100 of the population, therefore, stated in per cent this is 17.5 % of the population. 1 7.5 % -J- 5 3.5% = average per cent of increase per year, based on the population in 1900. PERCENTAGE AND APPLICATIONS 77 Example 2. In a certain machine f of the energy supplied to the machine is lost in friction and other resistances. What is the per cent of efficiency? If f of the loss of energy is in a certain part of the machine, what per cent of the total loss is in this part? Discussion. If in a machine it is known that -f of the energy expended is wasted in frictional and other resistances, we say that 40% is wasted, meaning that T 4 Ans. If the numerator and denominator are not each a perfect square, reduce the fraction to a decimal and then extract the square root as in Art. 81. Example 2. Find \/f Reducing to a decimal, |- = 0.28571428 . V0.28571428 = 0.5345. Hence, \/f = 0.5345 to four decimal places. It is worth noting here that the square root of -f may be found by extracting the square root of both numerator and denominator, and then dividing the square root of the num- erator by the square root of he denominator. This process would require two extractions of roots and one long division, and so make the work about three times what it is if the fraction is first reduced to a decimal and then the root extracted. 84. Short methods. Partly division. If it is required to extract the square root of a number to, say, five decimal 104 PRACTICAL MATHEMATICS places, making, say, seven figures in the root, the work may be shortened by extracting the root in the usual way till four figures are obtained, and then dividing the last remainder found by the corresponding trial divisor to obtain the last three figures of the root. In general, extract root till more than half the required number of figures are found, and then for the other figures of the root divide the remainder by the corresponding trial divisor. Process. 1 78. WOOWOO( 13.34 166 Ans. Example 1. Find \/178 to five decimal places. 23 263 78 69_ 900 789 2664 11100 10656 2668) The process may be contracted still further by using con- tracted division when dividing. 444000(166 2668 17720 16008 17120 16008 1112 Method by factoring. When the number of which the square root is to be extracted can be factored into two factors, one of which is a perfect square and the other the number 2, 3, 5, 6, or 7, a very useful short method may be obtained. For this purpose it is necessary first to have found the following square roots: A/2 = 1.4142, A/3 = 1.73205, A/5 = 2.23607, A/6 = 2.4494, \/7 = 2.6457. Of these the most useful are the roots of 2 and 3. Example 2. Find the A/32. 32= 16X2, so we may write A/32 =\/T6X A/2 = 4 XI. 4 142 = 5.6568. Ans. Example 3. Find \/125. A/1 25 = A/25 X A/5 = 5X2.236 = 11.180. Ans. POWERS AND ROOTS 105 86. Rule for square root. After carefully following through the solutions of the preceding examples, the following rule should be understood: RULE. (1) Begin at the decimal point and point off the whole number part and the decimal part into periods of two figures each. If there is an odd number of figures in the whole number part, the left-hand period will have only one figure. If there is an odd number of figures in the decimal part, annex a cipher so that the right-hand period shall contain two figures. (2) Find the greatest square in the left-hand period and place it under that period. The square root of this greatest square is the first figure of the required root. Subtract the greatest square from the left-hand period and bring down and unite with the remainder the next period of the number. This is the first remainder. (3) Take twice the root already found for a trial divisor, which write at the left of the remainder. Find how many times this trial divisor is contained in the remainder omitting the right-hand figure. This gives the next figure of the root, which place in the root and also at the right of the trial divisor, forming the true divisor. Multiply the true divisor by the figure last placed in the root and write the product under the remainder. Subtract and bring down and unite the next period in the number. This pro- cess is repeated for each figure of the root. (4) // at any time the trial divisor will not be contained in the corresponding remainder, place a cipher in the root and at the right of the trial divisor, bring down another period, and continue as before. (5) Point off in the root as many decimal figures as there are decimal periods in the number of which the root is extracted. 86, Cube root. The extraction of cube root is so seldom used that it is thought best to omit the usual consideration of it. It is found in a very simple manner by the use of loga- rithms, by which means any one root is as easily found as another. (See Art. 313.) EXERCISES 27 Find the square root of the following : 1. 516,961. Ans. 719. 2. 23,804,641. Ans. 4879. 1 1 If, I'UACTH'AL MA THEM A TICS 3. 0.3364. 4. 0.120409. 6. 1159.4025. 6. 2 to four decimals. 7. 786,432 to two decimals. 8. 7,326,456 to two decimals. 9. 3 to five decimal places. 10. 5 to three decimal places. 11. 6 to four decimal places. 12. 7 to five decimal places. 13. Wr 14. 27 -f- 156.25 to four decimals. Ans. 0.58. Ana. 0.347. Ann. 34.05. Ans. 1.4142. Ans. 886.81. ,4n*. 2706.74. Ans. 1.73205. Ans. 2.236. Ans. 2.4495. Ans. 2.64575. Ans. T V Ans. 0.4157. Suggestion. First perform the division, and then extract the root of the quotient. 15. I to four decimal places. Ans. 0.8819. In each of the exercises from 16 to 23, carry the root to five decimal places : 16. 143. Ans. 11.95826 20. 287. Ans. 16.94107. 17. 164. Ans. 12.80624. 21. 396. Ana. 19.89975. 18. 92. Ans. 9.59166. 22. 416. Ans. 20.39608. 19. 278. Ans. 16.67333. 23. 539. Ans. 23.21637. 24. Find the square roots of the following by short methods: (a) 28, (6) 72, (c) 288, (d) 75, (e) 147, (/) 192, (g) 432. Ans. (a) 5.2915, (6) 8.4852, (c) 16.971, (d) 8.6603, (e) 12.1244, if) 13.8564, (g) 20.7846. Fio. 22. Similar figures. 87. Similar figures. The following principles are useful in solving many problems: (1) The areas of similar figures are in the name ratio a,s the squares of their like dimensions. (2) The volumes of similar solids are in the same ratio as the cubes of their like dimensions. Similar figures are such as have the same shape. In Fig. 22 the following pairs are similar: (a) and (6); (c) and (d); (e) and (/); (g) and (h). POWERS AND ROOTS 107 EXERCISES 28 1. If the diameter of (a) is 6 in. and of (b) is 4 in., how many times as large as (b) is (a) ? Solution. Area of (a): area of (b) =6 2 :4 2 =36: 16 =2- Ans. 2. Find the ratio of areas of (e) to (/), if the shorter side of (e) is 9 ft. and of (/) 5 ft, Ans. 3.24. 3. If a round steel rod | in. in diameter, hanging vertically, will sup- port 12,000 lb., what will a rod | in. in diameter support? Ans. 36,750 lb. 4. Given that the electrical resistance is inversely in the same ratio as the areas of the cross sections of the conductors of the same material; find the ratio of the resistances of two copper wires of diameters | in. and 3 in. respectively. Ans. 64:9. 6. Two steam boilers of the same shape are respectively 12 ft. and 18 ft. long. Find the ratio of their surfaces. Ans. 4:9. 6. How many times as much gold leaf will it take to cover a ball 10 in. in diameter than to cover a ball 6 in. in diameter? Ans. 2 f 7. Two balls of steel are respectively 7 in. and 15 in. in diameter. The second is how many times as heavy as the first? Ans. 9.84 . 8. Which is the cheaper, oranges 2J in. in diameter at 30 cents a dozen or 3-in. oranges at 40 cents a dozen? What should the larger ones sell at to give the same value for the money as the smaller at 30 cents a dozen? Ans. The 3-in. oranges; 52 cents a dozen nearly. Suggestion. The price the 3-in. oranges should sell at is given by the proportion: (2|) 3 :3 3 =30:x. 9. Two balls of the same material are 10 in. and 3 in. in diameter respectively. If the smaller ball weighs 9 lb. what is the weight of the larger? Ans. 333^ lb. 10. The formula V = \/2gh gives the velocity V in feet per second a body will have after falling from a height h. Find the value of V for a stone that has fallen 400 ft. In the formula 0=32.2. Ans. 160.5 ft. nearly. Suggestion. As in this exercise, the evaluation of a formula often re- quires the extraction of a square root. The numbers that the letters stand for are put in place of the letters and we have V = A/2X32.2X400 = V25760 = 160.5 - . 11. The effective area of a chimney is given by the formula where ' = the effective area, and A=the actual area of the flue. Find the effective area if A =86 sq. in. If A =3.14 sq. ft. Ans. 85.44 sq. in.; 3.03 sq. ft. 12. When the pressure of water at the place of discharge is known, the rate of flow is given by the formula 108 PRACTICAL MATHEMATICS where F=velocity of discharge in feet per second, and P=pres8ure in pounds per square inch at the place of discharge. Find the rate of dis- charge if the pressure as given by a pressure gage is 50 Ib. per square inch. Ans. 85.98 ft. per second. IS. As in the last, find the velocity of discharge if the pressure is 200 Ib. per square inch. Compare the result with that of the preceding. Ans. 171.97 ft. per second PART TWO GEOMETRY CHAPTER X PLANE SURFACES. LINES AND ANGLES 88. In this and the following chapters are discussed some of the facts established in geometry, and some of their appli- cations to practical problems. The endeavor is to illustrate and make clear the principles and thus lay a broad foundation, rather than to follow narrow special lines. Many special problems, however, are given. From these the individual student can select those that are suited to his needs. There are many terms which, although quite familiar to the student, are used in geometry with such exactness as to require a careful definition or explanation. Point, line, angle, surface, and solid are such terms. Like all simple terms, such as number, space, and time, they are difficult to define; but it is hoped the explanations given will lead to a reasonable understanding of them. 89. Definitions. A material body, as, for example, a block of wood or an apple, occupies a definite portion of space. In geometry no attention is given to the substance of which the body is composed. It may be iron, stone, wood, or air, or it may be a vacuum. Geometry only considers the space occupied by the substance. This space is called a geometric solid or simply a solid. If one thinks of a brick, and then considers the brick removed and thinks of the space that the brick occupied, he has an illustration of a geometric solid. A solid has length, breadth, and thickness. A boundary face of a solid is called a surface. A surface has length and breadth but no thickness. 109 110 I'K. \ (' TIC A L MA THEM A TICK The boundary of a surface, or that which separates one part of a surface from an adjoining part, is called a line. A line has length only. That which separates one part of a line from an adjoining part is called a point. A point has neither length, breadth, nor thickness. It has position only. A point is read by naming the letter placed upon it. A line is read by naming the letters placed at its ends, or by naming B D Fio. 23. the single letter placed upon it. Capital letters are usually used at the ends of a line, while a small letter is placed upon a line. In Fig. 23(a), the line is read " the line AB" or simply "the line a." A straight line is a line having the same direction through- out its whole extent. See Fig. 23(a). A curved line is a line that is continually changing in direc- tion. See Fig. 23(6). A broken line is a line made up of connected straight lines. See Fig. 23 (c). Plane Surface Curved Surface Fin. 24. Parallel Lin*. If a surface is such that any two points in it can be con- nected by a straight line lying wholly in the surface, it is called a plane surface or simply a plane. A carpenter determines whether or not the surface of a PLANE SURFACES. LINES AND ANGLES board is a plane by laying the edge of his square or other straightedge on the surface in different positions, and observ- ing if the straightedge touches the surface at all points. A curved surface is a surface no part of which is a plane surface. Thus, the surface of a circular pipe and the surface of a ball are curved surfaces. Parallel lines are lines in the same plane and everywhere the same distance apart. In Fig. 24 are shown pairs of parallel lines. 90. Angles. Two straight lines which meet at a point form an angle. The idea of what an angle is, being a simple one, is hard to define. One should guard against thinking of the point where the two lines meet as the angle. This point is called the vertex of the angle. The two lines are called the sides of the angle. The difference in the direc- tions of the two lines forming the angle is the magnitude of the angle. For a further discussion of an angle see Art. 317. An angle is read by naming the letter at the vertex, or by naming the letters at the vertex and at the ends of the sides. FIG. 25. s 'I B 0) 3 ,-, Horizontal D Line 1 11111 i llllJ_ii_LLJJJ_LJ_l_[_r n P FIG. 26. When read in the latter way, the letter at the vertex must always come between the other two. Thus, the angle in Fig. 25 is read "the angle b," "the angle ABC," or ''the angle at B." If one straight line meets another so as to form equal angles, the angles are right angles, and the lines are perpendicular to each other. In Fig. 26 (a\ lines AB and CD are perpendicular to each other. 112 PRACTICAL MA THEM A TICK A vertical line or a plumb line is the line along which a string hangs when suspended at one end and weighted at the other. A horizontal line is a line that is perpendicular to a vertical line. Fig. 26(6). If a right angle is divided into 90 equal parts, each part is called a degree. It is usually written 1. An acute angle is an angle that is less than a right angle. An obtuse angle is an angle that is greater than a right angle and less than two right angles. See Fig. 26(c). D C E B Complementary Ansrles EB Supplementary Angles FIG. 27 Two angles whose sum is one right angle, or 90, are called complementary angles, and either one is said to be the com- plement of the other. Two angles whose sum is two right angles, or 180, are called supplementary angles, and either one is said to be the supplement of the other. SURFACES 91. Polygons. A polygon is a plane surface bounded by any number of straight lines. Any one of these lines is called D B FUJ. L'S. :i side. The point where two sides meet is called a vertex. The distance measured around the polygon, or the sum of the lengths of the sides, is called the perimeter of the polygon. PLANE SURFACES. LINES AND ANGLES 113 A triangle is a polygon having three sides. A quadrilateral is a polygon having four sides. A pentagon is a polygon having five sides. A hexagon is a polygon having six sides. An octagon is a polygon having eight sides. A regular polygon is one whose sides are all equal and whose angles are all equal. A diagonal is a line joining any two vertices not adjacent in a polygon. 92. Concerning triangles. A line drawn from any vertex of a triangle perpendicular to the opposite side and ending in it is called an altitude of the triangle. Since a triangle has three vertices, each triangle has three altitudes. The altitude ^ D FIG. 29. FIG. 30. may meet the opposite side, as CF in triangle ABC, Fig. 29; or the opposite side may have to be extended to meet it, as AD and BE, Fig. 29. A line drawn from any vertex of a triangle to the center of the opposite side is called a median. It is evident that in any triangle there are three medians. In Fig. 30, AD is a median. A line drawn through the vertex of an angle and dividing the angle into two equal parts is called the bisector of the angle. The bisector of an angle of a triangle is often taken as the length of the bisector of an angle of the triangle from the vertex to the opposite side. BE in Fig. 30 is the bisector of the angle ABC of the triangle. 1 1 4 I'K A CTICAL MA THEM A TICK It is evident that there are three bisectors of the angles in any triangle. 93. Concerning quadrilaterals. A parallelogram is a quad- rilateral whose opposite sides are parallel. See Fig. 31 (a). A rectangle is a parallelogram whose angles are right angles. See Fig. 31(6). A square is a rectangle whose sides are all equal. See Fig. 31(c). FIG. 31. A trapezoid is a quadrilateral with only two sides parallel. The parallel sides are called the bases. The altitude is the distance between the two bases. Fig. 31 (d) is a trapezoid; AB and DC are the bases, and EF is the altitude. The forms just discussed are very important, as any figure bounded by straight lines may be thought of as composed of rectangles and triangles. EXERCISES 29 In the following exercises use a ruler and a hard lead pencil. Letter all figures. 1. Draw two curved lines. Two broken lines. 2. Draw several parallel lines. 3. Draw a right angle. An acute angle. An obtuse angle. 4. Draw perpendicular lines. If two lines are perpendicular to each other is one of them vertical? Illustrate by a drawing. 6. Draw vertical and horizontal lines. Is a vertical line always perpendicular (o a horizontal line? 6. Estimate the size as nearly as you can and draw an angle of 45. Of 30. Of 60. Of 120. Of 135. Of 180. Which are acute angles? Which obtuse angles? 7. Draw two complementary angles. Two supplementary angles. 8. Draw a triangle. A quadrilateral. A pentagon. A hexagon. An octagon. A regular hexagon. 9. How many diagonals have each of the polygons of exercise 8? 10. What are the vertices of each polygon of exercise 8? What are the perimeters? PLANE SURFACES. LINES AND ANGLES 115 B 11. Draw a triangle having all its angles acute, and draw its three altitudes. 12. Draw a triangle having all its angles acute, and draw its three medians. Draw the three bisectors of its angles. 13. Draw triangles each having one obtuse angle, and follow the direc- tions of exercises 11 and 12. 14. Draw a rectangle. A square. A parallelogram. A trapezoid. A quadrilateral that is not any of these. Draw their altitudes. 15. Name objects in nature, or objects made by man that are of the forms asked for in the preceding exercises. AREAS OF POLYGONS 94. The rectangle. How to find the area of a rectangle is illustrated in Fig. 32. Suppose that this represents a rectangle whose length AD is 5 ft., and width AB is 4 ft. The rectangle is divided into small squares 1 ft. on a side, and so each represents 1 sq. ft. Since there are 4 rows of squares each containing 5 sq. ft., there are 4X5 sq. ft. =20 sq. ft. in the rectan- gle. What is said will also be true if . the lengths of the sides are frac- tional. This leads to the following: RULE. The area of a rectangle is equal to the product of its length and its width. Remark. The length and the width of the rectangle must be in the same unit before taking their product. The product is then square units of the same kind as the linear units. ( Thus, if the unit of length is the foot, the product will be square feet. 95. The parallelogram. A paral- lelogram and a rectangle, each hav- ing the same base and altitude, are equal in area. This is illustrated in Fig. 33. A BCD is the rectangle and ABEF is the parallelogram. The altitude BC is the same for each, and they have the same base, AB. Since the part BCE of the parallelogram may be cut off and fitted on ADF, it is evident that the parallelogram is just equal to the rec- tangle. Therefore, we have the following: D FIG. 32. E FIG. 33. 1 1 (i I'KA CTICAL MA THEM A TICK RULE. The area of a parallelogram is equal to the product of its base and its altitude. 96. Formulas. A rule stated in letters and signs is called a formula. It is a shorthand way of stating a rule. If A is used as an abbreviation for area, b for base, and a for altitude, the rule for the area of a rectangle or a paral- lelogram is given in the following formula: [I] A = ab. The form ab means altitude times base. Since the altitude times the base equals the area, by using well-known principles of division we have for the rectangle or parallelogram the following: RULE. (1) The altitude equals the area divided by the base. (2) The base equals the area divided by the altitude. These rules written as formulas are : [2] a = A-^b, [3] b = A^a. 97. The triangle. If a triangle and a parallelogram have the same base and have their altitudes equal, the triangle has half the area of the parallelogram. This is illustrated in Fig. 34. ABCD is the parallelogram. Q The diagonal BD divides it into two tri- angles ABD and BCD, which are equal. From this and the rule for the area of a parallelogram, it is clear that the fol- lowing is true : RULE. The area of any triangle is equal to one-half of its base times its altitude. If the area and either base or altitude of a triangle are given, the other dimension (altitude or base) is found by dividing twice the area by the given dimension. If A stands for the area, a for the altitude, and 6 for the base, we have these formulas for the triangle : [4] A = [5] a = 2A-^-b, [6] b=2A-a, PLANE SURFACES. LINES AND ANGLES 117 EXERCISES 30 1. Compute the areas of the following figures using the dimensions as given. 22 rd. FIG. 35. 2. If the sides only of a parallelogram are given can its area be found? 3. Draw two triangles and find their areas by drawing the three alti- tudes of each and measuring the sides and altitudes. 98. Area of a triangle when the three sides only are given. If a, b, and c stand for the three sides of a triangle; and if s stands for one-half the sum of a, b, and c, then the area A of the triangle is given by the formula: [7] A = a) (s-b) (s-c). This formula cannot well be derived here, but it is found in geometry. The area of the triangle can also be found by constructing it to scale, as explained later. The altitude can then be measured and the area be found by taking one-half the product of the base and the altitude. Since a formula is a rule stated in symbols, the above formula may be stated as the following rule for the area of a triangle when the three sides only are given : 118 PRACTICAL MA THEM A TICS HULK. Fiiul half the sum of the three sides. Subtract each side from this half sum. Take the continued product of the half fnim and the three differences. The square root of this product z.s the area of the triangle. This rule can be illustrated best by an example. Example. Find the area of a triangle with sides 40 rd., 28 rd., and 36 rd. Solution, a = 40, 6 = 28, c = 36. s = i(40+28+36) = 52. s-a = 52-40 = 12. s-6 = 52-28 = 24. s-c = 52-36 = 16. A = \/52X 12X24X16 = \/239,616 = 489.506. /.area = 489. 506- rd. 2 Ans. With very ordinary instruments this triangle can be con- structed to scale and measured, and the area found to within half a square rod of the computed area. 99. Area of trapezoid. A diagonal of a trapezoid divides it into two triangles which have the same altitude, and have as bases the two bases of the trape- zoid. Thus, in the trapezoid of Fig. 38, the diagonal AD divides the trapezoid into two triangles A CD and ADE. The area of A CD = $ of ACXa and area of ADE = \ of EDXa'. But a = o', hence the sum of the areas of the two triangles = $ (AC +ED)X a. Now the area of the trapezoid can evidently be found by finding the sum of the areas of the two triangles into which B Fir;. PLANE SURFACES. LINES AND ANGLES 119 it is divided; or what amounts to the same thing, by the fol- lowing: RULE. The area of a trapezoid equals one-half the sum of the two bases times the altitude. If B and b stand for the two bases and a for the altitude of the trapezoid, the formula is [8] A=i(B+b)Xa. Example. Find the area of a trapezoid whose lower base is 20 rd., upper base 14 rd., and altitude 9 rd. Solution. By formula [8], A = %(B-\-b)a. Putting the num- bers of the example in place of the letters of the formula, .'. area = 153 sq. rd. Ans. EXERCISES 31 1. Find the parts not given in the following exercises which refer to parallelograms: (1) Base 22| in. altitude 19 in. area . (2) Base - altitude 47 rd. area 426 rd. 2 (3) Base 33^ ft. altitude - - area 433f ft. 2 (4) Base altitude 102 in. area 9367 in. 2 Ans. (1) 427| in. 2 ; (2) 9* 3 7 rd.; (3) 13^ ft.; (4) 9U{ft in. 2. Find the number of acres in a farm 160 rd. long and 80 rd. wide. Ans. 80. 3. Find the number of square feet in a floor 16 ft. 8 in. by 13 ft. 6 in. Ans. 225. 4. Find the number of square meters in a rectangle 77 m. long and 5 Dm. wide. Ans. 3850. 5. A box 6 in. long, 4 in. wide, and 3 in. deep has six rectangular faces. Find the area of the surface of the box. Ans. 108 in. 2 6. Find the area of a triangle whose base is 25 ft. and whose altitude is 12 ft. 4 in. Ans. 154 ft. 2 7. How many acres are there in a triangular lot whose base is 432 ft. and altitude 320 ft.? Ans. 1.59-. 8. Find the number of hectares in a triangular field whose base is 196.8 m. and altitude 85 m. Ans. 0.8364. 9. Find the base of a triangle whose area is 20 acres and altitude 80 rd. Ans. 80 rd. 10. A rectangular field 48 rd. long contains 9 acres. Find the width. Ans. 30 rd. 11. If the perimeter of a rectangle is 96 ft. and the length is three times the breadth, find the area. Ans. 432 ft.'- 120 PRACTICAL MATHEMATICS 12. A rectangular garden 56 ft. long and 40 ft. wide has a path 6 ft. wide around it. Find the area of the path. Ans. 1296 ft. 1 13. A box of tin sheets for roofing, containing 112 sheets 14 in. by 20 in., will cover 170 ft. 1 What per cent of surface covered is allowed for joints and waste? Solution. Without allowing for joints and waste each box would 14X20X112 cover- ~i4l = 2175 sq.ft. 217$ sq. ft. 170 sq. ft. = 47J sq. ft. = allowance for joints and waste. 47J sq. ft. +170 sq. ft. =0.28+ =28 + %. 14. How many bricks each 9 in. by 4J in. by 1} in. will it take to pave a court 16 ft. by 18 ft., if bricks are laid flat? If laid on edge? Ans. flat 1024; edge 2634. 16. How many paving blocks each 4 in. by 4 in. by 10 in., placed on their sides, will it take to pave an alley 600 ft. long and 12 ft. 6 in. wide? Ans. 27,000. 16. What will be the expense of painting the walls and ceiling of a room 12 ft. 6 in. by 16 ft. and 10 ft. 4 in. high at 15 cents per square yard? Ans. $13.15. 17. Find the cost of sodding a lawn 31 ft. wide and 52 ft. long at 18 cents per square yard. Ana. $32.24. 18. Find the number of square feet in the floor of the room shown in Fig. 39. Ans. 277i ft. 1 Suggestion. Divide into rectangles and trapezoids. JT IG 3Q_ 19. At 15 cents per square foot, find the cost of building a cement walk 6 ft. wide, on two sides of a corner lot 33 ft. by 100 ft. Ans. $125.10. 20. Find the area of a trapezoid whose bases are 17 in. and 11 in. respectively and whose altitude is 13 in. Ans. 182 in. 1 21. Find the area of a triangle whose sides are 13 in., 15 in., and 21 in. Ans. 96.79- in.* 22. Find the area of a triangle whose sides are 54 in., 32 in., and 22 in. Ans. in. 1 23. Find the area of a triangle whose base is 27 in. and altitude 14 in. Ans. 189in. 24. Find the area of a board 14 ft. long and 18 in. wide at one end and 12 in. at the other. Ans. 17.5 ft. 1 26. Find the cost of painting both sides of a solid board fence 260 ft. long and 6 ft. high at 60 cents a square. How many gallons of paint will it take for two coats if 1 gallon will cover 250 sq. ft. two coats? (1 square = 100 sq. ft.) Ans. $18.72; 12$ gal. nearly. 26. How much did it cost to harvest a field of wheat 156 rd. by 76 rd., PLANE SURFACES. LINES AND ANGLES 121 if cutting and binding cost $1.50 per acre, setting up 25 cents an acre, and hauling $1.25 an acre? Ans. $222.30. 27. Find the area in acres of a farm which is represented on paper as a rectangle 3f in. by 102 in. on a scale of ?$ i n - to the rod. Ans. 63 A. 28. Find the area of Fig. 40(o). Ans. 23.592 in. 2 29. Find the area of Fig. 40(6). Ans. 6.02 in. 2 30. Find the area of Fig. 40(c). Ans. 8.625 in. 2 .S.40 FIG. 40. 31. Find the area of the footing for a column with a load of 168,000 Ib. if the safe bearing load of the soil is 4000 Ib. per square foot. Ans. 42 sq. ft. 32. How many square yards of plastering will be required for the four side walls of a hall 90 ft. long, 50 ft. wide, and 20 ft. high, with 4 doors 5J ft. by 10 ft., 14 windows 5 ft. by 11 ft., and a baseboard 9 in. high around the room? Find the cost at 40 cents per square yard. Find the contractor's profit at 20%. LUMBER 100. Measuring lumber. Lumber is measured in board measure. Timber used in framework is counted as lumber. Lumber and timber are sold by the 1000 ft. board measure. This is sometimes written 1000 ft. B.M., but more often it is indicated by the single letter M. 122 PRACTICAL MATHEMATICS One board foot is 12 in. square and 1 in. thick, and so con- tains one-twelfth of a cubic foot. The number of board feet in a stick of timber is the number of cubic feet times 12. The following rule may be used to find the number of board feet in a stick of timber: RULE. Take the product of the end dimensions in inches, divide by 12, and multiply the quotient by the length in feet. The student should make clear to himself the correctness of this rule. Example. Find the number of board feet in a stick of timber 6 in. by 8 in. and 14 ft. long. Solution. xl4 = 56ft. B.M. Ans. Lumber less than 1 in. is counted as if 1 in. thick in buying and selling. In widths a fraction of \ in. or more is counted as 1 in. Usually lumber is cut in lengths containing an even number of feet, as 12, 14, and 16 ft. Longer lengths than these are usually special, but classifications vary greatly. There are sixteen or more associations in America with specifications governing the classification of lumber, and these specifications differ more or less. Timber work is usually paid for at an agreed price per M, the timber to be measured in the work. 101. Estimations. There are various rules regarding the estimating of the amount of lumber required in a structure. In general, all that is necessary is to find the number of board feet in the lumber required and add a certain per cent for waste in cutting, matching, etc. Regardng this, the student can consult a handbook specially prepared for those in this line of work. 102. Shingles. Shingles are 16 in. or 18 in. in length, are counted as 4 in. wide, and put up in bunches of 250. The part of the shingle that is exposed when laid is said to be "laid to the weather." The part so exposed varies from 4 in. to 6 in. So a single shingle covers a space 4 in. wide and from 4 in. to 6 in. long. In laying shingles, the estimating is often made by the square, an area 10 ft. by 10 ft. or containing 100 sq. ft. PLANE SURFACES. LINES AND ANGLES 123 In stating the number of shingles, give the number so that only whole bunches will be required. Thus, do not give a number as 5550 but as 5750. The following table allows for waste and gives the number of square feet covered by a thousand shingles, and also the number of shingles required to cover a square, when laid at various distances to the weather. Inches to the Area covered by 1000 No. to cover weather shingles a square 4 100 sq. ft 1000 4i 110 sq. ft 910 4) 120 sq. ft 833 5 133 sq. ft 752 5| 145 sq. ft 690 6 157 sq. ft 637 EXERCISES 32 1. Find the number of feet of lumber it will take to build a tight board fence 5| ft. high and 70 ft. long, boards 1 in. thick and nailed at top and bottom to pieces of 2 in. by 4 in. stuff. (No waste allowed.) Ans. 478. 2. Find cost of lumber at $32.00 per M to build a walk 30 ft. long and 8 ft. wide; plank to be 2 in. thick and laid crosswise on 4 pieces of 4 in. by 4 in., running lengthwise. Ans. $20.48. 3. Find the amount of lumber to floor a room 30 ft. by 40 ft. with strips 3 in. wide, allowing for matching and 15% for waste. 4. Find how many shingles it will take to shingle a roof 36 ft. by 40 ft. if shingles are laid 4^ in. to the weather. (Use the table of Art. 102.) Solution. ~Tfj?r~ = 14.4 = number of squares. 833 XI 4. 4 = 11,995 = number of shingles required. /. 12,000 shingles must be bought. 6. How many board feet in 26 pieces of 2 in. by 4 in. by 14 ft. long, 20 pieces of 3 in. by 10 in. by 16ft. long? Ans. 1043. 6. What will it cost at $28 per M to cover the floor of a barn 32 ft. by 42 ft. with 2-in. plank? Ans. $75.26. 7. How many board feet are there in 3 sticks of timber 12 in. by 14 in. and 22 ft. long? Ans. 924. 8. Find the total cost of shingling the two sides of a roof each 18 ft. by 40 ft. Redwood shingles at $4.75 a thousand are used, and the laying, nails, etc., cost $1.90 per square. Shingles are to be laid 5 in. to the weather. (Use the table of Art. 102.) Ans. $79.61. 124 PR A CTICAL MA THEM A TICS 9. What docs the following cost at 25 cents a foot: 1 piece | in. by 6 in. by 10 ft. I piece } in. by 8 in. by 12 ft. 1 piece I in. by 18 in. by 4 ft. 2 pieces 1 in. by 6 in. by 8 ft ? Ant. IG.75. 10. Find the cost of the following bill of lumber if the quarter sawed is $90 per M and the common sawed is $65 per M : 2 pieces li in. X2J in. X12 ft. quarter sawed 2 pieces J in. X 8 in. X 12 ft. quarter sawed 1 piece | in. X2J in. X12 ft. quarter sawed 1 piece f in. X 2 in. X 12 ft. quarter sawed 5 pieces in. X 3 in. X 12 ft. quarter sawed 1 piece I in. XlO in. X 12 ft. quarter sawed 1 piece \ in. X 10 in. X 6 ft. common sawed 6 pieces i in. X 6 in.X 12 ft. common sawed 4 pieces i in. X 6 in. X12 ft. common sawed. An*. $0.18. Length 00 ; Roof extending '2 'at each end Fio. 41. 11. Fig. 41 is the end of a barn. Find the area of one end. Find the area of the roof. The rafters are placed 16 in. from center to center. Find the number of board feet in the rafters if made of 2 in. X 6 in. (Use 12-ft. stuff for short rafters.) Find number of feet of lumber to cover ends, sides and roof. Find how many shingles it will take for the roof if laid 4J in. to the weather. Ans. 1310 ft. 1 ; 3434J ft. 1 ; 2744; 8935; 28,750. 12. A ship builder gave $300 for a standing oak tree to make a long ship timber. The cost of felling, hewing, and hauling was $275. If the timber was 18 in. square and 98 ft. long, find the number of board feet in it and the cost per thousand feet. Ans. 2646; $217.31. PLANE SURFACES. LINES AND ANGLES 125 13. Find the number of board feet in the following list of framing tim- ber for a house: Girders 5 pieces 6 in. X 8 in. X20 ft. Sills 16 pieces 6 in. X 6 in. X 16 ft. First floor beams 45 pieces 3 in. XlO in. X28 ft. Second floor beams 45 pieces 3 in. X 8 in. X28 ft. Ribbons 16 pieces 1 in. X 8 in. X20 ft. Plates 32 pieces 2 in. X 4 in. X 16 ft. Outside wall studs 156 pieces 2 in. X 4 in. X20 ft. Inside wall studs 200 pieces 2 in. X 4 in. X 12 ft. Rafter studs 90 pieces 2 in. X 8 in. X24 ft. Collar beams 45 pieces 2 in. X 6 in. X 16 ft. Ans. 14,673. CHAPTER XI TRIANGLES THE RIGHT TRIANGLE 103. A right triangle is a triangle having one right angle. The side opposite the right angle is called the hypotenuse, and the sides about the right angle are called base and altitude, the base being the side the triangle is supposed to rest upon. The right triangle is of great importance as it is of very com- mon occurrence in practice. The solution of the right triangle depends upon the following relation es- tablished in geometry. The square formed on the hypotenuse is equal to the sum of the squares formed on the other two sides. This may be illustrated as in Fig. 42. AC is the hypotenuse and is 5 units in length. AB is the base, 4 units long. BC is the altitude, 3 units long. Here it is easily seen that the square on AC is equal to the sum of the squares on AB and BC. Hence AC 2 = AB 2 +BC 2 , or in general, if c stands for the hypotenuse, 6 for base, and a for altitude, then c 2 = a 2 +6 2 . From this are derived the three following formulas, by which any side can be found if the other two are known. Fio. 42. Example. Find the hypotenuse of a right triangle whose base is 14 ft. and altitude 16 ft. Solution. Using formula [9], c = \/a 2 +& 2 . = Vl6 2 +14 2 =\/452 = 21.26+ ft. Ana. 126 TRIANGLES 127 EXERCISES 33 In the following right triangles, solve for the parts named in the exer- cise: 1. a = 25, 6 = 16, find c and area. Ans. c = 29.68; area =200 square units. 2. c = 46, 6=30, find a and area. Ans. a = 34. 87 + ; area = 523. 05+ square units. 3. Area =2 acres, a = 15 rd. ; find b and c. Ans. b = 42.667- rd. ; c= 45.23- rd. 4. a = 16, c = 20, find b and area. Ans. b 12; area =96 square units. 5. Find length of the diagonal of a rectangle 16 ft. by 14 ft. Ans. 21.26ft. 6. Find the diagonal of a cube 9 ft. on an edge. Suggestion. In Fig. 44, the line marked D is of the cube. First find d and then D. Ans. 15.588+ ft. called the diagonal d=\/9 2 +9 2 =Vl62. D=\/162+9 2 =\/243. = A/8lX3=9\/3. 7. A man swims at right angles to the bank of a stream at the rate of 3.5 miles per hour. If the current is 7.5 miles per hour, find the rate the man is moving. Ans. 8.28 miles per hour. Suggestion. The rate the man is moving is the length of the hypote- nuse of a right triangle having a base =3. 5 mi. and an altitude = 7. 5 mi. 8. The diagonal of a rectangle is 130 and the altitude is 32. Find the area. Ans. 4032 square units. 9. What is the length of the longest line that can be drawn within a rectangular box 12 ft. by 4 ft. by 3 ft.? Ans. 13 ft. 10. The hypotenuse of a right triangle, with base and altitude equal, is 12 ft. Find the length of the base and altitude. Ans. 8.485+ ft. 11. The base of a triangle is 20 ft. and the altitude is 18 ft. What is the side of a square having the same area? Ans. 13.416+ ft. 12. The area of a rectangular lawn is 5525 m. 8 , and the length of one of its sides is 8.5 Dm. Find the length of its diagonal in meters to three decimal places. Ans. 107.005 m. 128 PRACTICAL MA Til KM A TICK 18. A steamer goes due north at the rate of 15 miles per hour, and another due west at 18 miles per hour. If both start from the same place, how far apart will they be in 6 hours? Ana. 140.58 + miles. 14. What is the length of the diagonal of a room 20 ft. by 16 ft. by 12 ft.? Ann. 28.284+ ft. 15. Find cost at $20 per M of roof boards on a third-pitch roof of a barn 45 ft. by 65 ft., if projections at ends and eaves are 2 ft. (In a third-pitch roof the distance of the ridge above the plate is one-third the width of the building.) Ans. $80.15. Suggestion. Distance of ridge above plate = J of 45 ft. = 15 ft. Length of rafters without projection = V22.5* + 15* = 27.04 ft. Total length of rafters =2 ft. +27.04 ft. =29.04 ft. Area of one side of roof = 29.04X69 = 2003.76 sq. ft. 16. How many thousand shingles will it take to cover the above roof, if shingles are laid 4J in. to the weather and a double row is put at the beginning on each side? (No allowance for waste.) Ans. 32,500 nearly. 17. In fitting a steam pipe to the form ABCD, Fig. 45, making a bend of 45, the fitter takes BC = CE + &CE. What is the error if CE = 18 in.? What is the correct length of CB, and what is the per cent of error by the fitter's method? Ans. Error, 0.0442- ; correct, C# = 25.4558 + in.; % of error, 0.17 + . 18. In cutting a rafter for a half-pitch roof a carpenter makes the length of the rafter AB = 1 ft. 5 in. for every foot there is in AC, Fig. 46. If AC =8 ft., find A B by this rule. What is the per cent of error by this method? Ans. Carpenter's method, 11 ft. 4 in.; correct, 11 ft. 3.76+ in.; error, 0.17+ %. 19. To find the diagonal of a square, multiply the side by 10, take away 1 % of this product, and divide the remainder by 7. Test the accuracy of this rule. Ans. 0.006 % too large. Solution. Take a square with a side of, say, 25 in. 10X25 =250 1% of 250 = 2.5 Remainder = 247.5 247.5 -J- 7 = 35.357 + in. - diagonally jrule. By formula for hypotenuse, diagonal = v/25*+ 25* = 35.355 + in. Hence error -35.357 in. -35.355 in. -0.002 in. 0.002 4- 35.355 = 0.006- % =per cent of error. TRIANGLES 129 It is evident that this rule is very accurate and is also easy of applica- tion. 20. Show that the following rules are correct. They are very useful in many problems connected with a square. (1) The diagonal of a square equals a side T of the square multiplied by \/2. (2) The side of a square equals one-half the diagonal multiplied by \/2. The number of decimal places used in \/2 will depend upon the degree of accuracy desired. Pipe fitters usually use \/2 = 1.41. It is often necessary to take three or more decimal places. \/2 = 1.4142136 to seven decimal places. 21. Use rule (1) in obtaining the correct values in exercises 17, 18, and 19. 22. What is the distance across the corners of a square nut that is 3s in. on a side? Use rule (1). FIG. 48. FIG. 49. Cap screw. 23. What must be the diameter of round stock so that a square bolt head If in. on a side may be milled from it? 24. Find the distance across the flats of the square head of a cap screw that may be milled from round stock 1| in. in diameter. Use rule (2) of exercise 20. Plan Section FIG. 50. 26. Fig. 50 shows a "scissors" roof truss with the lengths AB = B( = AC = 30 ft., CD = CF = 1S ft,, and CG = CE=8 ft, Find the lengths of NC and FG. Ans. NC = 25 ft. 11 f in. ; FG = 13 ft. 10 in. 130 J'KA CTICAL MA THEM A TICS 26. A smokestack is held in position by three guy wires that reach the ground 49 ft. from the foot of the stack. Find the length of a guy wire if they are fastened to the stack 70 ft. from the ground. Ans. 85.4+ ft. 27. An engine shaft is centered 9 ft. below and 3 ft. to the left of the center of a line shaft. Find the distance between the cen- ters of the two shafts. Ans. 9 ft. 5 j in. 28. The dimensional sketch, Fig. 51, shows plan and section of a roof. It has to be boarded. What will be the number of feet of boards required? Ans. 356. 29. In the gambrel roof shown in section in Fig. 52, find the lengths of rafters and parts not given. DB = 15 ft. 21 in.; C = 14 ft. 3f in. All to Fio. 52. Ans. AB = 15 ft. 71 in.; DB the nearest | in. SIMILAR TRIANGLES 104. Triangles that have the same shape are said to be similar. In Fig. 53 (a), ABC and ADE are similar. In (b) ABC and A'B'C' arc similar. Draw two triangles as in (a) and measure the sides a, b, c and a', b', c'. Then determine the ratios a :b, arc, b : c, a 1 : b', a' : c' and 6' : c'. Follow the same directions for the triangles in (6). Now compare the values of the ratios and notice whether or not they are equal. The results of the above should lead to the following: a : b = a' : b'; a : c = a' : c'; and b : c = b' : c'. TRIANGLES 131 Two triangles that have the angles of one equal respectively to the angles of the other are smilar. The sides about the equal angles in the similar triangles are called corresponding sides. Thus, c and c' are corresponding sides. Other corresponding sides are a and a', and b and &'. From the proportions given above, we arrive at the fol- lowing principle: Corresponding sides of similar triangles form a proportion. Example. To find the distance between the points P and Q on opposite banks of a stream, Fig. 54, where P is inaccessible. FIG. 54. Solution. As shown in the figure, measure distances AQ = 16 ft., AB = 10 ft., and BC = 60 ft. Because triangles AQB and APC are made similar we have the proportion AB :AQ = AC : AP .MO ft. : 16 ft. = 70 ft. :AP . . .AP = 16X70 I10 ,, = 112 ft. .'.PQ= 112 ft.- 16 ft. =96 ft. Ans. 105. Tapers. The man in the machine shop often finds it necessary to determine the taper per foot of a piece that is to be turned, in order that he may set his lathe properly. By the taper per foot is meant the decrease in diameter if the piece is 1 ft. long. In Fig. 55 (a), the taper is evidently 4 in. 3 in. = 1^ in, per foot. In (6) the taper is 2\ in. 2 in. \ in. in 4 in. Hence the taper per foot is 3 times as much or \\ in. 132 PRACTICAL MATHEMATICS If I stands for the length of tapered part in feet, t for the taper in inches in this part, and T for the taper in inches per foot, then the following proportion is true by similar triangles: The taper for the total length of the piece is evidently the taper per foot times the length in feet. Example. In Fig. 55 (c), what is the taper per foot? What would be the taper for total length of piece? Solution. Substituting in [12], T \ : 1 =\ : T. ..r=lXi-5- T 8 7 =l T in. Ans. 1J in.Xvf = 2.3 in. = taper if it were tapered the full length. r< 10 K 8 --- >j<-4->i Fro. 55. 106. Turning. In turning a piece in a lathe the taper is sometimes made by shifting the tailstock of the lathe. Since, when the piece is revolved, the same cut is made on all sides, it is necessary to set the tailstock over one-half of what the taper would be if the piece were tapered the full length. Thus, a piece 1 ft. long with a taper lj in. per foot requires the tail- stock to be set over \ of \\ in. = | in. If Z) = the large diameter and d the small diameter of the tapered portion, L the total length of the piece, and / the length of the tapered portion, then the offset x of the tail- stock is determined by the following formula: Example. A shaft 3 ft. long is to have a taper turned on one end 10 in. long, the large end of the taper being 4 in. in diameter and the small end 3$ in. Find the distance to set over the tailstock. Solution. x = -*X = 0.9in. Ans. TRIANGLES 133 EXERCISES 34 1. The following tapers per inch are what tapers per foot: 0.0026 in.; 0.0473 in.; 0.0379 in.? 2. How much will the tailstock need to be set over to give a taper of 1| in. per foot if the work is 1 ft. in length? If 8 in. in length? Am. & in.; f in. 3. The standard pipe thread taper is f in. per foot. How much is this per inch? Ans. -^ in. 4. Find the taper per foot to be used in turning a pulley with a 14 in. face crowned j% in. Ans. 0.32+ in. 5. If the crowning of a pulley is ->\ of the width of the face, find the taper per foot to be used in turning a pulley with a 10-in. face. Ans. I in. Taper pin. -Taper J in. per foot. Taper-pin reamer. Taper \ in. per foot. FIG. 56. 6. A taper-pin reamer has a taper of J- in. per foot. If the diameter of the small end is 0.398 in. and the length of the flutes is 5j in., find the diameter of the large end of the flutes. 7. A taper reamer has a taper of f in. per foot and the flutes are 3 in. long. If it is | in. in diameter at the large end, find the diameter at the small end. Ans. 0.5677 in. 8. Find the taper per foot of a taper reamer that has a diameter of f? in. at the large end of the flutes, and ff in. at the small end if the flutes are 2f in. long. Ans. f in. 9. A taper reamer has a taper of f in. per foot. If the diameter of the large end is 1^, find the diameter of the small end, the flutes being 3f in. long. Ans. fi in. 10. In Fig. 57, the timbers CB, DF, and EG are perpendicular to CA. From the given dimensions find the lengths of DF, EG, CF, DG, and AB. Ans. EG = 4 ft.; DG = 10.770 ft.; AB = 32.311 ft. Suggestion. CA :CB = DA: DF. Or 30: 12 = 20 : D/* 1 . 30 CF = v/io'^XS 1 = Vl64 = 12.800 134 PRACTICAL MA THEM A TICS 11. How much should the tailstock be offset to turn a taper on a piece of work 10 in. long, if the tapered portion is 4 J in. long and measures 1.275 in. in diameter at the large end and 0.845 in. at the small end? Ana. 0.5212 in. 12. What is the offset of the tail center for turning a taper 18 in. long on a bar 26 in. long, if the diameters at the ends of the taper are 3} in. and 2} in.? Ana. 0.27 +in. 10' D WE Fiu. 57. 13. To find the distance across the lake in Fig. 58, measure AB = 80 rd. ; AD =30 rd. ; DE = 25 rd., and find BC. Is it necessary to make right-angled triangles? Ans. 66 rd.; no. 14. Show how to find the height of a smokestack CD of Fig. 59, when the foot of the stack cannot be reached. Suggestion. On a level place measure from A to B in a line with C, and measure angles BAD and CBD. Suppose the line AB is 40 ft. and the angles are 40 and 60 respectively. Construct a figure to scale on paper and measure the line that corresponds to the smokestack. THE STEEL SQUARE 107. One of the most useful instruments known to man is the ordinary steel square or carpenter's square shown in Fig. 60. It is made in various sizes but the most common size is with the longer arm, called the body, blade, or stock, TRIANGLES 135 24 in. in length and 2 in. in breadth, and the shorter arm, called the tongue, 16 in. or 18 in. in length and 1 in. in breadth. Many books, having in some cases five or six hundred pages, have been written on the uses of the steel square. Here we wish only to call attention to the fact that the principles involved in using the steel square are mainly those involved in the solution of the right triangle and in similar triangles. One who understands the right triangle can devise many uses FIG. 60. FIG. 61. for the steel square, and can readily see the principles under- lying the various uses of this instrument given in the treatises on the steel square. Upon the ordinary steel square are found many figures, telling lengths of braces, board measures, etc. No attempt will be made here to explain these. Example. By use of a steel square, find the length of the hypotenuse of a right triangle that has a base of 8 in. and an altitude of 7 in. Solution. Measure the line drawn from the 8-in. mark on the blade to the 7-in. mark on the tongue. This measures about lOf in. which is near enough for most practical purposes. By the right triangle method the hypotenuse = -\/8~ 2 +V 2 = 10.63 +in. (See Fig. 61.) This method can readily be applied to find the lengths of braces supporting two pieces that are perpendicular to each other; to find rafter lengths, lengths of the parts of a trestle, etc. 108. Rafters and roofs. The run of a rafter is the distance measured on the horizontal from its lower end to a point under 130 PRACTICAL MATHEMATICS its upper end. The rise is the distance of the upper end above the lower end. In Fig. 62, AC is the run and CB the rise. The slant of a roof is usually told by stating the relation of the rise to the run. It is often given by stating the rise per foot of run; as 6 in. to 1 ft. Another way is to state what is known as the pitch of the roof. A roof is said to be half pitch, D quarter pitch, full pitch, etc., when the rise is \, }, 1, etc., times the full width of the building as represented in Fig. 62, where AD is the width of the building. The relation between rise and pitch is shown by the following table: 4 ft. rise is pitch. 6 ft. rise is \ pitch. 8 ft. rise is \ pitch. 10 ft. rise is VV pitch. 12 ft. rise is \ pitch. 15 ft. rise is f pitch. 18 ft. rise is f pitch. 12 ft. run to 12 ft. run to 12 ft. run to 12 ft. run to 12 ft. run to 12 ft. run to 12 ft. run to 12 ft. run to 24 ft. rise is full pitch. 109. Uses of the square. -The bevel or slant on the end of a brace or rafter, necessary to make it fit the part it rests against, can easily be marked by the square. LL Example \. Required to cut the lower end of a rafter that is to rest on the plate if the rise of the rafter is 8 ft. and the run 12 ft. Discussion. Place the square as shown in Fig. 63 and mark along the lower edge. This gives the proper slant. Marking along the tongue gives the slant for the upper end of the rafter. FKI. 03. TRIANdLEti 137 In placing the square on the stick it is only necessary to take the distances on the blade and tongue in the same ratio as the ratio of the run to the rise. In this case we could as well have taken 24 in. and 16 in. or 9 in. and 6 in. Example 2. Required to cut a rafter for a V-shaped roof on a building 12 ft. wide if the rise of the rafter is to be 4 ft. The rafter is to be made of a piece of 2 in. by 4 in. and half its width is to project 18 in. beyond the plate. FIG. 04. Discussion. Determine the slant for the plate end as de- scribed in example 1. Then place the square as shown in Fig. 64 so as to give a run of 24 in. to a rise of 16 in. The square is replaced with point A on C and this repeated as often as necessary to give a run of half the width of the building. In this case, it is necessary to place the square three times. In the last position, a mark along the tongue gives the slant of the upper end of the rafter and determines the length of the rafter. Any rafter can be cut in this way. EXERCISES 35 1. Show with a carpenter's square how to determine the length of a brace for a run of 4 ft. 6 in., and a rise of 3 ft. 6 in. Show how to cut bevels on ends. 2. Show how to mark the slants for the legs of the sawhorse shown in Fig. 65. Compute the length of legs. Ans. 29| in. nearly. 3. In Fig. 66 is a plan of one end of a roof on a house 18 ft. in width and 28 ft. long. CB, DE, etc., are common rafters; AB and NB are hip rafters; and FO, HI, etc., are jack rafters. Find the lengths to cut the several rafters if the roof is pitch. Show how to determine the slants 138 PRACTICAL MATHEMATICS at both ends of each. The rafters do not extend beyond the plates and are placed 1 ft. 6 in. from center to center. 4. Find lengths of the common, hip, and jack rafters for the roof of which Fig. 67 is the plan. It is J pitch, rafters 1 ft. 6 in. from center to center, and extend 2 ft. beyond the plates. Fio. 69. 6. Fig. 68 is a plan of the roof of a hexagonal tower. Find lengths of the rafters that end at the plates. Full pitch and rafters 1 ft. 6 in. between centers. Width of tower is 12 ft. TRIANGLES 139 6. Fig. 69 represents the plan of a roof of a house 20 ft. square, with a flat circular portion 8 ft. in diameter. If the circle is one-third the width of the building above the plates and the rafters 2 ft. between centers on the plates, find the length of each rafter and show how to cut slants. ISOSCELES AND EQUILATERAL TRIANGLES 110. Two other forms of triangles of common occurrence are isosceles and equilateral triangles. A triangle which has two equal sides is called an isosceles triangle. A triangle which has all of its sides equal is called an equi- lateral triangle. The following facts are proved in geometry. The student should satisfy himself that they are true by constructing the figures and measuring the parts. 111. The isosceles triangle. In Fig. 70 the isosceles triangle ABC has equal sides AC and BC. The angles A and B opposite the equal sides are equal. The line CD drawn bisecting the vertex angle, C, is per- pendicular to and bisects the base, AB. That is, AD = DB. It also divides the isosceles triangle into two equal right trian- gles, BDC and ADC. The line CD is then the bisector of the angle C, and also a median and an altitude of the triangle. The diagonal of a square divides the square into two equal right isosceles triangles. In these isosceles triangles each of the equal angles is 45. 112. The equilateral triangle. In Fig. 71, triangle ABC has its three sides equal and is an equilateral triangle. 140 PRACTICAL MA THEM A TICS The angles opposite the equal sides are equal, and therefore each angle equals 60. The line drawn from the vertex A and bisecting the angle is perpendicular to and bisects the opposite side BC. It also divides the equilateral triangle into two equal right triangles, ABD and ADC. Furthermore, each of the lines BE and CF divides the trian- gle in the same manner that it is divided by AD. Each of these lines then is a bisector of an angle, and also a median and an altitude of the equilateral triangle. E D FIG. 71. The point where these three lines meet is called the center of the equilateral triangle. It is one-third the distance from one side to the opposite vertex. That is, DO = \DA, It follows then that AO = 2DO, BO = 2EO, and CO = 2FO. Either of the triangles formed when an equilateral triangle is divided into two triangles by an altitude is a right triangle having acute angles of 30 and 60. This triangle is very important in practical work. It is readily seen that, in such a right triangle, the hypotenuse is twice the shortest side. That is, NR = 2MR. An altitude of an equilateral triangle equal* one-half of a side times \/3. As a formula, a = $s\/3 = i.sX1.732, where a = an altitude and s = a side. This is obtained by solving the right triangle AFC for FC. The student can easily carry through the work for any par- ticular value of a side. The following is the derivation in general and involves some algebra. Let s stand for one side of the equilateral triangle. Then /'( ' = VJC*-JF* = vV-($6>) 2 TRIANGLES 141 From this it follows that: A side of an equilateral triangle equals twice the altitude divided by \/3. As a formula, s = 2a-r- \/3- Since the area of any triangle is one-half its base times its altitude, it follows that : The area of an equilateral triangle equals the square of one-half a side times \/3. As a formula, A = (|s) 2 -v/3 = (^s) 2 X 1.732, where A is the area. 113. The regular hexagon. Another form often used in practice is the regular hexagon. From geometry we learn the following facts which will ap- pear true from a careful considera- tion of Fig. 72. The diagonals, drawn as shown, divide the hexa- gon into six equal equilateral triangles. The distance from the center to any vertex is the same as the length of a side. The area of the regular hexagon is equal to six times the area of an equilateral triangle with sides equal to the sides of the hexagon. The altitude NO may be found by solving the right triangle ANO; or may be found by taking AN times \/3. EXERCISES 36 1. Find the altitude of an isosceles triangle whose equal sides are 16 ft. and base 14 ft. Ans. 14.387+ ft. 2. Find the base of an isosceles triangle if equal sides are 18 in. and altitude 16.5 in. Ans. 14.387+ in. 3. Find the altitude of an equilateral triangle the sides of which are 12ft. Ans. 10.392+ ft. 4. In Fig. 72, find ON if AB is 10 ft. Ans. 8.660+ ft. 5. Find the area of an equilateral triangle 18 in. on a side. Ans. 140.3- in. 2 6. Find the area of an isosceles triangle whose base is 6 vn. and equal sides 9 in. . Ans. 25.456 in. 2 7. Compute the area of a regular hexagon one of whose sides is 5 ft. Am. 64.95+ ft. 2 8. An equilateral triangle has an area of 21.217 in. 2 What is the length of one side? Ans. 7 in. \\'2 PRACTICAL MATHEMATICS Solution. Ou page 141 it is stated that the area of an equilateral triangle equals the square of one-half a side times \ :;. Or A -da) 1 XI. 732. Then J *! .732. But A is given equal to 21.217. ' i -\/21.217 -5-1.732 -Vl2i25 -3.5. Or 8 = 2X3.5 = 7. 9. An isosceles triangle has a base 16 in. long and the equal sides 18 in. What is the area? Ans. 128.996+ in.' 10. Find the length of the steam pipe ABCD, Fig. 73, if AD = 6 ft., C = 16 in., and angle EEC = 30. Ans. 76.29- in. B E FIG. 73. Suggestion. ^ BC = 2EC = 2 X 16 in. = 32 in. " BE = \/3XEC = 1.732X16 = 27.71+ in. Increase along #C = 32 in. -27.71+ in. =4.29- in. Length of ABCD = 72 in. +4.29- in. =76.29- in. 11. The hypotenuse of a right triangle is 5 ft. and one side is 4 ft. Show that the equilateral triangle made on the hypotenuse is equal to the sum of the equilateral triangles made on the other two sides. Suggestion. Find the areas of the three triangles and show that the sum of the areas of the two smaller triangles equals the largest triangle. Use the formula A = (J) 2 X 1.732. Isosceles Right Triangle 80-60 Right Triangle Fio. 74. 12. Two very convenient forms of triangles used by draftsmen are right triangles made of celluloid or rubber; one a right isosceles triangle having the acute angles each 45, and one having acute angles of 30 and 60. If one of the equal sides of the isosceles right triangle is 6 in., find the hypotenuse. Ans. 8.485 in. 13. If the shortest side of the 30-60-right triangle is 4 in. find the other sides. Ans. 6.928 in. and 8 in. TRIANGLES 143 14. Using the draftsman's triangles, show how to construct the follow- ing angles: 15. 75, 105, 120, 135, 150. 15. A hexagonal nut for a Q-in. screw is lj in. across the flats. Find the diagonal, or the distance across the corners, of such a nut. Ans. 1.443 in. 16. What is the distance across the corners of a hexagonal nut thai is f in. on a side? What is the distance across the flats of the same nut? Ans. 1.5 in.; 1.299 in. 17. Show that the distance across the corners of a hexagonal nut is approximately 1.15 times the distance across the flats. Suggestion. It is readily seen that the distance across the corners is twice the side of an equilateral triangle whose altitude is one-half the distance across the flats. By Art. 112, s=2a +V1[. /. 2s = 4a^v^3 =2aXl.l5. But 2s = distance across corners, and 2a = distance across flats. 18. In a standard hexagonal bolt nut the distance across the flats is given by the formula F = 1.5Z) + |, where F is the distance across the flats and D the diameter of the body of the bolt. Find the distance across the flats and across the corners on a hexagonal nut for a bolt If in. in diameter. Ans. 2.75 in.; 3.1754 in. 19. To what diameter should a piece of stock be turned so that it may be milled to a hexagon and be If in. across the flats? Ans. 2.0207 in. 20. Find the size of round stock to make bolts with hexagonal heads and having bodies of the following diameters: f in., 1J in., li in., 1^ in. Ans. 1.0104 in.; 2.7424 in.; 2.0929 in.; 2.4177 in. Suggestion. Distance ocross flats, ^ = 1.50+1. When D = | in., F = 1.5X0.5+0.125 = 0.875 in. 0.875X2 Diameter of stock = distance across corners = ~7^= =1.0104 in. By the rule of exercise 17, distance across corners = 0.875 XL 15 = 1.006 in. SCREW THREADS 114. In the United States there are in use several different kinds of screw threads. Here we will consider: first, the 1 I! PRACTICAL MA THEM A TICS sharp V '-thread, or common V -thread; and second, the United States standard screw thread. Other kinds will be considered on page 437. 115. Sharp V-thread. The sharp V-thread, or common V-thread, is a thread having its sides at an angle of 60 to each other and perfectly sharp at the top and bottom. The objections urged against this thread are, first, that the top, being so sharp, is injured by the slightest accident; and second, that in the use of taps and dies the fine sharp edge is quickly lost, thus causing constant variation in fitting. V-Thread FIG. 76. FIG. 77. The common V-thread with a pitch of 1 in. has a depth equal to the altitude of an equilateral triangle 1 in. on a side. Hence its depth = %\/3 = 0.866 in. The root diameter equals the diameter of the bolt less twice the depth of the thread. We have then the following formulas: , fiAA 0.866 a = 0.866p = > [14] Dj = D-2d = D- 1.732 where p = pitch, rf = depth of thread, N = number of threads to the inch, D = diameter of bolt, and Di = root diameter. 116. United States standard thread. The United States standard screw thread (U. S. S.) has its sides also at an angle of 60 to each other but has its top cut off to the extent of $ its depth and the same amount filled in at its bottom, thus making the depth f that of the common V-thread of the same pitch. The distance / on the flat is J- its pitch. TRIANGLES 145 This thread is not so easily injured, the taps and dies retain their size longer, and bolts and screws having this thread are stronger and have a better appearance. ILS.Staniiard FIG. 78. For the U. S. S. screw thread we have the following formulas : _^ ~N' 0.6495 N 1.299 [15] D! = D-2d = D- The blank nut for a bolt is drilled with a tap drill the same size as the root diameter of the screw or bolt, or very nearly that size. Table V can be used in computations connected with screw threads. EXERCISES 37 1. Find the depths of common V-threads of the following number of threads to an inch: 10, 20, 5, 8, 40, 13. Ans. 0.0866 in.; 0.0433 in.; 0.1732 in.; 0.1083 in.; 0.0217 in.; 0.0666 in. 2. Find the depths of U. S. S. threads of same pitches as in exercise 1. Ans. 0.0649 in. ; 0.0325 in. ; 0.1299 in. ; 0.0812 in. ; 0.0162 in. ; 0.0500 in. 146 PRACTICAL MATHEMATICS 3. Find the root diameter of a screw of outer diameter i in. and 14 sharp V-threads to the inch. Ans. 0.3763 in. 4. Show that the depth of any sharp V-thread in inches is ^-\/3 divided by the number of threads per inch, or, what is the same thing, 0.866 divided by the number of threads per inch. 6. Check the depth of the sharp V-thread for five of the sizes given in Table V. 6. Using the formula, find the root diameter of the following common V-threads: 1 in. diameter and 20-pitch, f in.-16, in.-13, 1$ in.-6, 2ft in.-4J, 3J in.-3J. Ans. 0.1634 in.; 0.2668 in.; 0.3668 in.; 1.21 13 in.; 1.9276 in.; 2.9671 in. 7. Find the size of tap drill for a fs-in. 12-pitch sharp V-thread nut. Ans. 0.4182 in. 8. What is the tap drill size for the nut of a ft-in. 20-pitch common double-threaded nut? Ans. 0.4759 in. 9. Find the depth of the U. S. S. screw thread when there are 15 threads to the inch. 16 threads. Ans. 0.0433 in.; 0.0406 in. 10. Show that the depth in inches of any U. S. S. is |\/3 = 0.6495 divided by the number of threads per inch. 11. Using the diameter of the screw, check the depth of the U. S. S. thread for five of the sizes given in Table V. 12. Using the formula, find the root diameter of the following U. S. S. threads: f in.- 11, 1 in.-8, If in.-5. Ans. 0.507 in.; 0.838 in.; 1.490 in. 13. Check the root diameters for five sizes of screws found in Table V. CHAPTER XII CIRCLES 117. The importance of a geometrical form in the study of practical mathematics is determined, to a great extent, by the frequency of its occurrence in the applications. The circle occurs often, perhaps more frequently than any other geomet- ric form in applied mathematics. Wires, tanks, pipes, steam boilers, pillars, etc., involve the circle. In the present chapter will be considered the more useful facts about the circle, and some of their applications. Again, the student is recom- mended to select those parts that are most closely connected with his work or interests. 118. Definitions. A circle is a plane figure bounded by a curved line every point of which is the same distance from another point, called the center. The curved line is called the circumfer- ence. A line drawn through the center and terminating in the circumference is called a diameter. A line drawn from the center to the cir- c n J J- FlG - 80 - cumierence is called a radius. Any part of the circumference is called an arc. In Fig. 80, BC is an arc. If the arc equals ^^ of the circumference it is 1 of arc. There are then 360 of arc in one circumference. The straight line joining the ends of an arc is called a chord. In Fig. 80, DE is a chord. The chord is said to subtend its arc. The chord DE subtends the arc DmE. The area bounded by an arc and a chord is called a segment. In Fig. 80, the area DmE is a segment. 147 us PR A CTICAL MA Til EM A TICS The area bounded by two radii and an arc is called a sector. In Fig. 80, the area BOC is a sector. Circles are said to be concentric when they have a common center as in Fig. 81. D Fio. 81. Fio. 83. A polygon is inscribed in a circle when it is inside the circle and has its vertices on the circumference. The circle is then circumscribed about the polygon. The polygon ABCDE in Fig. 82 is inscribed in the circle O. A line is tangent to a circumference when it touches but does not cut through the circumference. In Fig. 83, AT is tangent to the circle O at the point T. The point T where it touches the circle is called the point of tangency. A polygon is circumscribed about a circle, or the circle is inscribed in a polygon, when the sides of the polygon are all tangent to the circle. In Fig. 84, the polygon ABCDE is circumscribed about the circle 0. A central angle is an angle with its vertex at the center of the circle. In Fig. 85, angle AOR is a central angle. CIRCLES 149 An inscribed angle is an angle with its vertex on the cir- cumference of the circle. In Fig. 85, angle CDE is an inscribed angle. An inscribed or a central angle is said to intercept the arc between its sides. The sides of the angle AOB intercept the arc AB, and the sides of the angle CDE intercept the arc CE. 119. Properties of the circle. The student should become familiar with the following properties, and satisfy himself that they are true by actual drawings and measurements. (1) In the same circle or in equal circles, chords that are the same distance from the center are equal. (2) A radius, drawn to the center of a chord, is perpendicular to the chord and bisects the arc which the chord subtends. In Fig. 86, the radius OC is drawn through the center of the chord AB. It is perpendicular to AB, and makes arc A (7 = arc CB. This appears true by measuring the parts in the drawing. A J3^ A- C FIG. 86. FIG. 88. (3) The angle at the center, that intercepts an arc, is double the inscribed angle that intercepts the same arc. In Fig. 87, the central angle ,40C = 60 , and the inscribed angle ABC is measured and found to equal 30. The student should draw several such figures and measure the angles. (See Art. 137 for measuring angles.) (4) The central angle has as many degrees in it as there are in the arc its sides intercept; and it is said that the central angle is measured by the arc its sides intercept. This is so because there are 4 right angles or 360 in the angles at the center, and the circumference also contains 360. (5) The inscribed angle has one-half as many degrees as the arc its sides intercept, and hence the inscribed angle is measured by one-half the arc its sides intercept. 150 I'ltA ('TIC A L MA Til KM A TICS (0) A radius drawn to the point of contact of a tangent is perpendicular to the tangent. In Fig. 88 ,OP is drawn to the point of contact of the tangent Ali. The angles oan be measured and found to be right angles. Hence the radius is perpendicular to the tangent. 120. The segment. In practical work it is often necessary to find the radius of the circle when we know the chord AK and the height of the segment DC of Fig. 89. If r stands for the radius of the circle, h for the height of the segment, and w for the length of the chord, we have the following formulas for finding r, h, and w: [17]h=r-Vr f -(|w) 2 , = 2\/h(2r-h). FIQ. 89. For the derivation of formula [16] the student is referred to exercise 35, page 306. It should be noticed that these formulas are found by applying the principles of the right triangle. Example. If the chord of the segment of a circle is 5 ft. 6 in. and the height of the segment is 10 in., find the radius of the circle. OQ2_1_ 1 M Solution. From formula [16], r = = 59.45 _ /\ lu .'.radius = 59.45 in. The method given in this article for finding the radius of the circle when the length of the chord and the height of the segment are known is used by street car trackmen as follows: A straightedge 10 ft. long is laid against the rail CIRCLES 151 on the inside of the curve. The distance from the center of the straightedge to the rail is measured. This is the height of the segment, or, as it is usually called, the "middle ordinate." The radius can now be found by the formula given. For the use of the practical man, tables are arranged which give the radius corresponding to any "middle ordinate" for the 10-ft. chord. 121. Relations between the diameter, radius, and cir- cumference. If the diameter and the circumference of a circle be measured, and the length of the circumference be divided by the length of the diameter the result will be nearly 3f. This value is the ratio of the circumference to the di- ameter of a circle, and cannot be expressed exactly in figures. In mathematics the ratio is represented by the Greek letter TT (pi). The exact numerical value of TT cannot be expressed. The value to four decimal places is 3.1416. (See Table II.) Because of this relation, if the diameter, the radius, or the circumference is known, the other two can be found. RULE. The radius equals one-half the diameter, or the diam- eter equals twice the radius. The circumference equals the diameter times 3.1416. The diameter equals the circumference divided by 3.1416. If r stands for the radius, d for the diameter, and C for the circumference, the rules are stated in the following formulas: [19] C=7rd. [20] d = C-^7r. [21] C = 27rr. [22] 2r = C-^7T. 122. Area of the circle The method of finding the area of a circle when the radius, diameter, or circumference is given is established in geometry. The following will show the reasonableness of the rules. In Fig. 90, suppose the half of the circle AnB is cut as indi- cated from the center nearly to the circumference, and then spread out as in (6). The length AnB of (6) is the half cir- cumference. Let the other half of the circle be cut in the same manner and fitted into the first half. It is evident that, if we make the number of the cuts large, the figure formed will lf>2 PRACTICAL MATHEMATICS be approximately a rectangle whose length is equal to one- half the circumference, and whose width is equal to the radius. We then have the following: n B (b) Fia. 90. RULE. To find the area of a circle, multiply one-half the circumference by the radius. This may be put in either of the following forms which are usually more convenient to use. RULE. The area of a circle equals IT times the square of the radius; or the area of a circle equals one-fourth of IT times the square of the diameter. If A stands for the area, C for the circumference, d for the diameter, and r for the radius, these rules are stated in the following formulas: [23] A = Cr. [24] A = 7rr 2 = 3.1416 Xr 2 . [25] A= i7rd 2 = 0.7854 Xd 2 . From formula [24], if the area of the circle is given, the radius equals the square root of the quotient when the area is divided by r. Or, in a formula, [26] r = From formula [25], we get [27] d = VA -r- ATT = VA -^ 0.7854. Example. Find the radius of a circle whose area is 28 ft. 2 Solution. Using formula [26] and putting in the numbers, = V28-f-3.14J6 = V8.9126 = 2.985. /. radius = 2.986 ft. CIRCLES 153 123. Area of a ring. In the ring, which is the area between the circumferences of two concentric circles, the area can be found by subtracting the area of the small circle from the area of the large circle. FIG. 91. If A and a, R and r stand for the areas and the radii respec- tively of the two circles, and A r for the area of the ring, then [28] A r =A-a=rR 2 -7rr 2 =7r(R 2 -r 2 )=7r(R+r)(R-r). This last is a very convenient formula to use. It may be stated in words as follows: RULE. To find the area of a ring, multiply the product of the sum and the difference of the two radii by TT. Example. Find the area of a ring of inner diameter 8 in. and outer diameter 12 in. Solution. Using formula [28] and putting in the numbers, A r = 3.1416(6+4)(6-4) = 3.1416X10X2 = 62.832. .'.area = 62. 832 in. 2 It should be noted that the rule holds even though the circles are not concentric, that is, the circles may be as in Fig. 91(6). 124. Area of a sector. The area of a sector of a circle is equal to that fractional part of the area of the whole circle that the angle of the sector is of 360. Thus, if the angle of the sector is 90, the area of the sector is -^VV of the area of the circle. Example. Find the area of a sector of 60 in a circle of radius 10 in. 154 PRACTICAL MA THEM A TICS Solution. In Fig. 92, the sector AOB has an angle of 60. Its area equals 6 c %X7rr 2 . If the radius is 10 in., the area of the sector is B A=iX3.1416X10 2 = 52.36in. 2 Ann. If 6 (the Greek letter theta) stands for the number of degrees in the angle of the sector, and the other letters the same as before, the area of the sector is given by the formula, Fio. 92. 125. Area of a segment. In Fig. 92, it is evident that the area of the segment ABD equals the area of the sector AOB minus the area of the triangle A OB. Since it requires a knowledge of trigonometry to find the area of a triangle when we have only two sides and an angle, or to find the area of a sector when the angle is unknown, we cannot usually find the area of a segment by geometry. (See page 434.) If the angle and the lines in the segment are measured, the area of, first, the sector, and second, the triangle, can be found. The difference be- tween these areas is the area of thp segment. Example 1. Find the area of a segment in a circle of radius 11 j in. and subtending an angle at the center of 105. Solution. The dimen- sions are as shown in Fig. 93, where the parts are constructed accurately to scale and measured. The area of sector OADB ***$%% of the area of the circle. .'.area of sector= 18$X3.1416X(11}) 2 = 115.97 in.-' Area of triangle OAB = \ X 18 X 6 J = 60.75 in. 2 Area of segment = area of sector area of triangle = 115.97 in. 2 -(50.75 in. - = 55.22 in. 2 Am. FIG. 93. CIRCLES 155 Many approximate rules are given to find the area of a segment. Perhaps the following are as good as any. [30(a)] A = fhw+~> 2w [30(b)] A = |h 2 ^-0.608. In these rules A is the area of the segment, h the height, and w the width, while r is the radius of the circle to which the segment belongs. If the height of the segment is less than one-tenth the radius of the circle, formula [30 (a)] may be shortened to A = %hw. Steam engineers often wish to find the area of a segment when the height is large compared with the radius, say, two-thirds of the radius. They then proceed as follows: In Fig. 94, let it be required to find the area of the segment CnD. Find the area of the half circle AnB, and then the area of the part ACDB considered as a rectan- gle. The area of the segment is roughly the difference be- tween these. Example 2. Find the area of the segment whose chord is 10 ft. and height 1.5 ft. Solution. By formula [30(a)]. 1 K3 = 10.17- ft. 2 FIG. 94. By formula [30 (b)], first finding r by formula [16], 2X1.5 -0.608 = 10.175 ft, 2 126. The ellipse. The ellipse is a figure bounded by a curved line such that the sum of the distances of any point in the boundary from two fixed points is constant, that is, always the same. Thus, in Fig. 95, any point P has the distances PF -\-PF' equal to the distances P'F+P'F', drawn from any other point P'. 156 PRACTICAL MATHEMATICS F and F' are the two fixed points and arc called the foci (singular focus). The point is the center of the ellipse. NA is the major axis, and MB is the minor axis. OA and OB are the semi-axes. If a stands for OA and b for OB, it has been proved that the area of the ellipse is given by the formula, [31] A=7rab. Example 1. Find the area Fio. 95. of an ellipse whose two axes are 30 ft. and 26 ft. respectively. Solution. Using formula [31] and putting in the values, 4=3.1416X15X13 = 612.612 ft. 2 While the area of an ellipse is easily found when the major and minor axes are given, the circumference, or perimeter, of the ellipse is determined with difficulty. Various approximate formulas are given for finding the circumference of an ellipse. If the ellipse is very nearly the shape of a circle, that is, if the major and minor axes are nearly equal, then [S2(a>] P=T(a+b), where P is the perimeter or circumference, a the semi-major axis, and b the semi-minor axis. When the ellipse differs considerably from a circle, that is, when there is considerable difference between the major and minor axes, either of the following rules may be used to good advantage : [32(b)] P=T[f(a+b)-yab], [32(c)] P = 7rV2(a 2 +b 2 ). The exact formula derived by the methods of higher mathe- matics may be stated in the following form : where e 0, This formula is not given with the inten- tion that it should be used, as the computation required is considerable. Example 2. Find the circumference of an ellipse whose major axis is 18 in. and whose minor axis is 6 in. CIRCLES 157 Solution. Here a = 9 and 6 = 3. By [32 (a)], P = 3.1416(9+3) =37.699 in. By [32(b)], P = 3. 1416[|(9+3) - \/9X3] = 40.225 in. By [32 (c)], P = 3.1416\/2(9 2 +3 2 ) =42.149 in. Formula [32 (b)] is the best to use when the two axes are not very nearly equal. EXERCISES 38 1. Using d for diameter, C for circumference, r for radius, and A for area of a circle, given the values in the first column to find those in the last two. (a) d = 75 ft. C =235.62 ft.; A =4417.86 ft, 2 (b) r=23ft. C = 144.51 ft.; A =1661.90 ft. 2 (c) d = 34.6 in. C = 108. 70 in.; A =940.25 in. 2 (d) d = 24.5 in. C = 76.97 in.; A =471. 44 in. 2 (e) C = 86.08 in. d = 27 A in.; A =589.65 in. 2 (f) (7 = 158.02 in. d = 50.3 in.; A = 1987.13 in. 2 (g) A = 1452.20 ft. 2 d = 43ft.; (7 = 135.09 ft. (h) A =27171. 6 ft. 2 r=93ft.; (7 = 584.34 ft. 2. Find the area of the cross section of a half-inch rod. Ans. 0.196+ in. By the cross section is meant the area of the end of the rod when cut square off. 3. Find the area of the ring enclosed between two circles, the outer 9 in. and the inner 8 in. in diameter. Ans. 13.352 in. 2 4. The inner and outer diameters of a ring are 9? and 10 in. respec- tively. Find the area of the ring? Ans. 7.658 in. z 5. Find the area of the ring in the cross section of a water main 40 in. in external diameter, if the iron is 1 in. thnk in the shell. Ans. 122.52+ in. 2 6. In an elliptical garden the longest diameter is 36 ft. and the shortest 22 ft. Find the area of the garden. Ans. C22.04- ft. 2 7. In a steel plate 3 ft. by 2^ ft. are 26 round holes, each If in. in diam- eter. Find the area of steel remaining. Ans. 1017.46+ in. 2 8. At the center of one side of a barn 40 ft. on a side a horse is tied by a rope 70 ft. in length. Find the area he can graze over in square rods. Ans. 43.27+ rd. 2 Suggestion. When the figure is drawn it is seen that the horse can graze over a half-circle having a radius of 70 ft., two quarter-circles having a radius of 50 ft., and two quarter-circles having a radius of 10 ft, 9. A 6-in. water pipe can carry how many times as much as an inch pipe? Solution. Area of 6-in. pipe =0.7854X6 2 in. 2 Area of 1-in. pipe =0.7854 XI 2 in. 2 Area of 6-in. pipe 0.7854X6 2 6 2 1-1 . . __ __ -IK 4 99 Area of 1-in. pipe 0.7854 XI 2 I 2 158 PRACTICAL MA THEM A TICS The quotient or the ratio of the areas of two circles can always be found by dividing the square of one diameter by the square of the other. The radii may be used instead of the diameters. The above is simply the principle that similar areas are in the same ratio as the squares of their like dimensions, applied to circles. 10. In putting up blower pipes, two circular pipes 11 in. and 14 in. in diameter respectively join and continue as a rectangular pipe 14 in. in width. Find the length of the cross-section of the rectangular pipe. Ana. 17.78+ in. 11. How many times the area of the cross section of a ^-in. wire is a half-inch wire? Ana. 64. 12. If an inch pipe will empty 2 barrels in 15 minutes, how many barrels will an 8-in. pipe empty in 24 hours? (Make no allowance for friction.) Ana. 12,288. 13. How many 3-in. steam pipes could open off from an 18-in. steam pipe? Ana. 36. 14. The diameter of the safety valve in a boiler is 3 in. Find the total pressure tending to raise the valve when the pressure of the steam is 120 Ib. per square inch. Ans. 848.23 Ib. 15. If the diameter of a piston is 30 in., find the total pressure on the piston when the pressure of steam is 100 Ib. per square inch. Ans. 70,686 Ib. 16. A circular sheet of steel 2 ft. in diameter increases in diameter by zJo when the temperature is increased by a certain amount, (a) Find the increase in the area of the sheet. (6) Find the per cent of increase in area. Ans. (a) 0.03149 ft.*; (b) 1%. 17. How many No. 20 B. and S. copper wires will have the same cross section area as one No. 00? (See Table VII.) Ans. 130.3 -. Fiu. 97. 18. A hot-air pipe 9 in. in diameter passes into a boot as shown in Fig. 96, and a rectangular pipe of same capacity passes upward from the boot. If the rectangular pipe is 4 in. wide, find its length in cross section. Ana. 15.9+ in. 19. Given two joining pipes 12 in. and 8 in. in diameter respectively, to find the diameter x of the continuation which has the same area. (See Fig. 97.) Ans. 14.4+ in. 20. Given an elliptical pipe of longest and shortest axes 16 in. and 1C CIRCLES 159 in. respectively, to find the diameter of the circular pipe having the same area of cross section. Ans. 12.65 in. 21. Show by means of the carpenter's square how to find the diameter of a circle having the same area as the sum of the areas of two given circles. Discussion. Suppose we take two circles 6 in. and 8 in. in diameter respectively. Lay off on one arm of the carpenter's square, as shown in Fig. 98, the diameter of the 6-in. circle and on the other arm the diameter of the 8-in. circle. The line joining the ends of these, or the hypotenuse of the right triangle, is the diameter of the circle having the same area as the sum of the areas of the two given circles. This is seen to be true in this particular case as follows: Area of 6-in. circle =6 2 X0.7854 in. 2 Areaof 8-in. circle =8 2 X 0.7854 in. 2 Sum of areas =(6 2 +8 2 )X 0.7854 in. 2 But 6 2 +8 2 = 10 2 , and if 6 in. and 8 in. are respectively the altitude and base of a right triangle then 10 in. is the hypotenuse. Hence the area of the circle equal to the sum is 10 2 X0.7854 in. 2 The diameter of this circle is evidently 10 in., which is the hypotenuse of the right triangle as drawn in the figure. A similar discussion would apply to any two circles. 22. Show by means of the carpenter's square how to find the diameter of a circle having the same area as the sum of the areas of any number of given circles. 23. If the drive wheels of a locomotive are 66 in. in diameter, find the number of revolutions per minute to go 40 miles per hour. Am. 203.7 + . FIG. 98. 40X5280X12 ;= 203.7. 60X66X3.1416 Solution. It is usual to work such problems as this by cancellation. Above the line are the numbers which give the inches in 40 miles. Below the line is 60, which we divide by to get the number of inches the train goes in 1 minute; and 66X3.1416, which is the number of inches in the circumference of the wheel. 24. Supposing that the driving wheels of a locomotive are 16 ft. in circumference, what number of revolutions must they make per minute so that the locomotive may attain a speed of 60 miles per hour? Ans. 330. 25. A locomotive wheel 5 ft. in diameter made 10,000 revolutions in a 100 1'RACTICAL MATHEMATICS distance of 24 miles. What distance was lost due to the slipping of the wheels? Ana. 5} miles. 26. If an arc of a circle is equal in length to the radius, what is the value of the central angle which it measures? Ann. 57.2958 . Solution. Since 2ir times the radius equals the circumference, and the entire circumference measures an angle of 360 at the center of the circle, the number of degrees = 360 -^2ir =57.2958 . 27. The radii of two circles are 2 ft. and 4 ft. The area of the second is how many times the first? Ans. 4. 28. The length of the circumference of a circle is 132 ft. Calculate the length of the diameter, the length of an arc of 40, and the area of a sector of 80. Ans. Arc 14.67- ft.; dia. 42.017- ft.; area 308. 1+ ft. 1 29. Find the weight of the iron hoops on a tank 15 ft. in diameter, there being 16 hoops weighing 3 Ib. per linear foot. Ans. 2261.9+ Ib. B 30. A regular hexagon, the perimeter of which is 42 ft., is inscribed in a circle. Find the area of the circle. Ans. 153.9+ ft. 2 31. What is the waste in cutting the largest possible circular plate from a piece of sheet steel 17 in. by 20 in.? Ans. 113.02 in. 2 32. Four of the largest possible equal sized pipes are enclosed in a box of square cross section 18 in. on an edge. What part of the space do the pipes occupy? Ans. 0.7854. 33. Find size of the box to enclose five 6-in. pipes, placed as in Fig. 99, and find the part the area of the pipes is of the area of the box. Ans. Box 12 in. by 16.392 in.; part occupied 0.7187+. Solution. AC = 12 in., AB = 2DM. DM = DN+NM, but DN =3 in. and .*. DM = 3 in. +5.196 in. =8.196 in. A AB = 2X8.196 in. = 16.392 in. /. area = 12 X 16.392 = 196.704 in. 2 Area of 5 circles =5X0. 7854 X6 2 = 141.372 in. 2 Part occupied by pipes = 141.372-^-196.704 =0.7187+. 34. If the diameter of a circle is 3 in., what is the length of an arc of 80? Ans. 2.0944 in. 36. The minute hand of a tower clock is 6 ft. long. What distance will the extremity move over in 36 minutes? Ans. 22 ft. 7.4+ in. 36. The maximum circumferential velocity of cast-iron flywheels is 80 ft. per second. Find the maximum number of revolutions per minute for a cast-iron flywheel 8 ft. in diameter. Ans. 191 nearly. 37. An emery wheel may have a circumferential velocity of 5500 ft. per minute. Find the number of revolutions per second an emery wheel 9 in. in diameter may make. Ans. 39 nearly. CIRCLES 161 38. The peripheral speed of a grindstone of strong grain should not exceed 47 ft. per second. Find the number of revolutions per minute a grindstone 3 ft. in diameter may turn. Ans. 299 nearly. 39. The area of a square is 49 sq. ft. Find the length of the circum- ference and the area of the circle inscribed in this square. Ans. 21.99+ ft.; 38.48+ ft. 2 40. Find the size of the largest square timber which can be cut from a log 24 in. in diameter. Ans. 16.97+ in. 41. A roller used in rolling a lawn is 6.5 ft. in circumference and 2.5 ft. wide. If the roller makes 10 revolutions in crossing the lawn once and must go up and back 12 times, what is the area of the lawn? Ans. 3900 ft. 2 42. Three circles are enclosed in an equilateral triangle. If the circles are 10 in. in radius, find the sides of the triangle. Ans. 54.64 in. A B I'lG. 100. Suggestion. The circles are as shown in Fig. 100. The triangle DEG = triangle DAH. Hence AH =GE = 10 X\/3 = 17.32 in., #7= the diameter of one of the circles =20 in., and 1C = AH = 17.32 in. 43. Using 4000 miles as radius of earth, find length in feet of one second of arc on the equator. Ans. 102.4 ft. Note. 1=60 minutes and 1 minute =60 seconds of arc. 44. Using 4000 miles as radius of earth, find the length in miles of arc of 1' (a) on the parallel of 45 north; (b) on the parallel of 60 north. Ans. (a) 0.823- miles; (b) 0.582- miles. Suggestion. For the parallel of 45 north, CB is the radius. But CB = OC since the triangle OCB is a right triangle with two equal angles. The relations are as shown in Fig. 101. 45. A bicycle is so geared that one revolution of the feet makes two revolutions of the wheels which are 28 in. in diameter How many revolutions per minute of the feet are necessary to go at the rate of 25 miles per hour? Suppose that the pneumatic tires are not well inflated, n 162 PRACTICAL MATHEMATICS what is the per cent of loss in distance made if the compression in the tire is i in.? An*. 150 + ; 3.6-%. Suggestion. If the compression of the pneumatic tire is J in., the wheel acts as if it were J in. less in radius. 46. What is the per cent of error in taking 4 times CA in Fig. 102 as the circumference of circle 0? Ans. 0.66 % too large. B FIG. 101. Fia. 102. 47. In the same circle, what is the per cent of error in taking rf "3 Q Q Q Q o Q e Q Q f" co S ri X ,-c ^ ^ S eb c? C 0) o o o p ' p * ' u. c O /rj ^ O CO O 9j ^^ C^l X s 3 3 c X t2 re CO CO co O5 o J^ - o'o ~ O Sig tn.Q c g Jl -27 7 " X i! CO X c; i - CO o 3 S J3.2 (N g -1- l> 2 M T tN X X c: (N 1C O -* o ~ / M 1 t C<3 tf V CO CO ^a oc OJ M M X 7i oc ~. CO OS i,' t> CO X -i cC ^ -< 1-1 -M CO * * i - 1 1 "^ c M _>. Triangle . . r -^ Pentagon. Hexagon . . Heptagon . Octagon. . Nonagon . Decagon . . Dodecagoi - o CO * SB 00 OS IN ^ O X "8 11)8 PRACTICAL MATHEMATICS (3) To find the radius of the circumscribing circle when a side of the polygon is given. Multiply the length of a side by a number chosen from column (5). This can be used to good advantage in drawing u regular polygon of a given side. Example. Construct a regular decagon having sides of 2 3 in. Radius of circumscribing circle = 2\ XI. 618 in. =4.045 in. With the compasses construct a circle of this radius. Then with the dividers open 2\ in. step around the circle, which should be divided into 10 parts. Connect these points succes- sively and the construction is complete. (4) To find the radius of the inscribed circle when a side of the polygon is given. Multiply the length of a side by a number chosen from column (6). (5) To find the length of the side of a polygon that can be inscribed in a circle of given radius. Multiply the given radius by a number chosen from column (7). Example. Construct a regular heptagon in a circle of 3-in. radius. A side of the polygon =3X0.8677 in. =2.6 in. With the dividers open 2.6 in. step around the circle, which should be divided into 7 equal parts. Connect these points successively and the construction is complete. EXERCISES 39 1. A round shaft is 3J in. in diameter. Find the length of the side of a triangular end that can be made on the shaft. Find the length of the side of a square end. Of a hexagonal end. Of an octagonal end. Ans. 2.81+ in.; 2.30- in.; If in.; 1.24+ in. Triangular Pentagonal Hexagonal Fio. 112. 2. Find the diameter of a circular shaft so that it may have a triangular end 2J in. on a side. A pentagonal end 1$ in. on a side. An octagonal end 1 1 in. on a side. Ans. 2.887- in.; 2.552- in.; 2.940- in. Suggestion. Use rule (3). CIRCLES 169 3. A square taper reamer is to be made which must ream 1| in. at the small end and If in. at the back end. What must be the distance on the flat face at each end? Ans. 0.795+ in.; 1.149+ in. 4. What is the diameter of the bearing that can be turned on a tri- angular shaft of side 2 in.? On a hexagonal shaft of side If in.? Suggestion. Use rule (4). Ans. 1.155 in. ; 2. 815 in. 6. Find the difference between the area of a circle of radius 5 in. and the area of the inscribed regular triangle. The inscribed regular penta- gon. Hexagon. Octagon. Decagon. Ans. 46.065+ in. 2 ; 19.10- in. 2 ; 13.59- in. 2 ; 7.94+ in. 2 ; 5.08- in. 2 Suggestion. Find the side of the inscribed polygon by rule (5) and its area by rule (1). 6. Find the radius and area of the largest circle that can be cut from a triangle every side of which is 4 ft. Ans. 1.155 ft.; 4.19 ft. 2 7. The triangular end on a round shaft is 1.7 in. on a side. Find the diameter of the shaft. Ans. 1.96+ in. 8. The area of a regular hexagon inscribed in a circle is 24\/3- Find the area of the circle and the length of the circumference. Ans. 50.266-; 25.133-. 9. A square end 0.875 in. on a side must be milled on a shaft. What is the diameter to which the shaft should be turned? Ans. 1.237+ in. 10. A pipe 10 in. in diameter is connected to a hexagonal pipe of the same area in cross section. Find the edge of the hexagon of the cross section of the hexagonal pipe. Ans. 5.50 in. TURNING AND DRILLING 128. Rules. The cutting speed of a tool is the rate at which it passes over the surface being cut. This applies to a lathe tool in turning a piece of work, such as a car axle; or to a drill used in making holes in a metal of any kind. The rate at which the tool can cut the metal without in- juring the tool depends upon the material in the tool, as well as upon the kind of work being turned. Since cutting speeds are usually given in feet per minute, the rate at which a tool is cutting can be found by the following : RULE. Multiply the circumference of the piece, or of the drill, in feet by the number of revolutions per minute. This gives the cutting speed in feet per minute. This applies to work turned in a lathe or to the drill in a drill press. It follows from the above that the number of revolutions, allowable per minute, is found by the following: 170 PRACTICAL MATHEMATICS RULE. Divide the cutting speed in feet per minute by the circumference of the work in feet. This gives the number of revolutions per minute. 129. Feed. The feed of a tool is the sideways motion given to the cutting tool. It is expressed in one of the follow- ing ways: (1) The feed is the part of an inch that the tool advances along the work for each revolution or stroke, as a feed of A inch. (2) The feed is the number of revolutions or strokes neces- sary to advance the tool 1 inch, as a feed of 20 turns to the inch. (3) The feed is the number of inches the tool advances in 1 minute, as the feed is f inch per minute. Thus, in turning a car axle, the shaving may be J in. wide, which means that the tool must advance that distance along the axle for every revolution of the axle. That is, it will take 4 turns of the axle to cover 1 in. of its length with the turning tool. 130. Cutting Speeds. Cutting speeds for carbon steel tools should be about 30 ft. per minute in steel, 35 ft. per minute in cast iron, and 60 to 100 ft. per minute in brass. The general rule is to run high-speed steel tools, in steel, about double, and in iron about three times the speed of the carbon-steel tool. The maximum speed given to any tool must be governed by the density and toughness of the material being cut, and by the way the tool "holds up." The feed of a drill should be from 0.004 in. to 0.01 in. per revolution. EXERCISES 40 1. In turning a brass rod 2 in. in diameter, what is the proper number of revolutions per minute if the cutting speed for brass is 100 ft. per minute? Ans. 191. 2. In turning a locomotive wheel 78 in. in diameter, what is the proper number of revolutions per minute, in order that the cutting speed may be 10 ft. per minute? Ans. 0.49 nearly. 3. In turning a tool-steel arbor, a carbon-steel turning tool is used. The cutting speed is 18 ft. per minute. How many revolutions per minute should the work make if the arbor is 3 in. in diameter? Ans. 22.9. 4. A J-in. drill, cutting cast iron, may cut at the rate of 40 ft. per min- ute. How many revolutions per minute can it make? Ans. 203.7. CIRCLES 171 6. In turning a car wheel 27 in. in diameter, it makes If revolutions per minute. What is the speed of the cutting tool? Ans. 12.37 ft. per minute. 6. How long will it take to turn off one layer from the surface of a car wheel 4 in. thick and 30 in. in diameter, if the cutting is 15 ft. per minute and the feed | in. ? Ans. 16f minutes nearly. 7. How long would it require to make one cut over the surface of a tool-steel arbor 2 in. in diameter and 10 in. in length, if the cutting speed is 18 ft. per minute and the feed of the cutting tool ^ in. per revolution of the work? Ans. 4.65+ minutes. 8. In Kent's Mechanical Engineer's Pocket-book are given the follow- ing formulas for finding results in cutting speed problems: Let d = the diameter of the rotating piece in inches, n=the number of revolutions per minute, and =the cutting speed in feet per minute, then irdn nna-ioj & 3.S2S , 3.82$ S= = 0.2618dn; = Show that these are true and apply them to the preceding exercises. 9. The diameter of a piece of cast iron to be turned is 7 in. If the lathe makes 22 revolutions per minute, what is the cutting speed? Ans. 40.3+ ft. per minute. 10. A piece of brass 4 in. in diameter is making 80 R. P. M. in a lathe. What is the cutting speed? Ans. 83.8 ft. per minute. 11. A wrought-iron shaft 2 in. in diameter and 30 in. long is turned at a cutting speed of 25 ft. per minute and a feed of 50 in. Find the time for turning the shaft. Ans. 25.1+ minutes. Solution. 30-5-3*5 = 1200 = number of revolutions. 2X3.1416X1200 _ 1QOC . =25. Io28 =number of mm. 1 2i X ^o 12. The cutting speed in a certain case must not exceed 40 ft. per minute. The piece to be turned is If in. in diameter. How many revolutions per minute can it make? Ans. 87.3. 13. Give to the nearest sixteenth of an inch the length of a f-in. steel rod that is turned per minute, if the cutting speed is 36 ft. per minute and the feed -^ in. Ans. 7^ in. 14. In turning a car wheel 3 ft. in diameter, the highest rate of speed allowable for the cutter is 40 ft. per minute. How many revolutions per hour can the wheel make? Ans. 254.6. 15. A car axle may be turned with the cutter moving 9 ft. per minute. If the axle is 4J in. in diameter, how many revolutions can it make per minute? Ans. 7.64. BELT PULLEYS AND GEAR WHEELS 131. The relation of size and speed of driving and driven gear wheels are the same as those of belt pulleys. In calcu- 172 PRACTICAL MATHEMATICS lating for gears we use the diameter of the pitch circle, or the number of teeth as may be necessary. A mechanic should be able to determine quickly and accu- rately the speed of any shaft or machine, and to find the size of a pulley in order that a shaft or machine may run at a desired speed. He should master the principles underlying the rules and formulas used as well as know how to use them. It is well then for the student to work many problems on pulley speeds before special formulas are taken up. This will help him to master the principles, and will make him independent of the formulas. It will also put him into position to derive the formulas. For a complete discussion of questions connected with belts and belting see any mechanical engineer's handbook. EXERCISES 41 1. A shaft having a pulley 6 in. in diameter makes 840 R. P. M. If this speed is to be reduced to 320 revolutions, what size of pulley should be used? Solution. If the 6-in. pulley makes 840 R. P. M., a point on the belt moves 6X3.1416X840 in. per minute. Then in order to make 320 6X3.1416X840 . R. P. M., the pulley must be ox~ ~ in. in circumference, and 6X3.1416X840 hence 320x3 1416 =1 ^* m< m diameter. 2. The pulley on the armature shaft of a dynamo is 4 in. in diameter. This is to be belted to a driving shaft which makes 500 revolutions per minute. The speed of the dynamo must be 1700 revolutions per minute. What must be the size of the pulley placed on the shaft? Ana. 13s in. in diameter. 3. A shaft has upon it two pulleys, each 8 in. in diameter. The speed of the shaft is 400 revolutions per minute. What must be the size of the pulleys of two machines if, when belted to the shaft, one of them has a speed of 300 revolutions per minute and the other 900? Ans. 10f in. and 35 in. in diameter. 4. The pulley on the headstock of a lathe is 3 in. in diameter. This is belted to an 8-in. pulley on a shaft that makes 420 revolutions per min- ute. At what rate will a block of wood placed in the chuck revolve? Ans. 1120 R. P. M. 6. If the wheels of an electric car are 2 ft., the axle cogwheel 8 in., and the cogwheel attached to the motor 12 in. in diameter, what must be the Hpeed of the motor to carry the car a mile in 5 minutes? Atu. 112.04+ R. P. M. CIRCLES 173 6. In two connected belt pulleys, or gear wheels, if D is the diameter of the driving wheel, d the diameter of the driven wheel, R the number of revolutions per minute of driver, and r the number of revolutions per minute of driven, find r in terms of D, d, and R. . , _^5. d Discussion. In Fig. 113, A is the driving pulley and B is the driven pulley. It is evident that, since the belt does not slip, a point on the circumfer- ence of B must move as far in a minute as a point on the circumference of A. Since A makes R revolutions per min- ute, a point on its circumference will move RirD units per minute. Similarly a point on the circumference of B will move rird units per minute. FIG. 113. i i RD or RD = rd and r = , d 7. In any system of pulleys or gears, the general rule holds: that the product of the diameters, or numbers of teeth, of the driving wheels and the number of revolutions per minute of the first driver must be equal to the product of the diameters, or the numbers of teeth, of the driven wheels and the number of revolutions per minute of the last driven wheel. As a formula this may be stated = RXDXD'XD"XD'"Xetc. dXd'Xd"Xd'"Xetc. where D, D', D", etc., are the diameters of the driving pulleys, d, d' d", etc., are the diameters of the driven pulleys, R is the R. P. M. of the first driver, and r is the R. P. M. of the last driven pulley. Show why this is true. 8. The number of revolu- tions the governor of a steam engine is intended to run is given by the builder. If the speed of the governor is 120 R. P. M., size of governor pulley 8 in., and the desired speed of the engine 90 R. P. M., find the diameter of the pulley to be put on the engine shaft to run the governor pulley. Ans. lOf in. 9. An endless knife runs on pulleys 48 in. in diameter as shown in Fig. 114, at a rate of 180 R. P. M. If the pulleys are decreased 18 in. in diameter, how many R. P. M. will they have to make to keep the knife traveling at the original speed? Knife 174 PRACTICAL MA Til EM A TICK Solution. 180X3.1416X48 in. = rate per minute. 3.141GX30 in. = circumference of reduced pulleys. 180X3.1410X48 'FTTo =288= number of R.P.M. of reduced pulleys. FIG. 115. FIG. 11G. Engine Shaft F UjD" Line Shaft Fio. 117. 10. Adapt the formula of exercise 7 to the following : A train of wheels consists of four wheels each 12 in. in diameter of pitch circle, and three CIRCLES 175 pinions 4 in., 4 in., and 3 in. in diameter respectively. The first three large wheels are the drivers and the first makes 36 revolutions per minute. Required the speed of the last wheel. Ans. 1296 R. P. M. 11. In the train of the preceding exercise, what is the speed of the first large wheel if the pinions are the drivers, the 3-in. pinion being the first driver and making 36 revolutions per minute? Ans. 1 R. P. M. 12. Pulleys are arranged as in Fig. 115. Pulley A makes 192 R. P. M., is the driver, and is 14 in. in diameter. Pulley B is 8 in. in diameter. Pulley C is 6 in. in diameter and is to make a required 1400 R. P. M. Find the diameter to make the pulley D, fastened to the same shaft as B, in order that C may have the desired number of revolutions per minute. Ans. 25 in. 13. Find the number of R. P. M. of the last gear shown in Fig. 116, if the gear having 84 teeth makes 36 R. P. M. 14. In Fig. 117, if a 160-in. pulley on the engine shaft drives a 60-in. pulley on the line shaft, and a 40-in. pulley on the line shaft drives an 18-in. pulley on the counter shaft, find the number of revolutions per minute of the counter shaft if the engine shaft runs at 80 R. P. M. Ans. 474 nearly. THE MIL 132. The circular mil. In most cases, electrical conductors have a circular cross section. We know that the area of a circle is found by the formula A = j-rrd 2 , which brings in the inconvenient factor fr, or 0.7854. In order to avoid this factor, a new unit has been adopted for commercial work. This unit is the circular mil (abbreviation C. M.) which is the area of a circle one mil, or 0.001 inch, in diameter. If A is the area in circular mils of any circle and d the di- ameter in mils, then, since a circle 1 mil in diameter has an area of 1 C. M., we have the proportion 1 =T Ad 2 for the areas of the circles are in the same ratio as the squares of the diameters. This proportion gives A =d 2 . This stated in words is the following: RULE. The area of a circle in circular mils is the square of the diameter in mils, or thousandths of an inch. Thus, a 0000 gage B. and S. wire is 0.46 in. =460 mils in diameter, and hence has an area of 460 2 =211,600 C. M. 176 PRACTICAL MATHEMATICS If the area in circular mils is given, the diameter in mils can evidently be found by taking the square root of the area, or 133. The square mil. The square mil is sometimes used, and is the area of a square 1 mil on a side. Since the area of a circle is 0.7854d 2 , it is seen that 0.7854 square mil = 1 C. M. EXERCISES 42 1. Find the number of circular mils in the area of B. and S. gage wires Nos. 40, 20 and 10. (See Table VII.) Ans. 9.88 + J 1021.5 + ; 10,383-. 2. How many square mils in a bar J in. by f in. in cross section? Ans. 187,500. 3. How many circular mils are equal to 20,000 square mils? Ans. 25,464.8-. 4. Find the diameter in mils and in inches of a circular rod having a cross section of 237,600 C. M. Ans. 487.4+ mils; 0.4874+ in. 5. In ordinary practice, trolley wire is or 00 B. and S. hard-drawn copper wire. What is the area of the cross section of each in circular mils? Ans. 105,535- C. M. or 133,076+ C. M. CHAPTER XIII GRAPHICAL METHODS ANGLES 134. Units. In measuring any magnitude, a unit of measure is necessary. In measuring length, there are various units, as the inch, foot, meter, and mile. Likewise in the measure- ment of angles, there are in use, as units, the right angle, the degree, and the radian. Right angle as unit. When using the right angle as a unit, we speak of an angle as such a part of, or as so many times, a right angle. The degree as a unit. The degree as a unit for measuring angles may be defined as the value of the angle formed by dividing a right angle into 90 equal parts. The degree is also used as a unit for measuring arcs. It is then defined as -$1-$ part of a circumference. In either case the degree is divided into 60 parts called minutes, and the minute into 60 parts called seconds. Degrees, minutes, and seconds of angle or arc are indicated by the signs , ', and ". For example, a measurement of 27 degrees, 47 minutes, 35 seconds is written thus: 27 47' 35". As already defined, if an angle has its vertex at the center of a circle and its sides formed by the radii of the circle, it is spoken of as an angle at the center of a circle. The number of degrees in the angle so placed is equal to the number of degrees in the arc of the circle intercepted between the sides of the angle. Thus, in Fig. 118, AOB is an angle at the center. The number of degrees in this angle equals the number of degrees in the arc AB. The angle AOB is said to be measured by the arc AB. 135. Circular measure, radian. The unit of circular meas- ure of angles is the radian. The radian is defined as the 177 178 PRACTICAL MA THEMATICS angle which at the center of a circle is measured by an arc equal in length to the radius of the circle. In Fig. US, arc ,4/? = rndius OA, hence angle AOB is one radian. Since a circumference is 2r times the radius, there are 2* arc lengths equal to the radius in a circumfer- ence; and hence 2w radians are meas- ured by the circumference of a circle, or 2ir radians = 360. From this, T radians = 360 -f- 2 = 180, and 1 radian = 180 -5- TT = 57.29578 - . Reduced to degrees, minutes, and seconds: 1 radian = 57 17' 44.8". In a similar manner, if 180 =T radians, then I=TT radians -5-180 = 0.01745+ radians. 136. The protractor. The protractor consists of a chcular or semicircular scale of convenient diameter. The circum- ference of this scale is divided into degrees, half degrees, and sometimes into quarter degrees, as shown in Fig. 119. The FII;. 118 divisions of the scale are numbered from to 180, beginning at each end. The sum of the two readings at any point is 180. This method of numbering enables one to measure an angle from either end of the protractor, or it enables one to set off angles in either a right- or a left-hand direction. GRAPHICAL METHODS 179 137. To measure an angle with protractor. Place the protractor over the angle to be measured, so that the point C is on the vertex of the angle, and either half of the side A B will fall upon one side of the angle. The other side of the angle should pass through some mark on the scale of the pro- tractor. The reading on the scale where this side of the angle crosses it is the measure of the angle in degrees. In work with the protractor, a hard pencil with sharp point should be used. 138. To lay off an angle with a protractor. Draw one side of the angle and locate the vertex. Place side AB of protractor on side drawn and point C on the vertex. Locate reading of value of angle required on scale of protractor, and connect this with the vertex. The degree of accuracy with which an angle can be laid off depends upon the instruments and the one who uses them, but largely upon the size of the protractor. It is well then to use a fairly large protractor, one 5 or 6 in. in diameter. EXERCISES 43 1. Draw three triangles and measure their angles. Find the sum of the angles of each triangle. Are their sums equal? 2. Draw a right triangle and measure the two acute angles. What is their sum? 3. Draw an equilateral triangle and measure the angles. What is their sum? 4. Draw an isosceles triangle and measure its angles. Are there two equal angles? 6. Construct angles of 26, 75 30', 106 45', and 146 15'. 6. Describe how an angle of If radians may be constructed. 7. How many radians in each of the angles: 27 45', 47 26', 109 30'? Arcs. 0.484 + ; 0.8279 -; 1.911 + . 8. What is the value in degrees and decimals of degrees of each of the angles: 2 radians, f radian, 1.75 radians? Ans. 114.5916-; 42.9718 + ; 100.2676 -i-. 9. Draw a line AB 2| in. long; at A lay off an angle of 115 30' and draw line AC f in. long; at B lay off an angle of 97 and draw line BD If in. long. Find length of CD by measuring. 10. Draw a parallelogram and measure its angles. \Vhat is their sum ? How do the angles compare? 11. Draw a> quadrilateral and measure its angles. What is their sum? 1 80 1'UA CTJCAL MA THEM A TICS ANGLES BY CHORDS 139. An angle can usually be laid off more accurately by means of a table of chords than by the protractor, and the method is often more convenient. While with the protractor it is not easy to measure more accurately than to half degrees, with the table of chords the angle may be laid off to tenths of a degree. In Table VI are given the lengths of the chords in a circle of radius 1, for central angles from to 89.9. 140. To find a chord length from the table. To find the length of the chord for an angle of, say, 27.6, find 20 in the left-hand column of the table; go to the right to column headed 7 and there find 0.467, the length of the chord for 27. Now add to 0.467 the number 0.010, found still further to the right in the column headed 0.6. This gives 0.477 as the length of the chord for 27.6. This means that, if the vertex of the angle is at the center of a circle of radius 1 in., the chord has a length of 0.477 in. For the length of the chord in a circle of any other radius, multiply this chord by the length of the radius of the given circle. Thus, for a circle of 6-in. radius, the chord for a central angle of 27.6 is 6X0.477 in. = 2.862 in. 141. To lay off an angle. How this can be done is best shown by an actual construction. Example. Lay off an angle of 48.3. From Table VI, the chord for 48.3 is 0.818. Draw a line OP, Fig. 120. i-'>. With as a center and radius 1 in., draw the arc AB. With A as a center and with radius 0.818 in., strike the arc cutting AB at C. Connect with C, and AOC is the angle 48.3 required. For this construction there are required a pair of compasses, a ruler divided into 0.01 in., and a sharp pencil. An angle larger than a right angle may be' laid off by first constructing a right angle, and then laying off the remainder adjacent to this. GRAPHICAL METHODS 181 The table can be conveniently used if a radius is taken 10 units in length. 142. To measure an angle. With the vertex of the angle as center and radius 1, draw an arc cutting both sides of the angle. Measure the chord drawn between these two points, and refer to the table of chords to find the angle. An angle larger than 90 may be measured by first laying off an angle of 90, and then measuring the remaining angle. EXERCISES 44 1. In a circle of radius 1 in., find the lengths of the chords for the following angles: (a) 43.8. Ans. 0.746 in. (e) 63 48'. Ans. 1.057 in. (b) 29.1. Ans. 0.503 in. (f) 10 40'. Ans. 0.186 in. (c) 13.3. Ans. 0.231 in. (g) 43 30'. Ans. 0.741 in. (d) 79.2. Ans. 1.275 in. (h) 78 15'. Ans. 1.263 in. 2. Construct the following angles: (a) 60, (b) 10 45', (c) 72.8, (d) 115 30', (e) 145.3, (f ) 39.9. 3. Draw the chords in a circle of 1 in. radius to form a regular polygon of 5 sides. (Each angle at the center is 72.) 4. Inscribe regular polygons of 6, 7, 8, 9, 10, and 15 sides in a circle of radius 3 in. Suggestion. Divide 360 by the number of sides to find the angle at the center of the circle determined by the side. From Table VI determine the length of the side, and space off the circumference using this length. If the number of sides is 9, the angle at center is 360 s-9 = 40. When the radius of the circle is 3, the side of the polygon is 2.052. AREAS, GRAPHICAL METHODS 143. Drawing to scale. If we wish to draw to scale the floor of a room 10 ft. by 20 ft., we may conveniently represent 1 ft. in the dimensions of the room by | in. in the drawing. Whatever dimension is measured in the drawing and given in eighths of an inch can be interpreted as feet when applied to the floor. The map of a country may be drawn on a scale of 50 miles to an inch, or any other convenient scale. 144. To construct a triangle having given two sides and the angle between these sides. Let the side AB = 10 ft., and AC = 8 ft., and the angle A between these sides be 47 45'. 182 PRACTICAL MATHEMATICS Choose a convenient scale, say, J in. for 1 ft. Draw line AB, Fig. 121, in length Y in.; lay off angle BAC=47 45'; make AC | in.; and draw CB. Then the triangle ACB is the required representation of the triangle. If it is required to find the area of the triangle, the measured length of the altitude DC in eighths of an inch times one-half of the base AB in the same unit, would give the area of the triangle in square feet. 145. To construct a triangle when given two angles and the side between these angles. Let the side AB = 30 ft., the angle A = 37, and the angle B = 49. Choose a scale, say, iV in. for 1 ft. Draw line AB, Fig. 122, in length f g in.; lay off angle A =37, and # = 49; and extend the sides of these angles till they meet. Then the triangle A CB is the required triangle represented on a scale of T V in. to 1 ft. The area of this triangle can easily be found by measuring the altitude DC and applying the rule for the area of a triangle. 146. To construct a triangle when the three sides are given. Let AB = 6Q rods, AC = 70 rods, and BC = 80 rods. Choose a scale of, say, 40 rods to the inch. Then 60 rods is represented by 1^ in., 70 rods by If in., and 80 rods by 2 in. Draw AB, Fig. 123, in length l in. With the compasses and a radius of If in. draw an arc with A as center. With B as center, and radius FIG. 123. 183 2 in., draw an arc to intersect this at C. Draw the lines AC and BC. Then triangle ABC is the required triangle. In a similar manner other shaped figures may be constructed to scale. In many cases these drawings may be divided into triangles, squares, and rectangles, which may be measured, and so the entire area of the figure be found. D n FIG. 124. Example. A piece of ground in the form of a quadri- lateral is represented by Fig. 124 to a scale of 40 rods to an inch. The area can be found by drawing the diagonal AC and the altitudes of the triangles A CD and ABC. The sum of the areas of these tri- angles equals the area of the quadrilateral ABGD. 147. Areas found by the use of squared paper. It is often convenient to find the area of an irregular figure by drawing it on squared paper (paper accu- rately ruled into small squares). The figure will usually be drawn to some scale that uses the side of one of the small squares as a unit. The method is most nearly accurate on irregular figures, and is liable to considerable error when the boundary has long straight lines, nearly parallel to the lines forming the squares. FIG. 125. 1M PRACTICAL MATHEMATICS As an illustration of the method, find the area of the circle in Fig. 125, if the circle is drawn on a scale of 1 in. to a side of the squares. First, determine how many squares are wholly within the circle. Second, count as whole squares the squares that are half or more than half within the circle, and neglect those squares that are less than half within the circle. Here it is most convenient to count the squares in one quar- ter of the circle, and then multiply by 4 to get the area of the whole circle. From the figure, it is seen that there are 30 whole squares in the quarter of the circle AOB. Counting the squares marked 1, gives 8 partial squares to be taken as whole squares. The squares marked are not counted. 30+8 = 38 squares for the quarter circle. 38X4 = 152 squares for the circle. .'. area of the circle is 152 in. 2 By formula [24], area = 3.1416X7 2 =154- in. 2 148. Other methods for approximating areas. (a) The planimeter is an instrument for estimating areas. There are several forms of this device; but, as instructions for its use are given with each instrument, it will not be described here. (6) The area, when very irregular, can often be estimated quite accurately by cutting full size or to scale out of card- board or sheet tin. Weigh on accurate scales the piece of tin or cardboard ; also weigh a square unit of the same material. Divide the weight of the piece by the weight of the square unit. The quotient is the number of square units in the figure. (c) Other methods will be found in Chapt. XXXIII. EXERCISES 46 1. The two sides of a triangle are 18 ft. and 24 ft., respectively, and include an angle of 98. Find the length of the other side, and find the area of the triangle by two separate sets of measurements, that is, draw two altitudes and use two sides as bases. GRAPHICAL METHODS 185 2. The base of a triangle is 10 ft., and the other two sides are 7 ft. and 5 ft. respectively. Find graphically the length of the altitude to the base. What is the area of the triangle? 3. Find the area of the triangle in exercise 2 by rule for the area of a triangle when the three sides are given. Ans. 16.25 ft. 2 4. In a triangle, one side is 28 ft. and the adjacent angles are 39 45' and 49 30'. Find the lengths of the other sides and the area of the triangle. 5. The angles of a triangle are 48, 78, and 54. Find the length of the side opposite the angle 78, if the side opposite 48 is 32 ft. 6. The two sides AB and BC of a triangle are 44.7 ft. and 96.8 ft., respectively, the angle ABC being 32. Find (a) the length of the perpendicular drawn from A to BC; (b) the area of the triangle ABC] (c) the angles at A and C. Ans. (a) 23.69 ft. ; 1 i < . < i Scale 5 I ,,,!,,,, , , , , . 1 , , , I 10 5 Vernier FIG. 141. scale, which is the vernier. The vernier is so divided that 10 divisions of its scale just equal 9 divisions of the graduated scale. If the mark on the vernier coincides with a division, say, the division as in Fig. 141, of the graduated scale, then the division 1 on the vernier stands at 0.9 on the scale; 2 on the vernier at 1.8 on the scale; and so on for the other divisions. If the vernier be moved along so that one of its divisions, as 4 in Fig. 142, coincides with a division of the scale, then the \ Scale / / 20 15 \i 1 i i i r ! i i 10 5 i ! i [ i i ! i , , ,\ N i i M Mil 10 ._ 5 . Vernier FIG. 142. division on the vernier just to the right or left of the coinciding division lacks 0.1 of a scale division of coinciding with a scale division. The next division of the vernier to the right or left lacks 0.2 of a scale division of coinciding with a scale division, and so on. In this case, the point on the vernier is removed 0.4 of a division to the left of a scale division. The reading then in Fig. 142, that is, the distance from the division on the scale to the division on the vernier, is 7.4. If the scale division is tenths of an inch, then the reading is 0.74 in. 192 PRACTICAL MA THEM A TICK If the vernier is moved to the left so that 6 on the vernier coincides with a division on the scale, then on the vernier is O.G of a scale division to the left of a scale division. It is evident that any number of divisions on a scale could be equal to one greater number of divisions on a vernier, and the readings could be made in a similar way. For instance, in instruments for measuring angles, if the scale divisions are to i, then a vernier with 30 divisions equaling 29 divisions of the scale will give a reading to ^V of \ or 1' of angle. 157. Micrometer with vernier. A micrometer that reads to thousandths of an inch may be made to read to ten-thou- sandths of an inch by putting a vernier on the barrel, so that 10 divisions on the vernier correspond to 9 divisions on the Fio. 143. thimble. There are eleven parallel lines on the sleeve occu- pying the same space as ten lines on the thimble. These lines are numbered 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. The difference between one of the ten spaces on the sleeve and one of the nine spaces on the thimble is ^V of a space on the thimble or m-Jinr inch in the micrometer reading. In Fig. 143(6), the third line from on the thimble coincides with the first line on the sleeve. The next two lines do not coincide by -fa of a space on the thimble, the next two marked 5 and 2 are i\ of a space apart, and so on. When the mi- crometer is opened the thimble is turned to the left and each space on the thimble represents mVu inch. Therefore, when the thimble is turned so that the lines 5 and 2 coincide the micrometer is opened Vo of TuVa inch or njXniTf inch. If the thimble be turned further, so that the line 10 coincides with GRAPHICAL METHODS 193 the line 7 on the sleeve as in (c) , the micrometer has been opened TTnhnj- inch- To read a micrometer graduated to ten-thousandths, note the thousandths as usual, then observe the number of divisions on the vernier until a line is reached which coincides with a line on the thimble. If it is the second line marked 1, add nrfanr; if the third marked 2, add T -jHhnr> etc. As an exercise give the different readings shown in the figure. CHAPTER XIV PRISMS 158. Definitions. Two planes are said to be parallel when they will not meet however far they be extended. That is, they are everywhere the same distance apart. A line is parallel to a plane when it will not meet the piano however far it may be extended. A line is perpendicular to a plane when it is perpendicular to every line of the plane that passes through its foot. A prism is a solid whose ends, or bases, are parallel polygons, and whose sides, or faces, are parallelograms. Fig. 144 is a prism. The bases ABCI) and EFGH are parallel polygons. The faces AEFB, BFGC, etc., are parallelo- grams. The lines AB, BC, etc., and EF , FG, etc., are base edges. The lines AE, BF, etc., are lateral edges. If the lateral edges are per- pendicular to the bases of the prism, the prism is a right prism. If the lateral edges are not per- pendicular to the bases, the prism is an oblique prism. In an oblique prism, the faces are parallelograms; but, in a right prism, they are rectangles. A prism is called triangular, square, rectangular, hexagonal, etc.; according as its bases are triangles, squares, rectangles, hexagons, etc. Fig. 145 is a right triangular prism. A cross section of a prism is a section that is perpendicular to the edges of the prism. In Fig. 145, MNO or either base is a cross section. The sum of the edges of the base is the perimeter of the base. 194 FIG. 144. PRISMS 195 The altitude of a prism is the perpendicular line between the two bases, as mn, Fig. 144. In a right prism, the altitude is the same length as an edge. In an oblique prism, this is not true. A right prism that has rectangles for bases is called a rec- tangular solid. The cube is a rectangular solid all of whose six faces are squares. 159. Surfaces. The right prisms are the forms of prisms that are met with usually in practical work. They are the ones considered here. The lateral area of a right prism is the area of its faces, not including the two bases. Since the faces are rectangles their areas can be easily found. As the area of each face is the product of its base by its altitude, the sum of the areas of the faces, or the lateral area of the prism is given by the following: RULE. The lateral area of a right prism equals the perimeter of the base times the altitude. The total area of the prism equals the lateral area plus the area of the two bases. Since the cube has six equal square faces, the total area is six times the square of an edge. If S stands for lateral area, T for total area, A for area of each base, p for perimeter of the base, h for altitude, and a for an edge, the rules are stated in the following formulas : [33] S=ph. [34] T = ph+2A. [35] T = 6a 2 , for the cube. [36] p = S-f-h. [37] h = S^p. Example. Find the total area of a right triangular prism, of altitude 20 ft., if the edges of the base are 2 ft., 3 ft., and 4ft. o JC FIG. 145. 196 PRACTICAL MATHEMATICS Solution. p = 2+3+4 = 9. By [7], A = Vs(s-a) (s-6) (s-c). s c=4J 4 = . .'. A = -V/475X 2.5 XI. 5X0.5 = \/K.4375 = 2.9047. By [34], r = pft+2A. /. 7 T = 9X20+2X2.9047= 185.81 -ft. 2 Ans. 160. Volumes. In Fig. 146, there are as many cubic inches on the base ABCD as there are square inches in its area, and since there are as many layers of cubic inches in the rectangular H a A / / / z / / / A \ / / / / / / / / /' -i / / / / 2 / / / '/< ' k i i '1 \'\ v/ / ' !,' f / / / / /I i-; '! !' 1 .' DA 7T& 7 r f / 1 > / / i -- -/-- "/" " -/ -t -f y- B FIG. 146. solid as there are inches in the altitude AE, the total number of cubic inches in the solid is found by multiplying the number of square inches in the base by the number of linear inches in the altitude. Any right prism can be taken in the same man- ner as the rectangular solid; hence the following: RULE. The volume of a right prism equals the area of the base times the altitude. If the prism is a rectangular solid this rule becomes: RULE. The volume of a rectangular solid equals the con- tinued product of the length, breadth, and height. If V stands for volume, and the other letters as before, we have these formulas: [38] V = Ah, for any prism. [39] V = a 8 , for the cube. [40] h = V-^A. [41] A = Vn-h. PRISMS 197 Example 1. One of the concrete pillars to hold up a floor in a concrete building has a cross section that is a regular hexagon. The dimensions are as shown in Fig. 147. Find its weight if concrete weighs 138 Ib. per cubic foot. Solution. The volume is found by formula [38] V = Ah, where h is 12 ft. and A the area of a hexagon with one side 8 in. A=6X4 2 Xl.732in. 2 = 6X4 2 X1.732 V = 144 12X6X4 2 X1.732 ft.' 144 ft. 1 1 \ .. w . , , 12X6X4 2 X1.732X138 1010 Weight = - = 19121b. 144 Example 2. How deep is a cistern in the form of a hexagonal prism to hold 100 bbl. if the base is 4 ft. on an edge? Solution. By formula [40], hV-r-A. F=100 bbl. = 100X31^X231 cu. in. A = area of hexagon with edge of 4 ft. Alt. of one triangle of hexagon = 2X1. 732 = 3.464 ft, Area of hexagon = 6X|(4X 3.464) =41 .568 sq. ft. 41.568 sq. ft. =41.568X144 sq. in. FIG. 147. EXERCISES 46 1. Find the volume of the following rectangular solids: (a) 8 ft. by 11 ft. by 32 ft. Ans. 2816 ft. 8 (b) 3 ft. by 4| ft. by 7f ft. Ans. 104f ft. 3 (c) 30 ft. 6 in. by 41 ft. 6 in. by 12 ft. Ans. 15,189ft. 3 (d) 17 ft. 2 in. by 19 ft. 3 in. by 3 ft. 7 in. to nearest 0.001 ft. 3 Ans. 1184.142+ ft. 3 (e) 3 ft. 4 in. by 14 ft. 7 in. by 11 ft. 11 in. to nearest 0.001 ft. 3 Ans. 579.282+ ft. 3 2. If the inside dimension of a cubical tank is 4 ft., find the number of barrels it will hold when full. Ans. 15.2 bbl. 3. Find the number of cubic yards of earth to be excavated in digging a cellar 40 ft. by 26 ft. by 7 ft. Ans. 269.63 -. 198 PRACTICAL MATHEMATICS FIG. 148. 4. A rectangular solid has the three dimensions: 5.25 ft., 17.23 ft,, and 4.062 ft. Find its capacity in bushels. Ant. 295.3 bu. 5. Find the weight of a block of granite (specific gravity 2.9) 6 ft. by 7 ft. by 1 J ft. An*. 11,418} Ib. 6. Find the weight of a solid cast-iron pillar in the form of a hexa- gonal prism 12 ft. high and 3 in. on edge of base. An*. 875 Ib. nearly. 7. A common brick is 2 in. by 4 in. by 8 in. Find the number of bricks in a pile 8J ft. by 4 ft. by 10 ft. (Use cancellation.) Ans. 9180. 8. How many rectangular solids 3 in. X4 in. X9 in. will fill a box 4 ft. X 6ft.X8ft? Ana. 3072. 9. One hundred eighty square feet of zinc are required for lining the bottom and sides of a cubical vessel. How many cubic feet of water will it hold? Ans. 216. 10. A box car that is 36 f ft. long and 8$ ft. wide, inside measurements, can be filled with wheat to a height of 5J ft. Find how many bushels of wheat it will hold if f cu. ft. are 1 bushel. Ans. 1317s. 11. How many cubic yards of soil will it take to fill in a lot 50 ft. by 100 ft., if it is to be raised 3 ft. in the rear end and gradually sloped to the front where it is to be 1J ft. deep? Ans. 416 J. Suggestion. The vertical cross section the long way of the lot is a trapezoid having parallel sides *"t~| fx. of lengths 3 ft. and \\ ft. respectively, and an altitude of 100 ft. 12. Find the number of cubic yards of crushed | rock to make a road one mile in length and of cross 5] section as shown in Fig. 148. Solution. The area of the vertical cross section can be found by considering it as two trapezoids each having parallel sides of 8 in. and 1 ft. respectively, and altitudes of 10 ft. u -2 * Area of cross section = (-f 1)10 = 16 J sq. ft. No. of cu. ft. of rock =5280X16? =88,000. No. of cu. yd. of rock =88,000-4-27 = 3259^ : . Any. 13. One cubic inch of steel weighs 0.29 Ib. An I-beam has a cross section as shown in Fig. 149, and a length of 22 ft. Find its weight. Ans. 1531+ Ib. 14. Find the weight of steel beams 10 ft. in length and of the cross sections given in Fig. 40, page 121. Am. 821 Ib.; 209.5 Ib.; 300.2 Ib. 15. Find what weight of load will be required to cover a surface 48 ft. long and 32 ft. wide, with lead fa in. thick, allowing 5% of weight for joints. Ann. 4 tons nearly. Fiu. 149. PRISMS 199 16. A rectangular box is made with |-in. sheet steel. Find weight of box (no allowance for corners) if it is 4 ft. by 3.4 ft. by 2.6 ft. (leu. in. = 0.29 lb.). Ans. 342.8 Ib. 17. What length must be cut from a bar of steel ^ in. by 1| in. in cross section in order to make 1 cu. ft.? Ans. 230.4 ft. 18. The base of a right prism is a triangle whose sides are 12 ft., 15 ft., and 17 ft., and its altitude is 8| ft. Find the lateral area. Ans. 374 ft. 2 19. Find the volume of the above prism. Ans. 745.872 ft. 3 Suggestion. Use formula [7] to find the area of the ba-se. 20. Find the volume of a cube whose diagonal is 8 in. Ans. 98.534+ in. 3 21. The cost of digging a ditch, including all expenses and profits, is estimated at 27 cents per cubic yard. Find the cost of digging a ditch 15 miles long, 10 ft. wide at the bottom, 20 ft. at the top, and 6 ft. deep. Ans. $71,280. 22. A river is 76 ft. wide and 12 ft. deep and flows at the rate of 3| miles per hour. How many cubic feet of water per minute passes a given point on the river? Ans. 280,896. Suggestion. The flow per minute is the volume of a prism of water of the cross section of the river and in length equal to the distance the water flows per minute. 23. A cistern 7 ft. long and 4 ft. broad holds 35 bbl. when three-fifths full; find the depth of the cistern. Find the cost of lining the bottom and sides with zinc at 5 cents per square foot. Ans. 8 ft. 9.27+ in.; $11.05. 24. An iron casting shrinks about | in. per linear foot in cooling down to 70 Fahrenheit. What is the shrinkage per cubic foot: Ans. 53.44 in. 3 Suggestion. The shrinkage per cubic foot is the difference between a cubic foot and the volume of a cube 11| in. on an edge. 25. A fall of | in. of rain is how many barrels per square rod? Ans. 4f. 26. A flow of 300 gallons per second will supply water for how deep a stream, if the stream is 4 ft. broad and flows 5 miles per hour? Ans. 16yf in. Suggestion. The computation in the form of cancellation is 300X231X60X60 . -I t> 3 2 4X5X5280X12X12 27. Find cost at 40 cents a pound for sheet copper to line bottom and sides of a cubical vessel 7 ft. on an edge, if the sheet copper weighs 12 oz. per square foot. How many barrels will the vessel hold? Ana. $73.50; 81.45+. 28. A stream, flowing 5 miles per hour, must be of how large cross section to supply 1? in. in depth of water per week for 160 acres of land? Ans. 28f in. 2 200 PRACTICAL MATHEMATICS 29. If a glass rod 1 in. long at centigrade is increased to 1.000008 in. at 1 centigrade, find the increase in volume of a cubic inch of glass when heated from to 1 centigrade. Ant. 0.000024+ in. 1 STONEWORK 161. Terms. A full description cannot be given here. For such the student is referred to a trade handbook. Stonework where the stones are broken with the hammer only is called rubble-work. If the stones are laid in courses it is called coursed rubble. When the stones showing in the out- side face of the wall are squared, the work is designated as ashlar. If all the stones of a course are of the same height, the work is called coursed ashlar. When the stones are of dif- ferent heights it is called broken ashlar. Ashlar work is both hammer-dressed and chisel-dressed. Any stonework where any other tool than the hammer is used for dressing is called cut- work. 162. Estimating cost of stonework. In estimating the cost of stonework, the custom varies greatly; we find it vary- ing even among the contractors of the same city. Cut-work is often measured by the number of square feet in the face of the wall. Rubble- work is almost universally measured by the perch, but the pereh used varies greatly. The legal perch of 24f cu. ft. is seldom used by stone masons. The perch of 16$ cu. ft. is the one most used. That of 25 cu. ft. or of 22 cu. ft. is sometimes used. Stonework on railroads is usually measured by the cubic yard. Openings, as a rule, are not deducted if containing less than 70 sq. ft. EXERCISES 47 1. Find the cost of laying a hammer-dressed ashlar wall, 45 ft. long, 6 ft. high and 2 ft. thick, at $2.75 a perch, using the 22-cu. ft. perch. Ana. $67.50. 2. Find the cost of making a rubble-work wall at $3.25 a perch, includ- ing all the material, under a building 25 ft. by 60 ft., the wall to be 30 in. thick and 8 ft. high. Use the 16J-cu.ft. perch. Ans. $630.30. 3. Find the cost of laying the stonework in two abutments for a bridge, each abutment to be 8 ft. high, 3J ft. thick, 20 ft. long at the bottom and PRISMS 201 15 ft. at the top. (The shape is a trapezoid.) The price for laying is $1.75 per cubic yard. Ans. $63.52. 4. Find the cost of building a sandstone rubble-work wall for a base- ment 36 ft. by 47 ft., 8 ft. high, and the wall 18 in. thick. (Use outside measurements and make no allowances for openings.) The stone costs $1.40 a perch (25 cu. ft.), and the labor of laying, including lime and sand, is $1.25 a perch. Ans. $211.15. BRICKWORK 163. Brick. The size of brick varies. In the United States, there is no legal standard. The common brick is approximately given as 8 in. by 4 in. by 2 in. In the New England States, they average about 7f in. by 3f in. by 2j in. In most of the western states, the common brick averages about 8| in. by 4 1 in. by 2^ in. The brick from the same lot may vary as much as -fo in., depending upon the degree to which they are burnt. The hard-burned bricks are the smaller. Walls made from these bricks are about 9, 13, 18, and 22 inches in thickness, that is, 1, 1^, 2, and 1\ bricks. Pressed bricks are usually larger than common bricks; the prevailing size is 8f in. by 4| in. by 2f in. 164. Estimating number, and cost of brickwork. This cannot be discussed in full here. The student is referred to an architect's and builder's handbook for the details. Plain walls are quite universally figured at 15 bricks to the square foot, outside measure, of 8- or 9-in. wall; 22^ bricks per square foot of 12- or 13-in. wall; 30 bricks per square foot of 16-, 17-, or 18-in. wall; and 1\ bricks for each additional 4 or 4| in. in thickness of wall. These figures are used without regard to the size of the bricks, the effect of the latter being taken into account in fixing the price per thousand. No deduction is made for openings of less than 80 sq. ft., and when deductions are made for larger openings the width is measured 2 ft. less than the actual width. Hollow walls are measured as if solid. For. chimney-breasts, pilasters, detached chimneys, and other forms the student is referred to a builder's handbook. Example. Find the cost of brickwork in the walls of a house 26 ft. by 34 ft., no cross walls, the basement walls to be 13 in. thick; the first story walls, 13 in. thick; second story walls, 9 in. thick; height of basement walls to top of first floor 202 PRACTICAL MATHEMATICS joists, 9 ft.; from first floor joists to top of second floor joists, 10 ft. 6 in.; from second floor joists to plate, 9 ft. (The chimneys, openings, and pressed brickwork are not considered as the method of estimating has not been given.) The cost of brick and laying, including lime, sand, scaffolding, etc., is $10 per thousand. Solution. Basement and first story walls: Girth of house = 2X26 ft. +2X34 ft. = 120 ft. Height of wall = 9 ft. + 10 ft. 6 in. = 19 ft. Thickness of wall is 13 in. and hence 22 brick are counted per square foot. Hence the number of bricks required is 120 X 19 X22 = 52,650. Similarly for the second story, 120X9X15=16,200. /.total number of bricks = 52,650+ 16,200 = 68,850. .'.cost = $10X68.850 = $688.50. Ans. EXERCISES 48 1. Find the cost at $9.50 per thousand to cover brick, material, and labor to build a brick wall on the front and one side of a corner lot 50 ft. by 100 ft. ; the wall to be 6J ft. high and a brick and a half thick, allow- ance being made for one opening 12 ft. in length (count it 2 ft. leas). Ans. $194.51. 2. The following is about the cost of furnishing and laying 1500 bricks, or one day's work : 1500 bricks at $6.00 per M, ~~ 1 barrel of lime at $1.00, ? 9 bushels of sand at 5 cents, x * 1 day's work for mason at $5.76, ! 1 day's work for helper at $3.32. 4'2 tf ^ Using this as a basis for estimating, find the cost of building the walls of an apartment house 25 Fio. 160. ffc by 54 f t ^ 41 ft nign j n f ront ftnc i 36 f t ^ tne rear. The walls are to be 13 in. thick and no allowances for open- ings. Ans. $1759.20. 3. Find the cost of common brick in the pier with a cross section as shown in Fig. 150, and a height of 12 ft. 6 in., at $7.00 per thousand. Count 20 brick to 1 cu. ft. Ans. $24.15. 4. How many enamel brick 4 in. by 8 in. are required to face a wall 30 ft. long and 12 ft. high, deducting for two windows 4 ft. by 8 ft. and one door 3ft. 4 in. by 10ft.? Ans. 1182. CHAPTER XV CYLINDERS 165. Definitions. A right circular cylinder, or a cylinder of revolution, is a solid formed by revolving a rectangle about one of its sides as an axis. Thus, in Fig. 151, the rectangle OABO' is revolved about OO' as an axis. This forms the cylinder as drawn. From this definition, the two bases are circles, and the lateral surface is a curved surface. The axis of the cylinder is the line 00' joining the centers of the bases. It is per- C' FIG. 151. pendicular to the bases, and hence it is equal to the altitude of the cylinder. The cross section of a cylinder is a section perpendicular to the axis. A cylinder is inscribed in a prism when its bases are in- scribed in the bases of the prism (see Fig. 152). The prism is then circumscribed about the cylinder. A cylinder is cir- cumscribed about a prism, or the prism is inscribed in the cylinder, when the bases of the prism are inscribed in the bases of the cylinder. 166. Area and volume. If the lateral surface of a right cylinder could be peeled off and spread out, it would form a rectangle of width equal to the altitude of the cylinder, 203 204 PRACTICAL MAT HEMATICS and of length equal to the circumference of the base of the cylinder. From this we get the following: RULE. The area of the lateral surface of a right cylinder equals the circumference of the base times the altitude. The total area equals the lateral area plus the area of the two bases. From a consideration similar to that for the prism, Art. 160, the volume of the cylinder is obtained by the following: RULE. The volume of a cylinder equals the area of the base times the altitude. The similarity of the cylinder and the prism is seen if the cylinder is thought of as a prism having a very great number of sides to the base. The above rules hold when the cylinder is not circular. Since the altitude times the cross section gives the volume, the altitude equals the volume divided by the area of the base or cross section. Also the area of the base equals the volume divided by the altitude. If S stands for lateral area, T for total area, A for area of base, and h for altitude, the rules given are stated in the following formulas: [42] S = Ch [43] V = Ah [44] h=V-rA = V^-irr 2 . [45] A = V-hh. 167. The hollow cylinder. The volume of a hollow cylinder may be found by subtracting the volume of the cylindrical hollow from the volume of the whole cylinder. If R is the radius of the cylinder, and r the radius of the hollow, then the volume of the hollow cylinder is as follows: [46] V = 7 rR%-7TT 2 h = Th(R 2 -r 2 )-7rh(R+r)(R-r). Example 1. Find the number of cubic inches of copper in a hollow cylinder 7 in. long, inner diameter 6 in. and outer diameter 8 in. Solution. By [46] , V = irh(R+r)(R-r). /. F = 3.1416X7(4+3)(4-3) = 153.938 in. 8 Example 2. Find the effective heating surface of a boiler of diameter 5 ft. and length 16 ft.; with 54 tubes 3| in. in CYLINDERS 205 diameter, assuming that the effective heating surface of the shell is one-half the total surface. Solution. By [42], effective heating surface of the shell = 145.299 ft. 2 Effective heating surface of 54 tubes /.total heating surf ace = 145.299 + 791. 683 = 937 ft. 2 nearly. Example 3. What is the weight of a cylindrical shaft of marble 3 ft. in circumference and 9 ft. high? Solution. r = f (3 -f-3.1416) =0.47746. By [43], F = 7rr 2 /i = 3.1416X0.47746 2 X9 = 6.4457 ft, 3 Weight = 2.7 X 62.5 X 6.4457 = 1087 Ib. Since 62.5 Ib. = weight of 1 cu. ft. of water, and 2.7 = specific gravity of marble. EXERCISES 49 1. If the radius of the base of a right circular cylinder is 5 in. and the altitude is 8 in., find the lateral area, the total area, and the volume. Ans. 251.33 in. 2 ; 408.4 in. 2 ; 628.3 in. 3 2. If r, d, h, A, S, and V have the meanings given in the preceding articles: (a) Given A =48 sq. in. and h = 16 in. ; find r, S, and F. (b) Given F = 4800 cu. ft. and A = 160 sq. ft.; find d and S. Ans. (a) r = 3.909 -in.; S = 393.0 -sq.in.; 7 = 768 cu. in. 3. Find the area of the rubbing surface in a steam cylinder 91J in. in diameter, the stroke of the piston being 6 ft. 8 in. Ans. 159.7 ft. 2 4. Find the total area and the volume of a cylinder whose radius is 7 ft, and whose altitude is 12 ft. Ans. 835.66 ft. 2 ; 1847+ ft. 3 5. Find the number of barrels each of the follow- ing cylindrical tanks will hold: (a) Diameter 5ft,, depth 5ft. Ans. 23.3. (b) Diameter 8 ft,, depth 9 ft, Ans. 107.4. (c) Diameter 20 ft., depth 19 ft. Ans. 1417.5. (d) Diameter 30 ft., depth 20 ft. Ans. 3357.3. Suggestion. Use 1 bbl. =4.211 cu. ft. 6. Find the number of cubic yards of earth removed in digging a tunnel 175 yd. long, if the cross section is a semicircle with a radius of 14 ft. Ans. 5986.5. 7. A peck measure is to be made 8 in. in diameter. it be? (See Fig. 153.) FIG. 153. How deep should Ans. 10.695 in. 206 PRACTICAL MATHEMATICS Suggestion. 1 pk. - J of 2150.42 cu. in. =537.605 cu. HI. Area of base -8* X0.7854 -50.2656 sq. in. Altitude volume -r-area of base. 8. A cylindrical oil tank 3 ft. in diameter and 10 ft. long will contain how ninny gallons of oil? Ans. 528.8. 9. The external diameter of a hollow cast-iron shaft is 18 in., and its internal diameter is 10 in. Calculate its weight if the length is 20 ft., and cast iron weighs 0.26 Ih. per cubic inch. Ans. 10,980 Ib. nearly. 10. Water is flowing at the rate of 10 miles per hour through a pipe 16 in. in diameter into a rectangular reservoir 197 yd. long and 87 yd. wide. Calculate the time in which the surface will be raised 3 in. Ana. 31.38 minutes. 10X5280 Suggestion. Ten miles per hour is =880 ft. per minute. 60 Now find the number of cubic feet of water that will flow through the 16-in. pipe in 1 minute. Then find the number of cubic feet required to fill the reservoir 3 in. The quotient found by dividing the required number of cubic feet by the flow per minute is the time in minutes. 11. In a table giving weights and sizes of square nuts for bolts, a nut 2 in. square and li in. thick, with a hole Ifa in. in diameter, has a given weight of 1.042 Ib. Use wrought iron and find this weight. 12. Find the length of steel wire in a coil, if its diameter is 0.025 in. and its weight is 50 Ib. (Use 1 cu. in. weighs 0.29 Ib.) Ans. 29,270 ft. nearly. 13. A cylindrical cistern is 6 ft. in diameter. Find the depth of the water when containing 10 barrels. .Ixx. 17. S7 in. 14. A nugget of gold is dropped into a cylinder of water and raises the surface of the water li in. If the cylinder is 2 in. in diameter and 25 grains of gold are worth $1, find the value of the nugget. Ans. $767. 16. In a table giving size, etc., of wrought-iron washers, a washer 3J in. in diameter with a hole 1 J in. in diameter is A in. thick. Find the number of these in a keg of 200 Ib. Ans. 582. 16. Find the per cent of error in the following FIG. l.vi. rule which applies to round bars, (a) for wrought iron, (b) for cast iron. Multiply the square of the diameter in inches by the length in feet, and that product by 2.6. The product will be the weight in pounds nearly. Ans. (a) 1.5-% too small; (b) 6.1 + % too large. 17. A conduit made of concrete has a cross section as shown in Fig. 154. How many cubic yards of concrete are used in making 500 yd. of this conduit? Ans. 1113.4 yds. CYLINDERS 207 18. In a table giving weights and areas in cross section of steel bars, a round steel bar f in. in diameter has its area given as 0.1104 in. 2 and weight 0.376 Ib. per linear foot. Verify these results if steel weighs 489.6 Ib. per cubic foot. 19. A water tank in a Pullman car has a vertical cross section as shown in Fig. 155, and a length of 52 in. Find its capacity in gallons. The arc is a part of a circle. Ans. 68.3 gal. 20. A rod of copper 8 in. long and 1 in. in diameter is drawn into wire of uniform thickness and 200 ft. long. Find the diameter of the wire. Ans. 0.0577 in. 21. The rain which falls on a house 22 ft. by 36 ft. is conducted to a cylindrical cistern 8 ft. in diameter. How great a fall of rain would it take to fill the cistern to a depth of 1\ ft.? Ans. 5.712 in. 22. Find the weight of 7 miles of |-in. copper wire, if copper weighs 0.319 Ib. per cubic inch. Ans. 1736.26 Ib. 23. Find the greatest tensile force a copper wire 0.25 in. in diameter can stand without breaking. (See Table VII.) Aris. Less than 1472.6 Ib. FIG. 155. By tensile force is meant a pulling force. The problem asks how great a weight the wire will hold when hanging vertically. 24. Calculate the size of a square wrought-iron bar to stand a pull of 43,000 Ib. (See Table VII.) Ans. 0.927 in. square. 25. What should be the diameter of a round cast-iron bar which is subjected to a tension of 30,000 Ib., if the pull on each square inch of cross section is 2400 Ib.? Ans. 3.989 in. 26. Find the pull per square inch necessary to break a rod 2| in. in diameter, which breaks with a load of 270,000 Ib. Ans. 55,000 Ib. nearly. 27. If a wrought-iron bar 2 in. by lj in. in cross section breaks under a load of 125,000 Ib., what load will break a wrought-iron rod 2 in. in diameter? Ans. 245,400 Ib. nearly. 28. A cast-iron bar has an elliptical cross section with axes 6 in. and 4 in. Find the pull per square inch of cross section under a total tensile load of 125,000 Ib. Ans. 6631 Ib. 29. A wrought-iron cylindrical rod 2000 ft. long and 1^ in. in diameter is suspended vertically from its upper end. What is the total pull at this end, and the pull per square inch of cross section? Ans. 11.87511).; 6720 Ib. JMS PRACTICAL MA THEM A TICS 1 Fio. 150. 30. Find the length of a wrought-iron bar, supported vertically at its upper end, that will just break under its own weight. Ans. 15,000 ft. 31. About what is the strength of an 18 gage B. and S. wrought-iron wire? (See Table VII.) An*. 63.8 Ib. 32. A trolley wire is copper and 00 gage B. and 8. Find the pull it will take to break the wire. Ans. 3130 Ib. 33. A cylinder to cool lard is 4 ft. in diameter and 9 ft. long and makes 4 R. P. M. As the cylinder revolves, the hot lard covers the surface I in. deep. How many pounds of lard will it cool in 1 hour if the specific gravity of lard is 0.9? Ans. 15,900 nearly. 34. If a tank 5 ft. in diameter and 10 ft. deep holds 10,000 Ib. of lard, what will be the depth of a tank of 2000 Ib. capacity if its diameter is 3 ft. ? If this tank has a jacket around it on the bottom and sides 3 in. from the surface of the tank, how many gallons of water will the space between the jacket and tank hold? Ans. 5J ft.; 124 nearly. 36. Find the volume of a wash boiler if the bottom is in the form of a rectangle with a semicircle at each end. The rectangle is 10 in. by 14 in., and the semicircles are on the smaller dimensions. The depth of the boiler is 16 in. .4ns. 15.14 gal. 36. Find the height of a 10-gallon wash boiler whose base is 10 in. wide with semicircular ends, the length of the straight part of the sides being 9i in. Ans. 13.5 in. 37. A certain handbook gives the following "rules of thumb" for finding the volume in gallons of a cylindrical tank: (1) V (in gal.) = (diameter in feet) 1 X5J X (height in feet). (2) V (in gal.) = (diameter in feet)*X of height in inches less 2% of same. Find per cent of error for each rule. Ans. (1) 0.003 + % too small; (2) 0.08 + % too large. Remark. Rule (2) is a quick rule to apply if the 2% is disregarded. This rule is in common use by many estimators. 38. In making the pattern of a teakettle with an elliptical bottom to hold 6 quarts, it is decided to have the bottom an ellipse with axes 10 in. and 7 in. Find the height. Ans. 6.30+ in. 39. How long a piece of copper will it take to make the body of the above teakettle? KTse Cir. =2x-J-') Ans. 26.70 in. 40. In drilling in soft steel, a IjViri. twist-drill makes 37 R. P. M. with a feed of ' in. Find the number of cubic inches cut away in 3J minute*. Ans. 3.1+. 41. A j 3 -in. twist-drill makes 310 R. P. M. with a feed of T*J in. Find the volume cut away in 31 minutes. Ans. 0.24 in. 3 CYLINDERS 209 42. In turning a steel shaft 6 in. in diameter and 4 ft. long, the cutting speed was 36 ft. per minute and the feed 0.125 in. Find the time re- quired for turning the shaft. If the depth of the cut was 0.05 in., find the amount of metal removed. Ans. 16 1 min. nearly; 45.2 in. 3 43. Find the number of pounds of cast iron turned off per hour in the following: (Consider the cutting as if on a plane.) Speed per min. Depth of cut Breadth of cut (a) 37.90 ft. 0.125 in. 0.015 in. Ans. 13.3. (b) 25.82 ft. 0.015 in. 0.125 in. Ans. 9.06. (c) 25.27ft. 0.048 in. 0.048 in. Ans. 10.9. 44. A steam chest cover is 42 in. by 24 in. How many steel studs 1 j in. in diameter should be used to hold the cover* if the steam pressure is 160 Ib. per square inch? The diameter of the bolts at the bottom of the thread is 1.065 in. Allow a stress of 11,000 Ib. per square inch. Ans. 16.5 bolts or 18 to make the number even. Solution. 42 X 24 X 160 Ib. = 161,280 Ib. total pressure. 161,280-^11,000 = 14.662 in. 2 =area in cross section of all bolts. 0. 7854 Xl.065 2 = 0.89082+ in. 2 = area of one bolt. 14.662 -7-0.89082 = 16.5-. .'. number of bolts is 18 to be even. 46. The flanges at the joining of two ends of flanged steam pipes 9 in. in inside diameter are bolted together by 12 bolts \ in. in diameter. If the pressure in the pipes is 200 Ib per square inch, find what each bolt must hold. How much is this per square inch cross section of the bolts? Suppose that bolts have 10 pitch U. S. S. thread. This makes the root diameter 0.620 in. (See Fig. 157.) Ans. 1060.3 Ib.; 3512+ Ib. 46. As in the last exercise, if the steam pipe is 18 in. in diam- FIG. 157. eter, and allowing the same pull per square inch of cross section of each bolt, find the number of bolts If in. in diameter at a joint of the pipe. (A li-in. bolt with 7 pitch U. S. S. thread is 0.940 in. in diameter at root of thread.) Ans. 22. 47. The following rule is often used to find the heating surface of any number of tubes in a steam boiler: Multiply the number of tubes by the diameter of one tube in inches, this product by its length in feet, and then by 0.2618. The final product is the number of sq. ft. of heating surface. Using this rule, what is the heating surface of 66 3-in. tubes each 18 ft. long? Does the rule give the correct result? Ans. 933 ft. 2 ; yes. 48. To find the water capacity of a horizontal tubular boiler, find f the volume of the shell and subtract from this the volume of all the tubes. Find the water capacity of a horizontal tubular boiler 185 ft. long, 66 in. ui diameter, with 72 3-in. tubes. Ans. 227.6+ ft. 3 49. The steam capacity of a horizontal boiler is often reckoned as 210 PRACTICAL MATHEMATICS one-third the volume of the shell. Find the steam capacity of a hori- zontal boiler 18 ft. long and 78 in. in diameter. Ans. 199 + ft. J 50. Use the following rule and find the heating surface of a boiler 12 ft. long, 5 ft. in diameter, and having 52 2J-in. tubes. Ana. 518.7 ft. 1 to 550.7 ft, 1 Rule. In finding the heating surface in a horizontal boiler, it is cus- tomary to take one-half to two-thirds of the lateral area of the shell, the lateral area of the tubes, one-half to two-thirds the area of the ends of the boiler, and subtract the areas of both ends of the tubes. 61. A steam boiler is 72 in. in diameter, 18 ft. long, and contains 70 tubes 4 in. in diameter. Find the heating surface, using one-half in the rule. Ans. 1505+ ft. 1 62. Find the steam capacity of a boiler 4 ft. in diameter and 16 ft. long, if the height of the segment occupied by the steam is 18 in.? Is this more or less and how much than one-third the total capacity of the boiler shell? Ans. 68.9 ft. J ; 1.9 ft. 1 more. Suggestion. Using [30(b)l as the most convenient, A =^X1.5 2 X\/ 4 - -0.608 = 4.305 ft. 1 o \ l.o Volume = 16 X4.305 ft. 3 = 68.9 ft.' 63. The cylinder of a pump is 6 in. in diameter, the length of stroke 8 in., and the number of strokes per minute 160. Find the flow in gallons per minute if the pump is double acting, that is, pumps the cylinder full rach stroke. Ans. 156.7 gal. 64. When the piston of a hand pump is 3 in. in diameter, and the sup- ply of water is drawn from a depth of 25 ft., what pressure is required on the handle 24 in. from the fulcrum when the piston rod is attached 3i in. from the fulcrum? Ans. 10.30 Ib. 66. The Cleveland Twist Drill Company records a test in which a IJ-in. "Paragon" high-speed drill removed 70.56 cu. in. of cast iron per minute. The penetration per minute was 57J in., the feed 1*0 in., and the drill made 575 H. P. M. Do these numbers agree p , K 66. What will be the weight of a cast-iron pipe 10 ft. long, 2 ft. in outer diameter, and 1 in. thick? (Use 0.26 Ib. per cubic inch.) Ans. 2254 Ib. 67. A tank car with a cylindrical tank 8 ft. in diameter and 34 ft. long will hold how many gallons? What weight of oil will it hold if the specific gravity of oil is 0.94? Ans. 12,784 gal.; 100,400 Ib. 68. Find the height of a cylindrical tank having a diameter of 30 in. in order that it may hold 4 barrels. Ans. 41 A in. 69. If a bar 1J in. in diameter weighs 6.01 Ib. per foot of length, what i the weight per foot of a bar 1 J in. square and of the same material? Ans. 7.65+ Ib. CYLINDERS 211 60. Find the weight of a hollow hexagonal bar 16 ft. long and weighing 0.28 Ib. per cubic inch. The cross section is a regular hexagon 1 \ in. on a side, with a circle \\ in. in diameter, at the center. (See Fig. 158.) Ans. 152ilb. 61. A cylindrical tank 22 ft. long and 6 ft. in diameter rests on its side in a horizontal position. Find the number of gallons of oil it will hold when the depth of the oil is 8 in. When 1 ft. 6 in. When 2 ft. 6 in. Use formula [30(b)l for finding the area of the segment. Ans. 282.5 gal.; 909.3 gal. 62. The segment in Fig. 159 is a counter-balance 5| in. thick. Find its weight if made of cast iron weighing 0.26 Ib. per cubic inch. Solution. Area of segment AnB=are& of sector AOBn area of triangle AOB. Area of sector AOn = |X84 2 XO. 7854 = 923.63 in. 2 Area of triangle AOB = |X42X21 X\/3 = 763.81 in. 2 Area of segment AnB =923.63 in. 2 -763.81 in. 2 = 159.82 in. 2 Volume of counter-balance =5^X159.82 in. 3 =879.0 in. 3 Weight of counter-balance =879.0X0.26 Ib. =228.5 Ib. Ans. 63. What is the diameter of a single round rod in order that it may be as strong as three rods having diameters of ^ in., 1 in., and 1J in. respectively? Ans. 1.87+ in. 64. Because the body of a bolt is greater in diameter than the threaded part, when the bolt is under strain the two parts will not stretch uniformly. For this reason the bolt is most liable to break where the threaded part joins the other part. To overcome this a p IG 159 hole is sometimes drilled from the center of the head to the beginning of the threaded part. This hole is made of such size that the cross-sectional area of the body is the same as that at the root of the thread. Find the diameter of the hole to be drilled in the following bolts in accordance with the preceding: (a) Diameter of bolt f in. with 10 U. S. S. threads to 1 in. (b) Diameter of bolt 1| in. with 5 U. S. S. threads to 1 in. Ans. (a) 0.422- in.; (b) 0.952 in. 65. In computing the safe working pressure for a steam boiler, a factor of safety of 5 is used. That is, the safe working pressure is 1 of the burst- ing pressure. The bursting pressure in pounds per square inch is given by the formula where P= bursting pressure in pounds per square inch, T tensile strength of boiler plate per square inch, I = thickness of boiler plate in inches, ? = radius of boiler in inches, and ft = a constant depending upon the riveting and is 0.56 for single 212 I'RA CTICAL MA THEM A TICS riveted boilers, 0.70 for double riveted boilers, and 0.88 for triple riveted boilers. Derive the formula. (See Fig. 160.) 66. Find the bursting and the safe working pressure for a double riveted boiler 66 in. in diameter, made of plate -fs in. thick, if tensile strength is 50,000 Ib. per square inch. Am. 331.4 lb.; 66.3 Ib. 67. What would be the bursting pressure in pounds per square inch of a wrought iron pipe having an inside diameter of 3 in. and a shell i in. thick ? Use a tensile strength of 40,000 lb. per square inch. Ans. 3333 J lb. 68. Holes are punched in sheets of metal by means of great pressure applied by a punch press. The pressure is usually reckoned at 60,000 lb. per square inch of surface cut over. For example, a hole 2 in. in circumference punched in a i-in. plate would require a pres- sure of 2Xi X60.000 lb., that is, the area of the cylindrical surface sheared off times 60,000 lb. Find the pressure necessary to punch a hole, having a diameter of i in., through a steel plate i in. thick. Ans. 11,781 lb. 69. Find the pressure necessary to punch at one blow a hole f in. in diameter and a rectangular hole J in. by J in. through a steel plate $ in. thick. .4ns. 130,686 lb. FIG. 1(51. 70. Find the blow necessary to cutout the corner squares and the holes in the box shown in Fig. 161. The dimensions are as given and the thickness of the sheet steel is & in. Ans. 17,678 lb. 71. Many small metal articles in common use are punched out of sheet metal and pressed into shape. The blank is usually cut so as to have the same area as the area of the finished article (Fig. 162). For example, the blank for a cylindrical box, having a diameter of 1 in. and a depth of 2 in., would have an area equal to the combined area of the sides and bottom of the box. Find the area and diameter of the blank for this box. CYLINDERS 213 Solution. Area = 3. 1416X(2) 2 +3.1416 X 1 X 2 = 7.06 sq. in. 89 Diameter of blank = V7.0686 ^0.7854 =3 in. Note. In shallow articles, as pail covers, the diameter of the blank is often found by adding twice the depth to the diameter of the top. FIG. 162. 72. Find the diameter of the blank for a pail cover whose diameter is 8 in. and depth f in. Work by both methods suggested above and com- pare results. 73. An aluminum cap for a paste bottle has the dimensions given in Fig. 163. Find the diameter of the blank from which it was pressed. Ans. 2.035 in. 74. A shoe-blacking box has a diameter of 3 in. and a depth of 1 in. Find the diameter of the blank from which it was pressed. Ans. 4.58 in. CHAPTER XVI PYRAMIDS, CONES, AND FRUSTUMS 168. Pyramid. A pyramid is a solid whose base is a polygon, and whose sides are triangles with their vertices at a common point, called the vertex or apex of the pyramid. A pyramid is triangular, square, hexagonal, etc., according as its base is a triangle, square, hexagon, etc. A right pyramid, or a regular pyramid, is a pyramid whose base is a regular polygon, and the sides or faces equal isosceles triangles. Fig. 164 is a regular pyramid with a square base. Fio. 164. In a regular pyramid the axis, or the line drawn from the ver- tex to the center of the base, is perpendicular to the base. This line is the altitude of the pyramid. In Fig. 164, OF is the altitude. The slant height of a right pyramid is the line drawn from the vertex to the center of one edge of the base. EF of Fig. 164 is the slant height, A lateral edge is the line in which two faces meet. BF of Fig. 164 is a lateral edge. 169. Cone. A circular cone is a solid whose base is a circle, and whose lateral surface tapers uniformly to a point, 214 PYRAMIDS, CONES, AND FRUSTUMS 215 called the vertex or apex. The axis of the cone is a straight line drawn from the vertex to the center of the base. A right circular cone is a cone whose base is a circle and whose axis is perpendicular to the base. In Fig. 165, F-ABC is a right circular cone. This might also be defined as a solid formed by a right triangle revolved about one of its legs as an axis. It may be called a cone of revolution. The altitude of a cone is the perpendicular line from the vertex to the base. The slant height is a straight line drawn from the vertex to the cir- cumference of the base. In Fig. 165, OF is the altitude, and CF the slant height. 170. Frustum. If the top of a pyramid or a cone is cut off by a plane parallel to the base, the remaining part is called a frustum of a pyramid or a cone. In Fig. 166, (n) and (b) are frustums. H G FIG. 166. The altitude of a frustum is the perpendicular between the bases, as NM of Fig. 166. The slant height of the frustum of a right pyramid or cone is the shortest line between the perimeters of the two bases. It is perpendicular to the edge of each base in the frustum of a right pyramid; and is, there- fore, the altitude of the trapezoids that form the faces of the frustum. In Fig. 16tJ(.o), 1>Q is the slant height. 216 PRACTICAL MATHEMATICS 171. Areas. The lateral area of a right pyramid is found by taking the sum of the areas of the triangles forming the faces of the pyramid. Since the altitudes of these triangles are each the slant height of the pyramid, they are equal. Because the base of a right pyramid is a regular polygon, the bases of the triangles are equal. We then have the following: RULE. The lateral area of a right pyramid or cone equals the perimeter of the base times one-half the slant height. The total area equals the lateral area plus the area of the base. Since the faces of the frustum of a pyramid are trapezoids, we have, by use of formula [8], the following: RULE. The lateral area of the frustum of a right pyramid or cone equals one-half the sum of the perimeters of the two bases times the slant height. The total area equals the lateral area plus the areas of the two bases. That these rules apply to the cone as well as to the pyramid may be seen by thinking of the cone as a pyramid with a very great number of sides to the base. Using S for lateral area, T for total area, h for altitude, 5 for slant height, p for perimeter (P and p for frustum), A for area of base (B and b for frustum), the rules may be written as the formulas: [47] S = ps, for pyramid or cone. [48] T = ^ps+A, for pyramid or cone. [49] S = i(P+p)s, for frustum. [50] T=i(P+p)s+B+b, for frustum. 172. Volumes. A particular case of the volume of a pyramid is seen as follows: The cube of Fig. 167 is divided into six equal pyra- mids with their vertices at the center of the cube. The volume of the cube equals the &rc&ABCD times the altitude PQ. Now the volume of one of the six pyramids, as 0-ABCD, equals $ of the Flo ! 07 volume of the cube, and hence equals the area of the base A BCD times of PO. RULE. The volume of a pyramid or a cone equals the area of the base times one-third the altitude. PYRAMIDS, CONES, AND FRUSTUMS 217 Which may be written as the formula: [51] V = iAh. The volume of the frustum of a pyramid or cone is best stated in the following formula : [52] y = J The volume of a frustum of a cone is usually more easily found by [53] V = i7rh(R 2 +r 2 +Rr), or [54] V = -iVh(D 2 +d 2 +Dd), where R and D are radius and diameter respectively of lower base, and r and d of upper base. It should be noted that the volume of a pyramid is one-third the volume of a prism of the same base and altitude, and that the FJG 1G8 cone bears a like relation to the cylinder. Example. Find the volume and the lateral area of a right cone of diameter 16 in. and altitude 12 in. Solution. A = 7rr 2 = 3. 1416 X8 2 = 201.0624. V=iAh = 3X201.0624X12 = 804.25 in. 3 s = \/12 2 +8 2 = 14.422, since the altitude, ra- dius, and slant height form a right triangle AOP of Fig. 168. S = ips = TITS = 3.1416X8X14.422 = 362.465 in. 2 EXERCISES 50 1. Find the volume of a right cone whose altitude is 8 in. and radius of base is 4.887 in. Ans. 200 in. 3 nearly. 2. Find the volume and total area of a cone whose radius of base is 6 in. and altitude 5.3 in. Ans. 199.8+ in. 3 ; 263.9 in. 2 3. Find the volume and lateral area of a cone whose altitude is 8 in. and radius 6 in. Ans. 301.6- in. 3 ; 188.5- in. 2 4. The circumference of the base of a conical church steeple is 35 ft. and the altitude is 73 ft. Find the lateral area. Ans. 1281 - ft. 2 Suggestion. Radius of base = 5(35-^3. 1416). The altitude and radius are the altitude and base of a right triangle of which the hypotenuse is the slant height of the steeple. 5. Find the weight of a conical casting of iron 8 in. in diameter and slant height 14 in. Ans. 58.4 Ib. '.'IS PR A CTICAI, MA THEM A TICS 6. A ji.-iil is 10 in. in diameter on the bottom and 12 in. on top. If tho slant height is 11 in., what is the number of square inches of tin in the pail? Ann. 458.67. 7. Find the total area and volume of a cone of revolution whose alti- tude is 12 ft., and the diameter of whose base is 10 ft. Arm. 282.74 ft. ; 314.16ft.' 8. The diameter of the top of a water pail is 12 in., the bottom is 10 in., and the altitude is 10 J in. How many quarts will the pail hold? Solution. By[64],F = i s X3.1416Xl0.5(12 1 -|-10 1 + 12XlO)=1000.6in. :i 1000.6 in. 3 -=-57.75 in. 3 = 17.33- = number of quarts. 9. Find the lateral edge, lateral area, and volume of a regular pyramid, each side of whose triangular base is 10 ft., and whose altitude is 18 ft. Ans. 18.90+ ft.; 273.45- ft. 1 ; 259.8+ ft. 3 10. A cone 12 in. in altitude and with cir- cular base 8 in. in diameter has a hole 2 in. in diameter bored through the center from apex to base. Find the volume of the part remaining. Ans. 169.646 in. 1 Suggestion. The part cut away, as shown in Fig. 169, consists of a cylinder 9 in. in altitude and a cone 3 in. in altitude. The height of the small cone can be found from the similar triangles AOP and BO'P in which we have the proportion AO:B(y=OP: O'P, or 4: 1 = 12: O'P. .: O'P =3. 11. Find the weight of a green fir log 215 ft. long, 4 ft. 6 in. in diameter at one end, and 20 in. in diameter at the other end, the specific gravity of fir being 0.78. Ans. 42 tons nearly. 12. Hard coal dumped in a pile lies at an angle of 30 with the hori- zontal. Estimate the number of tons in a pile of conical shape and 10 ft. high. Large egg size weighs 38 Ib. per cubic foot. (See Fig. 170.) Ans. 60 tons nearly. Suggestion. Radius of base of pile = 10X-s/3 by Art. 112. 13. Find the number of tons of large egg coal in a pile 20 ft. broad and 100 ft. long with cir- cular ends. Ans. 99 tons nearly. 14. A tank of reen forced concrete is 160 ft. long, 100 ft. wide, and 10ft. 6 in. deep outside dimensions. The side walls arc 8 in. thick at the top and 18 in. at the bottom, with the slope on the inside. The bottom is 6 in. thick. Find the number of cubic yards of cement in the tank and the capacity of the tank in barrels. Ans. 503 + ; 36,670 bbl. nearly. FIG. 170. PYRAMIDS, CONES, AND FRUSTUMS 219 15. Find the weight of a tapered brick stack of 10 ft. inside diameter, with a wall 4 ft. thick at the base, 1 ft. 6 in. at the top, and 175 ft. high. A cubic foot of brick weighs 112 Ib. Ans. 1095.5 tons. Suggestion. Find the volume of a frustum of a cone with lower base 18 ft. in diameter and upper base 13 ft. in diameter. Subtract from this the volume of the cylinder 10 ft. in diameter. 16. A cast-iron driver in the form of a frustum of a square pyramid is used on a pile-driving machine. Find the weight of the driver if it is 16 in. high, 10 in. square at the bottom, and 7 in. square at the top. Ans. 303.7 Ib. 17. A cast-iron cone pulley is 34 in. long. The diameter of one end is 12 in. and of the other end is 5 in. A circular hole 2 in. in diameter extends the length of the pulley. Find the weight of the pulley. Ans. 502.2 Ib. CHAPTER XVII THE SPHERE 173. Definitions. A sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within, called the center. A straight line passing through the center and ending in the surface is called a diameter. A line extending from the center to the surface is a radius. If the sphere is cut by a plane, the section is a circle. If the section is through the center of the sphere, it is called a great circle; if not through the center, it is called a small circle. The circumference of a sphere is the same as the circumference of a great circle. In Fig. 171, circles ACB and NCS arc great circles, and MER is a small circle. The parallels of latitude on the surface of the earth are small circles. The meridians of the earth all run through both the north and the south poles and, therefore, are great circles. 174. Area. The following is proved in geometry: RULE. The area of the surface of a sphere equals four times the area of a circle of the same radius. Or stated as a formula: [55] S = 47rr 2 = 7rd 2 , where S is the area of the surface of the sphere, r the radius, and d the diameter. The student may satisfy himself that this is true by winding evenly the surface of a ball with heavy cord, and then coiling the same cord into four circles of the same radius as the radius of the sphere. The rule can be derived from the fact that a sphere has the 220 THE SPHERE 221 same area as the lateral area of a cylinder having the same radius as the sphere, and an altitude equal to the diameter of the sphere. Area of lateral surface of cylinder = 2xr X 2r = 4irr 2 . 175. Volume. Geometry gives the following: RULE. The volume of a sphere equals the area of the surface times one-third of the radius. (a) FIG. 173. FIG. 172. Or stated as a formula: The reasonableness of this may be seen by thinking of the surface of the sphere as divided into a large number of small polygons. Let these be so small that they may be considered as planes. Now if we think of the sphere cut into pyramids having these polygons as bases, and having their vertices at the center of the sphere, as shown in Fig. 173, the volume of one of these small pyra- mids, represented in (a) of the figure, is found by [51] to be fr times the area of the small polygon. And the volume of all the small pyramids is equal to the whole surface of the sphere times |r. Hence V %Sr. 176. Zone and segment of sphere. A portion of the volume of a sphere included between two parallel planes is a segment of the sphere. If both the planes cut the surface of the sphere, the segment is a segment of two bases. In Fig. 174, the segment between the planes ABC and DEF is a segment of 222 I'RACTICAL MATHEMATICS two bases. The part of the sphere above DBF is a segment of one base. That portion of the surface of the sphere between two parallel planes is a zone. The altitude of the segment or zone is the perpendicular between the parallel planes. Thus, OQ is the altitude of the segment between the planes ABC and DBF. Here we can neither derive the rules for the area of a zone and the volume of a segment nor make them seem reasonable by any discussion. They are of some importance practically, especially the volume of the segment. RULE. The area of a zone is equal to the circumference of a great circle of the sphere times the altitude of the zone. Segment of One Ease Segment of Two Bases Fiu. 175. This rule is stated in the formula: [67] Z = 27rrh, where Z is the area of the zone, h the altitude, and r the radius of the sphere. It is readily seen from formula [57] that the area of any two zones on the same or equal spheres are to each other as their altitudes. It also follows that any zone is to the sur- face of the sphere as the altitude of the zone is to the diam- eter of the sphere. If a sphere is cut by parallel planes that are equal distances apart, as the planes cutting the sphere in Fig. 174, then the zones are all equal. Since the parallels of 30 north and south latitude are in planes that bisect the radii drawn to the north and south poles, then one-half of the surface of the earth is within 30 of (he equator. THE SPHERE 223 The volume of a spherical segment is given by the formula: [58] V = ih 7 r(r 1 2 +r 2 2 ) + i7rh 3 , where V is the volume, h the altitude, and n and r 2 the radii of the bases of the segment. If the segment has only one base, one of the radii is zero. Example 1. Find the surface, volume, and weight of a cast-iron ball of radius 12| in. Solution. By [55], S = 4*r* = 4 X 3.1416 X12.5 2 = 1963.5 in. 2 By [56], F = |Sr= 1X1963.5X12.5 = 8181. 25 in. 3 Since 1 in. 3 of cast iron weighs 0.26 lb., the weight = 0.26 Ib.X 8181. 25 = 2127. 125 lb. Example 2. A sphere 8 in. in radius is cut by two parallel planes, one passing 2 in. from the center, and the other 6 in. from the center. Find the area of the zone, and the volume of the segment between the two planes, if both planes are on the same side of the center. Solution. By [57], Z = 2irrh = 2X3.1416X8X4 = 201. 06 in. 2 ri = \ / (OB} 2 -(OW= VS^T 2 = A/60 ; r 2 = A/(OZ)) 2 - (OF) 2 = A/~8 2 -6 2 = A/28. By [58], V^hirW+rtf + brh* = |X4X3. 1416(60 + 28) + |X3. 1416 X4 3 = 586.43 in. 3 EXERCISES 51 1. Find the volume and the area of the surface of a sphere 6 ft. in diameter. Ans. 113.1 - ft. 3 ; 113.1- ft. 2 2. A sphere 4 in. in radius is cut from a cylinder 8 in. high and 8 in. in diameter. Find the volume cut away. Ans. 134.04+ in. 3 3. The radius of a sphere is 2 ft. Find the area of the surface and the volume. Ans. 50.266 -ft. 2 ; 33.51+ ft. 3 4. How much will a sphere of cast iron weigh if it is 3 in. in diameter, and if cast iron weighs 0.26 lb. per cubic inch? Ans. 3.675 lb. 5. A cubic foot of lead weighs 712 lb. Find the weight of a ball 3 in. in diameter. Ans. 5.825 lb. 6. How many square feet of tin will it take to roof a hemispherical dome 40 ft. in diameter? Ans. 2513+ ft. 2 7. Find how many acres of land on the surface of the earth, if one- fourth of the surface is land and the radius is 4000 miles. Ans. 32,170,000,000 nearly. PRACTICAL MA THEM A TICS 8. Find the volume of a cylinder 2 ft. in diameter and 2 ft. in altitude; of a sphere 2 ft. in diameter; and of a cone 2 ft. in diameter and 2 ft. in altitude. Compare the three volumes. (See Fig. 177.) 9. Find the volume of the segment between two parallel planes 6 in. apart that cut a sphere 12 in. in radius, if one plane passes 2 in. from the center. There are two cases: (a) when the center of the sphere lies out- side of the segment, and (b) when the center lies in the segment. Ans. (a) 2186.6- in.*; (b) 2638.9+ in. 1 10. An iron ball 3 in. in diameter has a coating of lead 1 in. thick. Find the volume of the iron, of the lead, and the weight of each. Ans. 14.137 in. 8 ; 51.313 in.;3.6T6- lb.;21.141- Ib. 11. A ball of lead 2 in. in diameter is pounded into a circular sheet 0.01 in. thick. How large in di- ameter is the sheet? Ans. 23 in. nearly. 12. A water tank, 6 ft. Fio. 177. in total length and 18 in. in diameter, is in the form of a circular cylinder with two hemispherical ends. Find its capacity in gallons. Ans. 72.7+ gal. 13. A hollow copper sphere used as a float weighs 10 oz. and is 5 in. in diameter. How heavy a weight will it support in water? Ans. Less than 27.9 oz. Suggestion. It will support a weight less than the weight of water displaced by the sphere minus the weight of the sphere. 14. A circular flower bed in a park is 25 ft. in diameter and is raised 2 ft. 6 in. in the center making a spherical segment. How many loads of dirt did it take to build it up if one load is 1 J cu. yd.? Ans. 15 J nearly. 15. In a practical hand- book the following rule is given as nearly correct. In fact, it is correct. The area of a flanged spherical seg- ment, a vertical section of which is shown in Fig. 178, is equal to the area of a circle of radius equal in length to the line drawn from the top of the seg- ment to the edge of the flange, that is, equal to a circle of radius AB. Find the area of a flanged segment having dimensions as given in Fig. 178. Work both by the rule and by using the formulas for area of a ring and of :i zone. Solution. By [16], radius, r, of sphere of which zone is a part FIG. 178. 5* +2* ! 2X2 = 7.25. THE SPHERE 225 By [57], 7 = 2X3.1416X7.25X2=91.1064 in. 2 By [28], area, A, of ring = 3.1416(8+5)(8-5) =122.5224 in. 3 91.1064 in. 2 + 122.5224 in. 2 =213.63- in. 2 = V68. FIG. 179. AB=- Area of circle having a radius =V<38 is 3. 1416 X (\/68) 2 = 213.63 -in. 2 , which is the same as the result by the first method. 16. Find the per cent of error in using the following rule: To find the weight of a cast-i/on ball mul- () tiply the cube of the diameter in inches by 0.1377, and the product is the weight in pounds. Ans. Rule is correct if 1 cu. in. of cast iron weighs 0.263 Ib. 17. Fig. 179 is the vertical cross section of a casting, the inner and outer "skins" being spherical zones. Find the weight of metal at 0.35 Ib. per cubic inch necessary to make the casting. Ans. 176 Ib. nearly. Suggestion. Find the difference of the volumes of the two segments. 18. A hemispherical cap of aluminum is 3 5 in. in diameter. Find the diameter of the blank from which it is pressed. Ans. 4.95 in. 19. Find the diameter of the blank if the cap in the preceding exercise has a flat ring ? in. wide around it. Ans. 5.70+ in. 20. Show that the volume of a round or button head of a machine screw is given by the formula V=irh where D is the diameter of the head and h the height of the head. Suggestion. In formula [68], But n = and r 2 =0. 21. Find the volume of the head of a round head machine screw if the diameter of the head is 0.731 in. and the height is 0.279 in. Ans. 0.070 cu. in. 22. The water tank shown in Fig. 180 con- sists of a cylinder with a hemisphere below. The diameter is 20 ft, and the height of the cylindrical part is 22 ft. Find the capacity of the tank in gallons. Ans. 67,369 gal. CHAPTER XVIII VARIOUS OTHER SOLIDS 177. Anchor ring. A ring formed of a cylinder bent into a circular form, as in Fig. 181, is called an anchor ring. The mean length of the rod in such a ring is the circumference of a circle of radius ON. Any cross section of such a ring will be a circle. Since the ring may be considered as a cylinder bent into circular form, the area of the surface is 2-irXONX circumference of a cross section. If ON = R, and the radius of the cross section NM is r, we have for the area the formula: [59] A = 27rRX27rr = 47T 2 Rr. The volume is the same as the volume of a cylinder with an altitude that is equal to the mean circumference of the ring, hence the following: [60] V = 27rRX7rr 2 = 27r 2 Rr 2 . These rules may be generalized so as to apply to any circular ring. In general, the area of the sur- face equals the perimeter of the cross section times the cir- cumference drawn through the center of gravity of the cross section. The volume equals the area of the cross section times the circumference drawn through the center of gravity of the cross section. EXERCISES 62 1. The cross section of a solid wrought-iron ring is a circle of 4 in. radius. The inner radius of the ring is 3 ft. Find the area of the surface and the volume of the ring. Ans. 6316.6 in.'; 12,633.2 in. 3 2. The cross section of the rim of a flywheel is a rectangle 6 in. by 8 in., the shorter dimension bring in the diameter of the wheel. The 220 VARIOUS OTHER SOLIDS 227 wheel is 22 ft. in outer diameter. Find the volume of the rim and its weight if of cast iron. Ans. 22.515 ft. 3 ; 10,132 Ib. 3. Find the weight of a cast-iron water main 12 ft. in length, 2 ft. in outer diameter, and 1 in. thick. Solve both by considering it as a ring and as a hollow cylinder. Ans. 2709.6 Ib. 4. Find the area of the surface and volume of a ring of outer diameter 10 in., made of round iron 1 in. in diameter. What is its weight at 0.28 Ib. per cubic inch? Ans. 88.83 in. 2 ; 22.207 in. 3 ; 6.218 Ib. 6. An anchor ring, 13 in. in outer diameter, of 1J in. round iron, has the same volume as what length of a bar 1 in. by If in. in cross section? Ans. 24.16 in. 6. Find the weight of an anchor ring of cast iron, outer diameter 3 ft., the iron being circular in cross section and 6 in. in diameter. (Use 450 Ib. per cubic foot.) Ans. 693.9 Ib. FIG. 182. 178. Prismatoids. A prismatoid is a solid whose bases are parallel polygons and whose faces are quadrilaterals or trian- gles. One base may be a point, in which case the prismatoid is a pyramid; or one base may be a line, in which case it is wedge-shaped. The rule for finding the volumes of prismatoids holds in very many cases when the faces become curved sur- faces and the prismatoid has become a cone, frustum of a cone, a cylinder, a sphere, a spindle of some kind, or one of various other forms that cannot well be described here. The rule is as follows: RULE. To find the volume of a prismatoid, add together the areas of the two bases and four times the area of a section midway between them and parallel to them, then multiply the sum by one-sixth the perpendicular between the bases. PR A CTICA L MA Til EM A TICK The rule may he stated in the formula: [61J V=Jh(B 1 +4M+B 2 ), where B\ and #2 are the two bases and M the mid-section. In Fig. 182 are given some forms to which the rule for the volume of a prismatoid will apply. The dimensions of the mid-section may be found by actually measuring the lines or by computing them. Example. By formula [61], find the volume of a frustum of a pyramid in which the bases are regular hexagons 10 in. and 6 in. on a side respectively and whose altitude is 18 in. (See Fig. 183.) h "->! Fro. 185. FIQ. 186. Solution. AB = $(1Q in. + 6 in.) =8 in. Area of lower base =5 2 X1.732X6 = #,. Area of upper base = 3 2 X 1.732 X 6 = B 2 . 4Xarea of mid-section =4X4 2 X1.732X6=43f. .'. #i+4M + 2 = 1018.416 in. 2 And F = JX 18X1018.416 = 3055.248 in. 3 EXERCISES 63 1. Use the rule for volume of prismatoids, and find the volumes called for in exercises 3, page 217; exercises 8 and 9, page 218. 2. Use formula [61] to find the volume of a hemisphere and check by the ordinary rule. VARIOUS OTHER SOLIDS 229 3. Use formula [61] to find the volume of the solid shown in Fig. 184. Ans. 3420 in. 3 4. Use formula [61] to find the volume of the solid shown in Fig. 185. Ans. 3360 in. 3 6. A concrete pier for a railway bridge has the dimensions shown in Fig. 186, the bases being rectangles with semicircles. Find the number of cubic yards of concrete in 21 such piers. Ans. 793.3+. 6. A railroad cut has the dimensions shown in Fig. 187, which shows the vertical section and three cross sections. Find the volume of the earth removed in cubic yards. Ans. 7972 1 yd. 3 PART THREE ALGEBRA CHAPTER XIX NOTATION AND DEFINITIONS 179. General remarks. In mathematics, the attempt is made to do certain things more easily and in less time than they can otherwise be done. In arithmetic, many processes were learned that saved time and labor. In performing these processes, certain signs and symbols were used to express the ideas. New signs and symbols were introduced as they were needed to express the new ideas that were involved. Thus, there were used the numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9; the letters of the alphabet; and various signs among which are +, , X, -^, ( ), and V- Algebra is a name applied to the continuation of arithmetic. The same signs and symbols are used in algebra as in arith- metic, and they have exactly the same meanings. From time to time as we proceed, we shall find it convenient to add to the symbols and signs, in order that we may express new ideas or perform new processes. We shall also expect that many new, simpler, and more powerful methods of procedure will be developed; in fact, this is the chief aim in continuing the study of mathematics. 189. Definite numbers. The numerals 0, 1, 2, 3, etc., have definite meanings. For instance, the symbol 4 represents the idea we call four. It may be 4 yards, 4 dollars, 4 pounds, or 4 of any other units; but, in any case, is a definite number. We have learned that the letter TT represents a definite number, the ratio of the circumference to the diameter of a circle. This cannot be expressed exactly by the numerals 1, 2, 3, etc.; but has, none the less, a fixed value. 181. General numbers. We have used the letter b to represent the number of units in the base of a triangle. Its 331 232 PRACTICAL MATHEMATICS value changed for different triangles, that is, it represented in a general way the length of the base of a triangle. Its value might be given as 10 ft., 6 in., or as any number of any sized units of length. Likewise, r represents the radius of a circle; but when it occurs in the formula A=7ir 2 , we do not think of a particular value for it. Such an idea as we rep- resent by b or by r cannot be represented by the numerals. The idea is a general number-idea. It is usually represented by a letter of the alphabet. In a particular discussion, the letter or letters used stand for the same value throughout the discussion. For example, when we are considering a particular circle the letter r repre- sents a definite length as 10 ft. 182. Signs. The signs +, , X, and -5- are signs of operation. They continue to have the same meaning as in arithmetic. As we have already seen, the sign X is not often expressed where a multiplication is indicated between numbers expressed by letters. The symbol () may be used instead; but usually no sign is expressed. Thus, aXb is written a-b or simply ab. Similarly, 2axy means 2 time a times x times y. The signs of grouping are the parentheses ( ), the brackets [ ], the braces { }, and the vinculum - . The first three are placed around the parts grouped, and the vinculum is usually placed over what is grouped. They all indicate the same thing; namely, that the parts enclosed are to be taken as a single quantity. Thus, 12 (10 4) indicates that 4 is to be subtracted from 10 and then the remainder is to be taken from 12. Hence 12 (10 4) =6. Exactly the same thing is indicated by 12 [10 4), 12 {10 4}, and 12-10~4. The vinculum is most frequently used with the radical sign. Thus, \/6425. 7 4 It is to be noted that in the form - A the horizontal lino 3+4 serves as a vinculum and as a sign of division. It thus per- forms three duties: first, indicates a division; second, binds together the numbers in the numerator; and third, binds together the numbers in the denominator. In performing the operations in a problem containing the NOTATION AND DEFINITIONS 233 signs of grouping, the operations within the grouping signs must be considered first. 183. Algebraic expression. An algebraic expression is any expression that represents a number by means of the signs and symbols of algebra. A numerical algebraic expression is one made up wholly of numerals and signs. A literal algebraic expression is one that contains letters. Thus, 14 + 13 (4+3) and 3ab 4cd are algebraic expressions; the first is numerical and the second is literal. The value of an algebraic expression is the number it represents. 184. Coefficient. If we have such an expression as 8abx, 8, a, b, and x are factors of the expression. Any one of these factors or the product of any two or more of them is called the coefficient of the remaining part. Thus, Sab may be considered the coefficient of x, or Sa the coefficient of bx', but usually, by the coefficient, we mean the numerical part only. It is then called the numerical coefficient. If no numerical part is expressed, 1 is understood. Thus, laxy is the same as axy. 185. Power, exponent. If all the factors in a product are equal as a-a-a-a, the product of the factors is called a power of one of them. The form a-a-a-a is usually written a 4 . The small number to the right and above indicates how many times a is taken as a factor. (See Arts. 74 and 75.) In the above power a is called the base and 4 the exponent. The exponent of a power is a number written at the right and a little above the base. When it is a positive whole number it shows how many times the base is to be taken as a factor. Thus, c 2 is read c square or c second power, and indicates that c is taken twice as a factor; c 3 is read c cube or c third power, and indicates that c is taken three times as a factor; c 4 is read c fourth power, and indi- cates that c is taken four times as a factor; c n is read c nth power or c ex- ponent n, and indicates that c is taken n times as a factor. When no exponent is written the exponent is understood to be 1. Thus, a is the same as a 1 . 234 PRACTICAL MATHEMATICS 186. A term in an algebraic expression is a part of the expression not separated by a plus or a minus sign. Thus, in 4ox+3c d, 4oz, 3c, and d are terms. It is convenient to have names for algebraic expressions having different numbers of terms. A monomial is an alge- braic expression consisting of one term. A binomial consists of two terms; and a trinomial consists of three terms. Any algebraic expression of two or more terms is called a poly- nomial or a multinomial. Terms that are exactly the same or differ only in their coefficients, are called like terms or similar terms. Terms that differ otherwise than in their coefficients, are unlike, or dissimilar terms. Thus, 60 ! z*, 7a'z 1 , and 16a j x* are like terms; while 6ox l , 7a f z*, and 16ar/2 are unlike terms. 187. Remarks. The object of the exercises of this chapter is to recall the meanings and uses of signs and symbols, and to fix in mind the new ideas that have been given here. The doing of these exercises must not be slighted by the student, for he must become familiar with the mathematical way of stating ideas in order that he may be prepared for the work that comes later. EXERCISES 64 Find the value of each of the following: 1. 3+6-2+4-1+7+2. 2. 7+2X6-3+8X2-2. 3. 8 + 16-5-2-4 + 14-5-7+21. 4. 75 + 10+25X6-3X3-15. 5. 150-5-6+4X25-75. 6. 150-5-5X6 + 10X9 + 17. Name the monomials in the following list, the binomials, the trinomials. Which of them are polynomials? 7. m'+2t>. 8. 7o6*c. 9. 4+9c+d. 10. ofe+ac+od. 11. a+4c+ in Ex. 7? Of a'b'c in Ex. 8? Of c in Ex. 8? Of a 6 in Ex. 12? Of r in Ex. 16? 23. Name the numerical coefficients in Exs. 9, 13, and 20. 24. Name the exponents in Exs. 8, 11, and 18. NOTATION AND DEFINITIONS 235 26. How many factors are there in Ex. 8? In Ex. 12? Write the following in algebraic symbols: 26. The result of adding 7 times a to 9 times b. 27. The result of subtracting 10 times a from 6 times n. 28. The product of 3 times a 3 times b square. 29. The product of 7 times a fourth power times n cube times d. 30. The product of the sum of a and b, times the difference found by subtracting b from a. Ans. (a-\-b)(a b). 31. The square of the sum of a, b, and c. Ans. (a+b+c) z . 32. The quotient of the sum of x and y divided by the difference when y is subtracted from x. 33. Express the product of m 3 and the sum of the two fractions, 2 f 2 31 divided by a and 3 divided by b. Ans. m 3 - + j- 1 (a o j 34. Express the product of the sum of a, 26, and 3c, and the sum of 3 & = 22, and c = 20. o+6+c 36+22+20 Substituting in s=- ~ s- --39. Substituting in \/s(s a) (s b) (s c}, we have N/39C39- 36)(39- 22)(39-20) = \/3779l = 194.4 - . Ans. FORMULAS AND TRANSLATIONS 239 EXERCISES 66 1. Express without exponents and find the value of the following: (1) 2 3 -4 2 - (2) 7 2 -2 2 -5. (3) 2 4 -5 3 -7 2 . Ans. 98,000. (4) 3 3 5 4 -7 2 . (5) 7-H 2 -5 3 . Ans. 105,875. (6) 3 4 -2 5 -13 2 . Ans. 438,048. If a = 2, 6 = 3, and c = 5, find the values of the following: 2. 3a 2 6 3 . Ans. 324. 3. 2a6 2 -c 2 . Ans. 11. 4. 3(a 2 +6 2 ). Ans. 33. 6. (a+6) 3 . Ans. 125. 6. (a+c-6) 3 . Ans. 64. 7. (c-a) 4 . Ans. 81. 8. a(a 2 +c 2 -6 2 ). Ans. 40. 9. a6 2 (c 2 -6 2 ). Ans. 288. 10. (o+6)(c-6). Ans. 10. 11. (c+6+a)(c-6+a). Ans. 40. 12. a 2 +2a&+6 2 . Ans. 25. 13. (a 2 +6 2 ) 2 -c 2 . Ans. 144. 14. c 3 -(a 2 +6 2 ). Ans. 112. 15. (a 2 +6 2 ) 2 -(c 2 -6 2 ). Ans. 153. 16. Find the value of the following when r = 18: (1) S = 47rr 2 . Ans. 4071.5+. (2) V = f^r 3 . Ans. 24,429+. 17. Find the value of the following when r = 7 and h = 9 : (1) S=2irrh. A ns. 395.84+. (2) V=irr z h. Ans. 1385. 4 + . If a = 1, b = 3, c = 5, and d = 0, find the numerical values of the following : 18. a 2 +26 2 +3c 2 +4d 2 . Ans. 94. 19. a 4 +4a'6+6a 2 6 2 -4a& 3 +6 4 . Ans. 40. 12a 3 -b 2 , 2c 2 20. oo + i L-> ri.t ' AfiS, O. 3a 2 a+o 2 5o 3 Suggestion. Remember that the lines in the fractions are vinculums and bind the terms in the numerators or denominators together, as well as indicate division. The multiplications, additions, and subtractions indicated in the numerators and denominators must be performed first. Thus, after substituting the values, 12XP-3 2 2X5 2 l+3 2 +5 3 = 3 50 135 3X1 1+3 2 5X3 3 3 + 10 135' If a = 1, b = 2, c = 3, d = 5, and e = 8, evaluate the following : 21. b 2 (a 2 +e 2 -c 2 ). Ans. 224. 22. (a 2 +6 2 +c 2 )(e 2 -d 2 -c 2 ). Ans. 420. _ 23. e-[v / e + l+2+e-v^e-4. Ans. 15. Evaluate the following when a = l, 6=2, c = 3, d = 4, and e~5: 24. abc' t +bcd 2 -dea 2 . Ans. 94. 25. e 4 +6e 2 6 2 +6 4 -4e 3 6-4e6 3 . Ans. 81. 26. - , , ,, . o , , i t -, ' Ans. o. a 3 +3a 2 o+3ao 2 +o 3 27. (a+6)(6+c)-(6+c)(c+d)+(c+d)(d+e). Ans. 43. 240 PRACTICAL MATHEMATICS 28. 29. 30. Xru. 72. 11. 7. 1, 6=2, c-3, and Ans. 6. 31. Evaluate (ac-M)Va*6c~-ffr 1 c/+c*ad-2, if a rf-0. 32. If A stands for the number of square units in the area of a circle, and r stands for the number of linear units in the radius, state in words the following formula: A = irr z . 33. In the above formula, can r be any number we please to make it? Can A1 If we make r = 5 in., can we then make A anything we please? Can T be any number we wish to make it ? Is T a general number ? Is A ? Isr? 34. Using A, a, and 6, state the following as a formula: The area of any triangle equals one-half of the base times the altitude. 35. Write a formula for finding the area of the cross section of a channel iron, using the letters given in Fig. 188. Ans. A = td+b(s+n). 36. Using the formula derived in the above exercise, find the areas of the cross sections of channel irons of the following dimensions: FIG. 188. FIG. 189. No. d t b n 8 Ans. A (1) 3 in. 0.17 in. 1.24 in. 0.38 in. 0.17in. 1.19 in. (2) 5 in. 0.19 in. 1.56 in. 0.45 in. 0.19 in. 1 . 95 in. (3) 8 in. 0.22 in. 2. 04 in. 56 in. . 22 in. 3.35 in. (4) 10 in. 0.24 in. 2. 36 in. 0.63 in. 0.24 in. 4.45 in. (5) 15 in. 0.40 in. 3. 00 in. 0.90 in. 0.40 in. 9.90 in. (6) 6 in. 0.41 in. 3. 15 in. 0.53 in. 0.46 in. 5.58 in. (7) 7 in. 0.45 in. 3. 00 in. 0.525 in. 0.475 in. 6. 15 in. (8) 10 in. 0.375 in. 3. 00 in. 0.469 in. 0.406 in. 6. 38 in. 37. Write a formula for finding the area of the cross section of an I-beam, using the letters as given in Fig. 189. Ans. A = dt+'2b(n+s). 38. Using the same letters for dimensions as in the last exercise, find the areas of the cross sections in the following I-beams: No. d I b s n Ans. A (1) 5in. 0.21in. 1 .395 in. 0.21 in. 0.44in. 2.86in. (2) 9 in. 029 in. 2. 02 in. 0.29 in. 0.627 in. 6 31 in.* (3) lf> in. 080 in. 2 80 in. 80 in. 1.27 in. 23 59 in.* 39. Using li for base, P for percentage, R for rate, and A for amount, write the formulas for determining the different things called for in percentage. FORMULAS AND TRANSLATIONS 241 40. If p stands for principal, i for interest, t for time in years, r for rate per cent, and a for amount, translate the following formulas into English : (1) i = prt. (2) a = prt+p. (3) r.=i+p. (4) r = -- pt 41. State in words the process of adding two fractions having a common denominator. State the same as a formula, using a and b for the numera- tors and d for the common denominator. Ans. -,+ -. = -. d d d 42. Do the same as in exercise 41, for subtracting fractions having a common denominator. 43. Translate the following into English: m a _ L _ c = a - y ( l C 9 % nXm _n ( b ' d be ^' dXm~d n + m n n nXm (3) = (4) -.Xni=-. d+m d d d 44. Write the formula that states that a times the sum of b, c, and d equals s. Ans. s=a(6+c+d). 46. In exercise 44, find s: (1) if a = 7, 6 = 6, c = 8 and d = 10; (2) if a = f , b = |, c = f , and d = 5 V . Ans. 168; l^j. 46. Write in algebraic symbols that if c, the cost, in dollars, of a harness, be increased by 5, the sum multiplied by 4 equals v, the cost, in dollars, of a horse. If c is 45, find the value of v in the formula. Ans. w = 4(c+5); 200. 47. Using bi and 6 2 for the bases and h for the altitude, state the following as a formula: The area of a trapezoid equals one-half the sum of the two bases times the altitude. Ans. A =5(^1+^2)^- 48. Using the formula of the preceding exercise, find the areas of the following trapezoids: (1) 61 =22.33 in., 6 2 =46.39 in., h =26.43 in. Ans. 908.13+ in. 2 (2) 6, =7.203 in., 6 2 = 5.826 in., h= 3.243 in. Ans. 21.127-in. 2 49. Write the formula stating that the area S of the surface of a sphere equals 4 times IT times the square of the radius r. Use the for- mula and find the area of the surface of a sphere 15 in. in radius. Ans. 2827.4+ in. 2 60. Write a formula stating that the volume F of a sphere equals | of IT times the cube of the radius r. Use this formula and find the volume of a sphere 12 in. in radius. Ans. 7328.2+ in. 3 51. Write a formula stating that the volume V of a rectangular solid equals the length I times the breadth 6 times the height h. Use this formula and find the number of cubic feet in a room 40 ft. by 30 ft. by 12 ft. Ans. 14,400. 62. Find the perimeter of each form in Fig. 190. Ans. (1) 4x; (2) 2(a+6); (3) 2(x+y)+46; (4) 2(b+c+d+y)+4x; (5) 2(a+6+d). 16 242 PRACTICAL MATHEMATICS 53. Find the area of each form in Fig. 190. An*. (1) x 1 y*; (2) ab-2xy; (3) xy-2ab; (4) 6c-y(2x+d) -cP; (5) ab-cd. 54. If a man is 35 years old, what was his age a years ago? What will it be 6 years from now? Ans. 35 a years; 35+6 years. 65. If x years was the age of a man a years ago, what is his age now? What will it be in c years from now? Ana. x+a years; x+a+c years. 66. Write in algebraic language: x diminished by a; y increased by b; j divided by n; the nth part of x; one ruth of a. 67. The sum of two numbers is a and one of the numbers is x, what is the other number V Express one with of the first plus one nth of the second number. x a-x Ans. a x; |- m n v V (1) *- x (2) ..~4 i i I 4--* >> If ] *-** t-*-* V i " i -*-x-^- y i I (4) FIG. 190. 68. If x is the first of two consecutive numbers, what is the second? What will represent each of three consecutive numbers, if x stands for the middle one? Ans. x + 1; x 1, x, and x + 1. 69. Represent three consecutive even numbers if x is the middle one. 60. If the length of a stick is a feet, how many inches is it in length? How many rods? How many miles? 61. If d is the number of dollars an article costs, what is the number of cents? Ans. lOOd. 62. If n room is a yards, 6 feet, and c inches long, how many inches in length is it? Ans. 36a + 126+c. 63. Given the product p and the multiplier m, find the multiplicand n. Ans. n= FORMULAS AND TRANSLATIONS 243 64. Given the dividend d and the quotient q, find the divisor 6. d Ans. b=-- Q 66. Given the divisor d, the quotient q, and the remainder r, find the dividend D. Ans. D=dq+r. 66. The difference between two numbers is n and the smaller one is a, what is the larger? 67. A number exceeds a by c, what is the number? Ans. a+c. 68. Write in symbols that a exceeds b as much as c is less than d. Ans. a b =d c. 69. Express in symbols that one-half of m equals the nth part of the f T i a-\-b-\-c sum of a, b, and c. Ans. \m= -- 70. If a cents is the price per quart for beans, what is the price per bushel? Ans. 32a cents. 71. The provisions that will keep a family of 9 persons 30 days will keep a family of 5 persons how many days? The provisions that will keep a family of a persons I days will keep 6 persons how many days? _. al Ans. 54; T- 72. If it takes m men d days to dig a ditch, how many days will it take md c men to dig it: Ans. -- C 73. How many pounds of sugar at c cents a pound will d dozen eggs at e cents a dozen buy? Ans. 74. A cubical tank e ft. on an edge will hold how many barrels, if one e 3 barrel equals 4.211 cu. ft.? Ans. . 4.^11 75. A box a ft. long, b ft. wide, and c ft. deep will hold how many bushels, if 1 bushel equals 2150.42 cu. in.? Ans. 76. A man earned $a per day and his son $6. How many dollars did they both earn in a month, if the man worked 26 days and the son 21 days? Ans. 26o+216. CHAPTER XXI POSITIVE AND NEGATIVE NUMBERS 191. Meaning of negative numbers. The degrees of tem- perature, indicated by the thermometer scale, are counted in two opposite directions from the zero point. We usually speak of a temperature as so many degrees above or below zero. In arithmetic we speak of temperature in this manner, but in algebra we seek some abbreviated form for stating the same thing. We might agree to use any convenient signs whatever. The signs + and have been generally adopted. The + sign, placed before the number of degrees, indicates a, temperature above zero; and the sign indicates a tempera- ture below zero. This use of these signs is different from the ordinary use in which they indicate addition and subtraction. Here they indicate the sense or direction in which the tempera- ture is measured or counted. Thus, +25 means 25 above zero, and 25 means 25 below zero. The number preceded by the + sign is called a positive number, and the one preceded by the - sign, a negative number. Two numbers so related, one positive and one negative, may be called relative numbers. The following are further examples of relative numbers. They will help to fix the negative number-idea in mind. Time is commonly measured forward and backward from a certain date. This might be shown by using the + and signs. Thus, 1918 A. D. might be written +1918, while 325 B. C. might 1>c written 325. A force acting in one direction and another in an opposite direction are designated as a + and a force respectively. Money gained and resources are + , and money lost and lia- bilities are . 192. Need of negative number. The necessity for extend- ing the number system so as to include negative numbers may be .seen from the following subtractions, where the min- 244 POSITIVE AND NEGATIVE NUMBERS 245 uend remains the same, but the subtrahend increases by steps of 1 as we pass from left to right. This causes the difference to diminish by steps of 1 from left to right. When the differ- ence becomes less than zero we indicate it by the sign placed before the number. 6666666 6 _3 _4 _5 J> _JT _8 __9 JO 3 2 1 0-1-2-3-4 etc. 193. Representation of negative and positive numbers. For convenience, the positive and negative numbers may be represented on a horizontal line; the +, or positive, numbers to the right of a certain point, called zero, and the , or nega- tive, numbers to the left of the zero point. This method of representing them is found very convenient in explaining addi- tion and subtraction. It should be carefully noted that toward the right is the t -6-5-4-3-2-1 + 1+2 + 3+4 + 5 + 6 ^+ i i i i i i I i i i i i i FIG. 191. positive direction and toward the left is the negative direction, no matter from what point we start. The idea of negative number is opposite to that of positive number. For example, if a man walks five miles east, or in the positive direction, and then five miles west, or in the nega- tive direction, he is at the starting-point. The negative distance has destroyed the opposite or positive. 194. Definitions. Positive and negative numbers, to- gether with zero, form the system called algebraic numbers. The absolute or numerical value of a number is the value which it has without reference to its sign. Thus, +5 and 5 have the same absolute value 5. The signs + and -- when used to show direction or sense are called signs of quality to distinguish them from the signs of operation used to indicate an addition or a subtraction. The sign +, used as a sign of quality, is usually omitted; but the sign , when a sign of quality, is expressed. To show that the sign is one of quality it is sometimes 246 PRACTICAL MATHEMATICS written with the number and enclosed in parentheses, as (3), (+4). (-5) + (+2) indicates that a -5 is to be added to a +2. 196. Remarks on numbers. What is the negative number and why do we need to trouble ourselves about it? The illustrations given above should help one to get the idea of negative number, and to see how it is forced upon us when we try to subtract a larger number from a smaller one. In mathematics when a new number-idea appears, the first thing to do is to represent it by a symbol; and, second, find a way of operating with it. That is, we must determine how to add, subtract, multiply, and divide such numbers. One of the first things we did in arithmetic was to determine methods of operating with positive whole numbers; then a little later, we did the same thing for the fractional numbers. In fact, much of the time spent in studying mathematics is spent in finding how to add, subtract, multiply, and divide numbers of different kinds, whole, positive, negative, fractional and combinations of these. It now remains to devise methods and rules for operating with algebraic numbers. This must be done in such a way that no old rules or principles shall be violated. The student is asked to consider carefully each step taken and to make every part seem reasonable. EXERCISES 56 1. A place is at 10 25' east longitude. What sign may be used instead of the word east to show this? 2. New York is 74 west longitude. Designate this by means of a sign. 3. Chicago is 41 45' north latitude, and Melbourne 37 south latitude. Designate these algebraically, that is, with signs. 4. A man overdraws his bank account. How may the banker indicate this on his books? 5. If the weight of a stone, weighing 100 lb., is written 4-100 lb., how may the upward lift of 500 lb. of a balloon be written? 6. What is meant by saying that a man is worth +$3000? By -$2000? 7. If a man has a -$200, a +$300, and a -$100, what is he worth? 8. A man was born in the year 35 and died in the year +48. How old waa he when he died? Ans. 83 years. POSITIVE AND NEGATIVE NUMBERS 247 9. If a man has assets of $3000, and we represent it by a line 3 in. long drawn to the right of 0, how could we represent a liability of $1500? 10. A man walks 10 miles east and then 6 miles west. Write this in symbols. 11. If a body is heated 37 and then cooled down 42, indicate the facts in symbols. 12. If 7 men came into a room and 8 men left the room, we could rep- resent the fact by saying that +7 and 8 men came into the room. A man received $25 and spent $21. Write this as money received. 13. Remembering the positive and negative directions, draw a line; locate a zero point; and, with a unit of 1 in., locate positive and negative numbers to the right and left of zero. Start at +5 and go +3, - 7, -4, +6, -10, +2, +1, and +4. Where do you stop? Ans. At the zero point CHAPTER XXII ADDITION AND SUBTRACTION 196. Definitions. The aggregate value of two or more algebraic numbers is called their algebraic sum. The process of finding this sum is called addition. 197. Addition of algebraic numbers. If we wish to add 3 to 4, we start with 4 and count 3 more, arriving at 7 which is the sum. If we consider the system of algebraic numbers arranged on a horizontal line, Fig. 192, we add a positive num- ber by starting with the number we wish to add to, and count- -10-9-8-7-6-5-4-3-2-1 0+1+2 +3*4+5 +6+7 +8+9+10 i i i i i i i i i i I i i i i i i i i i i FIG. 192. ing toward the right as many units as there are in the number added. Thus, in the above, we start at 4 and count toward the right to 7. To add +5 to 3, we start at 3 and proceed 5 units toward the right, arriving at +2. To add +3 to 7, we start at 7 and proceed 3 units toward the right, arriving at 4. Since adding 3 to +7 is the same as adding +7 to 3, the result of adding 3 to +7 is +4. In order to start with +7 and arrive at +4, we must move in the negative direction, or toward the left. Therefore, we conclude that to add a negative number, we go toward the left. Thus, to add 4 to +9, we start at +9 and go 4 units toward the left, arriving at +5. To add 7 to +2, we start at +2 and go 7 units toward the left, arriving at 5. To add 4 to 5, we start with 5 and proceed 4 units toward the left, arriving at 9. The above results arc given here: + 4 -:< -7 +7 +9 + 2 -5 + : * +1* + 3 -3 -4 -7 -4 + 7 +2 -4 +4 +5 -5 -9 248 ADDITION AND SUBTRACTION 249 198. Principles. A careful consideration of the above will disclose the following principles: (1) The algebraic sum of two numbers with like signs is the sum of their absolute values, with the common sign prefixed. (2) The algebraic sum of two numbers with unlike signs is the difference between their absolute values, with the sign of the one greater in absolute value prefixed. In adding three or more algebraic numbers, differing in signs, find the sum of the positive numbers, and then the sum of the negative numbers by principle (1), and then add these sums by principle (2). Thus, in finding the sum of +2, +10, -6, -3, -7, +9, we take +2 + 10+9 = +21 and (-6) + (-3) + (-7) = -16, then +21 + (-16) = +5, the sum. 199. Subtraction of algebraic numbers. Subtraction is the inverse of addition. If we are given one of two numbers and their sum, subtraction is the process of finding the other number. In arithmetic it is assumed that the minuend is always greater than the subtrahend. In the subtraction of algebraic numbers we not only may have the subtrahend larger than the minuend when the numbers are positive, but either or both subtrahend and minuend may be negative numbers. Since subtraction is the inverse of addition, if we consider the system of algebraic numbers arranged along the horizontal line as in Art. 197, we have the following principles: (1) Subtracting a positive number is equivalent to adding a numerically equal negative number. (2) Subtracting a negative number is equivalent to adding a numerically equal positive number. These may be combined in the following: RULE. Subtraction of algebraic numbers is performed by considering the sign of the subtrahend changed and proceeding as in addition of algebraic numbers. Applying the rule, we find the following algebraic differences : + 7 +4 -6 -3 -8 + 7 + 3 +6 -2 -7 _+3_ -2 +4 -2 -4 +4 -11 +9 250 PRACTICAL MATHEMATICS It should be carefully noted that 4 3 may be considered as a +4 minus a +3, or as a +4 plus a 3. It is this choice that causes more or less trouble to the beginner. If the number with its sign of quality is inclosed in paren- theses, we have, for example, (+4) + ( 6) (+7) ( 11) + (+7). This may also be written (+4) -(+6) -(4-7) + ( + ll) + (+7), which is the same as 4 6 7+11+7 where the signs indicate operations and all the numbers are positive. EXERCISES 57 Find the sum in exercises 1 to 7. 1. 7, -10, -13, 16, 25, -3. Ans. 22. 2. 27, 46, -100, -16, 17. Ans. -26. 3. 3, 16, -21, -1, 2, 1. Ans. 0. 4. !, -f, I, -i". An*. . 6. -3}, -7A, 3|, -A. Ans. -8|J. 6. 3.25, -7.16, -10.3, 14.1. Ans. -0.11. 7. 14.17, -16.19, -26.3. Ans. -28.32. 8. From 17.6 take -14.3. Ans. 31.9. 9. From -111 take -12. Ans. -99. 10. From -46 take 75. Ans. -121. 11. Is the absolute value of an algebraic number ever increased by subtraction? Illustrate. 12. Is the absolute value of an algebraic number ever decreased by addition? Illustrate. 13. How many degrees of latitude between places at +37 45' 17" and at -16 14' 53". Ans. 54 0' 10". 14. If a steamer is moving through still water at the rate of 20 miles per hour, and a man walks forward on the deck at the rate of 4 miles per hour, express the rate the man is moving with reference to the water. Suppose the man walks toward the stern of the boat at the same rate, how may the rate of the boat, and the rate of the man with reference to the boat, be expressed? 200. Addition and subtraction of literal algebraic expres- sions. We add 5 bushels, 8 bushels, 10 bushels and get 23 bushels. So we have 5 bu.+8 bu.+lO bu. = 23 bu. Similarly, 6d+4d+7d=17d, 4xy-\-7xy+8xy = 19xy, and In subtraction, we have 17o 5a = 12a and 46z V - 6z V - 40z V- We know that in arithmetic we cannot add or subtract ADDITION AND SUBTRACTION 251 unlike things; neither can we do so here. If we wish to add 3a to 26, we indicate the addition, thus, 3a+26. From these considerations, we have the following principle: Monomials which are alike, or similar, can be added or sub- tracted by adding or subtracting the coefficients. If the mono- mials are unlike, the operations can only be indicated. Examples of addition. (1) (2) (3) + 3abc Wxy 3 I7ab - Qabc + 3xy 3 3xy + 10a6c - 4xy 3 -4c 2 -16a6c - 7xy 3 +3a 2 - 3a6c -I2abc + 4xy 3 17a6-3:n/-4c 2 +3a 2 Examples of subtraction. (1) (2) (3) -21x 2 y Uab 3x z y 6c -24x 2 y 14a6 + 6c 201. Polynomials. The addition and subtraction of poly- nomials is similar to that of monomials. Write them so that like terms are in the same column, and combine the terms in each column as with monomials. Example of addition. Exam.ple of subtraction. - 32 17a:y 2 -14c 2 +4a + 7z IQx 2 - 5c 2 -8a 9z 7rn/ 2 -9c 2 +12a ax 2 + 4?/ 2 + 2z 202. Test or proof of results. It is very important that one should be able to test the results. The problems in addition or subtraction of literal algebraic expressions may be tested by substituting some definite values for the general numbers. Thus, if a = l, & = 1, and x = \ in the following example, the test is as given. Example. Test. - 7ab+ 422-3&Z - 7+ 4-3=- 6 - Sa&-10z 2 -4&z - 8-10-4= -22 lx 2 +6bx ~ 9 + 11+6= 8 -*>4ob+ 5x 2 - bx -24+ 5-1 = -20 252 PRACTICAL MATHEMATICS The test depends upon the fact that the letters used may have any values whatever. We could just as well take o = l, 6 = 2, x = 3. Of course, we usually choose values that make the computations as simple as possible. Ans. 24a. Ans. 3ab. Ana. -8x*. EXERCISES 68 Write the sums in exercises from 1 to 11. 1. 3a, 5a, 7 a, 9a. 2. 6a6, -4a6, 3a6, -2ab. 3. 9x, 10x, -14x, -13x*. 4. 14xV, -llxV, -16xV- 6. 2w+n, 6/n 4n, 7m 2n. 6. p+2q-r, 2p-3g. 7. a+46-6c, -26+5c. 8. 2x+3y 4z, x+yz. 9. 2x-|-3a-r-w, 2y-3a-m. 10. a-(-3c-17, 5a-6c + 14, 2a-5c-2. 11. 2a'-4cd, -8a 2 -7cd, -25a z + 16cd. 12. Subtract exercises 6 to 9 above. 13. From 3ox-4cd take 10ox-2cd. 14. From 17o6-2c take 3+c-/0+2) PRACTICAL MATHEMATICS 5. How many cents in a dollars? In 6 dimes? In c dollars and d dimes? Ans. lOOo; 106; lOOc+lOd 6. If x represents the number of bushels of apples bought, what was the price per bushel if $10 was the cost of all? 7. Find the cost per bushel of apples if 25 bushels at x dollars per bushel cost $10. State as an equation and solve. 8. A room is three times as long as it is wide and its perimeter is 96 ft. Find the length and width. Solution. (1) Let x = number of feet in width. (2) Then 3x = number of feet in length, (3) and z+i+3z +3z= perimeter. (4) Also 96= peri meter. (5) /. z+x+3z+3z = 96, by axiom (5). (6) Collecting terms, 8x = 96. (7) Dividing by 8, x = 12 = number of feet in width, and 3z = 36 = number of feet in the length. Note that in statements (3) and (4) are two expressions for the same thing, the perimeter. These two expressions put equal to each other in statement (5) give the equation to be solved. 9. In a company there are 64 persons, and the number of children is three times the number of adults. How many are there of each? Solution. (1) Let x = the number of adults. (2) Then 3x = the number of children, (3) and z+3x = number in the company. (4) Also 64 = number in the company. (5) .'. z+3* = 64. (6) 4x=64. (7) x = 16 = the number of the adults. (8) 3x = 48 = the number of the children. 10. If twice a number is added to six times the number the sum is 96. What is the number? Ans. 12. 11. A horse and wagon cost together $214. If the horse cost $76 more than the wagon, find cost of each. Ans. horse $145; wagon $69. 12. The second of three numbers is 4 times the first and the third ie 3 times the first. If the first and second are added and the third sub- tracted from the sum, the remainder is 60. Find the three numbers. Ans. 30; 120; 90. 13. The number of copies of a book sold doubled each year for three years, and in that time 36,750 copies were sold. How many were sold each year? Ans. 5250; 10,500; 21,000. 14. A rectangular lot is 30 rods longer than it is wide. Use w rods for the width and state in an equation that the perimeter is 260 rods. Solve this equation and find tho width and length of the field. Ans. 50 rd. ; 80 rd. EQUATIONS 263 15. The sum of two numbers is 300 and their difference is 200. What are the numbers? Solution. (1) Let x = the greater number. (2) Then 300 x = the lesser number, (3) and x- (300- x) = the difference. (4) Also 200= the difference. (5) /. x-(300-x) =200, by axiom (5). (6) Simplifying, x- 300+x = 200. (7) Transposing, x + x = 300 +200. (8) Collecting terms, 2x = 500. (9) Dividing by coefficient of x, x=250, the greater number. (10) 300-x=50, the lesser number. Test. The sum of 250 and 50 is 300, and the difference is 200. Hence the conditions of the problem are satisfied. 16. A man has $77 in $1 bills and $10 bills. How many bills has he if he has the same number of each? Ans. 14. Suggestion. Let x = number of each kind of bills. Then x = number of dollars represented in $1 bills, and lOx = number of dollars represented in $10 bills. .'. x + 10x=77. 17. If I have three times as many $5 bills as $2 bills and the amount of money in these bills is $85, how many of each kind of bills have I? Ans. 15; 5. 18. In my pocketbook are a certain number of silver dollars, twice as many quarters as dollars, and five times as many dimes as dollars. If the total amount of money is $4.00, find the number of coins of each kind. Ans. 2; 4; 10. 19. A shopper bought three articles. The second cost three times as much as the first and the third $3 more than the second. Find the cost of each if the total cost was $9. Ans. $ ; $2| ; $5f . 20. Three tanks hold a total of 24,500 gallons. The first holds 4500 gallons more than the second, and the second 2500 gallons more than the third. How many gallons does each hold? Ans. 12,000; 7500; 5000. 21. Of two candidates for the same office, the successful one received a majority of 265. How many votes did each receive, if the total number of votes cast was 6793V Ans. 3529; 3264. 22. Find the three consecutive even numbers whose sum is 216. Ans. 70; 72; 74. Suggestion. Let x = first number. Then x+2=second number, and x+4=third number. 23. Find the four consecutive odd numbers whose sum is 88. Ans. 19; 21; 23; 25. 264 PRACTICAL MATHEMATICS 24. Find the three consecutive numbers whose sum is 66. Ans. 21; 22; 23. 26. Divide $210 between A, B, and C so that B shall have $35 less than A and $20 more than C. Ans. $100; $65; $45. 26. One angle is the complement of another. If 14 is subtracted from the second and 14 added to the first, the first will be 44 larger than the second. Find the two angles. Ans. 53 and 37. Two angles are said to be complements of each other if their sum is 90. (See Art. 90.) 27. The difference between two angles is 14. Find the angles if they are complements of each other. Ans. 52 and 38. 28. A father and son earn $188 a month. If the son's wages were doubled, he would receive $62 less than his father. How much does the son receive? Ans. $42. 29. Three men, A, B, C, raised 4080 bushels of wheat. A raised three times as many bushels as B and 330 bushels more than C. How many- bushels did each raise? Ans. A, 1890 bu. ; B, 630 bu. ; C, 1560 bu. 30. The sum of four angles about a point is 360. The second is twice the first, the third three times the second, and the fourth is 10 greater than the first. Find the angles. Ans. 35, 70, 210, 45. CHAPTER XXIV MULTIPLICATION 212. Fundamental ideas. Multiplication of whole num- bers in arithmetic may be thought of as a shortened process of addition. For instance, 5X3 = 5+5 + 5 = 15. (1) We say that the multiplicand is used as an addend 1 as many times as there are units in the multiplier. This idea of multiplication must be enlarged in order to 8X3 include the multiplying by a fraction. Thus, 8Xf= . ', that is, we multiply by the numerator and divide by the denomi- nator of the multiplier. For multiplication in algebra we shall retain its arithmetical meaning when the multiplier is a positive whole number or fraction, but shall have to extend the meaning to include negative numbers. From the arithmetical meaning, since the multiplier is positive, we have, (-5)X(+3) = (-5) + (-5) + (-5) = -15. (2) Now we know that when we multiply two positive abstract numbers together, it does not matter which is used as the multiplier. If we assume that this principle holds when one of the numbers is negative, we have, (-5)X(+b) = (+3)X(-5) = -15. This gives us the following meaning for multiplication by a negative number: To multiply by a negative number is to multiply by its abso- lute value and then change the sign of the product. Thus, to multiply +5 by 3 we multiply +5 by +3 getting +15, and then change the sign of the result. That is, (+5)X(-3)= -15. (3) 1 An addend is one of the numbers to be added in an addition problem. 265 PRACTICAL MATHEMATICS Likewise, to multiply 5 by 3 we multiply 5 by +3 getting 15, and then change the sign of the result. That is, (-5)X(-3) = + 15. (4) In (1), (2), (3), and (4) we have examples of all the combina- tions possible of two algebraic numbers. 213. From the above considerations we see that in finding the product of two algebraic numbers: (1) The numerical part of the product is the product of the absolute values of the multiplicand and multiplier. (2) The sign of the product is plus when the signs of the multiplicand and multiplier are alike, and minus when their signs are unlike. This is called the law of signs in multiplication. It may be stated as follows: + x- = -, -x + = -. 214. Concrete illustration. For those who find the fore- going difficult to understand, the following may clear up matters. There is a machine shop employing laborers and appren- tices. The laborers are paid $15 per week, and the apprentices are charged $3 per week. Suppose that an increase in the number of men or dollars is positive, and a decrease in either is negative. Thus, a number of laborers or apprentices taken in will be called positive, and a number let go will be called negative; while the number of dollars received from an apprentice is positive and the number of dollars paid a laborer is negative. On these suppositions we have: (1) If apprentices are increased by 5, the amount of money is increased $15 per week. That is, (2) If apprentices are decreased by 5, the amount of money is decreased $15 per week. That is, MULTIPLICATION 267 (3) If laborers are increased by 5, the amount of money is decreased by $75 per week. That is, (-15)X(+5) = -75. (4) If laborers are decreased by 5, the amount of money is increased by $75 per week. That is, (-15)X(-5) = +75. From these considerations, we may deduce the same rules as already given. 215. Continued products. To find the product of three or more numbers, we find the product of the first two, and then multiply this product by the third, and so on till all the numbers have been used. By applying principles (1) and (2) of Art. 213, we obtain the following : (1) The product of an odd number of negative factors is negative. (2) The product of an even number of negative factors is positive. (3) The product of any number of positive factors is positive. Thus, (-2) (-2) (-2) (-2) (-2) = -32, while (-2X-2)(-2X-2)(-2)(-2) = The first one of these equals ( 2) 5 , and is then read "the fifth power of 2." The second is ( 2) 6 . It should be noted that such a form as ( 2) 5 does not mean the same as 2 5 , though they may be equal. The form 2 5 is read "minus 2 to the fifth power. 1 ' Thus, (-2) 2 = ^-2X-2)=4 = 2 2 , (-2) = (-2K-2)(-2) = -8= -2 3 , -2=- (2) (2) (2) (2) (2) = -32, (-3)*(-2)3 = (-3X-3)(-2)(-2)(-2) = -72, (4 2 X3')= 4-4-3-3 = 144. EXERCISES 64 1. Find the product of -7, -8, and +10. 4ns. 560. 2. Find the product of -2, -40, +75, and -60. 3. Find cube of -7, of +8, of -21. Jl.s PRACTICAL MATHEMATICS Find values of the following: 4. (-4)*. Ans. -1024. 5. (-8) 4 . Ana. 4096. 6. (-1)">. Ana. +\. 7. (-4)(-3). Ans. -576. 8. 2(+6)*(-6). Ana. 2592. If z = 2, j/= 3, z = 4, find value of: 9. *V- 4rw. 72. 10. xyz. Ana. 24. 11. j/'z. An. 108. 12. zV f - Ana. 576. IS. (6)(-5)(-2)t/>. ATM. -1620. 14. 15-(-2)-(-3)+4-(-5)-6. Ana. -30. 15. l-2-3+3-4-5 + (-5)-(-6)-(-7). Ana. -144. 16. 4-(-3)-5+2-(-3)-4+3-(-3)-4. Ans. -120. 17. (-2)-(-3)-(-4)-6-(-7). Ans. -2760. 18. (-3) J -3 + (-2) s -2 s . 4rw. -118. 19. (-2)-(-3) + (-4)-(-5) + (-5)-(-6). Ans. 56. 20. l-2-3-4-5-(-l)(-2)(-3)(-4)(-5). Ans. 240. 31. 4-5-6 7-(-4)(-5)(-6)(-7). Ans. 0. 216. Law of exponents. The law which applies to ex- ponents that are positive integers is derived from the defini- tion given in Art. 186. Since a 6 = a-a-a-a-o, and a 3 = a-a-a, then o 6 -a 3 = a-a-a-a-a-a-a-o = a 8 , and a 5 -a 3 = a 6+3 = a 8 . In general a n = a-a-a-o-- to n factors 1 and a m = a-a-a-a to ra factors, then a n -a m = a-a-a-a-a--' to (n+m) factors, and a n '(i m = a n+m . Similarly, when there are any number of factors we have LAW. The product of two or more powers of the same base is equal to that base affected with an exponent equal to the sum of the exponents of the powers. 217. To multiply a monomial by a monomial. Example. Multiply 14a 3 6 2 by 3a 4 6 3 . Process. Discussion. Since the multiplier is composed of 14a 3 6 2 the factors 3, a 4 , and 6 3 , the multiplicand may be multiplied by each successively. In each case the product for any one of these factors is obtained by multi- 1 A repetition of dots, as a, b, c,--. is the sign of continuation. It is read "and so on." MULTIPLICATION 269 plying a single factor in the multiplicand by it. We multiply by 3, by multiplying 14 by 3, which gives -42a 3 6 2 . This is multiplied by a 4 , by multiplying the a 3 by a 4 , which gives 42a 7 6 2 . This is multiplied by b 3 , by multiplying the 6 2 by b 3 , which gives 42a 7 6 5 , the answer. The multiplication is carried out by determining in the following order: (1) the sign of the product, (2) the coefficient of the product, (3) the letters of the product, (4) the exponents of these letters. Thus, in the above example the sign is + X = ; the coefficient is 14X3 = 42; the letters are a and b; and the exponents are, for a, 3+4 = 7, and for b, 2+3=5. This plan should be carefully followed by the beginner. 218. To multiply a polynomial by a monomial. RULE. The product is found by multiplying each term of the multiplicand by the multiplier, and taking the algebraic sum of these partial products. Example. Multiply 7 ax 3 - 21a6 4 - 3x 2 by 2a 2 6 3 z 4 . Process. 7ax 3 -21ab*-3x 2 - 42a 3 6 V - Explanation. The first term at the left of the product is obtained by multiplying the first term at the left of the multiplicand by the multiplier. The second and third terms in the product are obtained in a similar manner. In each of the multiplications we have a monomial by a monomial, which has been discussed in the previous article. 219. To multiply a polynomial by a polynomial. RULE. Multiply every term of the multiplicand by each term of the multiplier, write the like terms of the partial products under each other, and find the algebraic sum of the partial products. Example I . Multiply x 2 -\- 3xy 2y 2 by 2xy -2y*. Process. 2xy-2y z 2xy times (z 2 +3z?/ 2?/ 2 ) = 2x 3 y+Qx*y z 4xy 3 -2y 2 times (z 2 + ?>xy - 2?/ 2 ) = -2a?V- Adding these we get, 2x 3 y+4:x' 2 y 2 270 PRACTICAL MATHEMATICS Example 2. Multiply 3a 2 +36 2 +o& by &-2a l 6+a& 2 . Process. 3a 2 +36 2 +06 6-2a 2 b +ab 3a 2 & 3 +36 6 +a& 4 -6o 2 6 s -6a 4 6-2a6 2 a 2 b 3 +3afr 4 -2a 2 & 3 +36 5 +4a& 4 -6a 4 &+ a 3 6 2 220. Test. Problems in multiplication can be tested by substituting convenient numerical values for the letters. It is best to use values larger than 1, since with 1 the exponents are not tested, as any power of 1 is 1. Test of example 1, by letting x 1 and y = 2. z 2 +3xy-27/ 2 =4+12-8 = 8 2zy-2y* =8-8 =0 2x*y + 4s 2 2/ 2 - lOxy 3 + 4t/ 4 = 32 + 64 - 1 60 + 64 = 0. The work is probably correct if the product of the values of the two factors equals the value of the product. EXERCISES 65 Find the product of the following: 1. 10o6 2 and 3o'6. Ans. 30a 4 6. 2. 16n'x 5 and -2n'x t . 3. 4a 2 xV and 5x*y. Ans. -20a*xV- 4. -17iV&nd -3a'x 2 . 5. -5mn 2 . 6. x*y* and x*y*z*. 7. 3x*y s , x 2 , t/ 3 , and 4xy. Ans. 12x*j/ T . 8. 3PQ, Q, and 4P 2 Q. 9. ox 2 , Jo 2 x s , and 10. a", a 2B+1 , o 8 " +I , and o" +1 . Ans. a" n+ *. 11. (a 3 ) 2 or a 3 times a*. Ans. a*. 12. (m 2 *! 3 )^? ATM. m"n 12 . 13. (4a 2 6 14. (a n 6")'? Ans. 15. a s 6 2 c 18. x 4 j/ 2 z 17. 3 T FIG. 193. FIG. 194. 60. find the area of a rectangle 2x+5 ft. in length and x 6 ft. wide. Ana. 2x 2 -7x-30sq. ft. 61. Find the volume of a rectangular solid 2z 3 ft. wide, 7x 2 ft. long, and x +4 ft. deep. Ann. 14x 3 +31z* 94x+24 cu. ft. 62. Find the volume of a right circular cylinder if the altitude is h ft. and the radius 2A-4 ft. Ana. ir(4A 3 -16/i 2 + 16A) cu. ft. 1 Area = ab FIG. 195. (1) ac (2) be f f 1 t< a >|< b >] FIG. 196. 63. A hollow square has dimensions as shown in Fig. 193. Find its area. Ans. 3s*+28s+32. 64. A ring has dimensions as shown in Fig. 194. Find its area. Ans. M27a* + 150a+32). 221. Representation of products. If a is the number of units in the altitude in Fig. 195 and 6 is the number of units in the base, then the product ab is the number of square units in the area of the rectangle in the figure. Similarly, the two rectangles in Fig. 190 represent the product of (a+6) by c. The part marked (1) represents the partial pro- duct ac, and the part marked (2) repre- sents the partial product be. Fig. 1<)7 shows that (x+T/Xz+y) =x*+2xy+y-. The whole figure is a square x+y on a side. This is made up of a square x units on a side, having x~ square units; a square y Fio. 197. MULTIPLICATION 273 units on a side, having y 2 square units; and two rectangles each x units by y units, having xy square units each. Hence the whole figure contains x 2 -\-2xy +?/ 2 square units. EXERCISES 66 Find the product in each of the following, and show the product and the partial products by drawings. 1. (a+b+c)d. 2. (a+b+c)(d+e). 3. (a+6+c) 2 . 4. (x+y)(x-y). 5. (z-2/) 2 . 6. (a+b+c)(d+e+f). 222. Approximate products. By multiplication we get the formula (l+a)(l+fe) = l + a+6+a&. If in this formula a and 6 are small fractions, the product ab will be very small compared with a and 6. The value of (l+a)(l+6) will then be approximately l+a+6. Thus, if a = 0.05 and 6=0.02, the approximate value of (l+a)(l+6) = 1+0.05+0.02 = 1.07, while the exact value is 1.05X1.02 = 1.071. 1.071 -1.07 = 0.001 = difference. Example. If a = 0.01 and 6 = 0.02 find the per cent of error in the product (1 +a) (1 +6) if the term ab is disregarded. Solution. The approximate value of (1 +0.01) (1 +0.02) = 1 + 0.01 + 0.02=1.03. The exact value = 1+0.01 +0.02+0.0002= 1.0302. The error = 1.0302 -1.03 = 0.0002. The per cent of error is obtained by finding what per cent 0.0002 is of 1.0302. / . 0.0002 -T- 1 .0302 = 0.0002 - = 0.02 - %. Ans. EXERCISES 67 1. Find per cent of error if term ab is disregarded when a = 0.001 and 6=0.002. Ans. 0.0002-%. 2. Calculate the value of 1.002X1.05 to three decimal places by using the approximate method. Ans. 1.052. 3. To how many places can the product of 1.02X1.0024 be found by the approximate method? Ans. 4. 4. How much error per cent is there in assuming (l+a)(l+6) =l+a+& when a =0.003 and 6 = 0.005? Ans. 0.0015-%. 5. In assuming (l+a) 3 = l+3a, what is the per cent of error when a =0.0002? When a =0.002? When a = 0.02? When a = 0.2? Ans. 0.000012-%; 0.0012-%; 0.114-%; 7.41-%. 18 274 PRACTICAL MATHEMATICS 6. In measuring the radius of a circle of correct length 1 ft., there is an error of 1%. Find the per cent of error in the area of the circle deter- mined from the measured radius. Am. 2% nearly. 7. If in determining the area of a triangle by drawing to scale and measuring the base and altitude, the base is measured \\% and the altitude 2% too large respectively, find per cent of error in area, disre- garding the product term. Am. 3J%. 8. Find the approximate products of the following: (1)1.003X1.012. (4)0.97X0.98. (2)1.02X0.97. (5)0.996X0.997. (3) 1.004X0.998. (6) 0.985X0.996. 9. In measuring a rectangle the length is measured 2% too large and the width 3% too small. If the area is computed from these measure- ments, find the per cent of error in the area. Ans. 1% nearly, too small. 10. If, in measuring a triangle, the base is measured 1 % too large and the altitude 1J% too small, find the per cent of error in the area computed from these measurements. Ans. \% nearly, too large. 11. If the error in a number is 1%, what is the per cent of error in the square of the number? In the cube of the number? Am. 2%; 3% nearly. CHAPTER XXV DIVISION, SPECIAL PRODUCTS, AND FACTORS 223. Division. Division is the inverse of multiplication. That is, the quotient must be an expression that multiplied by the divisor will give the dividend. From the law of signs and the law of exponents in multi- plication we have the following : (1) In dividing, like signs give a positive, and unlike signs give a negative sign for the quotient. (2) In dividing powers of the same base, the exponent of the quotient equals the exponent of the dividend minus the exponent of the divisor. This applies, by definition of the positive integral exponent, only when the exponent of the dividend is larger than the exponent of the divisor. Thus, a 5 -f-a 3 = o 2 ; but when the exponents are equal and we subtract we obtain the exponent 0, which is meaningless. In dividing the same powers of the same base, the quotient is 1. Thus, a 3 -r-a 3 = l and, in general, a n -5-a n = l. 224. Division of one monomial by another. It is well for the beginner to carry out the work of a division in a regular order as in multiplication. (See Art. 217.) The steps in division are: (1) Determine the sign of the quotient. (2) Determine the coefficient. (3) Determine letters and exponents. Remember that in the process of division we divide where we multiply in multiplication, and we subtract exponents where we add exponents in multiplication. Example. Divide 25a 4 # 5 by 5a 2 x*. Process, carried out in steps. 275 276 PRACTICAL MATHEMATICS 25 -^ -5= -5, a 4 -i-a 2 = a 2 , 2 -2/ 2 +zy). 16. (7+ x +y)(7-x-y). 16. (3-x+y)(3+x+y). Suggestion. This may be written [(3+y) -x][(3+^) +x] = (3+?/) 2 -x 2 =etc. 17. (o-y-2)(o-y+z). 18. [4a-(*-2&)][4a+(-2&)]. 19. (Ha; 3 ?/-z 4 )(llx 3 ?/+2 4 ). 20. (30x 2 ?/ 3 +3a;2/)(30x 2 ?/ 3 -3x?/). 21. 95X105. 22. 995X1005. 23. 64+56. 24. 75X85. 26. 505X495. 26. 706X694. 232. Factors of the difference of two squares.^ From a consideration of the preceding it is easily seen that the difference of two squares can be factored into two binomial factors that are, respectively, the sum and the difference of the square roots of these squares. Example I. 4-a 2 =(2+a)(2-a). Example 2. 16a 4 -9?/= (4a 2 +3t/)(4a 2 -3?/). Example 3. (a+6) 2 -2 2 =(a+6 + 2)(a + fe-2). Example 4. a 2 -6 2 +26c-c 2 =a 2 -(6 2 -26c+c 2 ) =a 2 -(&- c) 2 =(a+6-c)(a-&+c). 282 PRACTICAL MATHEMATICS EXERCISES 75 Factor the following and test by multiplication : 1. 4-x. 2. 16-4y. 3. o-l. 4. l-9x. 5. Sla'-lGfc*. 6. 7* -5*. 7. 225-64a. 8. 36a'-496*. 9. 4a 1 6 10. 64a&-l(KW*. 11. 3>a<-2&. 12. 2* 3 4 t FIG. 198. A\< 40-ft FIG. 199. 39. A flagstaff, CD, Fig. 199, 75 ft. high, breaks at point B and end D strikes at A, a distance of 40 ft. from C. Find the length BD that was broken off. Ans. 48? ft. Suggestion. Let BD=x, then CB = 75x. 40. A piece of sheet iron containing 625 sq. in. is bent into a cylinder 9 in. in diameter. How high is the cylinder? What is its volume? Ans. 22.1 in.; 1406+ in. 3 41. At what rate simple interest will $75.00 amount to $106.50 in 6 years? Ans. 7%. 237. Equations solved by aid of factoring. The equations considered so far have reduced to a form in which a certain number of times the unknown equaled some number. Thus, 6z = 12 is such a form. They are called simple equations. All equations do not reduce to such a form as this. For instance, when the equation has been reduced, we may have an equation in which the square of the unknown equals some number. Thus, 2 = 5 is such a form. Such an equation is called a pure quadratic equation. Again, when the equation is simplified and reduced, we may have a form containing the square and the first power Jvs PRACTICAL MATHEMATICS of the unknown equaling some number. Thus, x 1 5x = 24 is such a form. Such an equation is called an affected quad- ratic equation. Some of these forms of equations, together with certain other forms, can be solved by the aid of factoring. Example 1. Solve the equation x 2 5x+6 = 0. Discussion. This equation puts the question: For what values of x does x 2 5x+6 equal zero? If we factor the expression in the first member we get (x 2)(x 3) =0. The question now is: For what values of x does the product (x-2)(x 3) have the value zero? We know that the product of two factors is zero if either, or both, factors are zero and not otherwise. Hence the product is zero, if x 2 = 0, or if x 3 = 0. Thus, the solution of x 2 5x+6 = depends upon the solution of the two simple equations, x 2 = and x 3 = 0. These give the values 2 and 3 for x. That these are the values of x may be tested by substituting each one separately in the equation x 2 -5x+6 = 0. Substituting x = 2, gives 4-10+6 = 0, or = 0. Substituting x = 3, gives 9-15+6 = 0, or = 0. The values of the unknown number that satisfy the equation, that is, answer the question, are called roots of the equation. A quadratic equation having one unknown letter always has two roots. Example 1. Solve the equation x 2 25 = 0. First solution. (1) Given equation, x 2 25 = 0. (2) Factoring, (x+5)(x-5) =0. (3) Putting each factor equal to zero, x+5 = and x 5 = 0. (4) Transposing, x = 5 and x = 5. Testforx=-5, 25-25 = 0. Test for x = 5, 25-25 = 0. Second solution. (1) Given equation, x 2 25 = 0. (2) Transposing, x 2 = 25. (3) Taking the square root of each member of the equation, x=5. EQUATIONS 289 Here the sign is read "plus or minus." It means that 5 is a plus as well as a minus quantity. It should be noted here that we are saying that 25 has the two square roots, +5 and 5. Either of these is the square root of 25, for (+5) 2 = 25 and also ( 5) 2 = 25. Hence both fulfill the definition of a square root, that is, one of the two equal factors into which a number may be divided. Any positive number has two square roots, one positive and one negative, both equal in absolute value. Example 3. Solve (3 + !) (3 -3) (23 -16) -0. Equating each factor to zero, & + l=0, 3-3 = 0, and 23- 16 = 0. Solving these, x= 1, x = 3, and re = 8. We have the following rule of procedure when solving an equation by the aid of factoring: RULE. (1) Simplify the equation as much as possible. (2) Transpose all terms to the first member of the equation. (3) Factor the expression in the first member. (4) Equate each factor to zero. (5) Solve each of these equations. EXERCISES 79 Solve the following by the aid of factoring: 1. (x-4)(x-3)=0. 2. (x-5)(x+6)=0. 3. (x-2)(x + l)(z+3)=0. 4. (x + l)(x-5)(x-3)=0. 6. (2x-l)(x+4)=0. 6. x(x-2)(3x+5)=0. 7. x 2 -16=0. 8. (z 2 -9)(x 2 -36)=0. 9. x 2 -16=48. Am. Sand -8. 10. x 2 -z = 56. Ans. Sand -7. 11. (2x + l)(x+3)=z 2 -9. Ans. -3 and -4. Suggestion. It is necessary first to clear of parentheses, transpose, and unite before factoring. Clearing of parentheses, 2x 2 +7x+3 = z 2 9. Transposing, 2x 2 -x 2 +7x+3+9 =0. Collecting terms, x 2 +7x + 12 = 0. Factoring, (x +3) (x +4) = 0. 12. If 24 is added to the square of a number, the sum equals 10 times the number. Find the number. Ans. 4 or 6. 13. If 78 be subtracted from the square of a number, the difference equals 7 times the number. Find the number. Ans. 13 or 6. 14. If 3 is added to a number, the square of the sum is 9 more than 13 times the number. What is the number? Ans. or 7. 19 290 PRACTICAL MATHEMATICS 16. A rectangle is 8 in. longer than it is wide. Find the dimensions, if the area is 240 sq. in. Ant. 12 in. by 20 in. 16. Find the dimensions of a rectangle that is 4 rods longer than it in wide if, when the length is increased by 6 rods and the width by 4 rods, the area is doubled. Ana. 10 rd. by 14 rd. 17. The base of a triangle is 3 in. longer than the altitude, and the area is 44 sq. in. Find the length of the base and the altitude. Ant. 11 in. and 8 in. 18. The altitude of a triangle is 3 times the base, and the area is 37 J sq. in. Find the length of the base and the altitude. An*. 5 in. and 15 in. 238. Formulas. A formula as given usually stands solved for one letter in terms of several others. For instance, formula [34] is T = ph+2A. Here T is stated in terms of p, h, and A. It often happens that one wishes to express, say, h in terms of T, p, and A. To do this it is only necessary to solve the formula as an equation, and find the value of the particular letter desired in terms of the others. Example. Solve the formula T = ph+2A for each of the other letters. Solution. Here there are three other letters than T, and we will solve for p, h, and A in turn. (1) Given equation, T = ph+2A. (2) Transposing, -p/i=-T+2A. (3) Dividing by the coefficient of p which is h, and indicating the division, since it cannot be performed, T-2A p 7 Ans. (4) Solving (2) for h, since it is properly transposed, , T-2A . h = Ans. P (5) To solve for A transpose (1), -2A = T+ph. rrt L (6) Dividing by the coefficient of A, A = Ans. EXERCISES 80 In the following, the numbers in the brackets are the numbers of the formulas as given in previous chapters, where their meaning can be found. The ability to do such problems as these is very important. 1. [32] A -rob. Solve for a and 6. Am. ~; ~ 2. [33] S = ph, Solve for p and A. Ant. |; EQUATIONS 291 e 3. [42] S = 2irrh. Solve for r. Ans. ^-^ V 4. [43] F = 7rr 2 /t. Solve for h. Ans. 2 - v 6. [46] V = irR z hnr*h. Solve for h. Ans. /p 2 _ 2) - 6. [69] A=47r 2 flr. Solve for R. Ans. - 47r 2 r ^ 7. [67] Z = 2irrh. Solve for r. Ans. -^-^ 8. Using the answer of exercise 7, find the radius of a sphere on which a zone of altitude 3 ft. has an area of 32 sq. ft. Ans. 1.698 ft. T IT 9. [35] 7 1 = 6a 2 . Solve for a 2 and then for a. Ans. -', x/- 10. Using the second answer of exercise 9, find the edge of a cube whose total surface area is 3258 ft. 2 Ans. 23.302 -f ft. 11. [60] 7 = 2ir 2 #r 2 . Solve for r. Ans. 12. [55] S = 47rr 2 . Solve for r. Ans. \- 13. Using the answer of exercise 12, find the radius of a sphere that has a surface area of 2756 ft. 2 Ans. 14.82- ft. /F 14. [43] 7=7rr 2 /i. Solve for r. Ans. A/^- \ TTrt 15. Using the answer of exercise 14, find the radius of a right circular cylinder whose altitude is 16 in. and volume 2674 in. 3 Ans. 7.294 in. 16. Disregarding the resistance of the air, v = \/2gh is a formula that gives the velocity in feet per second a body will have after falling from a height of h feet. Solve this for h and get a formula for the height to which a body will rise if thrown upward with , velocity of v feet per v 2 second. Ans. ft= r Suggestion. First square both members of the equation, which gives v*=2gh. 17. Use the formulas of exercise 16 and find (1) the velocity a stone will have after falling 125 ft., (2) the height to which a stone will go if thrown upward with a velocity of 200 ft. per second, (g = 32. 2.) Ans. v = 89.72+ ft. per second; h = 62 1.1 ft. 18. The formula vt = vo+32.2t gives the velocity that a falling body will have. t = time in seconds the body has been falling, z;o= velocity in feet per second the body has at the start, that is, V Q is the initial velocity, v t velocity in feet per second after t seconds. Solve for v and for t. Ans. t= ' 292 PRACTICAL MATHEMATICS 19. Using the formulas of exercise 18, find (1) the time for a falling body to have a velocity of 600 ft. per second if it started with a velocity of 40 ft. per second; (2) the initial velocity in order that a fulling body may have a velocity of 340 ft. per second after falling 5 seconds. Ana. 17.39+ seconds; 179 ft. per second. 20. Use the formula given in exercise 5 and find the height of a hollow cylinder having outer and inner radii of 14 in. and 10 in. respectively, that it may have a volume of 4.211 cu. ft. Ana. 24.13 in. 21. Given the formula V = far 2 h= 1.0472 r*h, for finding the volume of V I V a circular cone, solve for h and for r. Ans. L0472r ,; \fo472li' 22. Using the formulas of exercise 21, find (1) the altitude of a circular cone having a volume of 800 cu. in. and a radius of 8 in. ; (2) the radius of the base of a circular cone having a volume of 456 cu. in. and an altitude of 10 in. Ans. 11.94- in.; 6.6- in. 23. In the formula given in the answer to exercise 35, page 240, A =td+b(s-\-n). Solve this for t, d, b, 8, and n successively. Aru . t = A-b(s+n), s = A-td-bn d b 24. Using the formulas of exercise 23, find (1) t when A =3.35 in. 1 , 6=2.04 in., s=0.22 in., n=0.56 in., and d = 8 in.; (2) 6 when d = 10 in., t =0.24 in., n =0.63 in., s =0.24 in., and A =4.45 in. 1 ; (3) s when d = 5 in., <=0.19in., 6 = 1. 56 in., n = 0.45in., and A =1.95 in. 2 Ans. 0.22 in. ; 2.36 in. ; 0. 19 in. 25. In finding the area of a trapezoid, 2A = (bi+b t )h. Find 61 if A = 400 in. 2 , 6j = 15 in., and h = 20 in. Ans. 25 in. 26. Find the volume of a sphere that has a surface of 201.0624 sq. in., the formula for the volume of a sphere being V = Jur 3 . Use formula of exercise 12. Ans. 268.08 cu. in. 27. In reckoning simple interest, A=prt+p, where A = amount, p = principal, <=time in years, and r = rate per cent. Solve for each A Ap A p letter. Ans. P = I-T-T; * = i r = 1 +rt ' pr pt 28. Using the formulas of exercise 27, find (1) I when p =$250, r =6%, and A =$300; (2) r when *=3 years, p = $328, and A= $377.20; (3) p when A =$500, t=5 years, and r = 4%. Ans. 3J years; 5%; $416.67. CHAPTER XXVII FRACTIONS 239. The same names and terms are used when referring to fractions in algebra as are used in fractions in arithmetic, and these terms have the same meanings. The same principles are applied and the same operations performed as in arithmetic. Here only those processes will be considered that are neces- sary for the understanding of what follows. When in doubt about an operation in algebra, carry out a similar operation using numerical numbers, and from this determine what the operation with the algebraic expressions should be. REDUCTION OF A FRACTION TO ITS LOWEST TERMS 240. A fraction is in its lowest terms when there is no factor common to both numerator and denominator. Example 1. Reduce |~f- to its lowest terms. Process. - = -=. \.Zi\j p'p'o o Here each term of the fraction is factored and then the com- mon factors are cancelled. We handle an algebraic fraction in the same way. A'V 2^1 3 Example 2. Reduce 4 4 to it lowest terms. 6 *y 2 -x-y^.yy 2x 2 y ^ y- Example 3. Reduce ~TT~O~ ~~9 to its lowest terms. x 2 +2xy+y Process X * ~ ^ &^( x ~ I/) x ~ V Examvle 4 2 293 291 PRACTICAL MATHEMATICS EXERCISES 81 Reduce the following fractions to their lowest terms: 125 225 135 150' 28ax 4 35ax* 45a 5 cx* Ans. 5mr/ 2 141 2SS' 21a'x 24ax 4 ' Am. Ans. 3 Sax 12ax Ans. Ans. 3b 4 5cx*y 1 9. 10. 11. 12. 13. 14. 16. 16. 17. 18. 19. 20. 21. 22. 15acz 1296xVz 792zV 81zV ' 3ax* Am. 3a-i. Ans. 6y 4 88 Ans. - x(z-y)* a 2 -5a+6 a 2 -7a+10* n*+7n-30 n 2 -7n + 12 a(x-y) 3 Ans. a 2 +a-12 25-0* a 2 -! la +30* a-3 . - = a 5 n + 10 x+y 5a . : T a+4 5+a a(x+y)+c(x+y) x(a+c)+y(a+c) ax 8 2ax 8a ax 2 ax 6a , y~ x ~ z " x+y+z x-4 A na. x-3 REDUCTION OF FRACTIONS TO COMMON DENOMINATORS 241. In arithmetic, fractions are reduced to a least common denominator before adding, so in literal fractions we change the fractions to fractions having the lowest common denomi- nator before adding them. FRACTIONS 295 The lowest common denominator, L. C. D., is the lowest common multiple, L. C. M., of the denominators of the frac- tions. We must then first consider the finding of the L. C. M. of algebraic expressions. 242. Lowest common multiple. Example 1. Find the L. C. M. of 24, 32, and 40. First separate into prime factors, and then find a number which contains all the factors of each. Process. 24 = 2 3 -3. 32 = 2 5 . 40 = 2 3 -5. /.L. C. M. = 2 5 -3-5 = 480. Remark. The L. C. M. may also be found by the method of Art. 19, but that method is not as easily applied to algebraic expressions. Example 2. Find the L. C. M. of I2x 2 y, IQxy 3 , and 24x 3 y. Process. 12x 2 y = 2 2 -3-x 2 -y. :.L. C. The L. C. M. is found by taking each factor the greatest number of times it is found in any expression. Example 3. Find the L. C. M. of x-+2xy+y- and x 2 -y' 2 . Process. /.L. C. M. =(x+yY(x-y}. EXERCISES 82 Find the L. C. M. of the following: 1. 72, 288, 64. Ans. 576. 2. 576, 256, 128. Ans. 2304. 3. 5a 3 b 2 , 10a 2 b 3 , 25a 2 fc. Ans. 50a 3 6 3 . 4. 4c 2 , 2ab, 9cd 2 . Ans. 36a6c 2 d 2 . 6. x*-y z , x 2 -2xy+y 2 . Ans. (x-y) 2 (x+y). 6. x*-llx+3Q, z 2 -12z+35. Ans. (z-5)(z-6)(z-7). 7. a s -a& 2 , (a+6) 2 , (a-6) 2 . Ans. a(a+6) 2 (a-6) 2 . 8. a*b+ab 2 , a 2 +2ab+b*. Ans. ab(a+b)*. 9. x 2 +7x, x*+8x+7. Ans. x(x + l)(x+7). 10. o 2 +3a+2, a 2 -4, a 2 -!. Ans. a 4 -5a 2 +4. 243. Fractions having a L. C. D. Example 1. Change S, ^f> an d H to equivalent fractions having a L. C. D. 290 PRACTICAL MATHEMATICS The L. C. D. is found by the method of the preceding article to be 96. Now multiply both numerator and denominator of each fraction by such a number as will make the denomina- tor in each case 96. How is the multiplier in each case found ? 9 9X654 SS ' 16 ~ 16X6" 96' 24 24X4 96' 32 32X3 96' Example 2. Change ~^ y *+4 y -12' and y^y to frac ~ tions having a L. C. D. Process. By the preceding article the L. C. D. _ x-y(y+6) _ _ y-2 (y- vy 2v ~ The multiplier in each case is found by dividing the L. C. D. by the denominator of the fraction. The division is most easily performed by striking out those factors in the L. C. D. which are found in the denominator of the fraction considered. For this reason it is best not to multiply together the factors of the L. C. D. Thus, in the above the L. C. D. is left in the foim y(y 2)(y+6) during the process, instead of in the foHn y 3 +4y 2 12y. EXERCISES 83 Reduce the following to equivalent fractions having a L. C. D.: 7 9 41 . 42 J27 JB2 26' 52' 718' *' 156' 156' 156' JL 4 _ 5 _ A * 8q _ f) 4o' 6a 2 ' 12a 3 ' '' 12a v 12a 3 ' " _ 2 3 2a-26 3. rr> r Ant. ' a+b a-b " a'-b 1 a*-J . a / o* t __ ai+a 1 x*+ax a* x a x a x* a* 3 5 6. a*+3a+2 a *_2a-3 3o-9 5a + 10 Ans. (a+l)(o+2)(a-3) (a-r-l)(a+2)(o-3) 6. FRACTIONS 297 2x b 3x 2 56 2 24x(x+b) -6b(x+b) Ox 2 -lOb 2 Am " 12(x 2 -b 2 )' 12(x 2 -5 2 )' 12(x 2 -6 2 )' 12(x 2 -6 2 )' _ 7. a+x, -- . Ans. a 2 -x 2 a; . > - a x ax ax a 2 +x 2 a 2 -x 2 a 2 +2ax+x 2 a?+x z 8. a x. a+x, - Ans. > - > - a+x a+x a+x a+x ADDITION AND SUBTRACTION OF FRACTIONS 244. Fractions can be added or subtracted as in arithmetic, by first reducing them to fractions having a common denomi- nator, and then adding or subtracting the numerators. The result should then be reduced to its lowest terms. oc a ct^ ~\~ oc^ Example 1. Find the sum of - ', , and - a x a-\-x a z x* Process. L. C. D. =a 2 x 2 . x x(a-\-x) a x (a x) (a+x) a 2 a a(a x) _a z a-\-x (a-\-x)(a x) a 2 a 2 z 2 a 2 a; 2 Adding the numerators, the sum of the fractions is a a-\-2x 7 o v 4. i Example 2. From -r - take , a 2 ax a 2 a; 2 Process. L. C. D. =a(a-\-x}(a x) =a 3 ax 2 . (a+x) (a+x) _a 2 a 2 ax (a 2 ax) (a+x) a 3 ax 2 a+2x _ (a+2x)a _a 2 +2ax a 2 x 2 (a 2 x 2 )a a 3 ax 2 Subtracting the numerator of the second fraction from the 2 numerator of the first, the result is ax EXERCISES 84 1. to 8. Add the fractions in the preceding set of exercises, 9. Add f, |, I, |. Ans. 3J .... ...aa a a a a 10. Add , , , Ans. a. PRACTICAL MATHEMATICS a c-b c+b 6o+7c-b 12. Add - and o in > u inn , nT mr 13. Add - and --- Ant. - mn nr mr , x-2 . 3x+4 230z*- 14. Add lx-\ ,=-and 8xH = J -An. = -. - ox lor 4" 3o+12x 3x-3o 16. From 3x take -- = -- Ans. = -- o o 3a+2 4 , 7a6-106 12 -4a 17. From r take - r= -- Ans. -7 1/nn.c r; siita. r 00 3a-46 2a-6-c , 15a-4c a-46 81a-46 18. Combine = 5 1 T^ oi ' 84 .Sx-2.. 4x+5 16x+23 19. From 2xH = take 3x ^ Ans. - 76 42 ,34 7 7x-10 20. Add s ^, and -: -- r-=' Ans. - x 2 x 3 x 2 5x+6 x* 5x+6 111 1 o&*-a*-a*b-6' 21. Combine--^-^^-^^ Ans. ^^ 22. Combine t _ Q VonH i_i f~ -LQO' ^ ns ' 23. Combine 2 H f^f TTT' Ans. ; . 24. Combine - A x-6. * a ~ b b-a_ 21 -2a-2a6-46* r I~TI *- ^ i r~i *~ r* ATIS. ~~. a 2 6* a z -|-2a6-|-b 1 a 6 (a+6)*(a h) MULTIPLICATION AND DIVISION OF FRACTIONS 245. Multiplication of fractions. As in arithmetic, the prod- uct of two or more fractions is the product of their numerators divided by the product of their denominators. If we first cancel all factors common to both numerator and denominator, the result will be in its lowest terms when the multiplying is done. FRACTIONS 299 35 45 Example 1. Multiply ^ by Q-T- Process. ggX^p^X^^^- Here we cancel 7 and 9 in both numerator and denominator. 27 Example 2. Multiply 42 by ^o- Process. Example 3. Multiply * y J . 2 x 2 2xy+y* __ _ ^ __ g> " y> Process. -- __ X _ __ X 2 (z+T/Xz+y) (x-y)(x-y) x 3 x 3 246. Division of fractions. One fraction is divided by another by multiplying the reciprocal of the divisor by the dividend. The reciprocal of a number is 1 divided by that number. The reciprocal of a fraction is then the fraction inverted. c 6 Thus, the reciprocal of 4 is 5, of f is |, and of r is - _. ., x- llx 26, x 2 - Example. Divide -55- ^ by *.+ 26 a; 2 lSz + 65 Process. =--. x 2 -llx-26 x 2 -9a:+18 = (s-13)(a;+2) (a; -6) (a; -3) : ~ (x + 3) (* - 5) ~ x 2 - 2x - 15 EXERCISES 85 .,,,,., 25 , 48 1. Multiply^ by 2 25 5 2. Find product of 17X^X^- Am. 7^- ou Zu lo . 45 , 9 5 3. Divide gj- by j^- Ans. ^ . 3w . c m 4. Multiply by Ans. CX O X 300 PRACTICAL MATHEMATICS .Safe 16cx* 7d 5. Find product of x-^X^,- 6- Divide , , . by x 6+c u .i^xi|. -n. ..r y m y 13. S " IA 3x+y^4x 14 ' ~9~" ' T 16. ? 2o> ..llx'6 ... 22x' Sox 2c v/ A 1 TJJ Ai . .!">. Arw. 2 n-l'n+l ' 2n-l 2a+b 3a+2ft ; sr"*"-: r~r' ^ITW. 3a-26 ' 4a+b 9a-46 ... 5y* 21c.35c^ 3cy 17. Simplify sriX-; -- ^TTT^' A. -r 2 J 7a 3 4ax 7a'z 4a "5 x' Ails. -.' x 3 x 4 19. Multiply r v~by^ ^.rw. x^ k _. ., x^-Hx-lS, x'-12x-45 x+1 20 - Dlvlde 3 .2_4 3 ._ 45 - b y iFZr6x^27' ArW - xT5' X rrX rO X UX ** X T^O ~~ 1 oX f~ o(J X ~~\ OX ~f~ ^\) \ v/ -C I & , 3J "f" 1 ^^5^50" ' ^-6^^7"j X x^l' *' x^l' a 22. Multiply a 2 +2a6+6 2 by ^^- 2 ' Am. -^~~ 11 a 2 23. Multiply r by a -r- 1 1 6 a b-a Suggestion. r = -r L = r- a o ao ao ao a I o6 a* . . , 6 a , ab a* Now multiply -r- by j- FRACTIONS 301 24. Multiply +- by a Ans. 7. ' o a a a, b 25. Multiply 1 +- by 1 +r : As. Simplify the following: / . 8x \ / . 2x \ " ( V+ ^) * V T +x-3J 42 . x-y\ \x-2/ x+?// \x-?/ X+T// 26. Multiply2+^byl-^|. Ans. 27. Multiply 5-., by - +|- Ans. 2. 2x 2 +?/ 2 T/ 2x 28. Multiply ^rj- by -++- Ans. a. ac+afe+&c a 6 c 00 c x vx 3ax a 2 x 2 a 2 +ox 3a; 29. - X-rr-X~5 - ,-J-r rTT* AfW. -r a a; 4oy c 2 a: 2 bc+b.-v; 4i/ 30. -IK 3V 3, 31 - -r*yysy-y 4a+a x 2 -?/ 2 . 4+a . Wi ^ \o^/ t \o^ i /N o ./I /to. (x-7/) 2 (x+?/) 2 x+a ax?/ 2 xy+ay - -. s s - XA a 2 x 2 t a x . o+x . x 2 x . ^ ;; '. ~r~ r ~;^ r " A7 ~ " x 2 3x 4 x 2 x x 4 34 x 35 a 2 -a-6 5a+a 2 . a 2 +2a ** o 0,1- XN o i < <-. a ^./-v ^IftO. i- a; 2 -a;-20^ a; 2 -a;-2 . O-T-X ao. prr X , <-, F, ~ :~ "^r~* AW5. 3J. a 2 +6a-7 a& 2 +2a 2 b . a 3 -10a 2 +9a . ab-5b ~~2a+b~ a 2 +a-42 : a 2 -Tla+30* ' a-9 ' a+b 302 PRACTICAL MATHEMATICS I a \ / (z+y)\ / 3m \ 3m I ~ o , J I A*. ^ 45. _ 46. i_ 47. (x-x + l) ,+ + l' Xtw. 48. Simplify ^ - (^j ?* -ax(h -*) ? and get ^(3A* CHAPTER XXVIII EQUATIONS AND FORMULAS 247. Subject matter. In previous chapters a number of equations were presented for solution. These involved simpli- fications of various kinds, but were not what are called frac- tional equations; that is, equations in which fractions occur. In the present chapter, besides equations like those previously considered, will appear fractional equations. 248. Order of procedure. The main steps in the solution of a simple equation have already been given, but for the sake of clearness they are repeated here. (1) Simplify the equation; that is, free of signs of grouping, perform indicated operations of multiplication and division if possible, clear of fractions, etc. (2) Transpose all the terms containing the unknown to the first member and all other terms to the second member. (3) Collect terms. (4) Divide both members of the equation by the coefficient of the unknown. (5) Test the results by substituting each in the original equa- tion. 249. Clearing of fractions. A fraction is an indicated divi- sion, and usually a division that cannot be performed. Hence when an equation contains fractions, these must be removed by some other method than division. An equation can be cleared of fractions by multiplying both members of the equation by the lowest common denominator of all the fractions in the equation. Example 1. Solve f+f = 17 ^- o 8 1U 'Ts 2C 3 Solution. (1) Given equation, -+- = 17 O o 10 (2) Clearing of fractions by multiplying each term by the L. C. D . , 40, we have, Sx + 5x = 680 - 4x. 303 304 PRACTICAL MATHEMATICS (3) Transposing, 8z+5z+4x = (4) Collecting terms, 17z = 680 (5) Dividing by coefficient of x, x= 40. Test, f+f-17- or 13-13. jf> _ j D Example 2. In the equation S = -^ r solve for 7. o.zzu PI _ TO Solution. (1) Given equation, 10. |(x+2) = iU-3). ^n. -12. EQUATIONS AND FORMULAS 305 . 11. j --- v 3 =3. Ans. -5. .. 2-x 5x+21 12. = --- = =x+3. Ans. 2Jf. 2i O 13. (z + l) 2 +2(z+3) 2 =3o:(z+2)+35. Ans. 2. 8 = 0. Ans. -Of,. 15. An,l. 2 16. x-(3x-^^) = J(2z-57)-f. 4ns. 5. 17. %x+Q.25x-lx=x-3. Ans. 12. 0.36 0.09^-0.18 679 19. Given Sac 5c = 17; solve for c. Ans. ,. 3a 5 20. Given 4a6c-56c + 16=36c; solve (1) for a, (2) for 6, (3) for c. _26c-4 4 4 be 2cac' 2bab Solution for a. (1) Given equation, 4abc 5fec + 16 =36c (2) Transposing, 4a6c = 56c+36c 16. (3) Collecting terms, 4a6c = 86c 16. 8&C-16 26c-4 (4) Dividing by 4oc, a = jr j, Solution for b. (5) Transposing, 4a6c 5bc 3bc= 16. (6) Collecting terms, (4ac 8c)6 = 16. 16 4 (7) Dividing by 4ac 8c, 6=^ ^- = _ 21. Given (b-x)(b+2x) =6 2 -2x 2 -36+4; solve (1) for b, (2) for x. 4 4-36 22. (a+b)x + (a-b)x = a*. Ans. 23. |(a+.T)+l(2a+x)+i(3o+a!)=3a. 4ns. a. 24. 4(/+6+?/)+3(<+6 ?/) =?/, solvefor /. Am. b. The following are formulas that occur in physics, electricity, etc. 26. Given PD =TFD 1? solve for P. 4ns. > 26. Given F= > solve for TF, f, r, and V. Fgr WV* WV 2 fFJr Ans. -- ; -fj ; -r^ : * '- V 2 ' fr ' F^ ' 27. Given / = ^ ? solve for ^", 7?, r, and n. 20 n(E-lr) nE-IR IR AnS - n ' I ' In 'E-Ir 306 PRACTICAL MATHEMATICS En Solution for E. (1) Given equation, 7 ^ , (2) Clearing of fractions, IR+Inr=*En. IR+Inr_I(R+nr) n (3) Dividing by coefficient of E, E f 28. Given (1) 7-JL; (2) 7-^-; and (3) ; what value of n will make (1), (2), and (3) identical? Solve for n in (2) and (3). IR Ana. 1 ; -_ Ir E-Jr' E-IR 29. Given 7 =-H > solve for each letter in terms of the others. / P 9 p+q> L JP-. P-f Solution for p. (I) Given equation, 7 =-+- (2) Multiplying by L. C. V.Jpq, pq=fq+fp. (3) Transposing, pq fp=fq. (4) Collecting terms, (9/)p /? (5) Dividing by (q -/), P = ~Tf 30. If T. =-+ + / solve for k: and find the value of k if a = 19,000, A- a 6 = 90,000, and Z=3180. 31. Given = h solve for r. T TI TZ rl Ana. ;2G40. 32. Given E = RI+~, find n 33. Given nE = RI+~-, find 7=- m /t ,_ n m 34. Given R t = Ri(l+al) , solve for R , a, and t in terms of the other quantities. D D D D I? /t( /O *to **i *io Fio. 200. 35. Given 72 = radius of a circle, h = height of a segment, and W = length of the chord. If II' and h are known, find R. (See Art. 120.) Solution. In the figure, AB is the chord, DC is the height, and AO is the radius. ADO is a right triangle. Hence, AO'-DO'-IB" or EQUATIONS AND FORMULAS 307 Simplifying, R* - (R* - 2hR +h 2 ) = (~ . Simplifying, R*-R*+2hR-h 2 = * Transposing and collecting, 2A.R = ( - ] +h 2 . \ 2 / 2A 36. When a bar is balanced on a support F which is at a distance a from one end and b from the other, with weights m and n suspended from either end, we have found (see Art. 68) that m:n = b:a. This gives the equation, am = bn. A lever 10 ft. in length has a weight of 1000 Ib. on one end. Where must the support be placed so that a weight of 250 Ib. at the other end will make it balance ? Disregard the weight of the lever. FIG. 201. Solution. Let d =the distance from support to heavier weight. Then 10 d = distance from support to lighter weight. By proportion, 250 : 1 000 = d : 1 - d. .: 2500-250d = 1000d. Transposing, -250d - lOOOd = -2500. Collecting terms, - 1250d = - 2500. Dividing, d = 2=number of feet from the 1000-lb. weight to the support. The weight at one end of the lever in Fig. 201 has a tendency to pull that end down, that is, it tends to make the lever turn about the fulcrum in that direction. The amount of this turning effect is known as a moment, and is the product of the weight by the length of its lever arm. Thus am is the moment of the weight m acting on the lever arm of length a. Two moments that tend to turn in opposite directions balance each other when they are equal. Two such moments put equal to each other give the equation used in solving a problem in levers. 37. If a lever 16 ft. long is supported at a point 18 in. from one end, how heavy a weight can a man weighing 150 Ib. on the long part of the lever balance? Disregard the weight of the lever. Ans. 1450 Ib. 38. Two men are carrying a load, weighing 350 Ib., on a pole 10 ft. long. Find the weight carried by each if the weight is 4 ft. from one end of the pole. Disregard the weight of pole. Ans. 210 Ib. and 140 Ib. 39. Place a fulcrum under a lever 12 ft. long so that a force of 150 Ib. 308 PRACTICAL MATHEMATICS at one end will balance a weight of 1750 Ib. at the other end, if the lever weighs 10 Ib. per foot. Ans. Nearly 1 ft. 3 in. from the weight. When the weight of the lever is taken into consideration, the moment of each arm of the lever is the weight of that arm multiplied by half the length of the arm. Thus, in exercise 39: Let x =the length of the short arm in feet. Then 12 z=the length of the long arm in feet. 1750x = moment due to weight. 150(12 x) = moment due to force. $i'10x = moment due to short arm. }( 12 x)- 10(12 z) = moment due to long arm. The sum of the moments on one arm is equal to the sum on the other arm if the lever is in balance. Then the equation is 1750x+5x* = 150(12 -x) +5(12 -x) ! . . 1 l iy A 1 K*' ,, i ,, '/T 1 ' }< 12- >*<- -12-->{ '>!< 138 ' > V w FIG. 202. 40. In the arrangement of levers for a platform scale shown in Fig. 202, find the lengths of the arms of the lever resting on F 4 so that a weight of 1 Ib. at P will balance a load of 1000 Ib. at W. Ans. 4 in. and 40 in. In the following exercises, the numbers in the brackets are the numbers of the formulas as given in previous chapters, where their meaning can be found. 41. [30] A = Ihw. Solve for w. Ans. w = 42. [60] V='2v*Rr'. Solve for R. Ans. ^ 43. [46] V=vR*h-TT*h. Solve for r 1 and then for r. Ans. 2h V 44. A hollow cylindrical cast-iron pillar 12 ft. high nnd 10 in. in outside diameter is to weigh 1200 Ib. Find the diameter of the hollow. Ans. 7.7 in. EQUATIONS AND FORMULAS 309 Suggestion. Use the answer of exercise 43, and take cast iron at 450 Ib. per cubic foot. 2S 45. [47] S = /. Solve for s. Ans. 46. [48] T = \ps+A. Solve for p. Ans. 2( - T ~ A \ S 47. [49]-S = $(P+p)s. Solve for p. Ans. 2S ~ P *. S 48. [50] T = $(P+j))s+B+b. Solve for s. Ans. 49. [66] V = f TiT 3 . Solve for r. Ans. ., far-mx FT 2 60. Given F = TFZ-, solve for x. Ans. 7-5- Ff EP-pFf 61. Given E = , . -, solve for a;. Ans. ^ (P-ic)p' p rr' Cr 62. Given C =K , solve for r'. Ans. r+r 63. Given Q =K- ' , solve for ti and a. r+r" Kr C a. aKt z T-dQ dQ AnS " aKT 54. Given p 55. Given // = 1.600,000 -^^(1+0.004:^, solve for g(6i+62)-l,600,000(6i-&i) 6400(6! -62) 56. Find the side of a square room such that if its length and width are increased by 3 ft., the area is increased by 81 sq. ft. Ans. 12 ft. 67. Find the side of a square field such that if each dimension is de- creased by 10 rods, the field will contain 400 sq. rd. less. Ans. 25 rd. 68. A rectangle is 3 times as long as it is wide. If the width is increased by 4 in. and the length decreased by 5 in., the area is increased by 15 in. 2 Find the dimensions of the rectangle. Ans. 5 in. by 15 in. 59. The difference between the base and the altitude of a triangle is 6 in. and their sum is 36 in. Find the area of the triangle. Ans. 157| in. 2 60. The altitude of a triangle is 7 in. longer than the base. If the altitude is decreased by 4 in. and the base increased by 6 in., the area is increased by 25 in. 2 Find the altitude and base. Ans. 23 in. and 16 in. 61. Find the number such that we can take ^ of it away and still have i of 320. Ans. 576. 62. From what price can I deduct 33^% on goods which cost $3.20 per yard, and still make 20%? Ans. $5.76. 310 PRACTICAL MATHEMATICS 63. Find the principal that invested for 4} years at 3}% per annum will give an amount of $604.50. Ans. $600. Solution. Let p = number of dollars in principal. Then pX0.03$ X4} = 0.1575p = number of dollars in interest, and p+0. 1575p = number of dollars in amount. .'.p+0.1575p = 694.50. 1.1575p =694.50 p = 600. 64. Find the rate in order that an investment of $820 will amount to $912.25 in 3 yr. 9 mo. Ana. 3%. 66. The sum of $1100 is invested, part at 5% and part at 6% per annum. If the total annual income is $59, how much is invested at each rate? Ans. $700 and $400. Suggestion. Let x = the number of dollars invested at 5%, then 1100 x = number of dollars invested at 6%. ..0.05z+0.06(1100-z) =59. 66. The interest on $120 for a certain time at 6% is $16.56. Find the time. Ans. 2 yr. 3 mo. 18 da. 67. The interest on $584 for 2 yr. 8 mo. 7 da. at a certain rate per cent is $94.121$. Find the rate per cent. Ans. 6%. 68. A man saves $50 more than of his income, spends 3 times as much for living expenses as he saves, and spends the remainder which is $600 for rent. Find his income. Ans. $2400. 69. If air is a mixture of 4 parts nitrogen to 1 part of oxygen, how many cubic feet of each are there in a room 40 ft. by 30 ft. and 12 ft. high? Ans. 11,520; 2880. Suggestion. Let z=the number of cubic feet of oxygen in the room. Then 4x =the number of cubic feet of nitrogen in the room. 70. A room is -?$ as wide as it is long. If the length were decreased by 3 ft. and the width increased by 3 ft., the room would be square. Find the dimensions of the room. Ans. 20 ft. by 14 ft. 71. A room is $ as long as it is wide, and its perimeter is 70 ft. Find the area of the room. Ans. 300 ft. 1 72. A man made a journey of 560 miles, part at the rate of 40 miles per hour and part at 50 miles per hour. If it took him 13 hours, how far did he go at each rate? Ans. 360 mi. ; 200 mi. 73. A tree 189 ft. tall was broken into two pieces by falling. If | the length of the longer piece equals } the length of the shorter piece, how long is each piece? Ans. 100^ ft.; 88i? ft. 74. Sirloin steak costs 1$ times as much as round steak. Find the cost of each per pound if 3 Ib. of sirloin and 5 Ib. of round cost $2.28? Ans. 36 cents; 24 cents. 75. If 2 Ib. of butter cost as much as 5 Ib. of lard, and 4} Ib. of lard and 6 Ib. of butter cost $5.46, find the cost of each per pound. Ans. 70 cents; 28 cents. EQUATIONS AND FORMULAS 311 250. Thermometers. Two kinds of thermometers are in common use. The Fahrenheit, which is used for common purposes, has the freezing point marked 32, and the boiling point marked 212. The centigrade, which is used for all scientific purposes, has the freezing point marked and the boiling point marked 100. It is seen then that on the Fahren- heit scale there are 212 -32 = 180 between the freezing point and the boiling point; while on the centi- grade scale there are 100 in the same space. Hence 180 of the Fahren- heit scale = 100 of the centigrade scale. These relations are shown in Fig. 203. EXERCISES 87 1. If F stands for the number of degrees on the Fahrenheit scale and C for the number of degrees on the centigrade scale, find that (1) C = (F-32), (2) F = e s C+32. 2. A temperature of 176 Fahrenheit is what temperature centigrade? [Use (1) of exercise 1.] Ans. 80. 3. A reading of 24 centigrade is how many degrees Fahrenheit? Ans. 751. 4. Given that the following metals melt at the given temperatures in Fahrenheit scale, find the melting point of each in the centigrade scale: Wrought iron, 2822; steel, 2462; cast iron, 2210; silver, 1832; lead, 620; tin, 475. Ans. Wrought iron, 1550; steel, 1350; cast iron, 1210; silver, 1000; lead, 326|; tin, 246J. 6. 60 below Fahrenheit is what on the centigrade scale? Ans. 5H below 0. 6. At what temperature are the readings on the two thermometers the same? Ans. 40 below 0. Solution. Let x = the reading on each thermometer scale. Then by the formulas of exercise 1, 212F= 100C -100*0 ZI2 - F 210 - BOILING 3 200 - 100 ! 90 180 - ; i 80 170 5 : 160 - I 70 150 nO- eo ISO- i ^C ^ 60 110 : ! 100 : 40 T - 30 70 ^ 20 60 60 10 40 32 F 0C 0*0 82 F 30 FREEZING - 20 \ '10 > 10 20 -10 -20 30 -30 -40 , 40 FIG. 203. 312 PRACTICAL MATHEMATICS 5(z-32)-!x+32. 25x- 800* Six + 1440. 25x -Six -1440+800. -56x=2240. x=-40. .'. the reading is the same when temperature is 40. 7. With the oxyacetylene process of welding, the temperature of the flame is sometimes over 6000 Fahrenheit. What is this on the centi- grade scale? Ana. Over 3315. 261. Horse-power. The term horse-power was first used by James Watt, the inventor of the steam engine. He ascer- tained that a London draught horse was capable of doing work, for a short time, equivalent to lifting 33,000 Ib. 1 ft. high in 1 minute. This value was used by Watt in expressing the power of his engines, and has since been universally adopted in mechanics. The expression foot-pounds is used to denote the unit of work; It is equivalent to a force of 1 Ib. acting through a distance of 1 ft., or a force of ^ Ib. acting through a distance of 2 ft., or ^V Ib. through a distance of 10 ft., etc. Horse-power is the measure of the rate at which work is performed. One horse-power is equivalent to 33,000 Ib. lifted 1 ft. in 1 minute, or 1 Ib. lifted 550 ft. in 1 second. We say then that one horse-power equals 33,000 foot-pounds per minute, or 550 foot-pounds per second. Therefore, the horse-power of any machine can be found by dividing the number of foot-pounds of work done in 1 minute by 33,000, or Number of foot-pounds of work per minute Horse-power = - 3 pOO~ In electric-power machines where the watt is used, since 746 watts equal 1 horse-power, we have volts X amperes Horse-power = =-. 746 EXERCISES 88 1. What horse-power is necessary to raise a block of stone, weighing 3 tons, to the top of a wall 40 ft. high in 2 minutes? Solution. The number of foot-pounds per minute is 6000X40 and hence the A 6000X40 EQUATIONS AND FORMULAS 313 2. What horse-power is required to life an elevator, weighing 4 tons, to the top of a building 240 ft. high in 1J minutes? Ans. 39 h. p. nearly. 3. What horse-power is required to pump 30,000 barrels of water per hour to a height of 45 ft.? (Use 1 bbl. =4.211 cu. ft.) Ans. 179.4+ h. p. 4. The following formula gives the horse-power of a steam engine: PLAN 33,000' where H = indicated horse-power, P = mean effective pressure of the steam in pounds per square inch, L = length of stroke in feet, A =area of piston in square inches, and A r =number of strokes of piston (twice the number of revolutions) per minute. Solve the formula for each of the letters P, L, A, and N in terms of the others. 33,000// . 33,000/7 33,000/7 . 33,000/7 A Yi v t f - / i A /.... A/ .. . 'LAN ' ' PAN ' ' PLN PL A Remark. It is to be noticed that the formula 77 = 00 nnn has, in the oOjUlHJ numerator, an expression for the number of foot-pounds per minute. To see this, note that PA is the pressure on the piston in pounds. This times L, or PL A, is the foot-pounds for one stroke of the piston. Finally, multiplying this by the number of strokes per minute, N, gives PLAN, the number of foot-pounds per minute. 6. What horse-power will a steam engine with a cylinder 4 in. in diam- eter and a stroke of 6 in. develop at 300 R. P. M. of the crank, if the mean effective pressure is 95 lb.? Ans. 10.85+ h. p. Suggestion. Express in form of cancellation, 4X4X0.7854X95X1X600 33,000X2 cancel what you can and compute by slide rule if you wish. 6. Find the horse-power of an engine with a cylinder 10 in. in diameter, a stroke of 30 in., the crank making 96 R. P. M., and the mean effective pressure 120 lb. Ans. 137+ h. p. 7. Find the diameter of a cylinder to develop 95 h. p. with a stroke of 34 in., the crank making 110 R. P. M., the boiler pressure being 80 lb., and the mean effective pressure 65% of the boiler pressure. Ans. 11.1- in. Suggestion. Find the area of the piston by the formula of exercise 4, and then the diameter by [27]. 8. An engine is required to develop 50 h. p. with an average effective pressure of 46 lb. on a piston 13 in. in diameter, and a crank shaft speed of 100 R. P. M. Find the length of the stroke. Arts. 1.351+ ft. or 1 ft. 4, 3 , ; in. 314 PRACTICAL MATHEMATICS 9. Find the mean effective pressure on the piston of a steam engine with a cylinder 12 in. in diameter and a piston stroke of 18 in., if the num- ber of revolutions is 110 per minute and developed horse-power 40. Ana. 35.37 Ib. per square inch. 10. In gas engines it is not easy to determine the mean effective pres- sure, so the formula // = -^n cannot well be used. As a result of OO,UUU D*N experiments with engines used in automobiles, the formula H -s^r is .0 often used. Here D is the diameter of the cylinders in inches, and A' is the number of cylinders. Find //, if D =4} in. and N =6. Ans. 54.15. 11. In "Locomotive Data" of the Baldwin Locomotive Works, the following formula is given for the tractive power of a locomotive: C'SP ~D~' where C = diameter of cylinders in inches, S = stroke of piston in inches , P=mean effective pressure in pounds per square inch =85% of boiler pressure, D= diameter of driving wheels in inches, and T = tractive power in pounds. Solve for each of the letters C, S, P, and D in terms of the others. fDT ~_DT j>_DT n '^~' ' 12. Use the formula of exercise 11 and find T when C = 16 in., 5 = 22 in., and D = 64 in., if the boiler pressure is 160 Ib. per square inch. Am. 11,968 Ib. 13. To find the pressure on a lathe tool in turning steel multiply the area in square inches of a section of the chip cut by 230,000. In cutting cast iron use 168,000. What horse-power does it take to turn a 6-in. steel axle making 30 R. P. M., if the cut is 3*5 in. deep with a feed of \ in.? Ana. 1.3 nearly. 14. The power that is transmitted by a belt depends upon the pull of the belt and the rate at which it travels. The power may be given as a number of foot-pounds per minute or may be given as horse-power. Different makers of belts and writers on the subject give different values as the working strength for belts. The allowed pull for single belts is from 30 to 60 Ib. per inch of width of belt. For double belts it is from 60 to 100 Ib. per inch of width. Show that the following formula gives the horse-power transmitted by a belt: FWS 33,000' where // = horse-power, F = pull in pounds per inch of width, W = width in inches, 5 = speed of belt in feet per minute. EQUATIONS AND FORMULAS 315 Solve for each letter involved in the formula. 15. Find the horse-power that can be transmitted by a belt 14 in. wide, if the pull allowed per inch of width is 90 Ib. and the speed is 5000 ft. per minute. Ans. 190fr. 16. Find the horse-power transmitted by a single belt 6 in. wide, running over a pulley 16 in. in diameter, and making 350 R. P. M., if the pull of the belt is 45 Ib. per inch of width. Allow 2% for slipping. Ans. 11.85. 17. If a single belt 6 in. wide transmits 7 horse-power, find its speed per minute, allowing 35 Ib. pull per inch of width. Ans. 1100 ft. per minute. 18. Find the horse-power transmitted by a 10-in. double belt, run- ning over a 36-in. pulley, making 420 R. P. M. Use a pull of 75 Ib. per inch of width. Ans. 89.96. 19. Assuming a tension of 50 Ib. per inch of width for a single belt, and using D for diameter of pulley in inches, W for width of belt in inches, R for the number of revolutions per minute, and H for the horse-power transmitted, then rr^DRW ~ 2520 ' Use ir= ? r and derive this formula. Solve for W, D, and R in terms of the other letters. Remark. In the above formula it is assumed that the pulleys are prac- tically of the same size so that the arc of contact is 180. If the pulleys differ in diameter the arc of contact of the belt on the smaller pulley should be found and the following table used in finding the horse-power that can be transmitted. If angle of con- tact is . 90 100 110 120 130 140 150 160 170 Multiply by 0.65 0.70 0.75 0.79 0.83 0.87 0.91 0.94 0.97 20. Using the notation of the preceding exercise and 80 Ib. for the tension per inch of width for double belts, rj_DRW ~ 1575 ' Derive this formula if ir=V- Solve for D, R, and W in terms of the other letters. 21. Find the width of a single belt to transmit 3 horse-power when running over a pulley 15 in. in diameter, making 220 R. P. M. Allow a pull of 50 Ib. per inch of width. Ans. 2.3 in. 22. Find the number of R. P. M. a pulley 4 ft. in diameter must make that transmits 120 horse-power through a double belt 10 in. wide, having a pull of 80 Ib. per inch of width. Ans. 393 J. 316 PRACTICAL MATHEMATICS 23. How much work is done in lifting 150 Ib. from a mine 1100 ft. loop? How many horse-power would it take to lift this weight from the mine in 1 ! minutes? Ana. 3J horse-power. 252. Relation of resistance, electro-motive force, and current. Resistance, R, is measured in ohms ; voltage or elec- tro-motive force, E. M. F. or E, in volts; and current, 7, in amperes. The law connecting these is stated as follows: volts . E Amperes = -; or / = -= ohms R This law is Ohm's Law, and is the fundamental law for electrical work. It is an algebraic equation, and any one of the numbers can be found if the other two are given. Exercises 27, 28, 32, and 33, pages 305 and 306, are forms of this equation. EXERCISES 89 E E 1. Solve the equation J = p for E and R. Ans. E = IR; R**j- 2. In a certain circuit the voltage is measured and found to be 1.5 volts. If the total resistance is 12 ohms, what is the current in amperes? Ans. 0.125 ampere. 3. Find the number of amperes of current sent through a circuit of 20 ohms resistance by one Daniell's cell which has an E. M. F. of 1.03 volts. Ans. 0.0515. 4. Find the strength of current from 50 Daniell's cells united in series, assuming the E. M. F. of each cell to be 1.03 volts, the resistance within each cell 0.3 ohm, and the external resistance 25 ohms. Ans. 1.2875 amperes. Remark. The student not familiar with the meaning of "cells united in series" can note the fact that 50 cells in series have 50 times the E. M. F. of a single cell, and the resistance within the cells is 50 times that of one cell. The formula of exercise 27, page 305, can be used. 6. An electric bell has a resistance of 450 ohms and will not ring with a current of less than 0.06 ampere. Neglecting battery and line resistance, what is the smallest E. M. F. that will ring the bell? Ans. 27 volts. 6. If an electric car heater is supplied with 500 volts from the trolley, how great must its resistance be so that the current may not exceed 2.5 amperes? Ans. 200 ohms. 7. A certain wire has a resistance of 1 ohm for every 30 ft. of its length. What must be the E. M. F. in order that a current of 0.4 ampere may flow through 1 mile of the wire? Ans. 70.4 volts. EQUATIONS AND FORMULAS 317 253. Resistance of conductors. Consider the formula, K == K, a where R is resistance in ohms, I the length of the conductor in some unit, a the area of the cross section of the conductor in some unit, and k a constant value depending upon the material in the conductor and upon the units of length and cross section used. This formula is the fundamental formula in wiring calculations. In engineering practice, the length I is taken in feet; the area a is taken in circular mils, which equals the square of the diameter in thousandths of an inch when the conductor is circular (see Art. 132) ; and k is the resistance in ohms of 1 ft. length of the conductor having a cross section of one circular mil. That is, k is the mil-foot resistance of the conductor. The formula can then be written = -. a 2 If then it is required to find the resistance of any length of any size wire of any material, it is only necessary to know the mil-foot resistance for the -material, and substitute the values in the formula. The following table gives the mil-foot resistances for the materials named at zero degrees centigrade : Aluminum .......................... 17 . 5 ohms. Copper (commercial) ................. 9.6 ohms. German silver ....................... 125 . 7 ohms. Iron (pure) ......................... 58. 3 ohms. Iron (telegraph wire) ................. 90 . ohms. Platinum ........................... 54 . 3 ohms. Silver .............................. 9.1 ohms. Zinc ............................... 33.8 ohms. Example 1. What is the resistance of 2 miles of No. 12, B. and S. gage commercial copper wire? Solution. No. 12, B. and S. gage has an area of 6530 C. M. 2 miles =2X5280 ft. 2X5280X9.6 ,.R = -^-TC.?; ==15. 5 ohms. Ans. DOOU 318 PRACTICAL MATHEMATICS If it is required to determine the size of wire of a given length to have a given resistance, the formula is solved for d*. This M lk gives d 2 = ^- Example 2. Find the B. and S. gage of pure iron wire to have a resistance of 3 ohms per mile. Solution. Substituting in the formula, d* - 528 X58.3 _ 102)60g c M o d = V102.608 = 320.3 mils = 0.3203 in. The nearest to this size is No. 0, B. and S. which is 0.3249 in. in diameter. If it is required to find the length of wire of a given size to have a given resistance, the formula is solved for I. This gives Ra = *' Example 3. Find the length of a No. 20, B. and S. gage silver wire to have a resistance of 5 ohms. Solution. No. 20, B. and S. has an area of 1022 C. M. 5X1022 9.1 = 561. 5 ft. nearly. Ans. EXERCISES 90 1. Find the resistance of 340 ft. of No. 25, B. and S. gage German silver wire. Ans. 133.4 ohms. 2. Find the resistance of 20 miles of trolley wire, made of commercial copper, and No. 00, B. and S. Ans. 7.6 + ohms. 3. Find the resistance of 10 miles of the "third rail" conductor on an elevated railroad. The "third rail" is iron and has a cross section of 5.88 in. 1 Ans. 0.63 ohm nearly. 4. Find the resistance of 500 turns of No. 30, B. and S. gage silver wire, about a core j in. in diameter. Ans. 8.69+ ohms. 5. Find the B. and S. gage of commercial copper wire to give a resist- ance of 40 ohms per mile. Ans. No. 19 nearly. 6. Find the length of No. 16, B. and S. gage commercial copper wire to have a resistance of 20 ohms. Ans. 5381 ft. 7. The resistance of a certain commercial copper wire, 1 ft. long and 1 C. M. in cross section, is 10.79 ohms. A wire, 525 ft. long, has a cross section of 4117 C. M. ; what is its resistance? Ans. 1.376 ohms. CHAPTER XXIX EQUATIONS WITH MORE THAN ONE UNKNOWN 254. In previous chapters the equations solved have in- volved only one unknown letter. In the present chapter will be considered equations involving more than one unknown letter. These letters will occur to the first power only, or if squared, will be such as are easily handled. 255. Indeterminate equations. If we have the equation x y = 2, and try to solve it for the unknown numbers x and y, it is evident that we can get numerous pairs of values for these numbers. Thus, z = 3 and y = l is a pair of values that satisfy the equation; so are z = 4 and y = 2; x = 5 and ?/=3; x = 10 and y=8. Such an equation is called an indeterminate equation. Find pairs of values that satisfy the following: 1. x+y = lQ. 2. x-y = lQ. 3. 2x+y = 20. 4. 3z+2t/ = 40. 5. 7x-8y = 27. 6. 3x-8y = lQ. 1.6x-2y=-2. B.4x-8y=-lO. 256. Simultaneous equations. Take the two indeterminate equations (1) x y = 2, and (2) x+y = 12. Pairs of values that satisfy (1) are x = 3, y = l; #=4, y = 2; x = 5, y = 3; x = Q, y = 4; x = 7, y = 5; etc. Pairs of values that satisfy (2) are x = 2, 7/ = 10; x = 3, y = 9; x = Q, y = 6', x = 7, y = 5;x = 8, y = 4;etc. It is noticed that the pair of values, x = 7 and y = 5, is found in both sets of values; that is, these values satisfy both equa- tions. Such a set of two equations satisfied by a pair of values for the unknown is called a system of simultaneous equations. 319 320 PRACTICAL MATHEMATICS If, as in the system just considered, there is only one pair of values that satisfy both equations, the equations are called independent. If the equations are as in the set x+2y = 7 and 3x+6y = 15, where there is no pair of values alike for both, the equations are called inconsistent. If the equations are as in the set 2x+y = 10 and 6z = 30 3y, where every pair that satisfies one will satisfy the other, the equations are called equivalent. 257. Solution of independent equations. It is evidently not convenient to find a pair of values that will satisfy two inde- pendent equations, by writing down pairs of values for each equation, and then selecting the pair which is the same in each. It remains then to devise a way that is shorter. The different methods of solution that have been devised are alike in that an equation is obtained that has only one unknown. We say then that one unknown has been eliminated. There are three ways of eliminating an unknown. They are: (1) Elimination by adding or subtracting. (2) Elimination by substitution. (3) Elimination by comparison. 258. Elimination by adding or subtracting.- Example 1. Solve x y = 2 and x+y = 12 for x and y. Solution. (1) x y = 2. (2) z-h/ = 12. Add equations (1) and (2), first member to first member and second member to second member, and we have (3) 2z=14. .'.(4) z = 7. Substituting this value of x in (1) gives (5) 7-y = 2. .'. (6) y = 5. Hence the pair of values that will satisfy both equations is x = 7 and y = 5. Example 2. Solve 3z+2y = 21 and 7x 5v/ = 20 for JT and y. Solution. (1) 3x+2y = 2l. (2) 7z-5t/ = 20. Multiplying (1) by 7 gives (3) 21x+14j/= 147. Multiplying (2) by 3 gives (4) 21z-15y = 60. EQUATIONS WITH MORE THAN ONE UNKNOWN 321 Subtracting (4) from (3) = (5) 29y = 87. '. (6) y-3. Substituting in (1) gives (7) 3s +6 = 21. .'. (8) 3^ = 21-6. (9) 3z = 15. (10) a; = 5. Hence x = 5 and y = 3 are the required values. This way of eliminating one of the unknown numbers is called elimination by adding or subtracting. The following rule may be given for the process: RULE. (1) Multiply each equation, if necessary, by such a number as will make the absolute values of the coefficients of one of the unknown numbers the same in both of the resulting equations. (2) Add or subtract the corresponding members of the resulting equations so as to eliminate the unknown number having coefficients equal in absolute value. It should be noticed that we add when the coefficients are of opposite signs, and subtract when they are of like signs. 259. Elimination by substitution. The elimination can often be performed more easily by using a method called substitution. Consider the following system of equations: y = 42 7x and 3x y = 8. Solution. (1) y = 42 7x. (2) 3x-y = 8. Substituting the value of y from (1) into (2) gives (3) 3z-(42-7z)=8. (4) Zx-42+7x = 8. (5) 3z+7z = 42+8. (6) 10s = 50. (7) a; = 5. Substituting in (1) gives (8) y = 42-35. .'. (9)y = 7. Hence x = 5 and y = 7 are the values. This method of eliminating is called elimination by sub- stitution, and can generally be used to good advantage when one equation is much simpler in form than the other. The following rule may be given for the process: RULE. Solve one of the equations for the value of one of the unknown numbers, and substitute this value in place of that 21 322 PRACTICAL MATHEMATICS number in the other equation. This will give an equation with but one unknown number. 260. Elimination by comparison. Example. Solve z+4y = 21 and Zx y=ll. Solution. (1) x+4y = 21. (2) 3z-2/ = ll. Solving (1) for x gives (3) x = 2l4y. Solving (2) for x gives (4) a? = =-* o By axiom (5), equations (3) and (4) give (5) 21-4y-ii Clearing of fractions, (6) 63-12i/ = ll+y. (7)-12y-y-ll-63. (8) -13y=-52. (9) y-4. Substituting in (3) gives (10) x = 21 - 16. (11) z = 5. Hence x = 5 and y 4 are the values. This method of elimination is called elimination by com- parison. The following rule may be given for the process: RULE. Solve each of the equations for the value of one of the unknowns, and equate these values. This forms an equation having but one unknown. 261. Suggestions. Any method of elimination may be used. Usually one of the methods is shorter than the others. Elimination by adding or subtracting can usually be used to the best advantage. Free the equations of signs of grouping before eliminating. Usually clear of fractions before eliminating. Much practice and dealing with examples will help one to determine what method to use. EXERCISES 91 In exercises 1 to 10 eliminate in each by all three methods. 1. z+y**8 and x y=2. Ana. z = 5, y=3. 2. 2x+y = 10andz+2j/ = ll. Am. z=3, y = 4. 3. 4x-3y = 8 and 2z+y = 14. Ana. x-5, y**4. 4. 3x+2y=26and 5x-2y = 38. Ana. x=8,y = l. 5. 7x+y=42 and 3x-y=8. Ana. z5, y-7. EQUATIONS WITH MORE THAN ONE UNKNOWN 323 6. 8x+6y = 10 and 5x+2y = 1. Ans. x=l,y = 3. 7. 7x 9y = 13 and 5x+2?/ = 10. Ans. x = l*|, 2/=A- 8. 3x+5y=8 and 2x 3y = 12. Ans. x=4 T 8 5 , y= f. Solution. (1) 3x + 5y = 8. (2) 2x-3y = l2. Multiplying (1) by 2 gives (3) 6x + 10?/ = 16. Multiplying (2) by 3 gives (4) 6x- 9^ = 36. Subtracting (4) from (3) gives (5) 19?/= -20. Substituting in (1), x = 4j\,. 9. 5x+6y = 17 and 6x+5?/ = 16. Ans. x = l, y=2. 10. x-lly = l and llly-9x = 99. Ans. a; = 100, ?/ = 9. In exercises 11 to 14 draw the graphs of the equations, determine values from these, and solve equations by some method of elimination. (See Chapter XXXIII.) j.1. x~r~?/ = 17 niici x y == i. Ans. x 1^, t/ === o. 12. 3x+4y = 24 and 5x Qy = 2. Ans. x = 4,y = 3. 13. 5x+9?/ = 28and 7x+3y = 20. Ans. x = 2, y = 2. 14. 5x-2y = l, and 4x+5?/=47. Ans. x = 3, y = 7. Solve the following by some method of elimination : :c-j-l y-\-2 . x-\-y y~\~2 . _ Ans. x= -2, y=-2. ?_^_A ^ ?_^_1 3 6~2 a! 5~10~2' 18. ^+| = 5 and a-6 = 4. Ans. a = 10, 6 = 6. O <0 n + 1 3m 5 , n + 1 n m 19. ^Q-= 2 Tb~ = ~~8 Ans. m=3,n = lQ. f\f\ X U \ LJ ~ , *J** U J. A **M f~ f, f* jfl 1 1 Q n fl A *n a v *^ II s a\l. \J tillU j. _ V. yi Jto. X v, y a, 21. - ~ =4 and =+-~^=3. Ans. x = 14, y = 15. ft -xwl .XT/! o+c d 6 22.-+f=-rand --, = -, Ans.x= 7-7^-. 2/= rnr- a b ab c d cd ad+bc ad+bc Solution. (1) -+r=~T' 06 ab Clearing of fractions, (3) bx+ay = l. (4) dxcy = l. Multiplying (3) by c, (.5) lcx+acy=c. Multiplying (4) by a, (6) adx acy = a. Adding (5) and (6), (7) (ad+bc)x = a+c. - a+c 324 PRACTICAL MATHEMATICS Here it is beat to solve for y by eliminating x rather than to substitute the value of y in (1) or (2). 23. The sum of two numbers is 15 and their difference is 1. What are the numbers? Ana. 7 and 8. 24. Three times one number plus four times another number is ten, and four times the first plus the second is nine. What are the two numbers? An*. 2 and 1. 25. What is the fraction which equals | when 1 is added to the numerator, but equals i when 1 is added to the denominator? Ana. iV 26. Given *S = ry and T= jr, eliminate N and find the value of T j-vrir in terms of the remaining letters. Ana. T ' = -_ Suggestion. Solve each equation for N and eliminate by comparison. 27. The formulas of exercise 26 are used in lathe work. In these formulas, T is the time in minutes; S is the cutting speed in feet per minute; D is the diameter of the work in inches; N is the number of revolutions per minute; L is the length of the part to be turned in inches; F is the feed and is expressed as the numbers of turns to give a sidewise movement of 1 in. Thus, a feed of 16 means that each cut is ^ in. wide. Find the time to turn a piece 3 in. in diameter and 2 ft. long, if the feed is 20 and the cutting speed 15 ft. per minute. Ans. 25.13+ min. 28. In locating and boring holes in a drill jig, it is necessary to find the diameters of three circular disks tangent two and two, whose centers are at distances of 0.765 in., 0.710 in., and 0.850 in. Find the diameters of the three circles. FJG 204 Solution. The disks are placed as shown in Fig. 204. Let x, y, and z = the radii of the circles centered at A, B, and C respec- tively. Then (1) z+2/ = 0.850. (2) x+z = 0.710. (3) y+z = 0.765. Subtracting (2) from (1) = (4) yz =0.140. Adding (3) and (4) = (5) 2y = 0.905. (6) 77=0.4525. Substituting value of y in (1) gives x +0.4525 =0.850. /. i = 0.3975. Substituting value of y in (3) gives 0.4525+2 = 0.765. /. z=0.3125. The diameters of the circles are: at A, 0.795 in.; at B, 0.905 in.; at C, 0.625 in. EQUATIONS WITH MORE THAN ONE UNKNOWN 325 29. As in the last exercise, three holes are to be bored, the distance between whose centers shall be 0.650 in., 0.790 in., and 0.865 in. respec- tively. Find the diameters of the required disks. Ans. 0.725 in.; 0.575 in.; 1.005 in. 30. Three points A, B, and C are located as shown in Fig. 205. Three disks are centered at these points and tangent two and two. Find the diameters of the disks. Ans. At A, 0.8960 in.; at B, 0.9716 in.; at C, 0.9700 in. Suggestion. Let x, y, and 2= radii of circles centered at A, B, and C respectively. Then x+y= v / 0.680 2 +0.640 2 . x+2 = \/0.880 2 +0.310 2 . FIG. 205. 31. From geometry we know that the sum of the three angles of a triangle is 180. Find the three angles of a triangle if the sum of twice the first and the second is 90 more than the third, and the sum of the first and twice the third is 70 more than twice the second. Ans. 50; 60; 70. 32. A man has $98 in dollar bills, half-dollars, and quarters. Half of the dollars and f the half- dollars are worth $31, and f the half-dollars and -| the quarters are worth $10. How many pieces has he of each? Ans. 48 dollars; 70 half-dollars; 60 quarters. 33. In a factory where 700 men and women are employed, the average daily pay for the men is $4.25, and for the women $2.25. If $2415 is paid daily for labor, find the number of men and the number of women employed. Ans. 420 men ; 280 women. 34. A lever is balanced on a fulcrum with a weight of 40 Ib. at one end and 50 Ib. at the other. If 5 Ib. is added to the 40-lb. weight, the 50-lb. weight will have to be placed 1 ft. farther from the fulcrum to balance the lever. Find the lengths of the two arms of the lever at first. Ans. 10 ft. ; and 8 ft. Suggestion. Let z = the length in feet of the long arm, and ?/=the length in feet of the short arm. Then 40z = 507/, 36. A lever is balanced on a fulcrum with a weight Wi on one arm and Wi on the other. If p pounds are added to w\, w^ has to be moved over / feet to balance the lever. Find the lengths of the two arms of the lever. Ans. - and 326 PRACTICAL MATHEMATICS 36. A beam that is supported at its ends has a weight of 50 Ib. placed upon it so that it causes an increase in pressure on the support at one end of 20 Ib. It is also found that the same pressure is produced at this end by a weight of 60 Ib. placed 3 ft. farther from this end. How long is the beam? Ant. 45 ft. Suggestion. Consider the support where the increase is 20 Ib. as the fulcrum. Then there is a pressure of 30 Ib. at the other end. Let y = the length of the beam and x =the distance the 50-lb. weight is from the fulcrum. Then 30y 50x. Similarly, when the 60-lb. weight is applied, 40?/=60(z+3). 37. A glass full of water weighs 18 ounces. When the same glass is full of sulphuric acid of specific gravity 1.75 it weighs 27 ounces. Find the weight of the glass when empty. Ana. 6 ounces. 38. Given two grades of zinc ore, the first containing 45% of zinc, and the second 25% of zinc. Find how many pounds of each must be taken to make a mixture of 2000 Ib. containing 40% of zinc. Ans. 1500 Ib. of 45% ore and 500 Ib. of 25% ore. Suggestion. Let x= number of pounds of 45% ore, and y=number of pounds of 25% ore. Then z+y = 2000, and 0. 45* +0.25y = 0.40X2000. 39. The sum of the three sides of a triangle is 65 in. If the second side is 5 in. longer than the first and 7 in. shorter than the third, find the length of each side. Ans. 16 in.; 21 in. ; 28 in. EXPONENTS, POWERS, AND ROOTS 262. General statement. In previous chapters, we have used positive integral exponents. In Art. 185, a positive integral exponent was denned as showing how many times the base is to be used as a factor. The negative exponent, the fractional exponent, and the zero exponent are other kinds of exponents that occur in mathematics. They will be dealt with to some extent in the present chapter. Logarithms are exponents that will be considered in Chapter XXXIV. The few facts and theorems concerning exponents, given in this chapter, are not intended to be a complete treatment of the subject, but they are sufficient for what follows, and will give the proper viewpoint for logarithms. It should be carefully noted that the definition of a positive integral exponent cannot apply to an exponent that is negative, zero, or a fraction. For instance, in 8 2 , the 2 shows that 8 is taken twice as a factor; but in 8*, the ^ can in no sense show how many times 8 is taken as a factor. To say that 8 is taken | times as a factor is to make a meaningless statement. We shall therefore find it necessary to enlarge the definition of an exponent so as to make it include these new kinds of exponents. 263. Laws of exponents. The law of exponents in mul- tiplication (see Art. 216) has been stated and proved for positive integral exponents. It may be restated here in symbols as follows : a m -a n = a m+n . Likewise the law of exponents in division (see Art. 223) has been stated for positive integral exponents. A restatement in symbols is as follows: a m -^-a n = a m " n , where m is greater than n. When ra = n it is a n + a n = l. 327 328 PRACTICAL M AT HEMATICS In finding the power of a power the exponents are multiplied. That is, (a m ) n = o mn . This is found from the definition of a positive integral exponent. Thus, (a m ) n =a m a m a m - to n factors =a m + m """" to n terms a*". A numerical illustration is (a*) s =a*-a'-a'=o 18 . The power of a product is the same as the product of the powers of the factors. That is, (abc- -) n =a n b n c n - A numerical illustration is (2-34) 3 = 2-3 3 -4 3 . The power of a fraction equals the power of the numerator divided by the power of the denominator. fa\ m a* That is, ^ -~ 2\ * 2* 8 A numerical illustration is , \o/ o // we take the root of a power, we have the inverse of the opera- tion by which we get a mn , and have ^/a m = a m ^ n . A numerical illustration is -^/3*=3 44 - 2 = 3 2 = 9. The definition of a positive integral exponent gives us the right to state the above only when m-r-n is an integer. A summary of the six laws of exponents mentioned is: (1) a m a n = a m+n . (2) a m +a n = a m - n . (3) (a m ) n = a ron . (4) (a-b-c - - ) m = a m b m c m - - -. (G) Example 1. Find the fourth power of 3a z x 3 y 4 . Example 2. Find the cube of 2t/ 2 3 3 (a 5 ) 3 (b 4 ) 3 x* = 27a 9 b l *x* 2% 2 ) 3 8t/ Example 3. Find the cube root of 3 6 &V 2 . ;+ 3=3 2 6 3 a; 4 . EXPONENTS, POWERS, AND ROOTS 329 EXERCISES 92 Raise to the indicated powers, or find the root indicated. 1. (oV) 4 . Ans. a 8 ?/ 12 . 2. (2?/ 3 x 4 ) 6 . Ans. 64?/ 18 x 24 . 3. (3a 2 6) 3 . Ans. 27a 6 6 3 . 4. (-4a 2 x 6 ) 3 . Ans. -64a 6 x 18 . 6. (-3a 3 b 2 x) 4 . Ans. 81a 12 6 8 x 4 . 6. ( x 4 ?/ 6 ) 2 . Ans. x 8 ^ 12 . 7. (36 C ) 3 . Ans. S 3 ^ 30 . 8. (4a c x 2 ) 6 . Ans. 4 6 a 6c x 26 . 27x3 Ans. Ans. 1256 9 Ans ' 256c'x8' Ans. Ans. 7 x 4b 3 264. Zero exponent. If we assume that a m -;- a n = a m ~ n holds when m = n, we have a TO -r- a n = a m ~ n = a, for w n = Also a" 1 -s- a" = 1 by Art. 223, if m = n. :. a=i. Making the above assumption is the same as stating by definition: Any number other than zero affected by a zero ex- ponent equals one. Examples. (1) 0.8 = 1. (2) 100 = 1. (3) 0.01= 1. 265. Negative exponent. Assuming that the law a m +a n = a m- n holds when n is greater than m, we have 330 PRACTICAL MATHEMATICS a" Similarly a~" = o n By definition then, a number affected by a negative exponent equals 1 divided by the same number affected by a positive ex- ponent, equal in absolute value to the negative exponent. Examples. (1) 2 ~ 3 = o3 = o' (2) 4-> = = 4 3 64- 266. Fractional exponent. If we apply the law \s'a m = a m + n } when m and n have any values, we have a n . Also By definition then, a fractional exponent indicates a root. The denominator is the index of the root, and the numerator is the exponent of a power. m A form like a n means either -\/a m or (v^a) m - That is, the number a may be raised to the rath power and then the nth root taken, or the nth root may be taken first and then the result raised to the mth power. Thus, 8' = ^/8~' = ^64 = 4, or 8 J Examples. (1) 4 =\/4 = 2. (2) 64* = -^64 =4. (3) 32* EXERCISES 93 Find the values of the following: 1. 10. Ans. 1. 6. 16-*. Ans. \. 2. 4-' Ans. A. 7. 27>. Ans. 81. 3. 10'*. ATM. roooJinnnj. 8. 512~ J . Ans. &. 4. 8-'. Ans. ,},. 9. dh)-'. Ans. 25. 5. 16*. Am. 2. 10. 100*. Ans. 1000. EXPONENTS, POWERS, AND ROOTS 331 11. 1000-i. Ans. xta. 14. (4). Ans. 1. 12. 10-3 2 . Ans. 9. 16. 64~5 Ans. T1 fei. 13. (3)- 3 . Ans. 1. 16. 9*. Ans. 243. 17. Divide a~*x~* by cr 3 . Ans. ax~\ 18. Multiply a~ by a 8 . Ans. a*X 19. Multiply 3a* &* by 4a* 6*. Ans. 12a b*. 267. Exponents used in writing numbers. Often in engi- neering subjects we see such expressions as 4.25 X10 9 , or 726X10~ 8 . These forms are the result of an attempt to write certain expressions in a shortened form. Keeping in mind the meaning of an exponent, 4.25 X 10 9 = 4.25 X 1,000,000,000 = 4,250,000,000. This last number means the same as the given expression, but in some ways is not so convenient a form to handle, neither is it so easily compared with others of its kind as is the first form. Neither does the eye catch its value as quickly as it does that of the shortened form. The form 726 X 10~ 8 = 726 X ^ = 726 X i --- a 4a 2 4a 2 a . b I (o) Extracting roots, +K-=+A/ b , Ib 2 -4ac -b b 2 4ac This last form is a formula that can be used in solving any quadratic equation. In using the formula, care must be taken as regards the signs. They must be considered a part of the values of a, b, and c. Thus, in 3z 2 -2z-4 = 0, o = 3, b= -2, and c= -4. Example 2. Substitute in the formula of example 1, and find the value of x in 6z 2 +17o;+7 = 0. Solution, Here a = 6, b = 17, and c = 7. Substituting these values in the formula, -_ _ _j , ^6~ ~T2~ Example 3. By the formula, find the value of x in 6z 2 -|- 8z-30=0. Solution. Here a = 6, 6 = 8, and c= 30. _-8 + V8 2 -4-6(-30) = -828 338 PRACTICAL MATHEMATICS EXERCISES 98 Solve the following equations by any method. 1. z l -3x - -2. Ans. 1 or 2. 2. z+5x=-6. Ans. -2 or -3. 3. z'-ar-O. 4n*. 3 or -2. 4. t/*-2j/ = 168. An*. 14 or -12. 5. y l +2j/ = 120. Ana. 10 or -12. 6. j/ 2 -22j/=48. Ans. 24 or -2. 7. 5r 2 + 12r = 9. ATM. | or -3. 8. 2r 2 +5r=-2. 4ns. -J or -2. 9. 7a 2 +9ct = 10. ATM. ? or -2. 10. 7x*+2x = 32. An*. 2 or -2?. 11. F+i-;j =0. Ans. 2 or I V 12 '?+f = T Ans. 71 or -WL 96* 6 96 13. x*+4bx=-j-- Ans. - or ^' 14. Given S = Vt + $(7< 2 , solve for V, g, and <. 2S-gt\ 2S-2VI. -VVV*+2gg. 2t ' < 2 ' <; 15. Given !T = 27rr/i+2irr 2 , solve for r. Ans. - 16. Find three consecutive numbers, such that their sum is one-third of the product of the first two. Ans. 9, 10, 11 or 1, 0, 1. 17. Find two numbers, one of which is 5 times the other, and whose product is 4500. Ans. 150 and 30, or -150 and -30. 18. A walk containing 784 ft. 2 is to be built around a garden 50 by 49 ft. How wide must the walk be? Ans. 3 ft. 8 J in. nearly. 19. There are as many square feet in the surface of a certain sphere as there are cubic feet in its volume. Find its radius. Ans. 3 ft. Solution. Let r = number of feet in radius. Then 4nr* = number of square feet in surface, and 3irr 3 = number of cubic feet in volume. 4rr 8 -12irr J =0. Factoring, 4nr s (r-3) =0. ..4irr 2 =0 and r-3=0, or r = and r = 3. 20. Same as last exercise, but substitute cube for sphere and find the edge. Ans. 6 ft. 21. Find two numbers whose sum is 25 and whose product is 144. Ana. 9 and 16. 22. Divide 71 into two parte, the sum of the squares of which is 2561. Ans. 40 and 3L QUADRATIC EQUATIONS 339 23. If a train traveled 5 miles an hour faster, it would take 2 hours less to go 420 miles. Find the rate of the train. Ans. 30 miles an hour. 24. What is the radius of a circle whose area is doubled when the radius is increased by 2 ft.? Ans. 4.828 ft. 25. What is the diameter of a circle whose area is multiplied by 3 when the diameter is increased by 2 ft.? Ans. 2.732 ft. 26. The height of a right circular cylinder is 6 in., and the entire area is 100 in. 2 Find the radius of the base. Ans. 1.99+ in. Suggestion. Let r stand for radius of base. Then 2irr' 2 + l2wr is the entire area. 100 or r 2 +6r = -- = 15.915. 2ir 27. The differences between the hypotenuse and the two sides of a right triangle are 3 and 6 ft. respectively. Find the sides and the area of the triangle. Ans. 15 ft,, 12 ft., 9 ft., and 54 sq. ft. 28. The hypotenuse of a right triangle is 4 in. longer than one leg, and 8 in. longer than the other. Find the lengths of the three sides. Ans. 20 in.; 16 in.; 12 in. 29. The circumference of the hind wheel of a wagon is 5 ft. more than that of the front wheel. If the hind wheel makes 150 fewer revolutions than the front wheel in going one mile, find the circumference of each wheel. Ans. 16 ft. and 11 ft. 30. An aeroplane which is at an altitude of 1200 ft. and moving at the rate of 100 miles per hour in a northerly direction drops a bomb. Dis- regarding the resistance of the air, where will the bomb strike the ground? Ans. 1270 ft. north of starting point. Suggestion. Find the number of seconds it will take the bomb to reach the ground from the formula s = %gt~, where s=height of aeroplane in feet, g =32, and t =time in seconds. This gives t = 8.66 seconds. Then the bomb will strike as many feet north of the starting point as the aeroplane will travel in 8.66 seconds. CHAPTER XXXII VARIATION 276. General statement. We depend upon the ideas dealt with in variation for most of our physical laws and formulas. The meaning of many of the formulas already used will be made much clearer by considering their meaning from the viewpoint of variation. The relations considered in variation are those considered in ratio and proportion. In many ways, however, it will be found that the methods used in variation are more convenient than the methods of ratio and proportion. Familiarity with them gives the student another powerful mathematical instrument. For the principles of ratio and proportion used in the present chapter the student is referred to Chapter VIL 276. Constants and variables. A number whose value does not change is called a constant. In mathematical problems, certain constants occur that are always the same. The value of *, the ratio of the circumference to the diameter of a circle, is such a constant. There also occur other constants that do not change in the same problem, but which may have another value in a dif- ferent problem. The radius of a circle and the side of a square are such constants. A variable is a number that may take an unlimited number of values. For example, the number expressing a person's age, the height of a growing corn stalk, or the distance a moving train is from a station it has just left. 277. Direct variation. The idea of variation is a very com- mon one. Nearly everything is affected by its surroundings; that is, things vary according as something else varies. 340 VARIATION 341 The growth of a tree depends upon the amount of light it receives; the more light it receives, the faster it grows, if other conditions are favorable. In such a case we say its growth varies directly as the amount of light. The amount of pay a man gets varies directly as the time he works; that is, the longer he works the more pay he receives. Definition. If two numbers are so related that their ratio is constant, that is, if either increases the other increases, or if either decreases the other decreases, the two numbers are said to vary directly as each other. 278. Mathematical statement. Just as with many other ideas, there is a mathematical way of expressing the idea of variation. When the ideas are so expressed, they can be combined and handled according to the rules of mathematics. In this manner, new relations are seen and new results obtained. The sign <* means "varies as." If x and y are two variables that vary directly as each other, it may be written in the shorthand form x oc y. It may also be written in the form x = ky. This last is an equation stating that the ratio of x to y is a constant k. The equation is the form most often used. Example 1. If a train is moving away from a station at a uniform rate, express the relation between its distance d from the station and the time t since it left the station. Here evidently the distance d varies directly as the time t. The student should consider very carefully the meaning of the constant k in the different examples taken up. Here the k evidently represents the uniform rate per unit of time. If the time is in minutes, k stands for the rate per minute. This rate may be given in feet, rods, miles, or any other unit of length. The k then depends for its value upon the kind of units used, but this is not the only thing that may change the value of k. Example 2. If two numbers x and y vary directly as each other, and when x = 10, ?/ = 4, find x when i/ = 25. Solution. Since the relation between x and y is direct, x = ky. 342 PRACTICAL MATHEMATICS Substitute the values of x and y given, and .'. x = 2\y is the relation between the variables, and if ?/ = 25 we have z = 2^-25 = 62$. Ans. Such a relation as x = 2$y is often spoken of as a law. Example 3. The space passed through by a body falling freely from a distance above the ground varies as the square of the time. If s = the space in feet and < = time in seconds, write the law. Find the value of k, if the body will fall 402.5 feet in 5 seconds. Solution. Since the variation is directly as the square of the time, s = kt 2 . Substituting values of s and /, 402.5 = k -5 2 . .'. fc = 16.1. Hence the law or formula to find the distance in feet a body will fall in t seconds is s = 16. K 2 . This is usually written s = $gt 2 , where = 32.2. 279. Inverse variation. Consider a horizontal beam, resting at each end on a support, and having a weight at its midpoint. The size of the weight it will support depends upon its length ; but here the longer the beam, the less it will support. In such a case the variation is said to be indirect or inverse. The resistance to an electric current is less in a large wire than in a small one of the same material; the resistance varies inversely as the size of the wire in cross section. The intensity of the light from a lamp decreases as we go away from it. Here the variation is an inverse one, but the intensity of illumination is not one-half as much when the distance is doubled. It decreases inversely as the square of the distance; that is, the intensity is one-fourth as much at twice the distance. Definition. One number varies inversely as another when their product is constant. That is, if either increases, the other decreases, or if either decreases, the other increases. 280. Mathematical statement. The shorthand way of writing the fact that x varies inversely as y is x J The equation form is x = -> or xy = k. y VARIATION 343 Example 1. If x varies inversely as y, state the law and find the value of the constant if x = 10 when y = \ . Solution. xy = k is the mathematical statement of the variation. Substituting, lQ-^ = k. .' . k = 5. If this value of k is used, the law, or equation, becomes Example 2. If Z = TITO-J find y from the law of example 1. Substituting in the law, T -^y = 5. .'. ?/ = 500. Ans. 281. Joint variation. Definition. One number varies jointly as two or more other numbers when it varies directly as the product of the others. Thus, x varies jointly as u and v when x = kuv. A number may vary directly as one number and inversely as another. It then varies as the quotient of the first divided by the second. V Thus, if x varies directly as y and inversely as z, it is written x k- 2 EXERCISES 99 1. If x oc y and when a; = 20, y = QO, write the equation between x and y. Ans. 3x = y. 2. If b oc d and when 6 = 10, d = 15, find b when d =80. Ans. 53 \. 3. If x varies jointly as y and z, and when y = G and = 2, x = 120, find the constant. Find y when z = 200 and 3 = 15. Ans. fc = 10; y = \\. 4. The area A of a triangle varies jointly as the base & and the altitude a. Write the law if when a = 6 in. and 6 = 4 in., A =12 in. 2 What will be the area when the base is 25 in. and the altitude is 40 in. ? Ans. A=%ab; 500 in. 2 Remark. The law here is the well-known formula for the area of a triangle, but we have not supposed that we knew anything about the formula in working the example. 6. Similar figures 1 vary in areas as the squares of their like dimensions. The diameter of one circle is four times that of another. Using this principle, find the relation of their areas. (Similar figures are those that are alike in shape.) 1 The ideas used here are not new, but will help to show the intimate relation between variation and ratio and proportion. (See Art. 87.) 344 PRACTICAL MATHEMATICS Solution. Here the relation is best expressed as a proportion. Using A and a for the areas, and 4D and D for the diameters, we have Dividing both terms of the second ratio by D 1 , we have A:a = 16:l. /. A = 16a. 6. A grindstone when new is 48 in. in diameter. How large is it in diameter when } ground away? When i ground away? When J ground away? Arts. 41.57 in.; 33.94+ in.; 24 in. Suggestion. Let d\ in. = diameter when i ground away. Then since } remains, l:J=48 J :di 2 . Solving for di, we have di=41.57 . 7. To double the diameter of a circle has what effect on its area? To double the side of a regular hexagon has what effect on its area? Ans. Multiplies area by 4. 8. Similar solids vary in volumes as the cubes of their like dimensions. A water pail that is 10 in. across the top holds 12 quarts. Find the vol- ume of a similar pail that is 12 in. across the top. Ans. 20.736 quarts. 9. The number of vibrations made by a pendulum varies inversely as the square root of its length. A pendulum 39.1 in. long makes one vibration per second. How long must a pendulum be to make four vibrations per second? To make one vibration in 10 seconds? Ans. 2.44 in. + ; 325 ft. 10 in. k Solution. The law is n = ~7?> where n = number of vibrations per Vl second, and 1= length of pendulum in inches. To find k, put n = l and J = 39.1. To find the length of a pendulum to vibrate four times per second, put n=4. \/3jU , Z _39J_ 244 , To find the length of a pendulum to vibrate once in 10 seconds, put n = iV and / = 100X39. 1=3910 in. =325 ft. 10 in. Ans. 10 Vl 10. If a lever with a weight at each end is balanced on a fulcrum, the distances of the two weights from the fulcrum vary inversely as the weights. If two men of weights 160 Ib. and 190 Ib. respectively are bal- anced on the ends of a 10-ft. stick, what is the length from the fulcrum to each end? Ans. 5* and 4$ ft. 11. Could a 1-lb. weight balance a 100-lh. weight? How could they be placed on a 3-ft. bar? Ana. Yes. Fulcrum 2iVi ft. from 1-lb. weight. VARIATION 345 12. If /ocgg' and/ oc--, show that/ Temperature 45 45 45 45 43 4? 41 40 4? 51 57 5q 6? Hour P. M 1 2 3 4 5 6 7 8 q 10 11 1? Temperature 66 70 74 76 76 75 74 73 72 70 69 68 The change in temperature is quite easily seen from a study of this table, but it may be seen at a glance if we put the facts into a diagram as shown in Fig. 207. Here the hours are located on a horizontal line and the degrees on a line perpen- dicular to it. The reading for any hour is located so as to be above the hour and to the right of the degree of temperature. In this manner we can locate 24 points. Evidently if the temperature reading had been taken each minute instead of each hour, there would be determined 60 points in the same space in which we now have one. If we suppose no sudden change in temperature between the hourly readings, we may connect the points representing these by a line, and any point on this line will indicate the temperature for the corresponding time. Of course, if there had been a sudden rise and fall in temperature between the hours, it is not shown by this line. The oftener then we take the readings, the truer will the line indicate the changes. 1 The representation, made as in Fig. 207, is called a graph; 1 Cross-ruled paper, or paper ruled into squares of various sizes, can be obtained cheaply ; and, by using this paper, the points can be located with greater accuracy than on plain paper. 351 352 PRACTICAL MATHEMATICS and such a method of representing relations between numbers is called a graphical method. The use of the graph is of very wide application. At the weather bureaus there are thermometers with an attach- ment which automatically traces the graph of the temperature and time. Engineers make constant use of the graph in their 40 - 35 12128456789 10 11121 88456 789 10 11 12 A.M. Hours P.M. March. 21, 1917 FlO. 207. work. Laboratory data are put in the form of a graph. Graphs can be made for algebraic equations, and relations are thus clearly shown that otherwise would be difficult to see. Making the graph is often spoken of as plotting. 286. Definitions and terms used. Since we often wish to plot negative as well as positive numbers, it is necessary to give certain definitions and make certain assumptions which will now be explained. GRAPHICS 353 If in Fig. 208 OX and OY are drawn at right angles to each other, the position of any point, as P, may be located by measuring its distance from OY and from OX. These lengths, which in the figure are OA and OB, are called the coordinates of the point P. The length OA is called the x-cob'rdinate ; and OB, the y-coordinate. The two lines OX and OY are called the x-axis and the y-axis respectively. Together they are spoken of as the coordinate axes. The point where the two axes cross is called the origin. II III o -X IV FIG. 208. The rr-axis is called the axis of abscissas, and the y-axis, the axis of ordinates. The ^-coordinate and the ^-coordinate are also called the abscissa and ordinate respectively of the point. The coordinates are always measured from the origin. Any abscissa measured toward the right is positive, and measured toward the left is negative. Any ordinate measured upward is positive and downward is negative. The four parts into which the axes divide the plane are called quadrants. These are called the first, second, third, and fourth quadrants, and are numbered in Fig. 208 by the numerals I, II, III, IV. It is evident that, in the first quadrant, both coordinates are positive; in the second quadrant, the abscissa is negative and 354 PRACTICAL MA THEM A TICS the ordinate positive; in the third quadrant, both coordinates are negative; in the fourth quadrant, the abscissa is positive and the ordinate is negative. This is shown in the following table: Quadrant I II III IV Abscissa ... + f Ordinate. , + - 1 - 287. Plotting points. To plot a point is to locate it with reference to a set of coordinate axes. To plot a point whose y P (-2 ,5) r> P ( J ,4} tt _y / p (-4 _ > * I) P f 4 ' -a ) FIG. 209. x-coordinate is 3 and y-coordinate 4, first draw the axes (see Fig. 209), then choose a unit of measure and lay off OA 3 units to the right of 0. Through A draw a line parallel to the t/-axis. Now lay off OB, 4 units above 0, and through B draw a line parallel to the z-axis. The required point is located where these two lines meet. In the figure it is located as P(3, 4), which is the usual manner of writing the coordinates GRAPHICS 355 of a point. The abscissa is placed first; they are separated by a comma and inclosed in parentheses. It is read: "The point P whose coordinates are 3 and 4." In a similar manner the following points are located as given in the figure: P(-2, 5), P(-4, -3), P(4,-5). 3 PH / / / / j / 20 / / z / / / / j* X 1C. ^ LC 0" y Ib 50 G 7 7 3 b j Ye irs ) 1 2 3 FIG. 210. EXERCISES 101 1. Plot the following points: (-1, 6), (7-2), (0, 4), (4, 0), (-6, 0), (-3, -8), (-6,2), (0,0). 2. The temperatures read each hour for the 24 hours ending Jan. 26, 1918, at 2 p. M. were as follows: Above zero 11, 11, 10, 6, 8, 5, 4, 3, 2, 2, 1, 0, 0; below zero 1, 1, 2, 3, 2; above zero 1, 3, 6, 8, 10, 12. Plot these using the hours as abscissas and the temperatures as ordinates, and connect by a smooth curve. 3. The population of a town in 1850 was 16,000; in 1860, 16,300; in 1870, 16,850; in 1880, 17,800; in 1890, 19,100; and in 1900, 20,700. Plot a curve showing the variation in population, and estimate the probable population in 1875 and in 1907. Discussion. Draw the axes; lay off the years, beginning with 1850 at the origin, along the x-axis ; and lay off the population in thousands along the y-axis, beginning with 16,000 at the origin. The curve will then be as in Fig. 210. The population for any year is estimated by locating the year on the x-axis; drawing a perpendicular to the x-axis; and, from the intersection of this with the curve, drawing a parallel to the x-axis. The point where this parallel intersects the y-axis determines the population. 356 PRACTICAL MATHEMATICS The process by which we determine the coordinates of any point on the curve is called interpolating. 4. A company began to sink a mining shaft on June 4; on June 11 it had reached a depth of 30 ft. ; on June 18, 54 ft. ; on June 25, 70 ft. ; on July 2, 82 ft. ; on July 9, 100 ft. ; on July 16, 135 ft., and on July 23, 150ft. Plot a curve showing the progress of the work. 5. The population of the United States by decades was as follows. Plot and estimate the population for 1913 and 1919. Yrar Population Year Population 1790 3,929,214 1860 31,443,321 1800 5,308,433 1870 38,558,371 1810 7,229,881 1880 50,155,783 1820 9,663,822 1890 62,669,756 1830 12,806,020 1900 76,295,200 1840 17,069,453 1910 91,972,266 1850 23,191,876 6. Make a graph from which can be read the product of any number from -225 to 225 multiplied by 0.367. Suggestion. Plot the numbers from 225 to 225 as abscissas and the products as ordinates. 3 o 4 3 S 2 H 1 -J A' _ - _ - g- -rr ' _^ - 7~ -- -- ,_, ^-, -jT ^ ^ ^ ^-^ ^ ^ , / f A p 10 20 30 40 60 80 100 150 Distance in Feet Fio. 211. 7. Make a graph from which can be read the quotient of any number less in absolute value than 500 divided by 12.7. 8. A stone, falling from rest, falls through 4 ft. in J second, 16 ft. in 1 second, 36 ft. in 1 J seconds, 64 ft. in 2 seconds, 100 ft. in 2 seconds, and 144 ft. in 3 seconds. Find, by plotting a curve, how long a stone would be in falling 80 ft. ; and also how far it falls in 2} seconds. GRAPHICS 357 The graph for exercise 8 is as shown in Fig. 211. To find how long the stone would be in falling 80 ft., locate 80 ft. on the x-axis, draw a per- pendicular line AB, and through the point B on the curve draw a line parallel to the x-axis to intersect the y-axis. This point of intersection C determines the number of seconds it will take the stone to fall 80 ft. To find how far the stone will fall in 2f seconds, proceed in a similar man- ner, starting on the y-axis. 9. The football accidents for the years given are as follows: Year Deaths Injuries Year Deaths Injuries 1901 7 74 1910 22 499 1902 15 106 1911 11 178 1903 14 63 1912 13 1904 14 276 1913 14 1905 24 200 1914 12 1906 14 160 1915 16 1907 15 166 1916 19 1908 11 304 1917 12 1909 30 216 Plot two curves, using the years as abscissas and the deaths and injuries respectively as ordinates. State the conclusions that you can draw from the curves. 10. The monthly wages of a man for each of his first 13 years of work was as follows: $28, $30, $37.50, $45, $60, $65, $90, $95, $95, $137, $162, $190, and $210. Plot the curve showing the change. Estimate his salary for the four- teenth and fifteenth years. Can you be certain of his salaries for these years? Why? 11. A cyclist starts from a town at 8 A. M. and rides 4 hours at the uniform rate of 9 miles per hour. He then rests 1 hour and returns at the rate of 8 miles per hour. At 11 A. M. a second cyclist starts from the same town and rides over the same route at the rate of 6 miles per hour. Plot curves showing where they will meet. 12. (a) Reckon the simple interest at 6% on $100 for the several years 1, 2, 3, 10. Plot the years as abscissas and the interest as ordinates. Connect with a smooth curve from which may be read the interest on $100 at 6% for any time. (b) Reckon the compound interest at 6% on $100 for the several years 1, 2, 3, 10. Plot as in (a), using the same axes. 13. Plot the numbers 1, 4, 9, 16, 25, 36, etc., as abscissas and their square roots as corresponding ordinates. Make scale for ordinates 5 or 10 times that of abscissas. How can you determine roots of interven- ing numbers? (See discussion of exercise 3.) Find roots of 11, 14, 21, PRACTICAL MATHEMATICS 29, 42 from curve. How do these agree with computed values of roots of these numbers? 14. If T is the tensile strength, in tons per square inch of cross section, of steel containing X per cent of carbon, and we are given the following values, plot a curve to show, as accurately as the data will allow, the tensile strength of steel containing any percentage of carbon from 0. 1 to 1 per cent. What strength would you expect for 0.4 per cent of carbon? X 0.14 0.46 0.57 0.66 0.78 0.80 0.87 96 T 28.1 33.8 35.6 40 41.1 45.9 46.7 52.7 16. Plot the Fahrenheit thermometer scale on the z-axis and the centi- grade scale on the y-axis. Draw the line any point of which has as co- ordinates equivalent temperatures on the two scales. How can this graph be used to reduce from one scale to the other? Read from the graph the centigrade temperatures for the following Fahrenheit readings: 25, 90, 367, -40, 15. Give in Fahrenheit the following centigrade readings: 33, 76, 15, -46, -9. 16. In an experiment on the stretching of an iron rod the linear exten- sion, L in inches, for a load, W in pounds, was found to be as follows: W. .600 1100 1600 2100 2600 3100 3600 4100 4600 5100 ) i I n nt\r\ n rn o n r\OT Plot, choosing suitable distances on the z-axis for W, and on y-axis for L. Up to how great a load is the extension proportional to the load? That is, where does the curve change direction rapidly? Remark. While the curve is a straight line, the relation between L and W can be expressed in the language of variation by the equation W = kL. Or we may say that the straight line is the graph of a direct variation where the ratio is constant. 17. A stick of white pine 1 in. broad and 0.5 in. thick is supported at points 24 in. apart and loaded in the middle. The deflection, d in inches, for a load, W in pounds, is as follows: W o 5 8 18 ?8 38 48 58 63 68 j 69 70 d o 0088 14 035 056 077 099 1 ?? 1 35 1 685 1 765 1 85 1 Plot so as to show the deflection is proportional to the weight up to a certain point. 18. Plot the following data of the weight of railway locomotives. The weights are of the largest locomotives made each year. Plot time on the x-axis. GRAPHICS 359 Year 1898 1899 1900 1902 1904 1905 1909 1910 1911 Engine, pounds 230,000 232,000 250,300 259,800 287,240 334,000 425,900 440,000 616,000 Engine and tender, pounds 334,000 364,000 391,400 383,800 453,000 477,000 . 596,000 611,800 850,000 19. A rifle sighted to 1000 yd. rests upon a support 5 ft. from the ground and is fired. The height of the bullet above the support at the various distances is given in the following table: Distance in yards 1 100 200 300 400 500 600 700 800 900 1000 from firing point. J Vertical height 1 above support \ 7.3 11.2 15.0 18.5 21.0 23.3 25.0 22.5 16.5 in feet. Plot a representation of the path of the bullet. Show the ground level and the height of the support. Where does the bullet reach a height of 20 ft. ? 20. Plot the relation between centimeters and inches. Suggestion. Take the numbers of centimeters on the z-axis and the inches on the y-axis. Any point on the curve has as coordinates a num- ber of centimeters and a number of inches that are equivalent in length. 21. Plot a curve showing the relation between yards and meters. 22. Plot a curve showing the relation between pounds and kilos. 23. Plot a curve showing the relation between dollars and francs. Use 1 franc = 19.3 cents. 288. Graph of an equation. If we have an equation in two unknowns, as 3x+4y = l2, we can determine a number of pairs of values for x and y that will satisfy the equation. (See Art 255.) If we consider each of these pairs as the co- ordinates of a point, the value of x for the abscissa and y for the ordinate, then the graph determined by these points is called the graph of the equation or the curve of the equation. To plot the curve of 3z+4?/ = 12, determine the pairs of values: (0, 3), (4, 0), (2, 1J), (3, f),(5, -f), (8, -3), (-8, 9), 360 I'KACTH 'AL MA Til KM A TICS (4, 6), (12, 6). Plot these points and connect with a smooth line. The curve is shown in Fig. 212. Whenever the equation is of the first degree the graph will be a straight line. Since a straight line is determined by knowing two points on it, the graph can be drawn when two points only have been plotted. Usually the most convenient points to take are those where the line crosses the two axes. These two points are found by putting x = and finding y, and r A' s ^ s X, f- 8. 9 N ^ s X, t (. 4 g x S, s s, ), \) s (' (3 e .> ,(-. ) s S ,1 . II S '. ' \ s s, vf }.- i) % *s, s s 1 -f | s x s S N V, Fio. 212. then putting y = and finding x. The distances from the origin to these points are called the intercepts on the axes. If the line goes through the origin another point will be nec- essary. If the two points are near together the line will not be well determined unless the work is very accurate. 289. Simultaneous equations. If we plot the graphs of two simultaneous equations they will intersect at some point if they are not parallel. The coordinates of the point of intersection evidently satisfy both equations and are the values obtained by solving the equations as simultaneous. Example. Plot each of the following equations and solve them as simultaneous equations. (1) (2) 3z+2y = GRAPHICS 301 Pairs of values for first: (0, ), (1, |), (3, 0), (7, -1). Pairs of values for second: (0, 2), (2, -1), (4, -4), (-2, 5). These are plotted and the intersection is determined to be the point (1, |). Solving as simultaneous equations, we find re = 1 and y = \ as values of x and y. The plotting is shown in Fig. 213. FIG. 213. 290. The graph of an equation of any degree. The grapt of an equation of any degree in two variables (unknowns' can be plotted by taking values for one variable and deter- mining corresponding values for the other variable. Each pair of values determines a point, and a sufficient number of these points will determine the form of the curve. A graph of an equation of higher degree than the first is not a straight line. The graph of an equation of the second degree in two variables is a conic section, that is, the section of a cone. EXERCISES 102 Plot and solve the following systems of equations: = ll and and 3(52 ritACTH 'AL MA THEM A TICS 6. Plot the curve of the equation xy 1. This will give a curve from which can be read the reciprocals of any number. It is plotted by first finding a number of pairs of values for z and y which satisfy the equa- tion, and then plotting the points that have these pairs as coordinates. It should be noticed that when x is negative, y is negative, and when x is positive, y is positive. The pairs of values given m the table are found and plotted as shown in Fig. 214. 7. Plot the curve for j/ = x*, and thus find the curve from which can be read squares and square roots of numbers. Pairs of values x y A 16 1 8 j 4 4 2 1 1 2 4 3 1 6 i 8 i 16 A Fio. 214. 8. Plot the curve for y=x\ and thus find the curve from which cnn be read cubes and cube roots of numbers. 9. In simple interest, if p stands for principal, / for time, r for rate, and a for amount, then a = p(l+rO. If now particular numerical values are given to p and r, and if the different values of a be taken as ordinates, and the corresponding values of t as abscissas, then the graph of this equation may be drawn. Draw the graph. What line in the figure represents the principal? What feature in the graph depends upon the rate per cent? 10. With the same axes as used in exercise 9, draw the graph for which interest and time are the coordinates of points on the curve. GRAPHICS 363 11. Using the formula for the area of a circle A =irr 2 , plot the curve, using radii as abscissas and areas as oridnates, from which can be read the areas of circles of radii from to 6. 12. Follow directions similar to those in exercise 11 and plot the curve from which can be read the volumes of spheres. 13. If the temperature of a gas is constant, the pressure times the vol- ume remains constant. Using p for pressure and v for volume, plot so as to show their relative change when pv = 4. 291. Simpson's Rule. The area of the space included between a curve and a straight line can easily be found ap- proximately by the use of Simpson's Rule which may be stated as follows: Let AB, in Fig. 215, be the curve and CD the straight line. Divide the length CD into an even num- m o r FIG. 215. ber of equal parts, say 8, of length a, and erect the ordinates ho,hi,h 2 , , h s . Then the area of the figure CDBA will be given by the formula: It is to be noticed that the coefficients of the ordinates are alternately 4 and 2, excepting the first and the last. The greater the number of divisions made the more accurate, in general, will be the result. In words this may be stated in the following: RULE. Divide the base CD into an even number of equal parts, and measure the ordinate at each point of division. Add together the first and last ordinates, twice the sum of the other odd ordinates, and four times the sum of the even ordinates; multiply the sum by one-third the distance between consecutive ordinates. The result is the area inclosed (approximately). 364 PRACTICAL MATHEMATICS 292. The average ordinate rule. For approximate results the area between a curve and a straight line, base line, may be found as follows: RULE. Divide the base line into any number of equal parts; at the center of each of these parts draw ordinates. Take the average length of these ordinates and multiply by the length of the base line. The result is the area inclosed (approximately). In Fig. 215, mn, op, rs, etc., are the ordinates. A convenient way for adding the ordinates is to draw a line of indefinite length; then with the dividers measure the ordinates successively on this line. The total length can then be measured at once. This will avoid errors to some extent. 293. Area in a closed curve. Either of the methods given may be used in finding the area within a closed curve. Thus, FIG. 216. in Fig. 216, draw the two parallel tangents OY and MN, and draw OX perpendicular to these. Divide ON into any number of equal parts 1 (an even number for Simpson's Rule) and draw the ordinates AB, CD, etc. Call the several widths of the figure on these ordinates hi, hi, A s , etc. These widths can be used in Simpson's Rule to find the area within the closed curve. The widths mn, op, rs, etc., can be used in the average ordinate rule to find the area. 1 To divide a line into any number of equal parts see Art. 149. GRAPHICS 365 It should be noted that h and ^ 8 for this figure are each 0. 294. The steam indicator diagram. As a useful application of the discussion in the preceding articles, we will consider the steam indicator diagram. 1 The steam indicator is a mechanical device to attach to a steam engine to make a graphical representation of the steam pressure acting on the piston throughout the stroke. Knowing FIG. 217. the pressure, the indicated horse-power of the engine can be calculated from the formula given in exercise 4, page 313, rr H = 33^000' Where H indicated horse-power, P = mean effective pressure in pounds per square inch, L = length of stroke ;n feet, A = area of piston in square inches, N = the number of strokes per minute. The indicated horse -power is the power developed by the steam on the piston of the engine, without any deduction for friction. The effective horse-power is the actual available horse- power delivered to the belt or gearing, and is always less than the indicated horse-power, because the engine itself absorbs some power by the friction of its moving parts. The indicator diagram may be as given, Fig. 217. 1 For a full discussion of the steam indicator the student is referred to Peabody Manual of the Steam Engine Indicator. 366 PRACTICAL MATHEMATICS The width of a rectangle the same in length as this, and of the same area, would represent the mean effective pressure per square inch on the piston during the stroke. The diagram is always to a certain scale, which is known from the indicator. For instance, the scale might be 60 Ib. per inch in diagram. The mean effective pressure is then 00 Ib. multiplied by the average width of the indicator diagram. The average width of the diagram may be found by dividing the area by the length. The area can be found by Simpson's Rule or by the average ordinate rule. A convenient method for locating the ordinates is to place a common ruler as in Fig. 218, and locate 10 ordinates, the two FIG. 218. end ordinates being half as far from either end as the distance between the other ordinates. The average length of these ordinates multiplied by the scale, taken from the indicator, gives the mean effective pressure. Example. Taking the indicator diagram in Fig. 218, find the mean effective pressure of the steam if the scale is 30 Ib. to the inch. Find the horse-power of the engine if the diam- eter of the piston is 18 in., length of stroke 2\ ft., and number of revolutions 110 per minute. (The number of strokes of the piston is twice the number of revolutions.) Solution. Adding together the ten ordinates, we have 1.82+2.66+2.81 + 2.91+2.73+2.21 + 1.78+1.46+1.17+0.70 = 20.25. GRAPHICS 367 Since there are 10 ordinates, the mean is 20.25 -r- 10 = 2.025. Multiplying by the scale, we have 2.025X30 = 60.75 = pounds pressure per square inch, the answer to the first part. Area of piston = 3.1416X9 2 = 254.47 in. 2 t ^ u PLAN Formula for the horse-power is H = = 60.75, L = 2.5, A = 254.47, 60.75X2.5X254.47X220 EXERCISES 103 1. An indicator diagram has a length of 2 in. The ten ordinates, beginning at the left, are 0.70 in., 0.90 in., 0.97 in., 0.85 in., 0.67 in., 0.52 in., 0.42 in., 0.35 in., 0.23 in., 0.07 in. Draw a diagram that these ordinates will satisfy. If the indicator scale is 120 Ib. to the inch, find the mean effective pressure. Ans. 68.16 Ib. per in. 2 2. Find the indicated horse-power of an engine having the indicator diagram of exercise 1, if length of stroke is 3 ft., diameter of piston 23 in., and number of strokes 100 per minute. Ans. 257.4+ h. p. FIG. 219. 3. Draw a semicircle 2 in. in radius. Divide the diameter into four parts and find the area by Simpson's Rule. Divide the diameter into ten parts and find area by the same rule. Find the area by the formula. Compare the three results and state your conclusions. 4. Find the area of the ellipse in Fig. 219 by Simpson's Rule. Find the area by the formula A =irab, and compare the two results. 6. Find the area of the indicator diagram of Fig. 217 both by Simpson's Rule and by the average ordinate rule, using ten divisions. Compare the results. Ans. 4.46 in. 2 nearly. 6. If the indicator diagram of exercise 5 has a length of 3.73 in., and the scale is 50 Ib., find the mean effective pressure. If the stroke is 2 ft. 6 in., the cylinder 18 in. in diameter, and the number of revolutions per minute 80. find the indicated horse-power of the engine. Ans. 59.79 Ib. per in. 2 ; 184.4 + h. p. PART FOUR LOGARITHMS AND TRIGONOMETRY CHAPTE-R XXXIV LOGARITHMS 295. Uses. By the use of logarithms, the processes of mul- tiplication, division, raising to powers, and extracting roots of arithmetical numbers are much simplified. The process of multiplication becomes one of addition, that of division becomes one of subtraction, that of raising to a power be- comes one of simple multiplication, and that of extracting a root becomes one of simple division. Many calculations that are difficult or impossible by ordi- nary arithmetical methods are readily carried out by means of logarithms. For instance, by the help of logarithms a square root is more readily found than by ordinary methods, and any other root is found as easily as a square root. The value of a 17 number affected with any exponent, as 2.34 , can be com- puted easily by logarithms. 296. Exponents. For convenience the definitions and laws of exponents previously given are repeated here. Definitions. (1) a n = a-a-a - to n factors, n an integer. (2) <- = - (3) o = l. / A \ " Wl / (4) a* = a-\/a n . 24 369 370 PRACTICAL MATHEMATICS Laws. (1) a n -a m = a"" H ". (2) a n -s-a m =a n -" > . (3) (a-6-c- )" = a n 6"c n - (A\ f a V a " (4) (b) "V (5) (a n ) m =a nm . 297. Definitions and history. A logarithm of a number is the exponent by which the base must be affected to produce that number. The logarithms of all the positive numbers to a given base are called a system of logarithms, and the base is called the base of the system. Any base may be used in a system of logarithms; but the base 10 is commonly used because, as will be seen later, it makes a very convenient system of logarithms to work with. Logarithms were invented by John Napier of Scotland, who lived from 1550 to 1617. They were described by him in 1614. Napier used the number 2.71828 as a base. This base is still used in mathematics (see Art. 316) . Henry Briggs (1556 to 1631), professor at Gresham College, London, modified the new invention by using the base 10, and so made it more convenient for practical purposes. Because of the time they save and the help they give in performing difficult computations, logarithms may be consid- ered among the great inventions of the world. 298. Notation. If we take 2 as a base, we may write in the language of exponents, 2 4 = 16. In the language of logarithms, we may express the same idea by saying, the logarithm of 16 to the base 2 is 4. This is abbreviated and written thus Iog 2 16=4. Similarly, we have the following expressed in the language of exponents and in the language of logarithms: Language of Language of exponents logarithms 2 5 = 32. log, 32 = 5. 3 4 = 81. log, 81=4. 5 4 = 625. Iog 5 625=4. LOGARITHMS 371 Language of Language of exponents logarithms 8 3 = 512. logs 512 = 3. 4- 5 = 2. Iog 4 2 = 0.5. 8'= 4. logs 4 = . 64' = 4. Iog 64 4 = i 10 3 = 1000. Iog 10 1000 = 3. EXERCISES 104 Answer as many as you can orally. Cover the answers so that they will not be seen until the results are written. 1. Express the following in the language of logarithms: (1) 2 6 =64. (4) 16- 5 = 4. (7) 32- 4 =4. (2) 5 3 = 125. (5) 10 4 = 10,000. (8) 10 l - 3979 =25. (3) 7 3 = 343. (6) 125' =5. (9) 10 2 - 5465 = 352. Ans. (1) Iog 2 64=6. (4) logie 4 =0.5. (7) Iog 32 4=0.4. (2) logs 125=3. (5) logio 10,000 = 4. (8) logio 25 = 1.3979. (3) logy 343 =3. (6)logi 25 5 = i (9) logio 352=2.5465. 2. Express the following in the language of exponents: (1) log, 256=8. (4) logio 643 =2.8082. (2) loge 216=3. (5) Iog 10 429 = 2.6325. (3) logie 2=0.25. (6) logic 999 =2.9996. Ans. (1) 2 s =256. (4) 10 2 - 8082 =643. (2) 6 3 =216. (5) 10 2 - 6325 =429. (3) 16- 25 = 2. (6) 10 2 - 9996 = 999. 3. Find the logarithms of the following : (1) log; 49. (4) Iog 9 729. (7) Iog 8 2. (2) logs 243. (5) Iog 4 256. (8) Iog 16 64. (3) logs 3125. (6) log, 100,000. (9) log, 0.01. Ans. (1) 2; (2) 5; (3) 5; (4) 3; (5) 4; (6) 5; (7) *; (8) 4; (9) -2. 4. Find the value of x in the following : (1) logs x = 4. (5) logic x= -2. (9) Iog 25 z=i (2) logio z = 6. (6) logs x= -3. (10) logie z=f. (3) logie a; = i (7) logio x= -3. (ll)log m z = i (4) logio z=0. (8) logs x= |. (12) Iog 49 x = |. Ans. (1) 81; (2) 1,000,000; (3) 8; (4) 1; (5) 0.01; (6) ^ r ; (7) 0.001; (8) 16; (9) 125; (10) 64; (11) 25; (12) 343. 372 PRACTICAL MATHEMATICS 6. Find the value of x in the following : (1) log, 100-2. (5) log, 4-0.5. (9) log, 27-0.75. (2) log, 81 =4. (6) log, 4-0.25. (10) log, 2 = 0.125. (3) log, 512 = 3. (7) log, 16 ~J. (11) log* 49 = . (4) log, 1024 = 10. (8) log, 17-0.5. (12) log, 100 -J. Ana. (1) 10; (2) 3; (3) 8; (4) 2; (5) 16; (6) 256; (7) 32; (8) 289; (9) 81; (10) 256; (11) 343; (12) 1000. 299. Illustrative computations by means of exponents. It is very important that the fundamental ideas of logarithms shall be well understood. In this article are given examples illustrative of the use of exponents, or logarithms, in making computations in multiplication, division, raising to powers, and extracting roots. These computations are made by the help of the following table in which 2 is the base. The work is kept in the language of exponents but can be easily trans- lated into the language of logarithms. 2- 6 = & 2!=2 2" =256 2- & = & 2* =4 2 =512 2-' = ^ 2 3 = 8 2' = 1024 2~ 3 = t 2< = 16 2" = 2048 2- J = J 2 5 = 32 2 1 * = 4096 2-' = J 2 = 64 2" = 8192 2 =1 2 7 = 128 2 M = 16,384 Multiplication. Multiply 512 by 32. From the table, 512 = 2 9 , and32 = 2 6 . From the table, 2" = 16,384. .'. 512X32 = 16,384. Diinsion. Divide 512 by 4096. From the table, 512 = 2 9 , and 4096 = 2 12 . .'. 512-^4096 = 2* H-2 12 = 2- 3 . From the table, 2~ 3 = J- .'. 512 -h 4096 = I Raising to a power. Find the value of 16 s . From the table, 16 = 2 4 . .'. 16 3 = (2) 3 = 2". From the table, 2 12 =4096. .'. 16 s = 4090. 2. 256X64. 5. 32X256. 8. 2048 -=-256. 11. 16 3 . 3. 2048X8. 6. &X 16,384. 9. 16^-512. 12. 64 2 . 14. A/4096. 16,384X512 15. \/ 16, 384. 32X64 2048 20. (32X128)?- 512X8 21. (8192X32)1 LOGARITHMS 373 Extraction of a root. Find the value of ^4096. From the table, 4096 = 2 12 . From the table, 2 3 =8. .'. ^4096 = 8. EXERCISES 105 By the help of the table on page 372 find the value of the following. Do the work without using a pencil when possible. 1. 256X16. 4. 128X128. 7. 8192 -H 1024. 10. 8-^512. 13. Vl^384. ^j*?. !? 512 19. (4096)1 1024X64 / 512X256 \i 3 1256X1024 512X16,384' ' \1024 X8192/ \ 4096 2 300. Logarithms of any number. It is readily seen that, in the exercises of the last article, the numbers considered were all integral powers of 2. It is also readily seen that there are many numbers that cannot be expressed as integral powers of 2. The same thing is true for any base. Thus, for the base 3, we have as integral powers the numbers 3, 9, 27, 81, 243, etc., and no numbers except these between 3 and 243. We can, therefore, with any given base write integral log- arithms for only a small part of all possible numbers. That is, the logarithms of numbers to any given base are usually not integral. Thus, the logarithm of 95 to the base 3 is 4 and some fraction, because 95 is between 3 4 and 3 5 . What this fraction is cannot easily be determined. 301. Logarithms to the base 10. In what follows, if no base is stated, it is understood that the base 10 is used. When the base is 10 we evidently have the following: log 100,000 = 5. log 1 =0. log 10,000 = 4. log 0.1 -1. log 1000 = 3. log 0.01 =-2. log 100 = 2. log 0.001 =-3. log 10 = 1. log 0.0001 =-4. 374 PRACTICAL MATHEMATICS The logarithm of any number between 1000 and 10,000 is between 3 and 4, or it is 3 and some fraction. Between 100 and 1000 the logarithm is 2 plus a fraction. Between 0.01 and 0.1 the logarithm is 2 plus a fraction or 1 minus a fraction. In order that the fractional part of the logarithm may always be positive, we shall agree to take the logarithm so that the integral part only is negative. In general, the logarithm of a number consists of two parts, a whole number part and a fractional part. The whole number part is called the characteristic. The fractional part is called the mantissa. The mantissas of the positive numbers arranged in order are called a table of logarithms. The logarithm of 3467 consists of the characteristic 3 and some mantissa because 3467 lies between 1000 and 10,000. The logarithm of 59,436 is 4 plus a fraction because 59,436 lies between 10,000 and 100,000. The log 0.0236 is -2 plus a fraction because 0.0236 lies between 0.01 and 0.1. It is readily seen that multiplying a number by 10 increases its characteristic by 1. One of the great advantages in using the base 10 is that the characteristics can be determined by inspection. It is only necessary, then, to have the mantissas given in a table. 302. Rules for determining the characteristic. From what has been said in the last article, and from a further con- sideration of the table given there, the following rules are evident: (1) For whole numbers, the characteristic is one lefts than the number of whole number figures and is positive. (2) For decimals, the characteristic is one more than the num- ber of zeros immediately at the right of the decimal point and is negative. (3) In a number consisting of a whole number and a decimal, consider the whole number part and apply rule (1). Thus, the characteristics of the following are as given: of 326 is 2 by rule (1), of 37,265 is 4 by rule (1), of 0.046 is -2 by rule (2), of 0.000046 is -5 by rule (2), of 2.36 is by rule (3), and of 276.36 is 2 by rule (3). LOGARITHMS 375 EXERCISES 106 State the characteristics to the base 10 of the following: 846 3956 2.325 87,654 44.36 43,968 0.0123 9.3264 173.94 39.267 0.00492 0.0003967 4.7654 3333.3 0.4689 0.039643 303. The mantissa. The determination of the mantissa is more difficult than the determination of the characteristic. The mantissa is found from a table of logarithms. Tables of logarithms are made only by a great deal of work. They are spoken of as three place tables, four place tables, ten place tables, etc., according to the number of decimal places given in the mantissas. The degree of accuracy in computations made by logarithms depends upon the number of places in the table used, the more places in the table the greater the degree of accuracy. The tables generally used are those having from four to six places. The mantissa depends only upon the figures of the number, and not at all upon the decimal point. To illustrate this, consider the following: Suppose that we have given that log 867 = 2.9380. (1) This means that 10 2 - 9380 = 867. (2) Now 10 2 = 100. (3) Dividing (1) by (2), 10- 9380 = 8.67. (4) /. log 8.67 = 0.9380. Hence the mantissa of 8.67 is the same as the mantissa of 867. It is evident that if both members of (1) are multiplied or divided by any integral power of 10 the mantissa is unchanged. Hence the decimal point of a number can be moved as we please without the mantissa of the logarithm of the number being changed. This also illustrates the fact that a change in the position of the decimal point of a number changes the value of the characteristic of the logarithm of the number. 304. Tables. Upon examining a four place table of loga- rithms (see Table X), it is noticed that the first column has the letter N at the top. This is an abbreviation for number. The other columns have at their tops and bottoms the numbers 376 PRACTICAL MATHEMATICS 0, 1, 2, 3, 9. Any number consisting of three figures has its first two figures in the column headed N and its third figure at the top of another column. For instance, take the number 456; 45 is found in the column headed N and 6 at the top of another column. The columns, after the first, are made up of numbers con- sisting of four figures. These numbers are decimals, and are the mantissas of the logarithms of the numbers made up of the figures in the column headed N together with a figure from the head of another column. The difference between two consecutive mantissas is called the tabular difference, that is, the table difference. 305. To find the mantissa of a number. (1) When the number consists of three significant figures. Example. Find the mantissa of 347. From the manner in which the table is formed, the first two figures of 347 are found in the column headed N, and the third figure at the top of the page. The mantissa of 347 is found to the right of 34 and in the column headed 7. It is 0.5403. The mantissa of 3.47, 3470, or any number consisting of these figures in the same order is 0.5403. (2) When the number consists of one or two significant figures, the number is found in the column headed N, and the mantissa to the right in the column headed 0. Thus, the mantissa of 13 is 0.1139, and the mantissa of 4 is 0.6021; this is found to the right of 40. (3) When the number consists of four or more significant figures. Example 1. Find the mantissa of 7586. Since 7586 lies between 7580 and 7590, its mantissa must lie between the mantissas of 7580 and 7590. Mantissa of 7580 = 0.8797. Mantissa of 7590 = 0.8802. The difference between these mantissas is 0.0005, which is the tabular difference. Since an increase of 10 in the number increases the mantissa C.0005, an increase of 6 in the number will increase the mantissa 0.6 as much, or the increase is 0.0005 X 0.6 = 0.0003. Hence the mantissa of 7586 = 0.8797+0.0003 = 0.8800. LOGARITHMS 377 The process of determining the mantissa as above is called interpolation. As carried out, the mantissa is supposed to increase in a constant ratio between the values taken; however, this supposition is only approximately true. Example 2. Find the mantissa of 43,286. Mantissa of 43,200 = 0.6355. Mantissa of 43,300 = 0.6365, and the tabular difference = 0.0010. Since an increase of 100 in the number increases the mantissa 0.0010, an increase of 86 in the number increases the mantissa 0.0010X0.86 = 0.0009, to the nearest fourth decimal place. Hence the mantissa of 43,286 = 0.6355+0.0009 = 0.6364. The processes that have been given should seem reasonable ; but, since the process of finding a mantissa has to be performed so often, it is best to do it by rule. 306. Rules for finding the mantissa. (1) For a number consisting of three figures, find the first two figures of the number in the column headed N, and the third figure at the head of a column; then read the mantissa in the column under the last figure and at the right of the first two figures. (2) For a number consisting of one or two figures, find the number in the column headed N, and the mantissa opposite in the column headed 0. (3) For a number consisting of more than three figures, find the mantissa for the first three figures by rule (1), and add to this the product of the tabular difference by the remaining figures of the number considered as decimals. Illustrations. The mantissa of 243, 2.43, 0.0243, or any number consisting of these three figures in the same order, is found by rule (1) to be 0.3856. The mantissa of 25, 0.025, or any number consisting of these two figures in this order, is found by rule (2) to be 0.3979. The mantissa of 2364 is found by rule (3) to be 0.3736. Process. Mantissa of 2360 = 0.3729. Tabular difference is 0.0018. 0.0018 X 0.4 = 0.0007. Adding this to the mantissa of 2360 gives the mantissa of 2364 =0.3736. 307. Finding the logarithm of a number. In looking up the logarithm of a number, it is best to first determine the characteristic, and then the mantissa. 378 PRACTICAL MATHEMATICS Example 1. Find the logarithm of 236. By rule (1) for characteristic we find 2. By rule (1) for mantissa we find 0.3729. .'.log 236 = 2.3729. Example 2. Find the logarithm of 7326. Rule (1) for characteristic gives 3. Rule (3) for mantissa gives 0.8649. /.log 7326 = 3.8649. Example 3. Find the logarithm of 0.00037. Rule (2) for characteristic gives 4. Rule (2) for mantissa gives 0.5682. /.log 0.00037 = 4.5682. It is not permissible to place the minus sign before the characteristic in writing a negative logarithm, for this would indicate that both characteristic and mantissa are negative, and we have agreed that the mantissa shall always be positive. To overcome the difficulty, the negative sign is placed above the characteristic. Another way of writing the negative loga- rithm is to increase the characteristic by 10 and subtract 10 at the right of the mantissa. Thus, the logarithm of 0.00037 is written 4.5682 or 6.5682-10. EXERCISES 107 1. Study the following to fix in mind the meaning of characteristic and mantissa : log 4580=3.6609; that is, 4580 Id . 6809 log 458.0 = 2.6609; that is, 458.0 - in i, 8809 log 45.80 = 1.6609; that is, 45.80 - in i . 6809 log 4.580=0.6609; that is, 4.580 10 6809 log 0.4580=1.6609; that is, 0.4580 = 111 T .8809 log 0.0458=2.6609; that is, 0.0458 = 10 Z .8809 log 000458=3.6609; that is, 0.00458 = LO , .8609 2. Verify the following by the tables: (I) log 10 = 1.0000. (2) log 100=2.0000. (3) log 110 = 2.0414. (4) log 2=0.3010. (5) log 20 = 1.3010. (6) log 200 = 2.3010 (7) log 0.2 = 1.3010. (8) log 542 =2. 7340. (9) log 345 = 2. 5378. (10) log 5.07-0.7050. (II) log 78.5 = 1.8949. (12) log 0.981 = 1.9917. LOGARITHMS 379 (13) log 1054 = 3.0228. (14) log 1272=3.1045. (15) log 0.0165 =2.2175. (16) log 0.1906 = 1.2801. (17) log 21.09 = 1.3241. (18) log 0.09095=2.9588. (19) log 3.060 = 0.4857. (20) log 4.411 =0.6445. (21) log 07854=1.8951. (22) log 0.10125=1.0054. (23) log 54.657 = 1.7377. (24) log 0.09885 =2.9950. 308. To find the number corresponding to a logarithm. In nearly every problem involving logarithms it is not only necessary to find the logarithms of numbers, but the inverse process, that of finding a number corresponding to a logarithm, has to be performed. Since the decimal point in no way affects the mantissa, we can determine only the figures of the number from the mantissa. The decimal point has to be placed by the rules for determining the characteristic. (1) When the mantissa of the given logarithm is exactly given in the table. As an example, find the number having 2.8344 for a logarithm. Find in the table the mantissa 0.8344. To the left of this mantissa, in the column headed N, find the first two figures, 68, of the number, and at the head of the column in which the mantissa is found, find the third figure, 3, of the number. The number then consists of the figures 683, but we do not know where the decimal point is till we consider the characteristic. Since the characteristic is 2, there must be three figures at the left of the decimal point. Hence the number having 2.8344 for a logarithm is 683. This means that 10 2 ' 8344 =683. Notice that a change in the characteristic would change the position of the decimal point. Thus, the number corresponding to 4.8344 is 68,300; while the number corresponding to 2.8344 is 0.0683. (2) When the mantissa of the given logarithm is not exactly given in the table. As an example, find the number corre- sponding to the logarithm 3.4689. Find the mantissas 0.4683 and 0.4698, between which the given mantissa lies. The number corresponding to 3.4698 = 2950. The number corresponding to 3.4683 = 2940. That is, an increase in the mantissa of 0.0015 makes an increase of 10 in the corresponding number. The given mantissa is 0.0006 PRACTICAL MATHEMATICS larger than 0.4683. Then the required number is \ log.( T ) where t is time in seconds, C capacity, and V voltage. Compute the value of R if < = 120, F =123. V = 115.8, and C = 0.082. LOGARITHMS 385 Solution. In order that we may use the table of common logarithms, the formula may be written _ e t l C 2.3026 Substituting values, ff -1Q. X 12 A .082 ' 1 log 115.8) 2.3026(log 123 .-. R = 2.426 X10 10 . As other examples where the base of the natural system of logarithms occurs, consider the following: (1) In an alter- nating current circuit, the current i at any instant is given by the formula where / is the maximum current, R the resistance, L the coeffi- cient of self-induction, t the time in seconds, and e the base of the natural system of logarithms. (2) The work W done by a volume of gas, expanding at a constant temperature, from a volume F to a volume Vi is given by the following formula: Formulas that involve logarithms to the base e occur fre- quently in applications of calculus. EXERCISES 110 Use logarithms in evaluating the following : 1. 2. (0.543) 3 . (4.07) 3 . 3. (1.738) 3 . 4. (1.02) 5 . 6. GfWx) 3 . 6. Uj) 4 . 7. (0.1181)*. 8. (1381)*. 9. (1024) A. 18. (7.23) 7 - 23 . 19. Ans. 0.1601. Ans. 67.42. Ans. 5.248. Ans. 1.104. Ans. 0.004394. Ans. 2.868 X lO" 10 . Ans. 0.3436. Ans. 11.14. Ans. 128. 10. (1.4641)*. (0.00032)*. -v/2. 11. 12. 13. A/4. 14. A/3. 16. A/2^ 16. A/0.03 17. (4.56) 4 20. 48X64X11 52 2 X300 12 X. 31225X400,000' 25 Ans. 1.1. Ans. 0.2. Ans. 1.260. Ans. 1.414. Ans. 1.442. Ans. 1.149. Ans. 0.5023. Ans. 1012 Ans. 1,627,000. Ans. 0.2899. Ans. 0.5411. PRACTICAL MA THE MA TICK 21 J * \55X3.1416* 22. 50 X; 23. x / -431X96', \64X1500 24. (1.06) 4 3.8961 X. 6945 X. 01382 Ans. 1.622. Ana. 4.769. . 19.60. . 51.14. 4694 X. 00457 26. -C/. 0009657 -^70044784". / 7.61 X. 0593 U ' \ 1.307 /' 28. v/5i06.5 x .dobosHoaT. 29 /^_V ' W37J 30. (837.5 X. 0094325)'. 31. (.01)* -H ^7. .000561 6X^/424^65 Ans. 0.001743. Ans. 1.070. Ans. 0.4505. . 0.7945. 5.259. 32. 33. 34. (6.73)^6548 -\X72L83 .05287 Ans. 1.805. Ana. 0.0005228. Ans. 0.00001146. Ans. 188.2. An*. 1.023. v'TSTTXv'. 078359 P< __ T D 35. In the formula S = ^~> find R if 5 = 500, = 220, and 7 = 12. .'2-20 Ans. 9.17. Suggestion. Here, as in many exercises which follow, the computation can be more easily done without logarithms than with. Endeavor to use logarithms only when they are necessary or when they save time. 36. Using the formula for the horse-power of a steam engine, PLAN . H= 33000' (^ Art. 261): (1) Find H when P = 76.5, L = 2}, A =231.8, and # = 116. (2) Find N when H =52, P =49.12, L = 1.5, and A = 113.1. Ans. 140.3; 205.9. 37. Given TF = .0033XlO- 7 n, find H' when n = 75,000. Ans. 0.00002475. 38. If = E. M. F. in volts in moving conductor, L= length of con- ductor in centimeters, V = velocity in centimeters per second, B= the number of lines of force per square centimeter, we have the formula E^LVBIQ-*. Given V'--^-. -8000, and L -0.6X100, find . Ans. 0.072. 60 LOGARITHMS 387 39. Find the value of P from the following formula when A = 11, ~ A Kf = 1.71, and C = 1.3: p = 3 / ^ _ Ans. 0.6199. Suggestion. First find the logarithm of the expression under the radical, and then divide by 3. This will give the log P. 40. In measuring electrical resistance by a Wheatstone's bridge, the following data were taken : (1) R = 8, d = 539.7, a 2 =459. (2) R = Q, ^ =510.1, a 2 =488.4. Find the values of x from (1) and (2) by the formula !) 9.404; (2) 9.40. 41. In finding the diameter of a wrought-iron shaft that will transmit 90 horse-power when the number of revolutions is 100 per minute, using a factor of safety of 8, we have to find the diameter d from the formula 90~ = 68.5X / ennnn' Find the value of d. Ans. 3.591. Wgl 3 42. Find the value of M from the formula M = ., , 3 p> when gr = 980, TF = 75, Z = 50, 6=0.98178, d = 0.5680, and 5 = 0.01093. Ans. 11.69X10". Solution. log W = 1.8751 log 4 =0.6021 log g = 2.9912 log b =1.9920 log Z 3 = 5.0970 logd 3 =1.2629 log Num. = 9.9633 log B =2.0386 log Den. = 3.8956 log Den. =3.8956 log M = 12.0677 .'. M =1,169,000,000,000 = 11. 69X10". 43. Find the value of n from the formula n = - 2 4 , when g = 980, 1 = 28, 9 = 0.857, r = 0.477, L = 109.7, and m = 100. Ans. 2.476X10". 44. Use the same formula as in exercise 43, and find the value of n when L = 69.6, m = 10, gr = 980, Z = 28, = 1.1955, and r =0.317. Ans. 0.577X10". 45. If m = ar- 1 - 16 , find r when TO =2.263 and a = 0.4086. Ans. 0.2287. Solution. Solving for r, we have log a =1.6113 log m= 0.3547 log -=1.2566 388 PRACTICAL MA THE MA TICS It is now best to unite the characteristic and mantissa by algebraic addition. Thus, 1.2566= -1 +.2566 =-.7434. Now multiply by the exponent .862, and get -. 7434 X. 862= -.6408. This is next changed to the form of a logarithm as given in the table, that is, to a form where the mantissa is not negative. -.6408=1.3592. .MJ . _ = 1.3592. = 0.2287. Ana. 46. Given p = p c (2 \ _^- t ) y ~ 1 , find the value of p in terms of p u if -y : \V+IJ 1.41. FIG. 220. An*. p=0.527p . (For the meaning of this formula see Perry's Calculus, page 55.) 47. If an open tank kept full of water has a rectangular notch cut on one side as shown in Fig. 220, the number of cubic feet of water that will flow through * n * s notch P 61 " second is given by the formula: where Q = the amount of flow in cubic feet, c = a constant found by experiment, 6 = the width of the notch in feet, h =the depth of the notch in feet, and = 32.2. Find Q when A = 1J, b = 2, and c= 0.586. Am. 9.654. 48. Using the formula of the last exercise, find Q when h = I, b = 2.5, and c = 0.589. Aiis. 5.991. 49. If Q is the number of cubic feet that will flow through a V-shaped notch per second and h the height in feet of the water above the bottom of the notch, then Q hi If Q = 7.26 when h = 1 .5, find h when Q = 5.68. Ana. 1.360. Suggestion. If Q A* then Q = Kh*. Substituting values, 7.26 = A'1.5 I . /. A' = 7.26 -f- 1.5*. L T sing this value of K with Q =5.68, gives 5.68 = (7.26 -=-1.5')^. , _5.68X1.5 7.126 50. The following is an approximate formula for determining the num- ber of wires that can be enclosed in a pipe : N -0.907 /^-0.94\ '+3.7, LOGARITHMS 389 where N = the number of wires, D = the diameter of the enclosing pipe, d = the diameter of the wires. Solve this formula for D and find D =d [ 0.94 + 0.907 . 51. Use the formula of the preceding exercise and find the diameter of a casing to hold 100 wires each having a diameter of i in. Ans. 1.405 in. 52. The amount of a principal at compound interest for a certain time is given by the following formula: A=P(l+rY, where A=the amount, P = the principal, r = the rate per cent, and t =the time in years. Find the amount of $236 at compound interest for 14 years at 3 per cent. Ans. $356.50. 63. Find the amount of $3764 at compound interest for 21 years at 4J per cent. Ans. $9478. Remark. It should be noted that in such problems as those of exer- cises 52, 53, and the following, a four-place table of logarithms will not give results that can be relied upon when the exponent is large. For instance, if exercise 52 be computed by a six-place table of logarithms the amount is $356.97, and that of exercise 53 is $9486.07. Of course, all that is necessary to secure a desired degree of accuracy is to use a table of logarithms that has a sufficiently large number of decimal places. In life insurance' computations a ten-place table is often used. 54. The government Farm Loan Banks loan money to farmers for a period of years at 5 or 6 per cent, and the money is paid back in equal annual installments. The following table, in use by the United States government, gives the annual installment necessary to discharge a loan of $1000 with interest, the number of years and the rate of interest being indicated in the table. Table fcr Loan of $1000 Number of years 10 Annual ir at 5% $129.50 tstallment at 6% $135.87 102.96 87.18 78.23 72.65 68.97 66.46 15 96.34 20 80 . 24 25 70 . 95 30 65 . 05 35 61.07 40.. 58.28 Compute the installments in this table by using the following formula: Pr(l+r) 390 PRACTICAL MATHEMATICS where P~ the amount of the loan, r-=the rate per cent per annum, n = the number of annual payments, ;> = lh(> amount of each installment. Suggestion. If P = $1000, r = 6 %, and n = 10, then D 1000X.06X1.06 10 P= -M36.87. 65. The above formula will give the amount of each installment made at any equal intervals of time, in order to discharge a debt, if r divided by the number of intervals per year is used for the rate. Thus, if the install- ments are monthly, jo is the rate per month. Find the amount of the equal monthly payments to discharge a debt of $3000 in 5 years if the rate is 6%. Ans. $58.34. 06 Suggestion. Here P= 3000, n = 60, and rate per month is -.,, = 0.005. \i 66. Find the amount of the equal semi-annual installments to discharge a debt of $4500 in 7 years at 6%. Ans. $399.30. 67. If an indebtedness is paid in installments, the payments being equal and each including the interest to the date of the installment, then the number of installments necessary to pay the debt is given by solving the formula of exercise 54 for n and finding _logp-log(p-Pr) log(l-j-r) where P = the total indebtedness, p = the amount of one installment, and r = the rate per cent for the period between installments. How many semi-annual installments of $600 each will it take to dis- charge a debt of $7000 bearing 5% interest. Ans. 14. Suggestion. Here P = 7000, p = 600, and r= 0.025. Substituting in formula, log 600 -log (600 -175) log 1.025 _ 2.7782 -2.6284 0.1498 _ , , . ~~ "0.0107" 68. An indebtedness of $1500 is to be paid in installments of $15 p*>r month, each installment to cover the interest to that date. Find the number of installments if the interest is at 8 per cent per annum. Ans. 164.5. 69. An indebtedness of $1800 is paid in installments of $5.00 per week, each payment to cover the accrued interest to that date. How many payments will be required if the interest is at 5 per cent per annum, and 52 weeks are counted as one year? Ans. 461.5. 60. Find the radius of a sphere that contains 5263 cu. ft. Ans. 10.79 ft. LOGARITHMS 391 Suggestion. Take the formula for the volume of a sphere, V -Jar 3 , and solve it for r. This gives r = -C/ 61. In the equation y=log x, find the values of y corresponding to the values 0.5, 1, 2, 5, 10, 25, 50, and 100 of x. Choose a pair of coordi- nate axes and plot these pairs of points. This will give a curve that shows the relation between a number and its logarithm. (It will be best to choose a unit on the y-axis 5 or 10 times the length. of the unit on the z-axis.) 62. If the mixture in a gas engine expands without gain or loss of heat, it is found that the law of expansion is given by the equation pv 1 - 37 = C. Given that p = 188.2 when v = ll, find the value of C, then plot the curve of the equation using this value of C. This curve shows the pressure at any volume as the gas expands. Consider values of v from 11 to 23. Solution. Given pv 1 - 37 = C. Substituting p = 188.2 and w = ll, gives C = 188.2 Xll 1 - 37 . Computing by logarithms, log 188.2 = 2.2747 1.37 log 11 = 1.4267 log C = 3.7014 C = 5028. The formula is then pz; 1 - 37 = 5028 Now choose, say, six values of v, as given in the table, from 11 to 23 and compute the corresponding values of p. V 11 13 15 17 20 23 p. . 188.2 149.7 123.1 103.7 82.98 68.53 These values are plotted in Fig. 221 and a smooth curve is drawn through the points. From this curve any value of p corresponding to values of v from 11 to 23 can be read. Such curves when accurately plotted are of great value in engineering. 63. D is the diameter of a wrought-iron shaft to transmit an indicated horse-power H at N revolutions per minute. Given D= \~~\r > plot a curve showing the relation between D and H, from H = 10 to H = 80, when N is 100 revolutions per minute. From your curve find the diam- eters for horse-powers 35, 47, and 72. Remark. Many of the following exercises can be worked without the use of logarithms. Some will review ideas previously discussed. Use logarithms only where they are more convenient. 392 PRACTICAL MATHEMATICS 64. The tractive power of a locomotive is found by the formula: T DPtL w ' where D = the diameter of the cylinder in inches, P=the mean pressure of the steam in the cylinder in pound* per square inch, L = the length of the piston stroke in inches, W = the diameter of the driving-wheel in inches, and T = the tractive force upon the rails in pounds. r 200 150 100 CO 40 30 20 10 V \ \ \ \ ^ \ \ \ ' \ 5 \ . N \ ^ \ N ^ X \ V - 1 2 3 4 5 C 78 9 10 11 12 13 14 15 16 17 18 19 20 a 22 23 24 25 26 27 Values of v Fio. 221. Find the tractive force for the following data: (1) Z) = 16in., P = 901b., L = 45in., JT = 78in.; (2) Z) = 20 in., P = 100 lb., L=54in., T7 = 84 in. Arts. (1) 74,770; (2) 128,570. 66. To find the weight which a column of soft steel, with square bear- ings, will support per square inch of cross section we use the following formula: P = ~~(i2LF ^SoOOOr'' LOGARITHMS 393 where P is the weight in pounds; L, the length in feet; and r, the radius L of gyration in inches. Find P if - =5. Ans. 40,910 Ib. r 45000 45000 45000 Suggestion. P = 1 i (12L)' 144 /L\ * 144X5^ "T" oijrw-w-i_9 * "T" ocrvnn I / ' 36000r 2 ' 36000 \r/ 36000 d 66. In a column which is solid and circular in cross section. r = i 4 where d is the diameter of the cross section in inches. Find the weight such a column of soft steel will support per square inch of cross section, if L = 16 ft. and d = 6 in. Find the total weight this column will support. Ans. 30,920 Ib. per sq. in. ; 874,400 Ib. \/d 2 -\-di 2 67. In a hollow cylindrical column. r = ' where d and di are 4 the outer and inner diameters of the column respectively in inches. Find the weight the column in the preceding exercise will support per square inch of cross section and the total, if it is hollow and the shell is 1 in. thick. Ans. 34,220 Ib. per sq. in. ; 537,500 Ib. 68. For a column whose cross section is a regular hexagon, r = 0.264d, where d is the diameter of the circle inscribed in the hexagon. Find the weight a solid hexagonal pillar of soft steel and square bearings will support, if the length is 12 ft. and the edge of the base is 2 in. Ans. 276,900 Ib. Note. The safe load for the above columns is one-fourth or one-fifth of the value given. The formula used is Gordon's formula. For medium steel the 45,000 of the formula is changed to 50,000. 69. In any class of turbine, let P be the power of the waterfall; H, the height of the fall; and n. the rate of revolution. It is known that for a particular class of turbines of all sizes, n w = i s = ws is tw 71. Using the formula of the preceding exercise, find t when H = 50, w = 10 in,, and s =4000 ft. per minute. Ans. 41.25 Ib. 394 PRACTICAL MATHEMATICS 72. To determine the elevation of the outer rail in a curve in a railroad track, the following formula is used: where e = the elevation in feet, G = gage of the track in feet, F=* velocity of train in feet per second, and 72 = radius of curvature of the curve in feet. Find e for the following data, if G is 4 ft. 8i in., the standard gage (1) 72 = 5730 ft., (a) 7 = 20 mi. per hr., (b) F = 50 mi. per hr. (2) 72 = 2865 ft., (a) V = 15 mi. per hr., (b) F=40mi. per hr. (3) 72 = 716.8 ft., (a) F = 25 mi. per hr., (b) 7 = 50 mi. per hr. Ans. (1) (a) 0.02ft., (b) 0.14ft. (2) (a) 0.02ft., (b) 0.18ft. (3) (a) 0.27 ft., (b) 1.10 ft. 73. In long water pipes, when the diameter and length of the pipes are constant, that is, do not change, the amount of discharge varies as the square root of the head. How many times must the head H be in- creased to double the amount of discharge G? To make the discharge five times as much? Ans. 4; 25. Definition. The head is the distance the source of supply is above the point of discharge. 74. If the pipe is of such length and diameter that the discharge is 20 gallons per minute, what will it be if the head is doubled? Ans. 28.28+ gallons. 75. If when the head is 10 ft., the discharge through a certain pipe is 50 gallons per minute, what must be the head so that the same pipe may discharge 210 barrels per hour? Ans. 48. l>2 ft. 76. In long water pipes, when the lengths of the pipes are the same and when the head does not change, the amount of discharge varies directly as the square root of the fifth power of the diameter. Using D and d for the diameters, and G and g for the amounts of discharge, write this relation in the form of a proportion. Ans. G :0 = \/7J* :\/d*. 77. Write the above relation in the variation form both with and without the constant. What does the constant include in it? Would there be a different constant for each length of pipe? For each head? Ans. G = K\/D*', G a^/D*. 78. If the length and the head remain constant, what change in the discharge will be caused by a change in diameter from 3 in. to 4 in.? Ana. 2.05 times, nearly. Suggestion. Increased in the ratio -\/3* :\/4*. 79. A 3-in. pipe 100 ft. in length with a certain head discharges 110 gallons per minute; find the discharge from a 5-in. pipe of the same length and head. Ans. 394.5 gal. per minute. Suggestion. 1 1 : x \/3* ' \/5. 80. How many 1-in. pipes will it take to discharge the same amount LOGARITHMS 395 as one 6-in. pipe? Here we are considering long pipes so use N= where N = the number of small pipes, and D and d are the diameters of the large and the small pipes respectively. (See exercise 9, page 157.) Ans, 88.2. 81. In long water pipes, when the discharge and the length are con- stant, the head will be inversely as the fifth power of the diameter. Us- ing H and h for heads, and D and d for the diameters respectively, write this relation in the proportion form. Ans. H :h=d b :Z) 6 . 82. With a head of 4. 1 ft. and a length of 100 ft. a 3-in. pipe will dis- charge 95.4 gallons per minute; find the head so that a 2-in. pipe of the same length will discharge an equal amount. Ans. 31.13 ft. 83. Using H and D for head and diameter respectively, write the rela- tion given in exercise 81 in the variation form. Using this formula solve jf exercise 82. Ans. H = fz' 84. In long water pipes when the head and the diameter are constant the discharge will be inversely as the square roots of the lengths. Using G for discharge and L for length, state this in the variation form. In the proportion form. K ~ /- , Ans. G = ~7'> G'-a^Vl :\/L. 86. In a long water pipe of certain diameter and head, the discharge is 2000 gallons per minute. How many gallons will be discharged per minute under the same head and the same size of pipe if the length is doubled? If six times as long? Ans. 1414 gal.; 816.5 gal. 86. In long water pipes, when the discharge and the diameter are constant, the head varies directly as the length. Using the letters given in the preceding exercises, state this in the form of a proportion, and in the variation form. Ans. H:h=L:l; H KL. 87. The square of the initial velocity of a projectile in feet per second varies as the number of pounds of powder in the charge and inversely as the weight of the projectile. If 5 Ib. of a certain kind of powder will give a projectile weighing 10 Ib. an initial velocity of 1850 ft. per second, how great a velocity will 50 Ib. of powder give an 80-lb. projectile? Ans. 2068 ft. per second. Solution. Let p number of pounds of powder in charge, w= weight of projectile, and v = velocity in feet per second. Then t' 2 = fc-. w When p = 5 and w = 10, v = 1850. = fcjj, and & = When p = 50 and w = 80, 88. Using the same quality of powder as in the last, find the charge necessary to give a 1200-lb. projectile an initial velocity of 2100 ft. per second. Ans. 773 Ib. TRIGONOMETRY CHAPTER XXXV INTRODUCTION, ANGLES 316. Introductory. Each advance step in mathematics is an attempt to do something more easily than it could have been done before, or to accomplish something that was before impossible. We have seen that many problems could be worked more easily by algebra than by arithmetic, and that many other problems could be solved by algebra that could not be solved by methods of arithmetic. It was found that the area of a segment of a circle could not be obtained by geometry except in a few special cases; by methods of trigonometry, this area can be found in all cases where there are sufficient facts to do it by any means. By geometry, one side of a right triangle can be found if the other two sides are known; but there is no way by geometry of finding the acute angles when only the sides are known. By trigonometry, the angles as well as the sides can be found. Many such illustrations could be given in which trigonometry is a more powerful tool than either algebra or geometry. Trigonometry is based upon geometry, but makes use of the methods and machinery of algebra. While trigonometry can be applied at once to the solution of various practical problems, it is also of great assistance in other branches of mathematics. In the following chapters will l>e given some of the direct applications of the subject. 317. Angles. The definition of an angle as given in Art. 90 admits of a clear conception of small positive angles only. In trigonometry we wish to deal with negative as well as posi- tive angles, and these of any size whatever. We, therefore, need a more comprehensive definition of an angle. If a line is turned about a fixed point in the line and kept in the same plane, it is said to generate or sweep out an angle. 396 INTRODUCTION, ANGLES 397 The hand of a clock may be thought of as the line that is revolving and generating the angle. The size of the angle is determined by the amount of turning made by the line. If the line turns in a counter-clockwise direction, that is, opposite in direction to the hands of a clock, the angle described Initial Side y Initial Side FIG. 222. is called a positive angle. If the line turns in a clockwise direction, the angle described is called a negative angle. The position of the line at the start is called the initial line or side, and the final position is called the terminal line or side. A circular arrow drawn between the two lines and having its head in the terminal line shows the direction of turning and the size of the angle. 00 FIG. 223. In Fig. 222 (a), the line OX is imagined pinned at O and turning in a counter-clockwise direction to the position OP. The angle described is positive, and is read angle XOP, Notice that the initial line is read first. In Fig. 222(6), the line OX is thought of as turning in a clockwise direction, and so describing the negative angle XOP. PRACTICAL MATHEMATICS D It is evident that the idea of an angle given in this article allows it to be of any value whatever, positive or negative. Thus, an angle of 467 is one complete turn and 107. It is shown in Fig. 223(a). An angle of 229 is a turn of 229 in the negative, clockwise, direc- tion. It is shown in Fig. 223(6). An angle of 720 is two complete turns of the initial line. An angle of 3760 is ten complete turns and 160 more. 318. Location of angles, quadrants. For convenience in locating the angles, the agreement is made as for plotting (see Art. 286). Two lines, X'X &nd F'FofFig. 224, are drawn at right angles to each other. The directions of the lines and the location of the quadrants are as in the article referred to. If the positive direction of the x-axis is taken as the initial side, the angle is said to be in the first quadrant if its terminal side lies between OX and OF. It does not matter how many turns are made. Thus the angles of 40, 400, 760 all lie in the first quadrant. Similarly, if the terminal side lies between OF and OX', the angle is said to be in the second quadrant. If the terminal side lies between OX' and OF', the angle is said to be in the third quadrant. If the terminal side lies between OF' and OX, the angle is said to be in the fourth quadrant. Thus, angle XOA is in the first quadrant. Angle XOB is in the second quadrant. Angle XOC is in the third quadrant. Angle XOD is in the fourth quadrant. If the terminal side falls on OX, OY, OX', or OY' the angle is said to lie between two quadrants. 319. Measurement of angles. In Chap. XJJI, the three units for measuring angles are discussed. These units are the right angle, the degree, and the radian. INTRODUCTION, ANGLES 399 By definition, the radian is an angle of such size that when placed at the center of a circle its sides intercept an arc equal to the radius. It is found in the chapter referred to that 2-n- radians are measured by a whole circumference. This is illustrated in Fig. 225. From the definition it is evident that the size of the radian does not depend on the size of the circle. As before 1 radian = 57.29578- = 57 17' 44.8", and 1 = 0.017453 + radians. The measurement of an angle by the radian unit is often called circular measure or r-measure. 7T Since 2r radians = 360, -n radians == 180, ~ radians = 90, 7T ~ radians = 60, etc., it is often convenient to represent some o of the most frequently used angles by means of TT. In using circular measure, the word radian is usually omitted. 7T 7T Thus, we write TT, ~' T 3, 0.5, meaning in each case so many radians. To convert radians to degrees, multiply the number of radians by 1 on -or 57.29578-. 7T To convert degrees to radians, multiply the number of degrees by ~ or 0.017453 + . Example 1. Reduce 2.5 radians to degrees, minutes, and seconds. Solution. 1 radian = 57.29578, /. 2.5 radians = 2.5X57.29578 = 143.2394. To find the number of minutes, multiply the decimal part 3f the number of degrees by 60, .'. 0.2394 = 60X0.2394 = 14.364'. Likewise, 0.364' = 60 X0.364 = 21.8", .'. 2,5 radians = 143 14' 22". 400 PRACTICAL MATHEMATICS Example 2. Reduce 22 36' 30" to radians. Solution. First, change to degrees and decimals of degrees, 22 36' 30" = 22.6083+. 1 = 0.01 7453+ radians, /. 22.6083+ =22.6083X0.017453 = 0.3946- radians. 320. Relations between angle, arc, and radius. From the definition of the radian, it is evident that the number of radians in an angle at the center of a circle can be found by dividing the length of the arc its sides intercept by the length of the radius. arc That is, number of radians in angle = '- radius In Fig. 226, angle AOB (in radians) = -. pr-r- radius OA If 6 (the Greek letter theta) stands for the number of radians in an angle, r for the length of the radius of a circle, having its center at the vertex of the angle, and s for the length of the arc between the sides of the angle; then, Solving, first for s and then for r, r=s + d. These relations are important as they may be used in solving many practical problems. Example. The diameter of a gradu- ated circumference is 10 ft., and the graduations are 5 minutes of arc apart; find the distance, length of arc, between the graduations in fractions of an inch to three decimal places. Solution. By formula, s = rO. From tho example, r = 12X5 = 60in., and B = 0.01745XV% = 0.00145+. Substituting in the formula, s = 60X0.00145+ =0.087+. .'. length of 5' arc is 0.087+ in. 321. Railroad curves. In the United States it is customary to express the curvature of railroad tracks in degrees. The INTRODUCTION, ANGLES 401 degree of a curve is determined by the central angle which is subtended by a chord of 100 ft. Thus, in a circle, a 5-degree curve is one in which a 100-ft. chord subtends a central angle of 5 degrees. In curves commonly used, the error is slight if the arc is taken in place of the chord. Then, assuming that 1 radian = 57.30, the radius of a 1-degree curve is found by the formula for r of Art. 320. Thus, 57. oO Hence 1 degree of curvature gives a radius of 5730 ft. It follows that 5730 divided by the number of degrees in the curve gives the radius of the curve ; and 5730 divided by the number of feet in the radius gives the number of degrees in the curve. EXERCISES 111 1. In what quadrant is each of the following angles: 27, 436, 236, 4372, -46, -324, -90, -476, -2342, |TT, 3x, |TT, 4.3, 78.5, 82.3? Draw each angle. 2. Draw the terminal side of 237, and give value of a negative angle having the same terminal side. How many such angles are there? 3. Draw the following angles: 76, 25.6, 425, 5263, -25, -236, -146, -935. 4. Can a positive and negative angle each have its terminal side in the same position? Illustrate. 5. Express the following angles as some number of times x radians: 30, 45, 54, 60, 81, 90, 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360, 540, 720. . 7TTr3j7r7r97r:r27r3x5ir 7x STT 4x Sir Sir 7ir 6' 4' 10' 3' 20' 2' "3"' T' "6~' *"' IF' T' ~3' ~2' ~3' T' lllr O Q A ~7r~, Zir, OTT, 4ir. 6. Reduce the following to radians: (a) 47, (f) 135, (b) 75 30', (g) 120, (c) 16 43' 10", (h) 175 45' 40", (d) 125 46' 30", (i) 95 10' 10", (e) 62 40', (j) 127 41' 50". Ans. (a) .820 + ; (b) 1.318 -; (c) .292-; (d) 2.195 + ; (e) 1.094 -; (f) 2.356 + ; (g) 2.094 + ; (h) 3.068-; (i) 1.661 + ; (j) 2.229-. 26 402 PRACTICAL MATHEMATICS 7. Reduce the following (1) to degrees and decimals of degrees to four places, (2) to degrees, minutes, and seconds: (a) J, (d) 4.23, (g) 0.125, (b) fr, (e) 2.76, (h) 2.236, (c) ST, (f) UT, (i) 3.14159. An?., (a) 45. (d) 242 21' 40". (g) 79' 43". (b) 135. (e) 158 8' 11". (h) 128 6' 48". (c) 150. (f) 191 15" (i) 180. 8. How many radians are in each of the angles of a right triangle, if one of the acute angles is 36 47'? Ana. \v; .642-; .929-. 9. How many degrees in each of the angles of an isosceles triangle, if the angle at the vertex is \ir radians? Ana. 30, 75. 10. Two of the angles of a triangle are respectively f and of a radian. Find the number of radians and degrees in the third angle. Ans. 2.0749 radians = 118 52' 59". 11. If an angle of 126 at the center has an arc of 226 ft., find the radius of the circle. Solution. Use the formula r =8 + 0. = 126X0.017453=2.199. r = 226X2. 199 = 102.77. .'.radius is 102.77 ft. 12. A flywheel 20 ft. in diameter has an angular velocity of 3r per second. Find the rim velocity. Ans. 94.25 ft. per sec. 13. The circumferential speed generally advised by makers of emery wheels is 5500 ft. per minute. Find the angular velocity per second in radians for a 10-in. wheel. Ans. 220 radians per sec. Suggestion. Use the formula = s-t-r. 14. Solve similar problems for the velocities of the following: (a) Ohio grindstones, advised speed 2500 ft. per minute. (b) Huron grindstones, advised speed 3500 ft. per minute. (c) Wood, leather covered, polishing wheels, 7000 ft. per minute. (d) Walrus hide polishing wheels, 8000 ft. per minute. (e) Rag wheels, 7000 ft. per minute. (f) Hair brush wheels, 12,000 ft. per minute. 15. A flywheel of 4-ft. radius is revolving counter-clockwise with a circumferential velocity of 75 ft. per second. Find the angular velocity in radians per second. Ans. 18} radians. Solution. 16. A train is traveling on a curve of half a mile radius at the rate of 30 miles per hour. Through what angle does it go in 15 seconds? Ex- press the answer in both radians and degrees. Ana. 0.25 radian = 14 19' 26". Suggestion. Use the formula 6 =s -i-r, where s = i mile and r = J mile. INTRODUCTION, ANGLES 403 17. Find the radius of a circle in which an arc of 20 ft. measures an angle of 2.3 radians at the center. In this circle, find the angle at the center measured by an arc of 3 ft. 8 in. Ans. 8.70- ft.; 0.421+ radians. 18. Find the angular velocity per minute of the minute hand of a watch. Express in degrees and radians. Ans. 6 =0.1047+ radians. 19. A train of cars is going at the rate of 15 miles per hour on a curve of 600-ft. radius; find its angular velocity in radians per minute. Ans. 2.2 radians. 20. A flywheel 22 ft. in diameter is revolving with an angular velocity of 9 radians per second. Find the rate per minute a point on the cir- cumference is traveling. Ans. 5940 ft. per min. 21. Find the length of arc which, at the distance of 1 mile, will subtend an angle of 10' at the eye. An angle of 1". Ans. 15.36 ft. ; 0.0256 ft. 22. The radius of the earth's orbit, which is about 92,700,000 miles, subtends at the star Sirius an angle of about 0.4". Find the approximate distance of Sirius from the earth. Ans. 478 X 10 11 miles. 23. What radius has a 5-degree curve in a railroad track? A curve of 3 15'? If the radius of curvature is 4550 ft., what is the degree of curve? Ans. 1146 ft.; 1763+ ft.; H nearly. CHAPTER XXXVI TRIGONOMETRIC FUNCTIONS 322. Sine, cosine, and tangent of an acute angle. In geometry we learn of certain relations between the sides of a triangle; here we find that the angles are related to the sides in a certain way. These relations are very useful in the con- A struction of angles and in S^. solving various triangles and other figures. De^ruYt'ons. If an acute angle XOA, Fig. 227, is taken, and from P, any point X in OA, a perpendicular QP is ^ drawn to OX, a right triangle QOP is formed. It is evident because of similar triangles (see Art. 104) that the following ratios will not change no matter where the perpendicular QP may be drawn, so long as the angle XOA, or 6, does not change. QP The ratio ^p is called the sine of angle 6, written sin 6. The ratio ~p is called the cosine of angle 8, written cos 6. QP The ratio ~~ is called the tangent of angle 6, written tan 6. These lines can be measured, say, in inches, and numerical values of these ratios can then be found. 323. Ratios for any angles. The same ratios may be written for an angle in any quadrant. Thus, in Fig. 228, angle XOA is in the second quadrant. QP is the perpendicular drawn from any point in the terminal side to the z-axis. The ratios are sin XOA = -jtTp* cos XOA = pcp> tan XOA = ^ 404 TRIGONOMETRIC FUNCTIONS 405 As an exercise the student may write these ratios for the angles XOB and XOC. As will be seen in the next article, some of the ratios are positive and some are negative numbers. 324. General form for ratios. Let the distance from along the terminal side to the point chosen be called the distance, represented by r, and always considered positive. The length of the perpendicular to the z-axis is the ordinate IV s FIG of the point, positive if extending above and negative if below the x-axis, and is represented by y. The distance from O to the foot of the perpendicular is the abscissa of the point, positive if extending to the right of the origin and negative tor. PRACTICAL MATHEMATKs if to the left, and is represented by x. The ratios for any angle (see Fig. 229) may then be written as follows: ordinate y sin 6= = - distance r abscissa x cos 6= .. = distance r ordinate y tan 6 = , abscissa x The reciprocals of these ratios are often used and are named as follows: cosecant 6, abbreviated esc 6 = secant 6, abbreviated sec 6 = cotangent 6, abbreviated cot 6 = 1 = r sin 8 y 1 = r cos 0~x' 1 x tan0 y These six ratios are called trigonometric functions. They are of the greatest importance in trigonometry and must be learned so that they can be given at any time without hesitation. f A / B Fio. 230. 325. Acute angle in a right triangle. In the right triangle ABC, Fig. 230, sin A=- = cos B, cos A=- = sin B, tan A = c c Of O C C , = cot B, cot A = - = tan B, sec A = T = esc B. esc A = - =- sec B. b a b a These ratios for angle B will readily be seen if the triangle is placed in the position of Fig. 230(6). In working with right triangles, the following forms of the TRIGONOMETRIC FUNCTIONS 407 definitions of the trigonometric functions will be found con- venient for either acute angle: side opposite sin A (or sin J5) = cos A (or cos Z?) = hypotenuse side adjacent hypotenuse D , side opposite tan A (or tan B) -f-=- ,. > side adjacent hypotenuse esc A (or esc B) = -~- r > side opposite ^ hypotenuse sec A (or sec B) -^4 side adjacent , _,. side adjacent cot A (or cot B) = -rj- r side opposite A B As an exercise give rapidly the functions of the acute angles of the right triangles shown in Fig. 231. 326. Relation between the functions of an angle and the functions of its complement. In the right triangle, angle A + angle 5 = 90. That is, angle A and angle B are com- plements of each other. (See Art. 90.) In the previous article it is seen that sin A = cos B ; cos A = sin B ; tan A = cot B; etc. From this, we see FlG< 232 - that any function of an angle is equal to the co-function of the complement of the angle. For example, cos 30== sin 60, sec 22 = esc 68. 10S PRACTICAL MATHEMATICS 327. Trigonometric functions by construction and measure- ment. Let it be required to find the sine, cosine, and tangent of 40. Draw an angle XOA=40, Fig. 232. Take a convenient distance OP on the terminal side and draw the perpendicular QP. Then by measuring the lines we have the following: tan --- 84 ~OQ~.96~ The measurements are made in inches. As an exercise draw the angles and fill out the following table. Carry the results to two places of decimals. Angle tan Angle tan 10 50 15 55 20 60 25 65 30 70 35 75 40 80 45 85 ^ In the above table, compare the sine and cosine of 10 and 80, 15 and 75, 20 and 70, etc. As in Art. 326, it will be noticed that any function of any angle is the co-function of the complement of that angle. It follows that any function of an angle larger than 45 is a function of some angle that is less than 45. If then a table is made for the functions of all TRIGONOMETRIC FUNCTIONS 409 the angles from to 45, it can be used for finding the func- tions of the angles from 45 to 90 as well. Table XI is arranged in exactly this way. It includes the angles for every 10' from to 90. 328. Use of functions in constructing angles. Example 1. Construct an angle of 40. Construction, sin 40 = 0.64. By the help of this, the angle may be drawn as follows: Draw a straight line AB, Fig. 233. At some point Q erect a perpendicular QP 0.64 in. in length. With P as a center draw an arc with a radius of 1 in. cutting AB at 0. Draw OP. Angle QOP is 40, for sin QOP = 0.64 = sin 40. In constructing an angle of 40 all that is necessary is to make the sides QP and OP of such lengths that the ratio shall be 0.64. For this 1.28 in. and 2 in. could be used conveniently. Usually it is most convenient to make one of the sides unity. In making the construction, some other function of 40 could be used as well as the sine. Example 2. Construct an angle of 35 by using tan 35. Construction, tan 35 = 0.70. Draw a straight line MN, Fig. 234. At some point Q erect a perpendicular QP to MN 0.70 in. in length. Locate 0, making OQ=l in. Draw OP. Angle QOP is 35, for tan QOP = 0.70 = tan 35. EXERCISES 112 1. Construct by the use of sines, angles of 25, 65, 47, 53 20', and 25 30'. Suggestion. Use data from the table of Art. 327 or from Table XI. 2. Construct by use of cosines, angles of 20, 40, 17 20', and 67 40'. 3. Construct an angle 6 whose tangent is 2. Find the other functions of this angle. Ans. sin = .89; cos = .45; cot = .50; sec 6 =2.24; esc = 1.12. 410 PRACTICAL MATHEMATICS 4. Construct an angle whose cosine is .8. Find the other functions of this angle. Ans. sin 0-.6; tan 0-.75; cot 0-1.33; sec 0-1.25; esc 0-1.67. 5. Construct when (a) sec 0-3; (b) cot 0-5; (c) esc 0-2; (d) sin = .3. 329. Values of functions by computa- tion. From our knowledge of geometry we know the relations between the sides and the angles of right triangles when the 1 " acute angles are 30, 45, or 60. Thus, when the acute angles are 45 each, the two legs are equal; and if they are given the hypotenuse can be found. If one acute angle is 30 and the other 60, the shorter leg is one half the hypotenuse. (1) The 45 angle. Draw a right triangle ABC, Fig. 235, with the angle 4=45. Then the angle 5 = 45. Let ' = land (75 = 1. Then 45 = VlHT 2 = A/2. ' '-,= 0.707, sin 45 = cos 45 = V2 1 V2 tan 45 = 1. (2) The 60 and 30 angles. Draw a right triangle with the acute angles 60 and 30 respectively, Fig. 236. Let AC = 1, then AB = 2 and C5 = \/3- The functions can now readily be written: sin 60 = ^? = sin 30 = cos 60 = tan 60 = tan --- V3 330. Angles in other quadrants. In the second quadrant, the angle XOA = 120, Fig. 237, has the same numerical ratios between the abscissa, ordinate, and distance of the point P TRIGONOMETRIC FUNCTIONS 411 in the terminal side as the ratios between the abscissa, ordinate, and distance of the point P' in the terminal side of the angle ' = 60 in the first quadrant. Thus, using the values given in the figure, sin 120 = ~~ and sin 60 =~, cos 120 = - i and cos 60 = , tan 120 = - \/3 and tan 60 = Vs. y B FIG. 237. There are angles in the second, third, and fourth quadrants whose functions are connected in this way with the angles 30, 45, and 60 in the first quadrant. What are they? The functions of these angles can easily be written by remembering (1) that the distance is always positive; (2) that the ordinate is positive in the first and second quadrants and negative in the third and fourth; (3) that the abscissa is posi- tive in the first and fourth quadrants and negative in the second and third. (See Art. 286.) 331. Angles of 90, 180, 270, and 0. For an angle of 90, the ordinate equals the distance and the abscissa is zero. The functions are as follows : sin 90 =^ = 1, esc 90 = - = 1, r y cos 90 = - = 0, sec 90 = -= co r X tan 90 --=0, cot 90 = - = 0. X y 412 PR A CTICAL MA TH EM A TICS The symbol is read infinity, and here means that as the angle increases to 90, the tangent and the secant increase without limit. That the values of the functions are as given will readily be seen if one draws an angle XOP, Fig. 238, nearly equal to 90, and considers the values of the ratios as the angle changes to 90. For an angle of 180, the ordinate is zero and the abscissa and distance are equal. The functions are as follows: sin 180 = -=0, r cos 180 = *= -1, tan 180 = ^ = 0, esc 180= =co y sec 180 = -= -1. x cot 180 = -=. y These values may be found by considering angle XOR, Fig. 238. Y Fio. 238. Similarly the functions for 270 and are as follows: esc 270 sec 270 cot 270 C cscO secO cotO = 0, = 00, 1 A > sin 270 =-1, cos 270 =0, tan 270 = oo , sin =0, cos = 1, tanO =0, 332. The sine, cosine, and tangent of each of the angles mentioned in the previous articles are arranged in the follow- ing table. The student should carefully verify each result by drawing a figure and computing the ratios. Also express the TRIGONOMETRIC FUNCTIONS 413 results in decimals and compare with the values given in Table XI. Deg. Rad. sin cos tan Deg. Rad. sin cos tan 77T V3 1 1 210 IT -\ V3 30 7T 6 i V'3 2 1 V3 225 57T 4 1 ~V2 1 ~V2 1 45 IT 4 1 V2 1 V2 1 240 47T 3 V3 2 ~ 2 V/3 60 7T 3 IT 1 V3 270 2 -1 00 90 X 2 1 oo 300 STT 3 V3 2 -2 -V3 120 27T 3 V3 2 -1 -Vs 315 77T 4 1 1 V2 -1 135 STT 4 1 V2 1 V2 -1 330 UTT 6 i 2 V3 2 1 150 STT 6 ? V3 2 1 360 27T 1 180 IT -1 CHAPTER XXXVII TABLES AND THEIR USES 333. Nature of trigonometric functions. From what has been done with the trigonometric ratios, it is seen that they are abstract numbers and, in general, cannot be expressed exactly as decimals. For convenience in computing, these ratios are arranged in tables somewhat similar to the tables of logarithms. The ratios in these tables may be carried to any number of deci- mal places. The larger the number of decimal places, the more nearly accurate the computations with the tables will be. 334. Table of functions. In Table XI are arranged the natural and logarithmic functions of angles for every KX from to 90. The logarithms have 10 added when the characteristics are negative to avoid the writing of negative signs in the table. These logarithms are simply the logarithms of the natural functions that are in the adjoining columns, and are placed here for convenience. Since, as stated in Art. 327, each acute angle above 45 has as a function the co-function of an angle less than 45, each number in the table serves as the function of two different angles whose sum is 90. The angles less than 45 are found at the left of the page, and the names of the functions at the top of the page. The angles greater than 45 are found at the right of the page, and the names of the functions at the bottom of the page. 335. To find the function of an angle from the table. To find the function of an angle from the table we proceed much the same as with the table of logarithms. It can best be illustrated by examples. (A) When the angle is given in the table. Example 1. Find the tangent of 23 20'. Find the angle of 23 20' at the left of the page and read 0.4314 in the column headed natural tangent, /. tan 23 20' = 0.4314. 414 TABLES AND THEIR USES 415 Example 2. Find the cosine of 86 40'. Find 86 40' at the right of the page and read 0.0581 in the column with natural cosine at the bottom, .'. cos 86 40' = 0.0581. (B) When the angle is not given in the table. Example 3. Find sin 17 27'. Find sin 17 20' = 0.2979. Find the tabular difference between 0.2979 and the next ratio below. This difference is 0.0028. Since this difference is for 10', the difference for 7' is 0.7X0.0028 =0.0020, .*. sin 17 27' = 0.2979+0.0020 = 0.2999. Note that the interpolation is very similar to that in loga- rithms, and gives approximate results only. Example 4. Find tan 69 43.6'. Find tan 69 40' = 2.6985. Tabular difference for 10' = 0.0243. Difference for 3.6' = 0.0243X0.36 = 0.0087. .'. tan 69 43.6' = 2.7072. Example 5. Find cos 37 57.3'. Find cos 37 50' = 0.7898. Tabular difference for 10' = 0.0018. Difference for 7.3' = 0.0018X0.73 = 0.0013. .*. cos 37 57.3' = 0.7898 -0.0013 = 0.7885. It is to be noted that a subtraction is to be made when inter- polating in finding a cosine or a cotangent of an angle; while in finding a sine or a tangent, an addition is performed. This is because as the angle increases from to 90 the sine and the tangent increase, but the cosine and the cotangent decrease. 336. To find the angle corresponding to a function. (A) When the function is given in the table. Example 1. Find x if sin x = 0.2728. Find 0.2728 in the column labeled natural sine and read 15 50' in the column labeled angle. .'. x = 15 50'. 416 PRACTICAL MATHEMATICS (B) When the function is not given in the table. Example 2. Find x if tan x = 1.5725. In the column labeled natural tangent, find the ratio nearest 1.5725 and smaller. This is 1.5697 and is the tangent of 57 30'. The tabular difference is 0.0101. The difference be- tween 1.5697 and 1.5725 is 0.0028. Since a difference of 10' gives a difference in the ratio of 0.0101, it will take a difference of as many minutes to give a difference in the ratio of 0.002$ /. x = 57 30'+2.8' = 57 32.8'. Note again how similar the interpolating is to that in logarithms. Example 3. Find x if cos x = 0.7396. Since the cosine decreases as the angle increases, find the cosine nearest to .7396 but larger. This is cos 42 10' = 0.7412. Tabular difference =0.0020. Difference of 0.7412-0.7396 = 0.0016. ' = 8'. z = 42 10'+8'=4218'. Example 4. Find x if log sin x = 9.3762 - 10. From table log sin 13 40' = 9.3734. 9.3762-9.3734 = 0.0028. Tabular difference = 0.0052. MX 10' = 5.4'. ;. z = 1340'+5.4' = 13 45.4'. EXERCISES 113 1. Find the sine, cosine, and tangent of 40 10', 59 50', 76 30', and 5 40'. 2. Find the sine, cosine, and tangent of (a) 17 36', (b) 29 29', (c) 76 14', (d) 83 33', (e) 63 47'. Ana. (a) 0.3024 0.9532 (b) 0.3172 0.4922 0.8705 (c) 0.5654 0.9713 [ 0.9937 0.2380 (d) | 0.1123 (e) 4.0817 8.8468 0.8971 2.4418 2.0308 3. Find the angles having the following as sines: 0.5807, 0.2725, 0.4986, 0.9127, 0.0276. Ans. 35 30', 15 48.9', 29 54.4', 65 52.7', 1 34.8'. 4. Find the angles having cosines as follows: 0.3764, 0.8642, 0.9091, 0.4848, 0.0986. Ans. 67 53.3', 30 12.7', 24 37.5', 61, 84 20.5. TABLES AND THEIR USES 417 6. Find the angles having tangents as follows: 0.2256, 1.7624, 2.8427, 0.1111, 3, 0.6666. Ans. 12 42.9', 60 25.7', 70 37.1', 6 20.3', 71 33.9', 33 41.2'. 6. Find sin 34 40' and find the logarithm of this result from Table X. Find log sin 34 40' from Table XI and compare results. 7. Verify the following by the tables: log sin 56 35' = 9.9215; log tan 34 15.6' = 9.8332; log cos 27 55' =9.9462; log cos 19 53.4' =9. 9733; log sin 17 9' = 9.4696; log tan 75 56.8' =0.6015. 8. Find x in each of the following: (a) log cos z = 9.8236; (c) log tan x = 0.4293; (b) log sin x = 9.4737; (d) log cot x = 9.4236. Ans. (a) 48 13.6'; (b) 17 19'; (c) 69 35.3'; (d) 75 8.8'. 337. Evaluation of formulas. Formulas in various lines of work often contain trigonometric functions. As with other formulas, these can usually be evaluated with or without logarithms. Since logarithms are a very convenient and useful tool, they should be used whenever they can be used to advantage. To indicate the power of a trigonometric function the ex- ponent is placed before the symbol for the angle. Thus, sin 2 30 means the square of sin 30. The logarithms of trigonometric functions are found in Table XI and those of numbers in Table X. Example 1. Find \Xsin 47+tan 3 36. Solution. sin 47 =0.7314, from Table XI. log tan 36 = 9.8613 -10. /. log tan 3 36 = 9.5839 -10. /. tan 3 36 = 0.3836, from Table X. .". sin 47 + tan 3 36 = 0.7314 +0.3836 = 1.1150. log 1.1150 = 0.0473. log \/l. 1150 = 0.0158. .'. \/l. 1150 = 1.037. .'. \Xsin 47+tan 3 36 = 1.037. Ans. Example 2. Given x = , > find the value of x to O t7 ^.\J two decimal places. In an example like this it is agreed that 69 40' shall be changed to radians to give the number to divide by. Solution, tan 72 34' = 3. 1846. 27 418 PRACTICAL MATHEMATICS 69 40' = 69$ = 69X0.01745 radians = 1.216 radians. 3.1846^-1.216 = 2.62. /. z = 2.62. Ana. Logarithms could be used in solving this example. EXERCISES 114 1. Find the value of Vsin* 49 10'. Ans. 0.6583. 2. Find the value of \/tan 75 +56. Ans. 2.266. 3. Find the numerical value of r$ (s 2 < 2 ) tan 0, where r = 25.2, a = 90, *=49.6, and tf = 31 52'. Ans. 30,140. 4. Find the value of ae-* sin (,d+0),if a = 5, 6=200,c = 600, 0= -0.1745 radians, e =2.718, and t =0.001. Ans. 1.69. Suggestion, (ct + 0) =0.6-}- (-0.1745) =0.4255 radians. 0.4255X57.2957 =24.379 =24 22.7'. sin 45 56' 20" _ 6. Given x - on/ - > find the value of x to three decimal places. oO \J Suggestion. First change 36 20' to radians. Ans. 1.133. 6. Given x = > find x to four decimal places. _ Ans. 0.8689. 7. Evaluate \/o 2 +b 2 -2a6cos C, when a = 231, 6=357, and C = 55. ^rts.293.6. 8. Evaluate q sm B * m C , when a = 126, A =30, B = 72, C = 78. sin A Ans. 234.4. 9. Find the value of sin x cos y+ cos x sin j/, when z = 42 10', and y = 17 50'. Ans. 0.866. 10. Find the value of each of the following: (a) sin 2 20+cos 2 20. (c) sin 2 40+cos 2 40. (b) sin 2 30+cos 2 30. (d) sin 2 62 30'+cos 2 62 30'. Compare the results in the above and state conclusions. 11. The velocity t; of a body sliding a distance s down a smooth plane inclined at an angle

= 27 16'. Ans. 38.3 ft. per second. 12. If the resistance of the air is disregarded, the distance along a horiz^ ntal plane that a projectile will go is given by the formula d = - , where v is the velocity at which the body is projected in feet if per second, a the angle that the initial direction makee with the hori- zontal, and d the distance along the horizontal. The value of g may be TABLES AND THEIR USES 419 taken as 32. Find d if vis 800ft. per second and a is 5. Using the same velocity, find d when a. is 20, 30, 40, and 45. Ans. 3472 ft.; 12,856 ft.; 17,320 ft.; 19,696 ft.; 20,000 ft, 13. Disregarding the resistance of the air, the highest point reached by 4j2 gl ri 2 fy a projectile is given by the formula y= ~ Find the greatest ^9 height above the starting point reached by a projectile having an initial velocity of 2000 ft. per second, and having successive values for a of 5, 10, 20, 30, 45, 60, and 90. FIG. 240. 14. The height y that a projectile is after traversing a horizontal dis- tance x, when projected with a velocity v in a direction making an angle a with the horizontal, is given by the following formula : gx 2 yxt&a a- 2v 2 COS- a Find y when a; = 1000 yd., y = 2000 ft. per second, a = 5, and = 32. Ans. 226.2ft. 15. If the resistance of the air is disregarded, the greatest horizontal distance a projectile will go is found by making the initial direction at an angle of 45 with the horizontal. Find the greatest horizontal distance that a shell having an initial velocity of 2200 ft. per second can reach. Ans. 28.6+ miles. 16. If F is the force required to move a weight W up a plane inclined to the horizontal at an angle a, and M (Greek letter mu) the coefficient of friction, then Source sin a+M cos a cos a n sin d FIG. 241. Calculate F if TF = 800 lb., a = 30, and M = 0.2. Ans. 703 Ib. 17. In computing the illumination on a surface when the surface is not perpendicular to the rays of light from a source of light, the following formula is used: / 420 PRACTICAL MATHEMATICS where /-.'the illumination at the point on the surface in foot-candles / = the luminous intensity of the source in candles, rf = the distance in feet from the source of light, = 75. (2) d when 7 = 60, E = 0.25, and

= 45. Ans. (1) 0.1294; (2) 10ft.; (3) 362. 19. What do the formulas in exercise 17 become if ^>=0? That is, if the rays are normal (perpendicular) to the surface. Source 5 Horizontal Fio. 242. 20. To compute the illumination on a horizontal surface from a source of light at a given vertical distance from the surface the following formula is used: where Eh=the illumination in foot-candles at a point on the horizontal surface, / =the luminous intensity of the source in candles, h=ihe vertical distance in feet from the horizontal surface to the source of light, v> = the angle between the incident ray and a vertical line. Solve this formula for h and /, and obtain the following formulas: 21. By means of the preceding formulas compute: (l)^when 7=250, h = 12, and ^ = 55. (2) h when / = 100, E h = 65, and ? = 12. (3) / when E h =0.85, h - 8, and * = 37. Ans. (1) 0.3276; (2) 1.2 ft.; (3) 106.8. CHAPTER XXXVIII RIGHT TRIANGLES 338. Any triangle has three sides and three angles; these are called the six elements of the triangle. The angles are usually represented by the capital letters A, B, and C; the sides by the small letters a, b, and c, the side a being opposite angle A, side b opposite angle B, and side c opposite angle C. To solve a triangle is to find the values of the remaining elements when some of them are given. 339. Solving. A triangle may be solved in two ways : (1) By constructing the triangle from the known elements, and measuring the remaining elements with the ruler and the protractor. (2) By computing the remaining elements from those that are known. The first has already been done to some extent in Chapter XIII. The second has been done for some special triangles, as the right triangle, the isosceles triangle, and the equilateral triangle, in Chapter XI, but only for some of the elements, not including the angles. By trigonometry a triangle can always be solved when the facts given are sufficient for its construction; and not only can the sides be found, but the angles also. EXERCISES 115 If A, B, and C represent the angles of a triangle, and a, b, and c re- spectively the sides opposite these angles, construct carefully the follow- ing triangles, to scale if necessary, and measure the other elements. It is important that the student should carry out these carefully for it will help him to see when there are sufficient data for solution. 1. A =40, b =2 in., c = 2.5 in., find B, C, and a. 2. .4=50, C = 70, b = 2 in., find B, a, and c. 3. A =30, a = 10 ft., c = 15 ft., find B, C, and 6. 421 422 PRACTICAL MATHEMATICS Can more than one triangle be formed from the data in 3? 4. a = 20 ft., 6 - 15 ft, c = 12 ft., find A, B, and C. 5. A =40, =80, C-60, find a, b, and c. Can more than one triangle be formed from the data in 5? The following are right triangles and C is the right angle: 6. A =29, 6=2 in., find B, a, and c. 7. A =42, a = 4 in., find B, b, and c. 8. A =47, c = 3 in., find B, a, and c. 9. a =4 in., & = 6 in., find A, B, and c. 10. a -1.5 ft., c = 2.3 ft., find A, B, and 6. 340. The right triangle. The right triangle has already been solved when any two sides are known, but the angles were not found. The previous exercises from 6 to 10 should lead us to expect that we could find the other elements when any two are given, other than the right angle, and including at least one side. Geometry will not do this but trigonometry will. It is well to recall the following facts concerning the right triangle : (1) The hypotenuse is greater than either of the other two sides, and less than their sum. (2) The square of the hypotenuse is equal to the sum of the squares of the other two sides. (3) The sum of the two acute angles is 90, that is, the acute angles are complements of each other. (4) The greater side is opposite the greater angle, and the greater angle is opposite the greater side. An inspection of the problems of construction will show that all the possible sets of two given parts for the right triangle are included in the following cases: CASE I. Given an acute angle and a side not the hypotenuse. CASE II. Given an acute angle and the hypotenuse. CASE III. Given the hypotenuse and one other side. CASE IV. Given the two sides not the hypotenuse. 341. Directions for solving. To solve a right triangle it is necessary that two elements be given, at least one of which is a side. Each equation, as sin.A=-> contains three quantities. C When two of these are given the third can be found. These equations together with the facts from geometry: (1) that RIGHT TRIANGLES 423 the square of the hypotenuse equals the sum of the squares of the other two sides; and (2) that the sum of the two acute angles equals 90, enable one to solve any right triangle. These equations may be written thus : (1) sin A= a > (2) cos A=-, c (3) tan A =j-> (4) cot A=-, a (5) c 2 = (6) sin FIG. 243. (7) cos (8) tan (9) COt B = ^, (10) .4 +5 = 90. Notice that all of these except (5) and (10) are nothing but the definitions of the trigonometric ratios. 342. Case I. Given A and b, A and a, B and a, or B and b. Example. In a right triangle ,4=32 20' and & = 10 ft., find B, a, and c. Solution. First, construct the triangle carefully; second, write equations using two of the known elements and one of the unknown in each. Equations. Construction. (1) A + = 90, this gives B. (2) v-o.s A =-> this gives c. C (3) tan A j-> this gives a. Substituting in (1), 32 2 .-. B = 90 -32 20' =57 40'. 424 PRACTICAL MATHEMATICS Substituting in (2), cos 32 20' =- c .'. c = 10-!- cos 32 20' = 10 -K8450 = 11. 83 ft. Substituting in (3), tan 32 20' = y^ .'. a= lOXtan 32 20' = 10 X .6330 = 6.33 ft. This may be checked (a) by measuring the elements in the triangle constructed; (b) by using some other equation than the ones used in solving. Thus, substitute values in c 2 = o 2 +6 2 . 11.83 2 = 6.33 2 +10 2 . 139.95 = 39.97 + 100 = 139.97, which agrees closely. This solution has been carried through with natural func- tions. Logarithms could be used to advantage in performing the multiplications and divisions. Thus, in (2) c = - A cos A .'. log c =log b log cos A. log 10 = 1.0000 log cos 32 20' = 9.9268 - 10 log c= 1.0732 .'. c=11.83+ft., which agrees with the result ob- tained before. 343. Directions for solution of triangles. (1) Construct the triangle as accurately as possible with in- struments. This gives a clear idea of the relation of the parts, and will detect any serious blunder in computation. (2) Write dawn all the equations necessary to find the elements wanted. (3) Compute the elements by natural or logarithmic functions. (4) Check the work. (5) Strive for neatness and clearness in the work. RIGHT TRIANGLES 425 344. Case II. Given A and c or B and c. Example. Given A = 67 42.8' and c = 23.47 ft. Formulas. Construction. Q0. ' B = Q(r A. B (2) win A= , .' . d = csmA. (3)cosA=-j .'. b = ccosA. c Computation by logarithms. 6742.8 By (1), B = 90 -67 42.8' = 22 17.2'. By (2), a = 23.47 sin 67 42.8'. By (3), 6 =23.47 cos 67 42.8'. log 23.47 = 1.3705 log 23.47 = 1.3705 log sin 67 42.8' = 9.9663 log cos 67 42.8' = 9.5789 b FIG. 245. log a = 1.3368 /. o = 21.72 log 6 = 0.9494 .'. 6 = 8.90 Check. Usinga 2 = c 2 -6 2 =(c+6)(c-6). log (c+6) = 1.5101 log (c-6) = 1.1635 log (c 2 -6 2 ) =2.6736 = log a 2 . 345. Case III. Given c and a or c and b. Example. Given c = 35.62 ft. and a = 23.85 ft., find 6, A, B. Formulas. Construction. (1) sin A=- '. A =sin~ l (2) cos B = (3) tan A = 426 PRACTICAL MATHEMATICS Computation. log 23.85 = 1.3775 log 23.85 = 1.3775 log 35.62 = 1.5517 log tan A = 9.9549 log sin A = 9.8258 = log cos B log b = 1 .4226 .'. A =42 2.1' and 5= 47 57.9.' .'. 6 = 26.46. Check. log (c+6) =1.7930 log (c-6) =0.9619 log (c 2 -6 2 ) =2.7549 = log a 2 . 346. Remark on inverse functions. The form yl=sin~ l - c a " is read "A = the angle whose sine is - This is a convenient C way of expressing the fact, and allows the angle symbol to stand alone. The 1 that is in the position of an exponent is not a negative exponent in meaning. The form sin" 1 - is called an inverse trigonometric function. c It is also written arcsin - and invsin These forms are also c c . . . a ,< . a , ,,. . a read antisine - arcsine -> and inverse sine - c c c 347. Case IV. Given a and b. The equations are: (1) c 2 = o 2 (2) tan A =r> .'.A = tan -1 T- = -, /. B = tan~ 1 -- a a Check, sin A = -> and cos A =- c c EXERCISES 116 Solve the following right triangles and check each by making an accurate construction and by substituting into a formula not used in solving. 1. A =27 30', a = 14 in. ; find B, b, and c. An*. -62 30'; 6-7.288; r = 15.78. RIGHT TRIANGLES 427 2. B=46 25', a = 17 ft.; find A, b, and c. Ans. A =43 35'; 6 = 17.86; c =24.66. 3. A =75 26', 6 = 25 ft.; find B, a, and c. Ans. 5 = 14 34'; a = 96.2; c = 99.38. 4. 5 = 62 40', 6=2 ft.; find A, a, and c. 5. A = 17 50', c=47 yd.; find B, a, and b. 6. 5=53 20', c = 21 ft.; find A, a, and 6. 7. a = 2 ft., c = 3 ft.; find A, B, and b. 8. 6 =4 ft., c = 9 ft.; find A, B, and a. 9. a = 4.23 in., 6 = 7.23 in.; find A, B, and c. 10. a =27 in., 6 =20 in. ; find A, B, and c. 11. 5=29 45', c = 2.36 ft.; find A, a, and 6. 12. A =32 12', c = 8.23 in.; find B, a, and 6. 348. Orthogonal projection. If from a point P, Fig. 247 (a), a perpendicular PQ be drawn to any straight line RS, then the foot of the perpendicular Q is said to be the orthogonal projection or simply the projection of P upon RS. Olx^ I \^ \^_ ! A A s \ E 1 FIG. 247. The projection of a line segment upon a given straight line is the portion of the given line lying between the projections of the ends of the segment. In Fig. 247(6) and (c), CD is the projection of AB upon OX. In each case AE CD and AE = AB cos 9. Hence, if I is the length of the segment of the line projected, p the pro- jection, and Q the angle between the lines, then p = l cos 0. Similarly, the projection of AB upon a line OY that is per- pendicular to OX and in the same plane as OX and AB is p' = l sin 6. 349. Vectors. In physics and engineering, line segments are often used to represent quantities that have direction as well as magnitude. Velocities, accelerations, and forces are such quantities. 428 PRACTICAL MATHEMATICS R Q FIG. 248. For instance, a force of 100 Ib. acting in a northeasterly direction may be represented by a line, say, 10 in. long drawn in a northeasterly direction. The line is drawn so as to represent the force to some scale; here it is 10 Ib. to the inch. An arrowhead is put on one end of the line to show its direction. In Fig. 248, OP is a line representing a directed quantity. Such a line is called a vector. O is the beginning of the vector and P is the terminal. OQ is the projection of the vec- tor upon OX, and OR is the projection upon OY. OQ and OR are called compo- nents of the vector. As before, OQ = OP cos 6, and x OR = OP sin 6. Example I, Suppose that a weight W is resting on a rough horizontal table as shown in Fig. 249. Suppose that a force of 40 Ib. is acting on the weight and in the direction OP, making an angle of 20 with the horizontal; then the horizon- tal pull on the weight is OQ =40 cos 20 = 37.588 Ib., and the vertical lift on the weight is OR = 40 sin 20 = 13.68 Ib. Example 2. A car is moving up an incline, making an angle of 35 with the horizontal, at the rate of 26 ft. per second. What is its horizontal ve- locity? Its vertical ve- locity? Solution. Horizontal ve- locity = 26 cos 35 = 21 .3 ft. per second. Vertical ve- locity = 26 sin 35 = 14.9 ft . per second. EXERCISES 117 1. The line segment AB 17 in. long makes an angle of 33 with the line OX. Find the projection upon OX. Find its projection upon the line OY perpendicular to OX and in the same plane as OX and AB. Aiis. 14.258 in.; 9.258 in. 2. A steamer is moving in a southeasterly direction at the rate of 24 miles per hour. How fast is it moving in an easterly direction? In a southerly direction? Ana. 16.97 mi. per hr. in each. Fio. 249. RIGHT TRIANGLES 429 3. The eastward and northward components of the velocity of a ship are respectively 5.5 miles and 10.6 miles. Find the direction and the rate at which the ship is sailing. Ans. 11.94 mi. per hr., 27 25.4' east of north. 4. A roof is inclined at an angle of 33 30'. The wind strikes this horizontally with a force of 1800 pounds. Find the pressure perpendicu- lar to the roof. Ans. 993.4 Ib. 6. A roof 20 ft. by 25 ft. and inclined at an angle of 27 25' with the horizontal will shelter how large an area? Ans. 443.85 ft. 2 6. A hillside is on a slope of 16 and contains 5.2 acres. How much more is this than the projection of the hillside upon a horizontal plane? Ans. 0.2 acre. 7. Show in general that the projection of a plane area upon a fixed plane is equal to the given area times the cosine of the angle between the planes. 8. Show in general that the component of a force along any fixed line is equal to the magnitude of the force times the cosine of the angle between the direction of the force and the fixed line. 9. Two men are lifting a stone by means of ropes. The ropes are in the same vertical plane. One man pulls 85 Ib. in a direction 23 from the vertical and the other 105 Ib. in a direction 42 from the vertical. Determine the weight of the stone. Ans. 156.3 Ib. 350. Definitions. The angle of elevation is the angle between the line of sight and the horizontal plane through FIG. 250. FIG. 251. the eye when the object observed is above the horizontal plane. When the object observed is below the horizontal plane, the angle is called the angle of depression. Tims, if is the object observed by the eye at E, the angle XEO is the angle of elevation in Fig. 250, and the angle of depression in Fig. 251. Directions on the surface of the earth are often given by directions as located on the mariner's compass. As seen from 430 PR A CTICAL MA THEM A TICS Fig. 252, these directions are located with reference to the four cardinal points, north, south, east, and west. Directions are often spoken of as bearings. When greater exactness is required, the direction may lx^ given as a certain number of degrees from a cardinal point. Thus, a direction given north 10 east means a direction 10 oast of north; south 40 west means 40 west of south. FIG. 252. Mariner's compass. EXERCISES 118 1. If a vertical staff 20 ft. high casts a shadow 20 ft. long on level ground, find the angle of elevation of the sun. Ans. 37 33.8'. 2. How many degrees east of north is N.E.? N.N.E.? N. by E.? N.E. by N.? Ans. 45; 22*; 11}! 33J. 3. A flag staff 70 ft. high casts a shadow 40 ft. long. Find the angle of elevation of the sun above the horizon. Ans. 60 15.3'. 4. At 60 ft. from the base of a fir tree the angle of elevation of the top is 75. Find the height of the tree. Ans. 224 ft. nearly. 6. What is the inclination from the vertical of the face of a wall having a batter of J? Ans. 7 7.6'. A batter of | means that the wall slopes 1 ft. in a rise of 8 ft. 6. What is the angle of slope of a road bed that has a grade of 5 per cent? One with a grade of 0.25 per cent? Ans. 2 51.7'; 8.6'. 7. Find the angle between the rafter and the horizontal in the follow- ing pitch of roofs: two-thirds, half, third, fourth. Ans. 53 7.8'; 45; 33 41.3'; 26 33.9'. RIGHT TRIANGLES 431 8. Certain lots in a city are laid out by lines perpendicular to B Street and running through to A Street as shown in Fig. 253. Find the widths of the lots on A Street if the angle between the streets is 28 40'. Ans. 114.0 ft. 9. A man whose eyes are 5 ft. 6 in. above the ground stands on a level with, and 150 ft. distant from, the foot of a flag staff 72 ft. high. What angle does his line of sight when look- ing at the top of the staff make with the horizontal line from his eyes to the pole? Ans. 23 54.6'. 10. In surveying on the Lake Front in Chicago, measurements were taken as shown in Fig. 254. Find the distance on a straight line from A to E. Ans. 338.4 ft. 11. In an isosceles triangle one of the base angles is 48 20', and the i r FIG. 253. C\*~*> L *\B t ! 114.78-' E D Coal Pile FIG. 254. base is 18 in. Find the legs, vertical angle, and the altitude drawn to the base. Ans. Legs 13.54 in.; vert. ang. 83 20'; alt. 10.11 in. 12. The side of a regular pentagon (five-sided figure) is 12 in. Find the radius of the inscribed circle, and the area of the pentagon. Ans. Radius 8.258 in.; area 247.8 in. 2 Suggestion. Draw the pentagon and inscribe a circle as in Fig. 255. Angle = T2. Triangle AOB is isosceles. 6 We have tan 36 = - - = 6-r-tan 36 angle CAB -=40. the length of CP. Could this problem be solved as easily by geometry? 13. Find a side of the regular octa- gon circumscribed about a circle 20 ft. in diameter. Ans. 8.284 ft. 14. Given the right triangle ABC, with C the right angle, CB = 2Q ft., and Produce CB to P making angle CAP = 70, find Ans. 65.48 ft. 432 PR A CTICA L MA THE MA TICS 16. Find the shorter altitude and area of a parallelogram whose sides are 10 ft. and 25 ft., and the angle between the sides 75. Ana. Alt. 9.659 ft. ; area 241.5 ft. 1 16. Two points C and B are on opposite banks of a river. A line AC at right angles to CB is measured 40 rods long; the angle CAB is meas- ured and found to be 41 40'. Find the width of the stream. AHS. 35.00 rd. 17. Wishing to determine the width of a river, I observed a tree standing directly across on the bank. The angle of elevation of the top of the tree was 32. At 150 ft. back from this point and in the same direction from the tree the angle of elevation of the top of the tree was 21. Find the width of the river. Ans. 239 ft. nearly. Suggestion. Let a: = width of river, and y = height of tree. The rela- tions of the parts are as given in Fig. 256. D c (I)tan32-|- B (2)tan21 = ^L, Here are two equations and two unknown numbers. The solution of them will give the values of x and y. 18. Locate the centers of the holes B and C, Fig. 257, by finding the distance each is to the right and above the center O. The radius of the circle is 1.5 in. Compute cor- rect to three decimal places. Ans. B, 1.2135 in., 0.8817 in.; C, 0.4635 in., 1.4266 in. 19. A man surveying a mine measures a length AB = 220 ft. due east with a dip of 6 15'; then a length BC=325 ft. due south with a dip of 10 45'. How much lower is C than A1 Ans. 84.57 ft. 20. A building 80 ft. long by 60 ft. wide has a roof inclined at 36 with the horizontal. Find the area of the roof, and show that the result is the same whether the roof has a ridge or not. Ans. 5933 ft. 1 21. In the side of a hill that slopes upward at an angle of 32, a tunnel is bored sloping downward at an angle of 12 15' with the horizontal. How far is a point 115 ft. down the tunnel, below the surface of the hill? Ans. 94.63 ft. Fio. 257. RIGHT TRIANGLES 433 22. The angle of elevation of a balloon from a point due south of it is 60, and from another point 1 mile due west of the former the angle of elevation is 45. Find the height of the balloon. Ans. 1.225 miles. 23. From the top of a mountain 1050 ft. high two buildings are seen on a level plane and in a direct line from the foot of the mountain. The angle of depression of the first is 35 and of the second is 24. Find the distance between the buildings. Ans. 858.8 ft. Q FIG. 25S. 24. If R and r are the radii of two pulleys, D the distance between the centers, and L the length of the belt, show that when the belt is not crossed, Fig. 258, the length is given by the following formula where the angle is taken in radians: R-r L=2VD 2 -(R-r)*+Tr(R+r)+2(R-r) sin' 1 D 26. Show that when the belt is crossed, Fig. 259, the length is given by the following formula: L = R fr) 2 + (R +r) (* +2 sin- 1 T FKJ. 250. Note. These formulas would seldom be used in practice. An ap- proximate formula would be more convenient, or the length would be measured with a tape line. A rule often given for finding the length of uncrossed belts is: Add twice the distance between the centers of the shafts to half the sum of the circumferences of the two pulleys. 28 434 PRACTICAL MATHEMATICS 26. In exercise 24, given 7? = 18 in., r=8 in., and D = 12 ft., find the length of the belt by the formula. Find the length by the approximate rule. Am. 30.87- ft.; 30.81- ft. 27. Use the same values as given in exercise 26, and find by the formula the length of the belt when crossed. An*. 31.20 ft. 28. A belt connects two pulleys of diameters 6 ft. and 2 ft. respectively. If the distance between their centers is 15 ft., find the length of the belt, making no allowance for slack. A ns. 42.83 ft. 29. Two pulleys, of diameters 7 ft. and 2 ft. respectively, are connected by a crossed belt. If the centers of the pulleys are 16 ft. apart, find the length of the belt. An*. 47.41 ft. 30. A chord of 2 ft. is in a circle of radius three feet. Find the length of the arc the chord subtends and the number of degrees in it. An*. 2.038 ft.; 38 56.3'. Suggestion. In Fig. 260, the chord A B = 2 f t. and the radius OA = 3 ft. Triangle AOC is a right triangle. Angle AOC = \ angle AOB, and the central angle AOB has the same measure as the arc AnB. 31. Find the area of the sector AnBO in ex- ercise 30. Find the area of triangle AOB. Find the area of the segment ABn. An*. Sector 3.057 ft. 2 ; triangle 2.828 ft. 1 ; segment 0.229 ft. 2 32. Find the area of the segment whose FIG. 260. chord is 4 ft. in a circle of 5 ft. diameter. Ans. 2.794 ft. 1 33. Find the area of a segment whose chord is 6 ft. and height 2 ft. Ans. 8.67 ft. 1 34. In a circle of 60 in. radius, find the area of a segment having an angle of 63 15'. Find the length of the chord and the height of the seg- ment, take 3 of their product, and compare with the area found. Ans. 379m. 1 35. A cylindrical tank resting in a horizontal position is filled with water to within 10 in. of the top. Find the number of cubic feet of water in the tank. The tank is 10 ft. long and 4 ft. in diameter. Ans. 106.7 ft. 36. Compute the volume for each foot in the depth of a horizontal cylindrical oil tank of length 30 ft. and diameter 8 ft. Ans. 108.8 ft. 3 ; 294.8 ft. 3 ; 516.4 ft. 3 ; 754 ft. 3 ; 991.6 ft.'; 1213.2 ft.*; 1399.2 ft. 3 Note. In the same manner, the volume could be computed for, say, each i in. In this way a gage could be made for determining the quan- tity of oil in a tank. 37. The slope of the roof in Fig. 261 is 30. Find the angle which is the inclination to the horizontal of the line AB, drawn in the roof and making an angle of 35 with the line of greatest slope. An*. 24 11.1'. RIGHT TRIANGLES 435 38. Find the angle between the diagonal of a cube and one of the diag- onals of a face which meets it. Ans. 3-5 15.9'. 39. A hill has a slope of 32. A path leads up it making an angle of 45 with the line of greatest slope. Find the slope of the path. Ana. 22 0.3'. 40. Two set squares, whose sides are 3, 4, and 5 in., are placed as in Fig. 262 so that their 4-in. sides and right angles coincide, and the angle between the 3-in. sides is 50. Find the angle between the longest sides. Ans. 29 22.5'. 41. What size target at 30 ft, from the eye subtends the same angle as a target 4 ft. in di- ameter at 1000 yd.? Find the angle it subtends. Ans. 0.48 in. ; 4.6'. F IG . 261. 42. The description in a deed runs as follows: "Beginning at a stone, A, at the N.W. corner of lot 401; thence east 112 ft. to a stone, B; thence S. 36| W. 100ft.; thence west parallel with A B to the west line of said lot 401 ; thence north on west line of said lot to the place of beginning." Find the area of the land described. Ans. 6612.88 ft. 2 43. If the point of observation is at a distance of h feet above the surface of the earth, find the farthest distance that can be seen on the surface of the earth; that is, find the distance of the horizon. FIG. 262. FIG. 263. Discussion. In Fig. 263, let be the center of the earth, r the radius of the earth, and h the height of the point P above the surface; it is required to find the distance from the point P to the horizon at A. PRACTICAL MATHEMATICS (PA) 1 - (PQ)'-(QA)' - (r+A)-r -2rA+/i. 436 For points above the surface that are reached by man, h* ia very small as compared with 2rh. ' PA = \/2rh, approximately. Here PA, r, and h are in the same units. Now let h be in feet, and r and PA be in miles. Also let r = 3960 miles. Then PA = -y/ 2 X 3960 X ~ miles. We may then state the following approximate rules: The distance of the horizon in miles is approximately equal to the square root of $ times the height of the point of observation in feet. The height of the point of observation in feet is j> times the square of the distance of the horizon in miles. Definition. The angle APC = 6 is called the dip of the horizon. 44. Find the greatest distance at which the lamp of a lighthouse can be seen from the deck of a ship. The lamp is 85 ft. above the surface of the water and the deck of the ship 30 ft. above the surface. Ana. 18 mi. approx. 45. A cliff 2000 ft. high is on the sea shore ; how far away is the horizon ? What is the dip of the horizon? Ans. 54.8 mi. approx.; 47'. Fio. 265. Acme thread. 46. Find the radius of a circle circumscribed about a polygon of 128 sides if one side is 2 in. What is the difference between the circumference of the circle and the perimeter of the polygon? Ans. 40.81 in.; 0.417 in. RIGHT TRIANGLES 437 47. In Whitworth's English Standard screw threads, Fig. 264, the angle between the sides of the threads is 55. If the top of a thread is rounded off | of the height and the bottom filled in the same amount, find the depth to four decimal places of the threads of the following pitches: 1, 8, 14, and 26. Ans. 0.6403 in.; 0.0800 in.; 0.0457 in.; 0.0246 in. 48. In an acme thread, the angle between the sides of the threads is 29. When the pitch is P, the depth and the other dimensions are as shown in Figs. 265, 267. (a) Suppose that the top dimensions and the depth are given, find the dimensions at the bottom, (b) Find the dimensions at the top and bottom and the depth for an 8-pitch acme thread. Ans. (b) 0.0787 in.; 0.0463 in.; 0.0793 in.; 0.0457 in.; 0.0638 in. 49. In a worm thread, the angle between the sides is 29. The dimen- sions are as shown in Figs. 266, 267. (a) Suppose the top dimensions 14 Vi FIG. 266. Worm thread. and the depth are given, find the bottom dimensions, (b) Find the dimensions in a 7-pitch thread. (The above are the Brown and Sharp proportions.) What are the differences between a worm thread and an acme thread? Ans. (b) 0.0950 in.; 0.0479 in.; 0.0443 in.; 0.0981 in. 60. From a point A on a level with the base of a steeple the angle of elevation of the top of the steeple is 42 30'; from a point B 22 ft. directly over A the angle of elevation of the top is 36 45'. Find the height of the steeple and the distance of its base from A. Ans. 118.9 ft.; 129.8 ft. Worm Thread Acme Thread FIG. 267. U.S.S. Thread 61. A ship sailing due north observes two lighthouses in a line due west ; after an hour's sailing, the bearings of the lighthouses are observed to be southwest and south-southwest. If the distance between the lighthouses is 8 miles, at what rate is the ship sailing? Ans. 13.66 mi. per hour. 438 PRACTICAL MATHEMATICS 50 62. Show that R = - v-7y where R = the radius of curvature and D =* the degree of the curve. 361. Widening of pavements on curves. The tendency of a motorist to "cut the corners" is due to his unconscious desire to give the path of his car around a turn the longest possible radius. Many highway engineers recognize this tendency by widening the pavement on the inside of the curve as shown in Fig. 268. The practice adds much to the attractive appear- ance of the highway. If the pavement is the same width around the curve as on the tangents, the curved section appears narrower than the normal width ; whereas if the curved section is widened gradually to the mid-point G of the turn, the pave- ment appears to have a uniform width all the way around. In order that the part added may fit the curve properly it is necessary to have the curve of the inner edge a true arc of a circle, tangent to the edge of the straightaway sections, and therefore it must start before the point E of the curve is reached. The part added may be easily staked out on the ground with transit and tape, by means of data derived from the radius r, the central angle 6 of the curve, and the width w. In practice the width w is taken from 2 ft. to 8 ft. according to the value of r. The width added can be readily computed when values for r, w, and 8 are given. Referring to the figure, derive the following formulas: x r sec $0 r = cos . r. x-\-w = r' sec \d r' = .'. r' = cos x+w (x-\-w) cos sec $0 1 1 cos $0 t = r tan $0. ' = r' tan i0. RIGHT TRIANGLES 439 Area added =BFCEAG = BPAO'-FPEC-BGAO'. BPAO' = r't r . FPEC = FPEO -FCEO = rt - 360 .'. area added =r't'- * - = r't'-rt- 6 360 7r(r'+r)(r' r). Exercise. Find the number of square feet in the area added if r =300 ft., w=4 ft., and = 100. Ans. 1395 ft. 2 352. Spirals. If a line is drawn around a circular cylinder so that it advances a certain distance along the cylinder for each revolution, the curve thus formed is a spiral or a helix. If a piece of paper is cut as shown in (a) Fig. 269, and lines AB and CD drawn, this piece of paper can be rolled into the D B D B,C (a) FIG. 269. cylinder (6) Fig. 269, where the lines AB and CD of (a) form the spiral running from A to D of (6). The advance along the cylinder for each turn of the spiral is the lead of the spiral, or the spiral lead. In Fig. 269, AC is the lead. It is customary to give the lead of the spiral as so many inches per one turn. For instance, a spiral that advances 8 in. in one turn is called an 8-in. spiral. 440 PRACTICAL MATHEMATICS The angle a that the spiral makes with an element of the cylinder is the angle of the spiral. It is seen that _ circumference of cylinder lead of the spiral C'R or, in (a) Fig. 269, tan a .= In setting milling machines for cutting spirals such as worms, spiral gears, counter bores, and twist drills, it is often necessary to know the angle of the spiral. To find the angle of a spiral or for the cutters in cutting a spiral, make a drawing as shown in Fig. 270; the angle C being a right angle, CB the circumference, and AC the lead. Angle A is the angle required, and may be measured with a fin protractor, or it may be found by finding tan A = -j~ and using the table of tangents. For ready reference the following rules are given: Lead FlO. 270. ANGLE. Divide the circumference of the spiral by the lead (advance to one turn) , and the quotient is the tangent of the angle of the spiral. LEAD. Divide the circumference of the spiral by the tangent of the angle, and the quotient is the lead of the spiral. CIRCUMFERENCE. Multiply the tangent of the angle by the lead of the spiral, and the product will be the circumference. When applying calculations to spiral gears, the angle is reckoned at the pitch circumference. EXERCISES 119 1. Find the angle of the spirals in the following twist drills: (1) Diameter of drill & in., lead 2.92 in. Ans. 18 35.2'. (2) Diameter of drill 1 i in., lead 9.33 in. Am. 20 44.8'. (3) Diameter of drill tf in., lead 7.29 in. Ans. 19 C 13.3'. RIGHT TRIANGLES 441 2. Find the angle of the spiral thread on a double-threaded worm of pitch diameter 3| in. and having three threads in 2 in. Ans. 83 5.1'. 3. In a tower the outer diameter of a winding stairway is 12 ft. Find the angle of the spiral formed by the outer end of the steps if the stairway makes one turn in ascending 18 ft. Find the angle of the spiral formed by the inner ends of the steps if the steps are 4 ft. long. Ans. 64 28.7'; 34 55.1'. 4. The piece shown in Fig. 271 has a length of 85 in. and a diameter of | in. If the spiral grooves make a half turn in the length of the piece, find the angle of the spiral. Ans. 15 40'. 6. Find the angle for setting cutters in cutting the following spirals: (a) Diameter jin. and lead 2.78 in.; (b) f in. and 7.62 in.; (c) 2 in. and 10,37 in.; (d) 1 in. and 22.5 in. Ans. (a) 15 46.7'; (b) 17 10.9'; (c) 31 13.1'; (d) 6 59.2'. FIG. 271. 6. A cylinder 2 in. in diameter is to have spiral grooves making angles of 20 with the center line of the cylinder. What will be the lead of the spiral? Ans. 17.26 in. 7. Find the length of one turn of a spiral around a cylinder 3 in. in diameter if the lead of the spiral is 9 in. Ans. 13.03 in. 8. Show that the length of any spiral is given by formula L=n\/C 2 +/ 2 , where L= length of spiral, n=the number of turns of the spiral, C = cir- cumference of cylinder the spiral is on, and Z=lead of the spiral. 9. Find the length of a spiral making 20 turns in 8 in. on a cylinder 3.5 in. in diameter. Ans. 220.0 in. CHAPTER XXXIX RELATIONS BETWEEN RATIOS, AND PLOTTING 353. Relations between the ratios of an angle and the ratios of its complement In Art. 327 it was pointed out that the function of an acute angle is equal to the co-function of its complement. This gives the following relations. It can easily be shown that these relations hold for any value of the angle 6. [62] sin (90 -6)= cos 6. [63] cos (90 -6)= sin 6. [64] tan (90 -6)= cot 6. [66] cot (90 -6)= tan 0. [66] sec (90 -8)= esc 6. [67] esc (90 -6)= sec 6. As previously explained, it is these relations that make a table of trigonometric functions do double duty. Example, sin 60 = sin (90 - 30) = cos 30. 354. Relations between the ratios of an angle and the ratios of its supplement. The following relations between the ratios of an angle and its supplement are true for any value of 0. They are convenient when one wishes to find by means of the tables the functions of angles lying between 90 and 180. [68] sin (180 -6)= sin 6. [69] cos (180-6) = -cos 6. [70] tan (180 -6) = -tan 6. [71] cot (180 -6) = -cot 6. [72] sec (180-6) = -sec 6. [73] esc (180-e)=csc 6. Proof. In Fig. 272, angle XOP = 6 is any angle, and angle XOQ = 18Q-8. From any point in the terminal side of XOP, as B, draw the perpendicular AB to the z-axis; and from D, any point in the 442 RELATIONS BETWEEN RATIOS, AND PLOTTING 443 terminal side of XOQ, draw the perpendicular CD to the z-axis. The right triangles OAB and OCD are similar. Also OA, AB, OB, CD, and OD are positive, while OC is negative. Then sin (180 - 0) = gg = ' AJso cos = sn Examplel. sin 130 15' = sin (180-49 45') = sin 49 45' = 0.7633. Example 2. sos 160 = cos (180 - 20) = -cos 20 = -0.9397. FIG. 272. 355. Relations between ratios of an angle 6 and 90 +6. The following relations connect the ratios of any angle 6 and 90 + 0. They are also convenient to use in finding, by means of the tables, the functions of angles lying between 90 and 180. [74] sin (90 +6)= cos 0. [75] cos (90 +6) = -sin 0. [76] tan (90 +6) = -cot 6. [77] cot (90+e) = -tan 6. [78] sec (90+6) = -csc 6. [79] esc (90 +0)= sec 6. Proof. In Fig. 273, angle XOP = 8 and angle XOQ = 90+ e. From any point in the terminal side of each draw a perpen- Ill PRACTICAL MATHEMATICS- I in ilar to the z-axis. The right triangles AOB and OCD thus formed are similar, and have all their sides positive except OC. Then sin (90+0) = ^ = ~ = cos d. (JL) (Jo Also tan (90+0) == = -cot d, Example 1. sin 130 15'=sin (90+40 15')= cos 40 15' = 0.7633. Example 2. tan 116 20' = tan (90+26 20') = -cot 26 20' = -2.0204. Fio. 273. 366. Relations between the ratios of an angle and the ratios of its negative. The following relations connect the ratios of positive and negative angles, and are true for any value of 0. [80] sin (-0) = -sin 6. [81] cos (-6) =cos 6. [82] tan (-6) = -tan 6. [83] cot (-6) = -cot 6. [84] sec (-0) = sec 0. [86] esc (-6) = esc tt. Proof. In Fig. 274, angle XOP = 0, and angle XOQ=-d. From any point in the terminal side of each draw a per- pendicular to the x-axis. The right triangles OA B and OAC RELATIONS BETWEEN RATIOS, AND PLOTTING 445 thus formed are similar, and have all their sides positive but AC, which is negative. ml / n\ ^r At) Then sin ( 0) = -^ = -=-= = sm 6. UL Uo . M OA OA And cos ( 0)=7T7; = pr^ = cos0. t/C UJD Also COt ( 6)= TTr rn = ~ c t &' AC/ AD Example 1. sin (-30) = -sin 30= -0.5. Example 2. tan ( - 46 10') = - tan 46 10' = - 1 .0416. EXERCISES 120 Draw a figure in each case and prove the following: 1. sin (90 -0) =cos 6. 7. esc (180 -0) =csc. 0. 2. tan (90 -0) =cot 0. 8. cos (90 + 0) = -sin 0. 3. sec (90 -0) =csc 0. 9. cot (90 + 0) = -tan 0. 4. tan (180 -0) = -tan 0. 10. sec (90 + 0) = -esc 0. 6. cot (180 -0) = -cot 0. 11. tan(-0) = -tan 0. 6. sec (180-0) = -sec 0. 12. sec (-0) =sec. 0. By means of the tables find the following: 13. sin 140. 16. cos (-49). 19. sin 159 40'. 14. cos 150. 17. tan (-17). 20. cos (-117 30')- 15. tan 170. 18. cot (-125). 21. tan (-156 10'). Ans. Ex. 14. -0.8660. Ex. 17. -0.3057. Ex. 20. -0.4617. 357. Relations between the ratios of any angle. As stated in Art. 324, [86] sin 8 = or esc = - - esc sin 6 [87] cos = or sec 6 = -. sec 6 cos 6 [88] tan == or cot 6 = cot 6 tan 6 By definition, [89] versedsine 0, abbreviated vers 6 = 1 cos 6. [90] coversedsine 0, abbreviated covers = 1 sin 6. The truth of the following can easily be proved when is any angle. 446 PRACTICAL MATHEMATICS [91] sin'O + cos 2 6 = 1. [92] sec 2 6 = l+tan 2 e. [93] csc 2 = l+cot 2 6. [96, cot .- Proo/ o/ [91]. By definition (see Fig. 275), 7/ 2 X 2 sin 2 = and cos 2 5 = -= r 2 r 2 Hence adding, sin 2 0+cos 2 8 = ^+^ = V * But t/ 2 -|-x* = r 2 because the triangle OCZ? is a right triangle. FIG. 275. r 2 Hence sin 2 0+cos 2 = = 1. r 2 Proo/ of [92]. By definition, sec 2 = , and tan 2 = x 2 x 2 r 2 ?/ 2 /. sec 2 6 = 1+ tan 2 B because -, = 1 + ^- = X 2 X 2 EXERCISES 121 1. Prove [93], [94], and [96] by using the definitions. 2. From [91] derive (a) sin 6 = \/l cos* 0, (b) cos 9 8. Solve [92] for sec and tan 0. __ An*, sec 0= -v/l+tan 2 0; tan -v/sec* 1. 4. Solve [93] for esc 6 and cot 6. __ Ans. esc 0= \/l-r-cot 8 0; cot 0= f-y/csc 1 0-1. RELATIONS BETWEEN RATIOS, AND PLOTTING 447 5. From the relations already given, write five other relations between the ratios. 6. Prove the following relations : (1) cos 6 tan 0=sin 6. (3) sec 6 cot 0=csc 6. (2) sin cot 8 = cos 6. (4) tan 6 esc 8 = sec 0. Use the formulas and find the other functions : 7. When sin = %. 8. When cos = f . 9. When tan 0=3. 10. Find the following from Table XI : (1) sin 122 U 20', (2) cos 110 44', (3) tan 163 15', (4) cot 172 50', (5) cos (-42 16'), (6) tan (-21 49'). Ans. (1) 0.8450; (2) -0.3540; (3) -0.3010; (4) -7.9530; (5; 0.7400; (6) -0.4003. 368. Plotting the sine curve. In Chapter XXXIII the plotting of the curve of an algebraic equation was considered. The trigonometric equations can be represented by curves plotted in a similar manner. y 30 15 60 90 120135'l50\lSO 210 240 270 300 330 360 FIG. 276. Example. Plot the curve for y = sin x. Choose suitable angles for values of x and take the values for y from the table of Art. 332 or from Table XI. Values of x, 30 45 60 90 120 135 150 180. Values of y, 0.5 0.7 0.87 1 0.87 0.7 0.5 0. Values of x, 210 225 240 270 300 315 330 360. Values of y, -0.5 -0.7 -0.87 -1-0.87-0.7-0.5 0. Choose a convenient unit on the z-axis and y-axis and plot the points as shown in Fig. 276, using the angle as abscissa and the sine as ordinate. The curve in Fig. 276 is called the sine curve. It extends its PRACTICAL MATHEMATICS Y Fio. 277. 3r y = tan z- Fio. 278. Fio. 279. RELATIONS BETWEEN RATIOS, AND PLOTTING 449 both ways indefinitely, repeating the part given. That is, if values between 360 and 720 be taken the curve will be found to be of the same shape. The same will be true for values of the angle less than 0. A curve that repeats itself in this way is called a periodic curve. The equation giving rise to it is called a periodic function. The relative lengths of the units representing x and y may be changed, and the curve thus drawn out or shortened. However, the curve will still be of the same general form. 359. Curves for cosine, tangent, cotangent, secant, and cosecant. By choosing suitable angles for values of x and determining the corresponding values of y, the coordinates of points are obtained which, when plotted, give the curves of Figs. 277, 278, and 279. EXERCISES 122 1. Plot y = sin x, using various units for x and y. 2. Plot y=cos x from x = to x = 360 and get form of Fig. 277. 3. Plot y=tan x and y=cot x from x=0 to z = 360 and get forms of Fig. 278. 4. Plot y =sin z+cos x from x = to x 180. 5. Plot y =cos 2 x sin 2 x from x = Q to = |?r. 00 6. Plot 7/=cos -1 T from x=0 to a; = 3. 7. Plot y =sec x and y = csc x from x =0 to x = 360 and get the forms of Fig. 279. 360. Projections of a point having circular motion. Example 1. A point P, Fig. 280, moves around a vertical circle of radius 3 in. in a counter-clockwise direction. It starts with the radius in a horizontal position and moves with an angular velocity of one revolution in 10 seconds. Plot a curve showing the distance from the center of the projec- tion of P on the vertical diameter at any time. Discussion. Describe a circle with center and radius 3 in., to scale if desired. In this circle the point P starts at A. After 1 second OP has turned to the position of OPi through an angle of 36 = 0.6283 radians, after 2 seconds to the position OP % through an angle of 72 = 1.2566 radians, and so on to the positions OP 5 , OP 4 , OP 10 . 29 1.-.0 PR A CTICAL MA THEM A TICS The points NI, Ni, etc., are the projections of PI, P 8 , etc., on the vertical diameter. Produce the horizontal diameter OA, and lay off the seconds on this to some scale. For each second plot a point whose ordinate is the corre- sponding distance of TV from 0. Fractions of seconds may be taken and more points located if desired. Connect these points, and we have a curve for which any ordinate y is the distance from the center O of the projection of P on the vertical diameter at the time t, represented by the abscissa of the point. It is to be noted that the ordinate of the curve increases from to +3, decreases from +3 to and from to 3, f " . j D r \ P ^ s P / N 2 / \ / Ai // 1\ V i \ A \ h \ _ Y \ ^ A ; 'lice iu S< '. cond ' s \ I * j i / V? E ,/ \ / * X X \ ^-ji / P P 1 FIG. 280. and increases from 3 to 0. The curve is made up of two like-shaped parts, one below and one above the horizontal line. Together they form a cycle of the curve. It is evident that, for the second and each successive revo- lution, the curve repeats itself, that is, is a periodic curve. Since OP turns through 36 per second, the angle AOP = 36< = 0.6283J radians. Because of the right triangle OPN, y =ON = OP sin 36J = 3 sin 36<. Or, in general, yr sin wt, where o> (Greek letter omega) is the angle turned through in one second, that is, the angular velocity. It is then the equation of the curve. It is readily seen that if a straight line of length r, such as a crank, starts in a horizontal position when t = 0, and revolves in a vertical plane around one end at a uniform angular velocity RELATIONS BETWEEN RATIOS, AND PLOTTING 451 u> per second, the projection of the moving end on a vertical straight line has a motion which is represented by the equation y = r sin coi. Similarly, the projection of the moving end on a horizontal straight line is given by the ordinates of the curve whose equation is y = r cos o)t. If OP represents the crank of an engine with a connecting rod which is very long compared with OP, the motion of the cross-head is represented approximately by that of the point .V, and, hence, by the equation y = r sin cot. A simple periodic motion like the above is sometimes repre- sented by the equation y = r sin cot and sometimes by y = r Y P, '45 .5 FIG. 281. cos ut. The first, if the time is measured from the instant when OP is in the line of the simple periodic motion, and second, if measured from the instant when OP is perpendicular to the line of the simple periodic motion. If the time is counted from some other instant than the above, we have y = r sin (o^+a), where a is the angle that OP makes with the line OA at the instant t is counted from. Example 2. A crank OP, Fig. 281, of length 2 ft. starts from a position making an angle a = 40 with the horizontal line OA when t = 0. It rotates in a vertical plane in the posi- tive direction at the rate of 2 revolutions per second. Plot a curve showing the projection of P on a vertical diameter. Discussion. As before, draw a circle of 2-ft. radius to some scale. Extend the horizontal axis to the right, and represent the part of a second necessary for one revolution to some suitable scale. 452 PRACTICAL MATHEMATICS When < = 0, OP starts from the position OPo making an angle .4OP = a = 40. OP turns through 720 in 1 sec., or 36 = 0.0283 radians in 0.05 sec. Here we plot a point for every 0.05 sec. Any other convenient fraction of a second could be chosen. The position of the free end of the crank for each 0.05 sec. until a complete revolution is made is located by Po, PI, P 2 , PS . These are at intervals of 36 starting with 40. At any time t, the position of OP makes with OA an angle AOP=o)t+a. And the distance from O to the projection of P at N is ON = OP sin (w*+a), or O# =2 sin (720<+a). The curve of this example is periodic. It repeats itself for each complete revolution of the crank. It follows that, in general, the equation of the curve is 2/=rsin (ut+a). 361. Sine curves of different frequency. The frequency refers to the number of cycles of the periodic curve in a unit of time. Thus, if a crank makes 4 revolutions in 1 second, the curve showing the motion makes 4 cycles in 1 second; while the curve for a crank that makes 2 revolutions in 1 second makes 2 cycles in 1 second. The frequency of the first is 4, of the second 2. The angular velocity w divided by 360 when measured in de- grees, or by 2ir when measured in radians, gives the frequency f t The time necessary for one cycle is called the period. It is evident that the period T is given by the formula Consider the following equations: Equation Angle at Angular Frequency Period velocity u u r _ ? T / ~2r ^~w (1) y = r sin}/ \t 4 0.0796 12.566 (2) y = r sin* t 1 0.1592 6.283 (3) y = r sin2t 2t 2 0.3183 3.1416 RELATIONS BETWEEN RATIOS, AND PLOTTING 453 If r is taken as 1 and the three curves plotted, they are as in Fig. 282. It is to be noted that the frequency of a sine curve varies directly y = B in y 2 t = 8 in 1J = sin 2 t -1 - FIG. 282. as the value of u, that is, as the coefficient of t in the angle ut. The number a> is sometimes called the frequency factor. 362. Variation in the amplitude of sine curves. The am- -1 2 / \ y*= h sin e - I \ n \ y= sin a . y = 2 sin - \ W= 2 sin 3^- \ \J y i \./ --.....x v FIG. 283. plitude of a sine curve is the difference between its greatest and least ordinate. In the previous article all of the curves plotted have the 454 PRACTICAL MATHEMATICS same amplitude, and it will be remembered that the radius in each case was taken equal to unity. Consider the curves of the equations: (1) y = \ sin 0, (3) y = 2 sin 0, (2) 2/ = sin 0, (4) y = 2 sin 36. These curves are plotted in Fig. 2&3. It is to be noted that the amplitude of a sine curve varies directly as the coefficient of sin 6, that is, as the value of r in the generating circle. EXERCISES 123 1. A crank 18 in. long starts in a horizontal position and turns in a positive direction in a vertical plane at the rate of 0.7854 radian per second. The projection of the moving end of the crank upon a vertical line oscillates with a simple periodic motion. Construct a curve whose ordinates show the distance of the projection from the center of its path at any time. 2. Write the equation of the curve in the previous exercise. Find the value of the ordinate when t =0.5. When t =2.3. Ans. y = 18 sin(0.7854<); 6.89 in.; 17.50 in. 3. A crank 8 in. long starts in a position making an angle of 55 with the horizontal, and rotates in a positive direction at the rate of 20 revolu- tions per minute. Draw a curve to show the projection of the moving end of the crank on a vertical line. 4. Write the equation of the curve in the last exercise. Find the value of y when t = 1.5 seconds. Ans. t/ = 8 sin (120<+55); -6.55 in. Plot the curves that represent the following motions: 5. y =3.5 sin (35* +36). 6. y = 12sin (1.88* +0.44). 8. Plot y=r sin - and i/ = r sin (-<+ -Ion the same axes. Notice that the highest points are separated by the constant angle - Such 4 curves are said to be out of phase. The difference in phase is stated in time or as an angle; in the latter case it is called the phase angle. 9. Plot (1) y=r sin ^t, (2) y = r sin (f-^) ' and (3) y =r cos ^t, all on the same axes. What is the difference in phase between (1) and (2)? 10. Plot on the same set of axes: (1) y=r sin rt, (2) y=r sin ^t, (3) y-rsin ^(. 11. Plot on the same set of axes: (1) //=40ain 9, (2) /=30sin 9, (3) y=20sin 29. CHAPTER XL TRIGONOMETRIC RATIOS OF MORE THAN ONE ANGLE 363. In the present chapter will be given a number of formu- las of very great use in more advanced subjects of mathematics and in their applications. These formulas are given for reference. But little attempt will be made here to prove them or illustrate their uses. 364. Functions of the sum or difference of two angles. The following formulas express the functions of the sum or the dif- ference of two angles in terms of the functions of the separate angles. These formulas are sometimes called the addition and subtraction formulas of trigonometry. [96] sin (A+B)=sin A cos B+cos A sin B. [97] sin (A B)=sin A cos B-cos A sin B. [98] cos (A+B)=cos A cos B sin A sin B. [99] cos (A-B)=cos A cos B+sin A sin B. MAAI A (K \-o\ tan A + tan B [100] tan FIG. 284. tan A tan B Proof of [96] for A<90, <90, and (A+5)<90. In Fig. 284, angle XOQ = A and angle QOP = B. Hence angle XOP = A+B. Draw NP, QP, QR, and MQ perpendicular respectively to OX, OQ, NP, and OX. Then angle RPQ = A. NP MQ+RPMQ RP 455 450 PRACTICAL MATHEMATICS But MQ = OQ sin A, and RP = QP cos A, from the right triangles. Putting these values in place of MQ and RP, OQ sin A . QP cos A .". sin (A+Z?) = sin A cos B-j-cos A sin /?. Formulas [97] to [101] may be proved in a similar manner. Example 1. Given the functions of 45 and 30, find the value of sin 15. Solution, sin 15 = sin (45 -30) = sin 45 cos 30 -cos 45 sin 30 = 0.7071X0.8660-0.7071X0.5=0.2588. Example 2. Test formula [100] by using A = 16 and B=27. Substituting these values in the formula, tan a6+27}- Jan_16 +tan_27 ~l-tanl6tan27 6 ' Putting in the values of tan 16 and tan 27 from Table XI, 0.2867+0.5095 = r^o7286Txo:5095 = a9325 - 366. Functions of twice an angle and half an angle. By putting B = A in the formulas for the sum in Art. 364, the following formulas are obtained: [102] sin 2A 2 sin A cos A. [103] cos 2A = cos 2 A -sin 2 A = 1-2 sin 2 A = 2 cos 2 A - 1. ,104] taa 2A = These formulas are used to express the function of any angle in terms of the functions of one-half that angle. From the above may be derived the following formulas, which are used to express the functions of any angle in terms of the functions of twice that angle. MAKI ,*-cos 2A [106] sm A= H Mnei /1-f COS 2A [106] cos A-+-J \ & MA _, /I cos 2A sin 2A 1 cos 2A [107J tan A= \/ 1 'l-f-cos2A 1+cos 2A sin 2A TRIGONOMETRIC RA TIOS OF MORE THAN ONE ANGLE 457 Example 1. Test formula [102] by using A =20. Substituting in the formula, sin 40 = 2 sin 20 cos 20 = 2X0.3420X0.9397 =0.6428. Example 2. Test formula [106] by using A = 35. Substituting in the formula, /I + cos 70 /I +0.3420 cos 35 = A/ ~ =A/ ^ =0.81915. 366. Formulas for changing products td sums or differ- ences, and sums and differences to products. The following four formulas are convenient for expressing a product of two functions as the sum or difference of two functions. [108] sin A cosB = | sin (A+B)+| sin (A-B). [109] cos A sin B = | sin (A+B)- sin (A-B). [110] cosAcosB = | cos (A+B) + ^ cos (A-B). [Ill] sin A sin B= -| cos (A+B) + cos (A-B). The following four formulas express a sum or a difference as a product. They are often convenient when working with logarithms. [112] sin A+sin B = 2 sin | (A+B) cos | (A-B). [113] sin A-sin B = 2 cos \ (A+B) sin \ (A-B). [114] cosA+cosB = 2 cos (A+B) cos HA B). [115] cos A cos B = -2 sin \ (A+B) sin \ (A-B). EXERCISES 124 1. Find sin 90, cos 90, and tan 90, using the addition formulas and 90 = 60 +30. 2. Find sin (A +B) if sin A = and sin B = jj, both being acute angles. Find cos (A+B). Ans. 1; 0. 3. Find cos (A B) if cos A = A and cos 5={|, both being acute angles. Find sin (A B). Ans. iff; iJ|. 4. Find tan (A +B) if tan A =2 and tan B =3, both being acute angles. What is the value of (.4 +5)? Ans. - 1 ; 135. 6. Given the functions of 30 and 45, find sin 75, cos 75, and tan 75. 6. Take values from Table XI for the functions of 25 and 18, and find sin 43, cos 43, sin 7, and tan 7. 7. Take values from Table XI for the functions of 15 and find sin 30, cos 30, and tan 30. 8. Using the functions of 45, find sin 22, cos 22, and tan 22^- 9. Given sin 20 = , find tan and cot 6. Ans. ; 3. 458 PRACTICAL MATHEMATICS 10. Show that COB 50 +cos 30 = 2 cos 40 cos 0. 11. Show that sin 70 sin 30-2 cos 50 sin 20. 12. Show that cos 125 -cos 75 = -2 sin 100 sin 26. 13. Express sin 7o+sin 5a as a product. Ana. 2 sin 6a cos a. 14. Express cos 46+cos 28 as a product. Ann. 2 cos 37 cos 9 15. Show that sin 575 cos 927+cos 575 sin 927 -sin 1602. . tan327+tan_846 16. Show that ^^3270^8^0 = tan 1173 . . -TT j A i_ i . . 11* 15*- 9r OT 17. Innd the value of sin cos --- htan r cot -75-- o 4 4 i Ann. -0.3536. 18. Find the value of ll-\ . 19r . / 4\ . / 7r\ ~ T ) 8m T + C08 (~ 37 8m \ ~"37 ' 19. If sin a = $ and a is in III quadrant, and cos = } and ft is in 1 quadrant, find the value of sin (a+0), sin (a 0), and tan (a+0). Ans. -0.7618; 0.5475; 1.1760. . sin 47+sin 17 20. Show that ._ -r=5=tan 32. cos 47 +cos 17 ,, . sin 3+sin 5a 21. Show that =cota. cos 3a cos 5a ,, , sin 67 sin 23 22. Show that C0867 o +cos23 o=tan 22. 23. In a right triangle cos a - and cos /3 = - Show that c c ] c b . . i- ^ Ic a. and sin i/3 = \/~o~~ 24.,If a is less than 360, in what quadrants may a be if sin i a is nega tive? Positive? If tan \a is negative? Positive? If cot $a is nega tive? Positive? If sec i a is negative? Positive? CHAPTER XLT SOLUTION OF OBLIQUE TRIANGLES 367. Cases. A triangle that is not a right triangle is called an oblique triangle. The student should now reread Arts. 338 and 339 and the exercises following them. It is evident, then, that an oblique triangle can be solved when there are three elements given at least one of which is a side. There arise the four following cases: CASE I. When any side and any two angles are given. CASE II. When any two sides and the angle opposite one of them are given. CASE III. When any two sides and the angle included between them are given. CASE IV. When the three sides are given. The oblique triangle can be divided into right triangles by drawing convenient altitudes, and so be solved by methods already given in the chapter on right triangles. It is, however, usually a saving of time to solve them by means of formulas derived especially for that purpose. The simpler of these formulas will now be derived. 368. The law of sines. The law is: In any triangle, the sides are proportional to the sines of the opposite angles. Proof. Let ABC in Fig. 285 be any triangle. Draw the altitude h from B to the side A C. Because of the right triangles, in either (a) or (6), h c sin A and h = a sin C, 459 400 PR A CTICAL MA THEM A TICS .'. a sin C = c sin A. Dividing both members of this by sin C sin A, a c_ sin A sin C Similarly by drawing an altitude from C, - ' sm A sin B b c [116] .'. a tions: sin A sin B sin C From the law of sines there may be written the three equa- a b a c b c sin A sin B' sin A sin C' sin B sin (f any one of which involves four elements of a triangle. It is evident that if any three of the elements involved in one of the equations are given, the remaining element can be found. Thus, if in - -r = - 5' A, B, and 6 are given, then solving sin A sm B for a, a = sin b sin A sin B 369. The law of cosines. The law is: In any triangle, the square of any side equals the sum of the squares of the other two sides minus twice, the product of these two sides times the cosine of the angle between them. O) b D C A m. 2JSC. Proof. In Fig. 280, either (a) or (6), draw (ho altitude h from B to the side AC. Let AD = m and DC = n. Because of the right triangles, in either (a) or (6), a 2 = A 2 +n 2 . In (a), n = b m = b c cos A. In (b), n = m b = c cos A 6. In either case, n 2 = 6 2 26c cos A-f-c 2 cos 2 A. In either (a) or (6), h* = c* sin 2 A. SOLUTION OF OBLIQUE TRIANGLES 461 Substituting these values for h 2 and n' 2 in, a 2 = a 2 = c 2 sin 2 A+b 2 2bc cos A+c 2 cos 2 A. :. a 2 = & 2 +c 2 (sin 2 A+cos 2 A)-2bc cos A. [117] .'. a 2 = b 2 +c 2 -2bc cos A. Similarly there may be obtained: [118] b 2 =a 2 +c 2 -2accosB. [119] c 2 = a 2 +b 2 -2abcosC. Solving these for cos A, cos B, and cos C respectively: [120] cos A = [1211 cos B = [122] cos C = b 2 +c 2_a2 2bc 2ac a 2 +b 2 -c 2 2ab By these formulas the values of the angles of a triangle can be computed when the sides are known. 370. Directions for solving. It will be noticed that each of the formulas from the law of sines and the law of cosines involves four elements of the triangle. In solving a triangle, select a formula that involves three known elements besides the element that is to be found, solve for the unknown element in terms of the known, substitute the numer- ical values, and evaluate. The work may be checked (a) by making a careful construc- tion, and (b) by using some other formula than is used in solving. 371. Case I, a side and two angles given. Example. Given a -45, 5 = 36 17', and C = 83 32'; to find 6, c, and A. Formulas. (1) A=--180-(B+C). Construction. (2) sin A sin B' /.& = sin A sin C' ' a sin B sin A a sin C sin A Computation by natural functions. A = 180 -(36 17'+83 32')=60 C a sin B 45 X. 5918 11' c sin A a sin C sin A .8676 45 X. 9937 .8676 = 30.70. = 51.54. FIG. 287. 462 1>K A CT1CA L M A Til KM A TICS Computation by logarithms. log 45= 1.6532 Iogsin3617' = 9.7722 11.4254 log sin 60 11'= 9.9384 log 6 = 1.4870 . 6 = 30.69 log 45= 1.6532 log sin 83 32'= 9.9972 11.6504 log sin 60 11'= 9.9384 log c= 1.7120 /. c = 51.53 Check by using the law of cosines. 372. Case II, two sides and an angle opposite one of them given. With these parts given it is possible (a) that there is only one solution, that is, that only one triangle can be found; (b) that there are two solutions, that is, that there are two different triangles that fulfill the conditions; (c) that there is no solution, that is, that no triangle exists that will fulfill the conditions. Whether there is one solution, two solutions, or no solution can readily be determined by making a careful construction of the triangle from the given parts. Example 1. Given a = 15, c=10, and A =40 30'; to find 6, B, and C. The construction shows that there is only one solution. Formulas. Construction. .'. sin C = sin A sin C c sin A (2) = 180-(A+C). a b a sin B Fio. 288. sin A sin B' sin A Example 2. Given a = 20, c = 25, and A =52 40'; to find 6, B, and C. From the construction it is seen that there are two triangles, ABC and ABC', that fulfill the conditions. (1) SOLUTION OF OBLIQUE TRIANGLES Formulas. Construction, a c 463 .'. sin C = c sin A sin A sin C (2) C" = 180-C. (3) 5 = 180-(A+C). (4) B'=ABC' a b 7 a sin B a (5) (6) sin A sin J5 sin A sin 5' .'.6' = sin A a sin 5 sin A Example 3. Given a = 12, c = 20, and A =62 20'; to find 6, B, and C. FK} 28g From the construction, Fig. 290, it is evident that the side a is not long enough to form a triangle. Hence there is no solution. FIG. 290. 373. Case III, two sides and the angle between them given. Example. Given 6=45.2, a = 56.7, and C=47 45'; to find c, A, and B. Construction. B Formulas. (1) c=V 2 +6 2 -2a6 cos C. C (2) = sin .'. sm A= sin A sin a sin C .'. sin B sin 5 sin C b sin C 464 PRACTICAL MA THEM A TICS Computation. c = V56.7 2 +45.2 2 -2X56.7X45.2X0.6723 - 42.56. n . =80 26'. C/iecfc. A+5+C = 180. 80 26' +51 49'+47 45' = 180. Logarithms cannot be used conveniently with formulas such as (1) above. Formulas in which logarithms can be used for the solution of examples tinder Case III can be derived from the law of sines. They are numbers 123, 124, and 125 on page 473. 374. Case IV, three sides given. Example. Given a =10, 6 = 12, and c=15; to find A, B, and C. (1) cos A = (2) cos B = (3) cosC = Formulas. 6 2 +c 2 -a 2 Construction. 2bc a 2 +c 2 -b* 2oc 2ab Computation. 12 2 +15 2 -10 2 -2XT2X-15- = - 7472 ' R 10 2 +15 2 -12 2 cos B = ^TTTTTT-r^r- = 0.6033, cos C = 2X10X15 10 2 +12 2 -15 2 = 0.0792, Fm. 292. A =41 39'. # = 52 53.5'. C = 85 27.6'. 2X10X12 Check. ^+B+C = 180. 41 39' + 52 53.5'+85 27.6' = 180 0.1'. The above formulas are not suitable for work with loga- rithms; but formulas can be derived from them that lend themselves readily to logarithmic work. These formulas are given on page 473, numbers 126 to 134 inclusive. SOLUTION OF OBLIQUE TRIANGLES 465 EXERCISES 126 1. From the law of sines, find 6 in terms of a, A, and B. Find C in terms of 6, c, and B. . ^_g sin B . n _ . _ t c sin B Ans. u ; j j C sin r sin A With the following three elements given, find those remaining and check the results when the answers are not given. 2. A =44 6.5'; 5 = 57 42.5'; a =4.23 in. 3. A =48 39.2'; C = 115 23.8'; a = 14.83 in. Ans. 5 = 15 57'; 6 =5.428 in.; c = 17.85 in. 4. 5=30 36.8'; C = 107 15.5'; 6 = 144. 6. C = 44 17.3'; 6 = 14.33; c = 13.67. Ans. 5 = 47 2.5'; A =88 40.2'; a = 19.57; ' = 132 57.5'; A'=2 45.2'; a' =0.940. 6. A =53 16.5'; c = 25.64; a =31.4. 7. a = 79.8; 6=46.7; 5 = 23 19.6'. 8. 6 = 17;c = 16; A =47 16.4'. 9. a =99.4; c = 90.4; B = 11 7.8'. Ans. A =110 20.4'; C = 58 31.8'; 6 = 20.5. 10. a = 21; 6=24; c = 31. 11. a = 61.52; 6 = 81.74; c = 75.34. Ans. A =45 53.4'; 5 = 72 33.2'; C = 61 33.4'. 12. a = 2.46; 6=3.5; c = 4.2. 13. To find the distance AB through the swamp, Fig. 293, the following data were measured: a = 120 rd., 6 = 146 rd., and C = 41 25'. Compute the distance AB. Ans. 97.14 rd. 14. Compute the inaccessible distance AB, Fig. 294, from the measured data 6 =450 ft., A =82 30', and C = 67 42'. Ans. 837.2 ft. 15. Two points, P and Q, Fig. 295, are on opposite sides of a stream and invisible from each other on account of an island in the stream. A straight line AB Is run through Q, and the following measurements taken: AQ = 75Q ft., QB = 562 ft., angle AP=47 28.6', and angle ' = 57 45'. Compute QP. Ans. 852 ft. FIG. 293. FIG. 294. 16. From a point on a horizontal plane the angle of elevation of the top of a hill is 23 46' ; and a tower 45 ft. high standing on the top of the hill subtends an angle of 5 16'. Find the height of the hill. Ans. 173 ft. 30 466 PRACTICAL MATHEMATICS 17. Two observers, A and B, 100 rd. apart on a horizontal plane ob- serve at the same instant an aviator. His angle of elevation at A is 68 25 and at B 55 58.2'. The angles in the horizontal plane made by the projections of the lines of sight with the line AB are 43 27' at A and 23 45' at B. Find the height of the aviator. Ans. 1820 ft. 18. B is 42 miles from A in a direction W. 22 N., and C is 58 miles from A in a direction E. 73 N. What is the position of C relative to Bl Ana. 68.6 mi. E. 35 21.1' N. 376. Resultant of forces. If two forces, represented by the vectors PQ and PS, Fig. 296, act upon a body at point P, then the combined effect of these forces is the same as that of the force, represented by the vector PR, where PR is the diagonal of the parallelogram of which PQ and PS Fio. 295. are t wo sides. The force PR is called the s R resultant of the forces PQ and PS. The resultant of any number of forces is a single force that will produce the same effect as the combined effect of all the given forces. FIG. 290. The resultant of any number of given forces can bd found by find- ing the resultant of any two of the given forces, then the resultant of a third force and the first resultant, continuing till all the forces are used. Thus, if a, 6, c, and d, Fig. 297, are four forces acting at the point P, r\ is the resultant of a and b, r 2 the resultant of r\ and c, and r 3 the resultant of r 2 and d. Therefore r 3 is the resultant of a, b, c, and d. 376. Computation of a resultant. If two forces act at right angles to each other, their resultant is evidently equal Fio. 297. SOLUTION OF OBLIQUE TRIANGLES 467 in magnitude to the square root of the sum of the squares of their respective magnitudes. Thus, in Fig. 298, r=\/a z -\-b 2 . Also the direction cxf r can be found, for tan QPR = If two forces act at any angle to each other, their resultant can be found by using the law of cosines, Art 369, and is equal in magnitude to the square root of the sum of the squares of their respective magnitudes increased by twice their product times the cosine of the angle between the two forces. Thus, in Fig. 299, r = \/a 2 -}-b 2 2ab cos

, r = \/a 2 +& 2 -4-2a6 cos 0. The angle between the resultant and either force can be found by the law of sines, Art. 368. Velocities can be combined in exactly the same manner as forces. EXERCISES 126 1. Given two forces of 40 Ib. and 60 lb., acting at an angle of 30. Find the magnitude of their resultant. Ans. 67.66+ lb. 2. Given a force of 500 lb. acting toward the east and a force of 700 lb. acting northeast. Find the resultant in magnitude and direction. Ans. A force of 1111.3 lb. acting 26 26.9' N. of E. 3. An aeroplane, which is at an altitude of 1600 ft. and moving at the rate of 100 miles per hour in a direction due east, drops a bomb. Dis- regarding the resistance of the air, where will the bomb strike the ground if during its fall it is acted upon by a wind of 40 miles an hour from a direction 30 east of south? .4ns. 1279 ft. 23 24' N. of E. of point where bomb was dropped. Solution. To find the number of seconds it is in falling, use equation of exercise 30, page 339, \gP = 1600, where g = 32. This gives t = 10, the time in seconds. 468 PRACTICAL MATHEMATICS It would move east from the starting point as far as the aeroplane travels in 10 seconds if the wind is not considered. 100X5280X10 60X60 During 10 seconds the wind would carry the bomb 40X5280X10 60X60 The resultant of these displacements is _ V / 1467 1 +586.7*+2Xl467X586.7Xcos 120 = 1279 ft By the sine law the direction is found to be 23 24' N. of E. 4. An automobile is traveling W. 36 N. at 27 miles per hour, and the wind is blowing from the N. E. at 30 miles per hour. What velocity and direction does the wind appear to have to the chauffeur? Ana. 37.09 mi. per hr. from N. 59' W. 5. A train is running at 30 miles per hour in a direction W. 35 S., and the engine leaves a steam track in the direction E. 10 N. The wind is known to be blowing from the N. E. ; find its velocity. Ans. 22 mi. per hr. 6. In a river flowing due south at the rate of 4 miles per hour, a boat is drifted by a wind blowing from the southwest at the rate of 15 miles per hour. Determine the position of the boat after 40 minutes if resist- ance reduces the effect of the wind 70 per cent. Ans. 2. 19 mi. E.14 26' S. TABLES I. SUMMARY OF FORMULAS. II. USEFUL NUMBERS. III. DECIMAL AND FRACTIONAL PARTS OF AN INCH. IV. ENGLISH INCHES INTO MILLIMETERS. V. U. S. STANDARD AND SHARP V-THREADS. VI. CHORDS OF ANGLES IN CIRCLES OF RADIUS UNITY. VII. STANDARD GAGES FOR WIRE AND SHEET METALS. VIII. SPECIFIC GRAVITIES AND WEIGHTS OF SUBSTANCES. IX. STRENGTH OF MATERIALS. X. FOUR-PLACE TABLE OF LOGARITHMS. XI. TABLE OF NATURAL AND LOGARITHMIC SINES, COSINES, TANGENTS, AND COTANGENTS OF ANGLES DIFFERING BY TEN MINUTES. 469 TABLE I Summary of Formulas 11] A = ab, rectangle, parallelogram. [2] a = A-5-6, rectangle, parallelogram. [3] b = A-j-a, rectangle, parallelogram. [4] A = $ab, triangle. [5] a =-2A -j-b, triangle. [6] b = 2A-5-a, triangle. [7] A = \/s(8 a)(s b)(s c), where = i(a+b+c), triangle. [8] A = JCB+b) Xa, trapezoid. [9] c=\/a 2 +6 2 , right triangle. [10] a =VC 2 b 2 , right triangle. [11] b = \/c 2 a 2 , right triangle. [12] I : 1 =t : T, tapers. D-d^L [13] X "~ rt ^ 7 J tflpcrs. t I [14] D, 1.732 , _ , = D \r ' snarp V-tnreads. 1 2oHo u -..j t u. o. >3. inn'iicis. [16] r (iw) 2 +A 2 n _ f _. ,_ ^yT ) segment ot circle. [17] h =r \/r 2 (Jtc) 2 , segment of circle. [18] w = 2\/h(2r h), segment of circle. [19] C =rd, circle. [20] d = C-:-5r, circle. [21] C = 2nr, circle. [22] 2r = C-f-ir, circle. [23] A = ^Cr, circle. [24] A = nT 2 , circle. [25] A = \-ird* = 0.7854rf 2 , circle. [26] r = \/A-J-T, circle. [27] d = \/A -j- JTT, circle. [28] A r = A-a=*R*-irr* = ir(R l -r*)=*(R+r)(R-r), ring. [29] A = g^Xr*, sector. I30(a)] A /t 1 = Aw+o segment. I30(b)] A = JA* Vy- 0.608, segment. 131] A = Tab, ellipse. [32(a)J P T(a-f-b), ellipse. 470 TABLES 471 [32(c)] P=7r\/2(a 2 +b 2 ), ellipse. [33] S = ph, prism. [34] T = ph+2A, prism. [35] r = 6o 2 , cube. [36] p=S + h, prism. [37] h=S + p, prism. [38] V = Ah, prism. [39] F = a 3 , cube. [40] h = V +A, prism. [41] A = F-T-/I, prism. [42] S = Ch =Trdh=2irrh, cylinder. [43] V = Ah = wr 2 h = \Trd*h, cylinder. [44] h = V -T- A = V -s-irr 2 , cylinder. [45] A = V -v- h, cylinder. [46] V =-irR 2 h irr^h =irh(R 2 -r 2 ) =vh(R +r) (R-r), cylinder. [47] S = %ps, pyramid or cone. [48] T = %ps-{-A, pyramid or cone. [49] S = i(P-fp)Xs, frustum. [50] T = l(P+p)Xs+B+b, frustum. [51] V = lAh, pyramid or cone. [52] V = %h(B+b+\/Bb), frustum of pyramid or cone. [53] V = %irh(R+r 2 +Rr), frustum of cone. [54] V=&Trh(D 2 +d 2 +Dd), frustum of cone. [55] S = 4irr 2 , sphere. [56] V \Sr = |7ir 3 = gird 3 , sphere. [57] Z =2irrh, area of zone. [58] F = 57r/t(r 2 i+i" 2 2)+g7r/i 3 , volume of segment. [59] A =2wRX2Trr = 4TT 2 Rr, solid ring. [60] F = 27r#X7rr 2 =27r 2 .Rr 2 , solid ring. [61] V = h(B l +4M+B 2 '), prismatoid. [62] sin (90 -0) =cos 6. [63] cos (90 0) =sin 6. [64] tan (90 -0)= cot d. [65] cot (90- e)= tan d. [66] sec (90 -0) =csc 0. [67] esc (90 -0)= sec 9. [68] sin (180 -6} =sin 9. [69] cos (180 (?) = -cos 0. [70] tan ( 180 -0) = - tan 0. [71] cot (180 0) = -cot 0. [72] sec (180 -0) = -sec 0. [73] esc (180-\ tan A +tan B 1 1 1 1 1 ( ^. f~ ,/J ) ^ s ~ ^* [101] tan (A B) tan A ~ tan B 1+tanAtanB [102] sin 2A =2 sin A cos A. [103] cos 2A=cos 2 A -sin 2 A =12 sin 2 A =2 cos 2 A-l [104] tan o , _ 2 tan A tall t\ - 1 tan 2 A [105] /I -cos 2A |n ^ _I_ .4 I sin A : "\/ 9 ' ^ [106] /I +cos 2A f*r\a a <4- * / - ^ [107] 11 -cos 2A sin 2 A 1 -cos 2 A ' V 1 +cos 2 A : " 1 +cos 2 A sin 2 A [108] sin A cos B = i sin (A +B) +} sin (A -B). [109] cos A sin B = i sin (A +B) i sin (A B). [110] cos A cos B = ^ cos (A +B) + i cos ( A B). [111] sin A sin B = cos (A +B) + J cos (A B). [112] sin A+sin B=2sin i(A+B) cos J(A-B). [113] sin A sin B =2 cos j(A +B) sin (A B). 1114] cos A +cos B =2 cos \(A +B) cos $(A B). 1115] cos A cos B = 2 sin i(A +B) sin j(A B). TABLES 473 sin A sin B sin C [117] a 2 = b 2 +c 2 -26ccos4. [118] 6 2 =o 2 +c 2 -2accos B. [119] c 2 =a 2 +6 2 2ab cos C. 7,2_|_,.2_, 7 2 [120] cosA= + . ' Zbc /7 2_l_ r 2_7,2 [1211 ...B-tjiLj-L [122] cos C = " 2a6 Formulas for solving Case III, oblique triangles, by logarithms: [123J tan \(B-C) =j- tan |~C [124] tan $(A-C) =^- tan l(A+C). a i~c [125] tan hA -5)= tan Formulas for solving Case IV, oblique triangles, by logarithms : a+b+c Let s = [130] cos \B = [131] cos |C = Let r 1R-- /(-&). , /(8-q)(-b)(-c) -\- p- [132] tan *A=- s a [133] tan |/? = -%- s o [134] tanJC = -^-- s c [135] Area of a triangle = 5 bcsin A = \ac sin B = %ab sin C = \/s(s a)(s 6) (s c). Other formulas for any angle. tan 6 1 A/1 +tan T VHhcoV- 474 PRACTICAL MATHEMATICS 1 cote \/l+tan0 Vl+cot f 1 BCC OM " [1371 cos 0~ Vl-sm'0 U88 , tan g. -JEL V 1 sin* [139] cot [140] sec = [141] csc0-- - == sin 6 V 1 cos 2 Vaec*0-l [142] sin 3A =3 sin A -4 sin 3 A. [143] cos ZA =4 cos 3 A -3 cos A. 3 tan A tan 3 A [144] tan 34 = - TABLE II Useful Numbers 1 cu. ft. of water weighs 62.5 Ib. (approx.) = 1000 oz. 1 gal. of water weighs 8 Ib. (approx.). 1 atmosphere pressure = 14.7 Ib. per sq. in. =2116 Ib. per sq. ft. 1 atmosphere pressure = 760 mm. of mercury. A column of water 2.3 ft. high =a pressure of 1 Ib. per sq. in. 1 gal. =231 cu. in. (by law of Congress). 1 cu. ft. =7J gal. (approx.) or better 7.48 gal. 1 cu. ft. = j bu. (approx.). 1 bbl. =4.211- cu. ft. (approx.). 1 bu. =2150.42 cu. in. (by law of Congress) = 1.24446- cu. ft. 1 bu. = J cu. ft. (approx.). 1 perch =24} cu. ft. but usually taken 25 cu. ft. 1 in. =25.4001 mm. (approx.). 1 ft. =-30.4801 cm. 1 m. =39.37 in. (by law of Congress). 1 lh. (avoirdupois) = 7000 grains (by law of Congress). TABLES 1 Ib. (troy or apothecaries) = 5760 grains. 1 gram = 15.432 grains. 1 kg. =2.20462 Ib. (avoirdupois). I liter = 1.05668 qt. (liquid) =0.90808 qt. (dry). 1 qt. (liquid) =946.358 cc. =0.946358 liter, or cu. dm. 1 qt. (dry) =1101.228 cc. =1.101228 liters, or cu. dm. TT= 3. 14159265358979 + = 3.1416 = fB = 3f (all approx.). 1 radian = 57 17' 44.8" = 57.2957795+. 1 = 0.01745329+ radian. Base of Napierian logarithms =e =2.718281828 . log 10 e= 0.43429448 log JO =2. 30258509 . 1 horse-power second = 550 foot-pounds. 1 horse-power minute = 33,000 foot-pounds. V2 = 1.4142136. V3 = 1.7320508. \/5 = 2.2360680. \/6 = 2. 4494897. ^2 = 1.2599210. -^3 = 1.4422496. 475 TABLE III Decimal and Fractional Equivalents of Parts of an Inch 8ths 32nds 64ths 1 = . 125 1 = . 03125 1 = . 015625 33 =.515625 2 =.250 3 = .09375 3 =.046875 35 =.546875 3 = . 375 5 = .15625 5 =.078125 37 =.578125 4 = . 500 7 =.21875 7 = . 109375 39 = . 609375 5 = . 625 9 =.28125 9 = . 140625 41 = . 640625 6 = . 750 11 = .34375 11 = .171875 43 = . 671875 7 = . 875 13 =.40625 13 =.203125 45 =.703 125 16ths 15 = . 46875 15 = . 234375 47 = . 734375 1 = . 0625 17 =.53125 17 = . 265625 49 = . 765625 3 =.1875 19 = . 59375 19 = . 296875 51 = . 796875 5 =.3125 21 = . 65625 21 = .328125 53 = . 828125 7 = .4375 23 =.71875 23 = . 359375 55 = . 859375 9 = . 5625 25 =.78125 25 = . 390625 57 = . 890625 11 = .6875 27 = . 84375 27 =.421875 59 = . 921875 13 = .8125 29 =.90625 29 = .453125 61 = .953125 15 = .9375 31 = .96875 31 = .484375 63 = .984375 476 I'll A C'TICAL MA THEM A TICS ffl : *f O> >O O O H cp CN t- CN O5 CO O5 ^f Cft 1C O CO ' CO C$ h- CO OO I > * + O W a A c HH J3 w ^ (J .^ . _____ . . g > "cog,beobed6r!g5gQerJggg'-!^^.^. ll3M3IOKa . --.-- ~r~~~r,, s ' I H2 "cooooocoio6ooco?oooc5coQOn!M=eoSr;?2?S? g -a o CO CO CO CO ^t* ^* ^* ^& *^ *^ "O *O __ , _ * -.^J.,-^ \. - J," " "'.."_. O'* oo e o o ^ oo ci i o ** oo M o o^t" oo cNloo * OO~CN e ^5 ^5 ^^ ^? ^O ^^ I s * C^ t^ CO QC ^t* O^ ^* C_? ^O ^^ ^3 ^^ ^** ^^ ^- CO CC ^* o^ *o t 1 ^ ^5 01 *o i*^ ^3 ^^ *^ t^* ^b co *o oc ^^ co *o oc ^2 re *^ ^c c fl O ,' _.--..---.,-._,*,. .^^*-_.~^ TABLES 477 TABLE V U. S. Standard and Sharp V-threads Diameter of screw Threads per inch Depth U. S. S. Depth sharp V Root dia. U. S. S. Root dia. sharp V i 20 . 03247 .04330 .1850 .1634 tV 18 .03608 .04811 .2403 .2163 t 16 . 04059 .05412 .2936 .2668 A 14 .04639 .06178 .3447 .3139 * 13 . 04996 . 06661 .4001 .3668 A 12 .05412 .07216 .4542 .4182 f 11 . 05905 .07873 .5069 .4675 i 10 .06495 .08660 .6201 .5768 1 9 .07216 .09622 .7307 .6826 i 8 .08119 . 10825 .8376 .7835 li 7 .09277 .12371 .9394 .8776 H 7 .09277 .12371 1.0644 1.0026 H 6 . 10825 . 14433 1.1585 1.0863 U 6 . 10825 . 14433 1 . 2835 1.2113 l! 5 .11809 . 15745 1 . 3888 1.3101 if 5 . 12990 . 17325 .1 . 4902 1 . 4035 U 5 . 12990 . 17325 1.6152 1 . 5285 2 4J . 14433 . 19244 1.7113 1.6151 2J 4 . 14433 . 19244 1.9613 1.8651 2J 4 . 16238 .21650 2.1752 2.0670 2f 4 . 16238 .21650 2.4252 2.3170 3 3J . 18557 . 24742 2.6288 2.5052 3} 3J . 18557 . 24742 2.8788 2.7552 3^ 31 . 19985 . 26647 3.1003 2.9671 3| 3 .21666 . 28866 3.3167 3.1727 4 3 . 21666 .28866 3.5667 3.4227 4J 21 .2259 .3012 3 . 7982 3.6476 4* 21 .2362 .3149 4.0276 3.8712 4f 2f .2474 .3299 4.2551 4.0901 5 2-1 .2598 .3465 4.4804 4.3070 51 21 .2598 .3465 4.7304 4.5500 5* 2| .2735 .3647 4.9530 4.7707 51 2f . 2735 .3647 5.2030 5.0207 6 2J .2887 .3849 5.4226 5.2302 478 PRACTICAL MATHEMATICS > 3 H S m .S as CO CO 1C .-. - - CO CN CN d s SB _ _ _ CO CO M CN 1 J B ! g. CN CN CN CN -H ~ O O OS 3 o v - 8. O O O O O OS OS 00 00 Is S * g* 05 OS CS 00 00 CO t^o 1 1 t^r^^ !> CO CO COCO.O J B I M 1C 1C 1C 1C 1C >C 1 tc S CO CO CO CO CO CO CO CO CO s 2 CN CN CN CN CN CN i-H 1 r-l 1 - t^. O 1-H 1C CO O 1-1 CO 1C r^ os ic CO CN 00 CO 00 CS CO CN CN CO t^ O 1 *o 00 O CO * Tf -H 00 1-1 CO ^ * CO O S"H l>- 00 OS CO C! OS i 1C CO ^H CN CO . be 1. CN cp r^ CN OS CO i-i CN * 1C l>- ^J CO C5 1C Tj< 1C t- 1-1 CN CO 1 1 1 1 1-H .& 3 1 1 U5 N- H CO S3 CO I s - OS 1C 00 -H t^ 1C O CN CO 00 "** i a s - O CO CO t>- T*< 1-H O CN -^ ggs CO 00 O CO - c d aj 05 09 -w bD Lengths W CN CO OS iC CN OS O CN CO 00 CO CN CO CO CS c p 10 -1< OS CN O -* CO -3 * 3 ^ O CO CN CO 1-H t* t^ O 1C CN I>- -H -H CO *S ^ s O GJ ^ - h- ^< < ssi * Q ~* CO O CO 1C t^. 00 1C < C35 ^i O OS O 1-1 CN FH 1-H 1 1 u ^ 5 & SS 00 Tf 1C iC CO oo 58 5*0*0 111 rt i-l H b b o o r o a 23 O Q CO -f C to t>- <* t^ ' TABLES 479 TABLE VII Standard Gages for Wire and Sheet Metals Diameter or thickness given in decimals of an inch Number of gage Birmingham wire gage American, Brown and Sharp (B. and S.) United States standard plate iron steel British Imperial American Steel and Wire Co. 3000000 .5 .5 000000 .46875 .464 00000 .4375 .432 0000 .454 .46 .40625 .400 .3938 000 .425 .409642 .375 .372 .3625 00 .380 .364796 .34375 .348 .3310 .340 .324861 .3125 .324 .3065 1 .300 . 289297 .28125 .300 .2830 2 .284 .257627 .265625 .276 .2625 3 .259 . 229423 .25 .252 .2437 4 .238 .204307 .234375 .232 .2253 5 .220 . 181940 .21875 .212 .2070 6 .203 . 162023 .203125 .192 .1920 7 .180 . 144285 .1875 .176 .1770 8 .165 . 128490 .171875 .160 .1620 9 .148 .114423 . 15625 .144 .1483 10 .134 .101897 . 140625 .128 .1350 11 .120 . 090742 .125 .116 .1205 12 .109 . 080808 . 109375 .104 .1055 13 .095 .071962 .09375 .092 .0915 14 .083 . 064084 .078125 .080 .0800 15 .072 . 057068 .0703125 .072 .0720 16 .065 .050821 .0625 .064 .0625 17 .058 . 045257 .05625 .056 .0540 18 .049 .040303 .05 .048 .0475 19 .042 .035890 .04375 .040 .0410 20 .035 .031961 .0375 .036 .0348 21 .032 .028462 .034375 .032 .03175 22 .028 . 025346 .03125 .028 .0286 23 .025 .022572 .028125 .024 .0258 24 .022 .020101 .025 .022 .0230 25 .020 .017900 .021875 .020 .0204 26 .018 .015941 .01875 .018 .0181 27 .016 .014195 .0171875 .0164 .0173 28 .014 .012641 .015625 .0148 .0162 29 .013 .011257 .0140625 .0136 .0150 30 .012 .010025 .0125 .0124 .0140 31 .010 .008928 .0109375 .0116 .0132 32 .009 .007950 .01015625 .0108 ' .0128 33 .008 .007080 .009375 .0100 .0118 34 .007 . 006305 .00859375 .0092 .0104 35 .005 .005615 .0078125 .0084 .0095 36 .004 .005000 .00703125 .0076 .0090 37 .004453 .006640625 .0068 38 .003965 .00625 .0060 39 .003531 .0052 40 .003144 .0048 480 PRACTICAL MATHEMATICS TABLE VIII Specific Gravities and Weights of Substances .Vamc of substance Pounds per en. in. Pounds per cu. ft. Specific gravity Air 0.0795 Aluminum 162 2 6 Anthracite coal, broken. . . . 52 to 60 Antimony 418 6 7 Asphaltum 87.3 1 4 Beech wood 46 .73 Birch wood . . 41 65 Brass, cast (copper and zinc) .... 506 8 1 Brass, rolled 525 8 4 Brick, common 125 Brick, pressed . ... 150 Chalk 156 2 5 Coal, bituminous, broken 47 to 56 Coke, loose 23 to 32 Corundum . . 3 9 Copper, cast 542 8 . 6 to 8 8 Copper, rolled .319. 555 8 . 8 to 9 Cork 15 24 Ebony wood 76 1 23 Elm wood 35 .56 Flint 162 2 6 Glass 186 2 5 to 3 45 Gold .695 19 3 Granite 170 2.56 to2.88 Hickory wood . . 53 85 Ice 57.5 .92 Iron, cast. .26 450 6.7 to 7.4 Iron, wrought. . . .28 480 7.69 Lead .412 712 11 42 Marble .... 168.7 2.7 Maple wood 49 79 Mercury. . .49 13.6 Nickel .318 8.8 Oak wood, red .... 46 .73 to .75 Pine wood white . 28 .45 Pine wood yellow . . 38 .61 Platinum . . .... 21.5 Quartz 165 2.65 Silver .379 655 10.5 TABLES 481 Specific Gravities and Weights of Substances Continued Name of substance Pounds per cu. in. Pounds per cu. ft. Specific gravity Steel .29 490 459 438 62.417 62 . 355 59.7 7.85 7 . 2 to 7 . 5 6 . 8 to 7 . 2 1 Tin Zinc . . Water, distilled Water, distilled Water, distilled at 32 F. . at 62 F at 212 F Specific gravities referred to air : Air 1 Oxygen 1.11 Hydrogen . 07 Chlorine gas 2 . 44 TABLE IX Strength of Materials Ultimate tensile strength Ultimate com- pressive strength Coefficient of linear expansion Pounds per square inch Pounds per square inch For 1 Fahrenheit Hard steel 100,000 120,000 0000065 Structural steel 60,000 60,000 . 0000065 Wrought iron 50,000 50,000 0000067 Cast iron 20,000 90,000 0000062 Copper ... . .... 30,000 . 0000089 Timber, with grain Concrete 8,000 to 25,000 300 4,000 to 12,000 3,000 0.0000028 0000055 Granite. ... . . . . 11,000 0000050 Brick 3,000 0000050 In the column of ultimate tensile strengths are given the pulls neces- sary to break a rod of one square inch cross section of the given material. In the column of ultimate compressive strengths are given the weights necessary to cause a support of one square inch cross section to give way under the pressure. In the column of coefficient of linear expansion are given the fractional parts of their length, bars of the different materials will increase when the temperature rises one degree Fahrenheit. 31 482 PRACTICAL MATHEMATICS TABLE X. COMMON LOGARITHMS t) , . . 1 1 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 1* 0792 0828 0864 0899 0934 ii'.tii'.i 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 .1584 1614 1644 1673 1703 1732 16 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 ITU 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 10 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 UM 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 80 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 6185 5198 5211 5224 5237 5250 5263 5276 5380 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 36 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 40 6021 6031 6O42 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 fC'Sl 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 i.l. ->1 6464 6474 6484 6493 6503 6513 6522 46 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 BOH 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 60 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 61 7070 7084 7093 7101 7110 7118 7126 7135 7143 7152 62 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 63 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 64 7324 7332 7340 7348 7350 7364 7372 7380 7388 7396 N. 1 2 3 4 9 6 7 8 9 TABLES TABLE X. COMMON LOGARITHMS Continued 483 N. 1 2 3 4 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 66 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 N. 1 2 3 4 5 6 7 8 9 PRACTICAL MATHEMATICS TABLE XI. TRIGONOMETRIC FUNCTIONS Angle* Sine* Cosines Tangents Cotangents AngW Nat. Log. Nat. Log. Nat. Log. Nat. Log. O e OO' .0000 a, 1 . 0000 OOOO .0000 OP CD 00 90 00' 10 .0029 7.4637 1.0000 0000 .0029 7.4637 343.77 2.5363 50 20 .0058 7048 1.0000 OOOO .0058 7648 171.89 2352 40 30 .0087 9408 1.0000 OOOO .0087 9409 114.59 0591 30 40 .0116 8.0658 .9999 OOOO .0116 8.0658 85.940 1.9842 20 50 .0145 1627 .9900 OOOO .0145 1627 68 . 750 8373 10 100' .0175 8.2419 .90M 9.9999 .0175 8.2419 57.290 1.7581 89 00' 10 .0204 3088 . 9998 9999 .0204 3089 49.104 6911 50 20 .0233 3668 .9997 9999 .0233 3669 42.964 6331 40 30 .0262 4179 .9997 9999 .0262 4181 38.188 5819 30 40 .0291 4637 . 9996 9998 .0291 4638 34 . 368 5362 20 50 . 0320 5050 .9995 9998 .0320 5053 31.242 4947 10 2 00' .0349 8.5428 .9994 9.9997 .0349 8.5431 28.636 1.4569 88 00' 10 .0378 5770 .9993 9997 .0378 5779 26.432 4221 50 20 .0407 6097 .9992 9996 .0407 6101 24 . 542 3800 40 30 .0430 6397 .9990 9996 .0437 6401 22.904 3599 30 40 .0465 6077 .9989 9995 .0466 6682 21.470 3318 20 60 .0494 6940 .9988 9995 .0495 6945 20.206 3055 10 3 00' .0523 8.7188 .9986 9.9994 .0524 8.7194 19.081 1.2806 87 00' 10 .0552 7423 .9985 9993 .0553 7429 18.075 2571 50 20 .0581 7645 .9983 9993 .0582 7652 17.169 2348 40 30 .0610 7857 .9981 9992 .0612 7865 16.350 2135 30 40 .0640 8059 .9980 9991 .0641 8067 15.605 1933 20 60 .0669 8251 .9978 9990 .0670 8261 14.924 1739 10 400 / .0098 8.8436 .9976 9.9989 .0699 8.8440 14.301 1.1554 86 00' 10 .0727 8613 .9974 9989 .0729 8024 13.727 1376 50 20 .0756 8783 .9971 9988 .0758 8795 i 13.197 1205 i 40 30 .0785 8946 .9909 9987 .0787 8960 12.700 1040 1 30 40 .0814 9104 . 9967 9980 .0810 9118 12.251 0882 20 50 .0843 9256 .9964 9985 .0840 9272 11.820 0728 10 5 00' .0872 8.9403 .9902 9.9983 .0875 8.9420 11.430 1.0680 85 00' 10 .0901 9545 .9959 9982 .0904 9503 11.059 0437 50 20 .0929 9682 .9957 9981 .0934 9701 10.712 0299 40 30 .0958 9816 .9954 9980 .0963 9836 10.385 0164 30 40 .0987 9945 .9951 9979 .0992 9966 10.078 0034 20 60 .1016 9.0070 .9948 -9977 .1022 9.0093 9.7882 0.9907 10 6 00' .1045 9.0192 .9945 9.9976 .1051 9.0216 9.5144 0.9784 84 00' 10 .1074 0311 .9942 9975 . 1080 0336 9.2553 9664 50 20 .1103 0426 .9939 9973 .1110 0453 9.0098 9547 40 30 .1132 0539 .9936 9972 .1139 0567 8.7769 9433 30 40 .1161 0648 .9932 9971 .1169 0678 8.5555 9322 20 50 . 1 190 0755 .9929 9969 .1198 0786 8.3450 9214 10 7 00' .1219 9.0859 .9925 9.9968 .1228 9.0891 8.1443 0.9109 83 OO 7 10 .1248 0961 .9922 9966 .1257 0995 7.9530 -9006 50 20 .1276 1060 .9918 9964 .1287 1096 7.7704 8904 40 30 .1305 1157 .9914 9963 .1317 1194 7.5958 8806 30 40 .1334 1252 .9911 9961 .1346 1291 7.4287 8709 20 50 . 1363 1345 .9907 9959 .1376 1385 7.2687 8615 10 800 / .1392 9.1436 .9903 9.9958 .1405 9.1478 7.1154 0.8522 82 00' 10 .1421 1525 .9899 9956 . 1435 1509 0.9082 8431 50 20 .1449 1612 .9894 9954 .1465 1658 0.8269 8342 40 30 . 1478 1697 . 9890 9952 .1495 1745 6.0912 8255 30 40 .1507 1781 .9886 9950 .1524 1831 6.5600 8169 20 50 .1536 1863 .9881 9948 .1554 1915 6.4348 8085 10 9 00' .1504 9.1943 i .9877 9.9940 .1584 9 1997 6.3138 0.8003 81 00' Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angle* Cosines Sines Cotangents Tangents Angles TABLES 485 TABLE XL TRIGONOMETRIC FUNCTIONS Continued Angles Sines Cosines Tangents Cotangents Angles Nat. Log. Nat. Log. Nat. Log. Nat. Log. 9 00' .1564 9.1943 .9877 9.9946 .1584 9.1997 6.3138 0.8003 81 00' 10 .1593 2022 .9872 9944 . 1614 2078 6.1970 7922 50 20 . 1622 2100 .9868 9942 .1644 2158 6.0844 7842 40 30 .1650 2176 . 9863 9940 . 1673 2236 5.9758 7764 30 40 .1679 2251 .9858 9938 . 1703 2313 5.8708 7687 20 50 . 1708 2324 .9853 9936 . 1733 2389 5.7694 . 7611 10 10 00' .1736 9.2397 .9848 9.9934 .1763 9.2463 5.6713 0.7537 80 00' 10 , .1765 2468 .9843 9931 . 1793 2536 5.5764 7464 50 20 . 1794 2538 . 9838 9929 . 1823 2609 5.4845 7391 40 30 . 1822 2606 .9833 9927 .1853 2680 5.3955 7320 30 40 . 1851 2674 .9827 9924 . 1883 2750 5.3093 7250 20 50 . 1880 2740 .9822 9922 .1914 2819 5.2257 7181 10 11 00' .1908 9.2806 .9816 9.9919 .1944 9.2887 5.1446 0.7113 79 00' 10 . 1937 2870 .9811 9917 . 1974 2953 5.0658 7047 50 20 . 1965 2934 .9805 9914 . 2004 3020 4.9894 6980 40 30 . 1994 2997 .9799 9912 .2035 3085 4.9152 6915 30 40 . 2022 3058 . 9793 9909 .2065 3149 4.8430 6851 20 50 .2051 3119 . 9787 9907 .2095 3212 4.7729 6788 10 12 00' .2079 9.3179 .9781 9.9904 .2126 9.3275 4.7046 0.6725 78 00' 10 .2108 3238 .9775 9901 .2156 3336 4 . 6382 6664 50 20 .2136 3296 .9769 9899 .2186 3397 4 . 5736 6603 40 30 .2164 3353 .9763 9896 .2217 3458 4.5107 6542 30 40 .2193 3410 .9757 9893 .2247 3517 4.4494 6483 20 50 .2221 3466 . 9750 9890 .2278 3576 4.3897 6424 10 13 00' .2250 9.3521 .9744 9.9887 .2309 9.3634 4.3315 0.6366 77 00' 10 .2278 3575 .9737 9884 .2339 3691 4 . 2747 6309 50 20 . 2306 3629 .9730 9881 .2370 3748 4.2193 6252 40 30 .2334 3682 .9724 9878 . 2401 3804 4.1653 6196 30 40 .2363 3734 .9717 9875 . 2432 3859 4.1126 6141 20 50 .2391 3786 .9710 9872 .2462 3914 4.0611 6086 10 14 00' .2419 9.3837 .9703 9.9869 .2493 9.3968 4.0108 0.6032 76 00' 10 .2447 3887 .9696 9866 .2524 4021 3.9617 5979 50 20 . 2476 3937 .9689 9863 . 2555 4074 3.9136 5926 40 30 . 2504 3986 .9681 9859 .2586 4127 3 . 8667 5873 30 40 . 2532 4035 . 9674 9856 .2617 4178 3.8208 5822 20 50 . 2560 4083 .9667 9853 .2648 4230 3.7760 5770 10 15 00' .2588 9.4130 .9659 9.9849 .2679 9.4281 3.7321 0.5719 75 00' 10 .2616 4177 .9652 9846 .2711 4331 3.6891 5669 50 20 . 2644 4223 . 9644 9843 .2742 4381 3.6470 5619 40 30 . 2672 4269 .9636 9839 . 2773 4430 3 . 6059 5570 30 40 .2700 4314 .9628 9836 . 2805 4479 3.5656 5521 20 50 .2728 4359 .9621 9832 . 2836 4527 3.5261 5473 10 16 00' .2756 9.4403 .9613 9.9828 .2867 9.4575 3.4874 0.5425 74 00' 10 .2784 4447 .9605 9825 . 2899 4622 3.4495 5378 50 20 .2812 4491 .9596 9821 .2931 4669 3.4124 5331 40 30 . 2840 4533 .9588 9817 .2962 4716 3 . 3759 5284 30 40 .2868 4576 .9580 9814 .2994 4762 3.3402 5238 20 50 .2896 4618 .9572 9810 .3026 4808 3.3052 5192 10 17 00' .2924 9.4659 .9563 9.9806 .3057 9.4853 3.2709 0.5147 73 00' 10 . 2952 4700 .9555 9802 . 3089 4898 3.2371 5102 50 20 .2979 4741 .9546 9798 .3121 4943 3.2041 5057 40 30 .3007 4781 .9537 9794 .3153 4987 3.1716 5013 30 40 .3035 4821 .9528 9790 .3185 5031 3.1397 4969 20 50 3062 4861 .9520 9786 .3217 5075 3.1084 4925 10 18 00' .3090 9.4900 .9511 9.9782 .3249 9.5118 3.0777 0.4882 72 00' Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angles Cosines Sines Cotangents Tangents Angles 486 PR A CTICAL MA THEM A TICS TABLE XI. TRIGONOMETRIC FUNCTIONS Continued Angle* Sine* Cosines Tangents Cotangents Angles Nat. Log. Nat. Log. Nat. Log. Nat. Log. 18 00' .3090 9.4900 .9511 0.9782 .3249 9.5118 3.0777 0.4882 72 00' 10 .3118 4939 .9502 9778 .3281 6161 3.0475 4839 60 20 .3145 4977 .9492 9774 .3314 6203 3.0178 4797 40 30 .3173 6015 .9483 9770 .3346 5245 S.M87 4755 30 40 .3201 5052 .9474 9765 .3378 5287 2.9600 4713 20 60 .3228 5090 .9465 9761 .3411 5329 2.9319 4671 10 19 00' .3256 9.5126 .9455 9.9757 .3443 9.5370 2.9042 0.4630 71 00' 10 .3283 5163 .9446 9752 .3476 5411 2.8770 4589 50 20 .3311 5199 .9436 9748 .3508 5451 2.8502 4549 40 30 .3338 5235 .9426 9743 .3541 5491 2.8239 4509 30 40 .3365 5270 .9417 9739 .3574 5531 2.7980 4469 20 60 .3393 5306 .9407 9734 .3607 5571 2 . 7725 4429 10 20 00' .3420 9.5341 .9397 9.9730 .3640 9.5611 2.7475 0.4389 70 0 I/BRANCH, >S mxS