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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. PA«K
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. PAc»
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 0° 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 0 and 90° + 0 443
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 0 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 0 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 0 = 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 _0 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, Jg2-, 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, 46j
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 -j7^ 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. TV5V Ans. f- 10. ffi- Ans. H-
11. fil?- Ans.lt- 12. r9|g- Arw. H-
13. Jg|3- Ans. $J. 14. AUf. Ans. f
15. if|8- Ans. 1TV 16. mi' -^ns. rfflfc-
17. TViWo- ^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 0 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 ^, ^ i¥o> and 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 2V to lOOths. 8. f , g, i, and | to 120ths.
9. if, A, |, and f to 208ths. Ans. i§f, Hf , HI, Hf •
10. |, T5r, H, and & to 396ths. ylns. HI, HS, Ml, sVr Change the following to fractions having a L. C. D.
11. -| and |- Ans. T50, /„• 12. | and T\- Ans. f f, 4^- 13. fV and if ^^s- io, IS- 14. T93 and T\- ^ns. Ml, Hf • 15. | and fr 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, 4P0, 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 + T52 + 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 T4r from i9i-
Since like numbers can be subtracted we can subtract 4 elevenths fiom 9 elevenths and have the remainder 5 elevenths. This may be written -i9r~~T4r='iV Ans.
Example 2. Subtract T7T 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 <n t
3 TT~ff tt~~T8' A-ns-
Example 3. 7f - 3| = what? Solution.
In this case the fractional part of the 71 = 710
subtrahend is less than that of the 03—09°
minuend. The fractional parts of mixed — -
numbers are reduced to fractions having the L. C. D., the fractional parts subtracted, and then the whole numbers.
Example 4. 7\ — 3f = what?
In this case the fractional part of the subtrahend is greater than the frac- Solution.
tional part of the minuend. The frac- 7^=7f =6f tions are changed to fractions having 3f = 3f =3|- the L. C. D. as before. It is then Ans. 3f
noticed that the fraction | in the sub- trahend is larger than f- in the minuend and so cannot be subtracted from it. To overcome this difficulty we take 1 from the 7 and change it to sixths. This gives 6f instead of 7f . The subtraction is then made as before.
RULE. To find the difference between two fractions having a common denominator, find the difference of the numerators and write it over the common denominator. If the fractions do not have a L. C. D. reduce them to such before subtracting. If the numbers are mixed numbers, subtract the fractional parts and then the whole numbers.
20 PRACTICAL MATHEMATICS
EXERCISES 7
Subtract the following and give the results in their simplest forms.
1. 1-i- 2- l-l 3. A-i-
4- ft-A 6- J-A- 6. 2J-|.
7. 8}-f 8. 7-4S- 9. 9i-l$-
10. ff-rir ^w- 2|f- 11. M-A' An*. TH
12. U-fr Ans. HI- 13. 4|-1/0. Ana. 2^
14. 8i-2|- Ans. 6/0- 15. 9f-3$- ^rw. 5Jf-
16. 463J 17. 346| 18. 461?
146& 146J 145|
19. 469ft 20. 192ft 21. 229 J
21 & 142A 163g
22. 230| 23. 117§ 24. 403ft
103f 96| 2311
Simplify the following, that is, do the operations indicated:
25. 12| +28|-15|- Ans. 25|g- 28. 4?-?-|+6f Ans.
26. 5f+2i-3l- Ans. 4J- 29. 14+6i-9f- Ans. 10$
27. 4J-21+1J- Ans. 3gg- 30. | + 13-(6J-|)+}- Ans. 8JJ-
The parentheses indicate that the enclosed operations must be per- formed first. Thus, in the above, f must be subtracted from 6J before they are subtracted from 13.
31. ?+17+ft-(6 + 9j)- Ans. 21-
32. 4|+3f+6i-(H + H> Ans. 10JJ-
33. 7f+6J-2f+?+2ft- Ans. 14\\-
34. 3i+41 + l?-(?+22»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 6i7o 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 T3^ 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*.
OrT\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 X3§X9X412,
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 ,5j 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. A7sXl6. Ans. IfJ. 40. 7V,X96. Ans. 6T83.
41. fiiX96. Ans. 62£. 42. 28T52X14. Ans. 397«.
43. 816§X17. 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. TV
54. fXfXlfXl=what? Ans. f.
55. AX2jX7f X29gX3|=what? Ans. 7£f-
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 56/0 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 X1/- 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 j3o 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 $2i7(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) $2457; (j) 3095^.
DIVISION OF FRACTIONS
26. Division of a fraction by an integer. — Example 1. Divide 4 by 4.
(1) Since to divide by 4 is to find one of the 4 equal parts and to get \ of a number is to find one of the 4 equal parts, we have
f-s-4 = i off =&• Ans.
(2) Or, using the principle that multiplying the denominator of a fraction divides the value of the fraction, we have
(3) In division of fractions, we can often divide the numera- tor and thus divide the fraction. Example 2. Divide Yi4 by 25.
925 925^-25 37
TT -IT =TT=3*'
This may be written W^-25 = WX-sVHi =3^- Ans.
COMMON FRACTIONS 27
RULE. To divide a fraction by an integer, divide the numerator, or multiply the denominator, of the fraction by the integer; or multiply the fraction by 1 over the integer.
27. Division by a fraction. — Example 1. Divide 6 by f
(1) If we reduce 6 to thirds, we may divide the numerators, since then the numbers will both be thirds, and so be like numbers.
6 -5-f= -^-5-1 = 18 thirds ^-2 thirds =9. Ans.
(2) Since there are 3 times ^ in 1, and | as many times |» there are f times 3, or f times f in 1. Now -f is f in- verted. Hence we can find how many times the fraction | is contained in 1 by inverting the fraction, f will be contained 6 times as many times in 6 as it is contained in 1.
.-. 6-hf = 6Xf = 9. Ans.
Definition. The reciprocal of a number is 1 divided by that number. Thus, f is the reciprocal of f. f is the recipro- cal of f or 4.
Example 2. Divide c§ by £|.
49 _i_ 1 4 — 49V39 — 21— O 1 65 • 39" ~ (55 A 1¥~ "17 — ^1 IT
Example 3. Divide 4| by 3|.
First reducing each to improper fractions, we have 4*-3-3$=¥-¥ = 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.
ljV -5-H.
11. *
j-j.
2. \\-
-3.
7.
_W"X12.
12. f-
LA-
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 § + } =TV of the patch in one hour. They can hoe the entire patch in l-=-i7j = 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 $15?
Ans. 10. 48. 49. 60.
51. 100 -4| 52.
53. 31| -I 54.
In the following five exercises, reduce all to fractions, take reciprocals of each divisor and cancel. 62. 2^X31 Xr4rX2r^-20f.
Process.
2 Xa+20f-xxxx-. Ans.
Ans. ^V.
55. 3}i-^7|f.
A -MO 295
/ins. 5^7.
Ans. |f.
56. T&r-TriT.
Ans. 2f4f.
Ans. !TvV
57. lOH-5-lrf*.
Ans. 10.
Ans. 20 1$.
58. 51-^3-
Ans. 24|.
Ans. T95.
59. 7^3MJ.
Ans. Iff.
Ans. 18.
60. 104T3TH-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-l15a+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»0.
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: <i = 14$, 6 = 16?, c = 33j, and d = 27§; find the values of the fractions expressed by the letters.
86. --. Ans. 161. 88. - An..
c — o —
COMMON FRACTIONS
31
90. An alloy is composed of 92 Ib. of copper, 17 Ib. of zinc, and 7 Ib. of tin. What part of the alloy is of each metal?
Ans. || copper; ^^ zinc; T|? tin.
91. Three men did a piece of work in 26 \ days for which they received $344£. What was the average pay per day for each man?
Ans. $4f
29. Pitch and lead of screw threads. — The pitch of a screw thread is the distance from the center of the top of one thread to the center of the top of the next.
The lead of a screw thread is the distance the screw will move forward in a nut for each complete turn of the screw.
Single Threaded
H -- L- --H
Double Threaded
Triple Threaded
L = Lead, P = Pitch
FIG. 2.
For a single-threaded screw the pitch and the lead arc equal; but for a double-threaded screw, the lead is twice the pitch; and for a triple-threaded screw, the lead is three times the pitch.
In a single-threaded screw, there is only one thread run- ning around the screw; in a double-threaded screw there are two threads running side by side around the screw; and a triple-threaded screw has three threads side by side running around the screw.
The above statements are made clearer by reference to the figure.
32
PRACTICAL MA THEM A TICS
92. According to the Franklin Institute standards for the dimensions of bolts and nuts, a |-in. bolt has 11 threads per inch. What is the pitch? The lead if single-threaded? How many full turns of the nut will it take to advance the bolt 2J in.? Ans. jt in.; ft in.; 24}.
93. A 4i-in. bolt has 2 j threads per inch. What is the pitch? What is the lead if triple-threaded? Ans. i*j in.; 1ft in.
94. In a special threaded screw for a screw-power stump puller, the screw is double-threaded with a pitch of H in- How many turns of the nut are required to lift the stump 4J ft.? Ans. 37 ft.
30. The micrometer. — The screw is used in very many mechanical devices. Many of these will be used in illustrative problems in later chapters. The use of the screw in measuring small distances where great accuracy is required, is illustrated in the ordinary micrometer shown in Fig. 3.
A — Frame
B — Anvil
C — Spindle or Screw
D — Sleeve or Barrel
E— Thimble
FIG. 3. — Micrometer.
The object to be measured is placed between the anvil, B, and the spindle, C. The spindle has a thread cut 40 to the inch on the part inside the sleeve, D. The thimble, E, is outside the sleeve and turns the spindle. The thimble has a beveled end that is divided into 25 equal divisions. The sleeve is graduated into divisions each ^V °f an inch, every fourth of which is marked 1, 2, 3, etc. The numbered marks then rep- resent tenths of an inch.
If the thimble is turned through one of its graduations, the spindle is evidently advanced -5*5 of ^V in. = T^TV in. The vernier, which will be described later (Art. 167) enables one to read to -jV of one of the graduations on the thimble and hence makes it possible to measure -fa of TTfSv in. =TT^T) in.
The readings of a micrometer are usually stated in decimals of an inch.
COMMON FRACTIONS
33
95. Find the distance the spindle advances when the thimble makes 7 full turns and 17 divisions. Ans. t\fo8o or 0.192 in.
96. What is the measurement when the reading on the sleeve is 5 graduations and on the thimble 14? Ans. ^ftj- or 0.139 in.
31. Screw gearing. — Spiral or screw gearing is often used where it is desired to reduce the speed greatly. The teeth are arranged in the same manner as the threads of a screw. A screw wheel may have one tooth or any number of t,eeth. A one-toothed wheel corresponds to a one-threaded screw, a many-toothed wheel to a many-threaded screw.
In Fig. 4, the upper wheel has 12 teeth, and corresponds to a 12-threaded screw; while the lower wheel has 36 teeth, and corresponds to a 36-threaded screw. Here the smaller wheel makes three revolu- tions while the larger is revolving once. If the smaller had but one tooth or was single threaded, it would make 36 turns to one of the larger wheel.
When the number of threads or teeth on the smaller wheel is few, the smaller wheel is called a worm, and the larger the worm wheel.
FIG. 4.— Righthand spiral gears.
97. If a worm is 2-pitch and single-threaded, how many inches will it cause the circumference of the worm wheel to advance for one revolu- tion of the worm? How many inches if double-threaded? If triple- threaded? If 6-threaded? (2-pitch means 2 threads to the inch.)
98. If a single-threaded worm, making 20 revolutions per second, turns a worm wheel having 54 teeth, Fig. 5, how many revolutions per minute will the worm wheel make? Ans. 22 1.
99. In turning a piece in a turning lathe the distance the cutting tool advances along the piece for each revolution is called the feed. The feed is usually given as a fraction of an inch. How long will it take to turn a piece 2J ft. long, if it makes 17 revolutions per minute, and the feed is ^ in.? Ans. 28A minutes.
100. How many turns per minute is a piece making if 4 ft. of length is turned in 75 minutes when the feed is -fa in.? Ans. 6||.
101. What feed is necessary to run a cut of 45 in. in 10 minutes at 36 revolutions per minute? Arts. | in.
3
34 PRACTICAL MATHEMATICS
102. In planing an aluminum casting of width 10) in. and length 12} in., find the time required if cutting speed is 40 ft. per minute, return speed 80 ft. per minute, and feed /j in. Ans. 27 minutes nearly.
Siiggestion. The cutting tool cuts a strip V* in. wide each time across. Or it takes 64 times across to plane a strip one inch wide. If the cut is
Lcfthand single thread. Righthand double thread
FIG. 5. — Worm and worm wheel.
made the long way of the casting it takes 64X10J =672 cuts. The re- turn strokes take one-half the time of the cutting strokes. No allowance is made for over run in the strokes.
103. A A-in. twist-drill has a speed of 130 revolutions per minute (R. P. M.) when cutting steel. How long will it take to drill through a f-in. plate if 120 revolutions are required to drill 1 in.?
Ans. \\ minutes.
FIG. 6.— Twist drill.
104. A lA-in. drill makes 66 R. P. M. in iron and has a feed of sV in. How long will it take to drill 20 holes through a $-in. plate if i minute is allowed for setting for each hole? Ans. 28i*r minutes.
105. In drilling through mild steel 1 f in. thick, a hole 1 in. in diameter is drilled in If minutes; find the distance drilled per minute.
Ans. \\ in.
106. A A-in. drill can make 320 R. P. M. in brass. If 120 turns are made to drill 1 in., find how many holes can be drilled in a i-in. plate in 1 hour if J the time is used in setting the drill. Ans. 192.
107. The following rule is often used to find the weight of round steel and wrought iron: Square the diameter in inches and multiply by |,
COMMON FRACTIONS 35
the product is the weight in pounds of 1 ft. of the bar. (The product of a number multiplied by itself is the square of the number.) Using this rule find the weights of round bars of steel of the following dimensions:
(1) Diameter f in., length 12| ft. Ans. 25jf Ib.
(2) Diameter 2f in., length 2J ft. Ans. 41& Ib.
(3) Diameter 8| in., length 1| ft. Ans. 326 f Ib.
108. Three pipes can empty a reservoir in 6, 5, and 4 hours respectively. How long will it take them to empty it if running together?
Ans. If? hours.
Solution. Since the first pipe can empty the reservoir in 6 hours, it can empty f of the reservoir in 1 hour. Reasoning in the same manner, the second pipe can empty g of it in 1 hour, and the third pipe I of it in 1 hour. Hence, the three pipes can empty £+i+l = i? of the reservoir in 1 hour. To empty the whole reservoir, it will take all the pipes to- gether, l-r-|J=l|? hours.
109. In the preceding exercise, how long will it take to empty the reservoir if it is full to begin with, and the first two pipes are emptying out of and the third emptying into the reservoir? Ans. 8| hours.
110. Three pipes open from a tank. The first alone can empty it in 6 hours, the second in 4 hours, and the third in 3 hours. How many hours will it take to empty the tank if the pipes are all running together?
Ans. 1|.
111. In the previous exercise the tank is full to begin with and the first and third pipes are emptying out of, and the second emptying into the tank. How long will it take to empty the tank? Ans. 4 hours.
112. If A can do a piece of work in 9 hours and B can do the same work in 6 hours, how long will it take them if working together? What part can each do in 1 hour? What part of the work can both together do in 1 hour? Ans. 3§ hours; $ and £; -f$.
113. If A can do a piece of work in 10 J days and B can do the same work in 81 days, what part of the work can they both do in one day? In how many days can they do the work if working together?
Ans. A; 4f.
114. If one gang of men can do a piece of work in 20 days, and another gang can do the same piece of work in 25 days, how long will it take both gangs working together to do the work? Ans. 11| days.
115. A contractor is to grade a street in 30 days. Fifteen men work on the grading for 20 days and do one-half of the work. How many men must work for the next 10 days to finish the grading? Ans. 30.
116. In 10 days, 15 men do -fa of a piece of work. How many men will it take to finish the work in 15 days?
117. How many turns must be made with a triple-threaded screw having 4j threads to the inch to have it advance a distance of 3 in.?
118. The lead screw on the table of a milling machine has a double thread with a pitch of J in. How many inches per minute is the feed if the lead screw is making 4 R. P. M.?
36 PRACTICAL MATHEMATICS
119. A man who owns $ of a city block valued at $140,000, gold 11, of his sharp. What is the value of the part he has left?
Ann. $15,000.
120. A merchant bought a stock of goods for $7426.50 and sold \ of it at an advance of \ the cost, J of it at an advance of i the cost, and the remainder at a loss of 1*5 the cost. Did he gain or lost* and how much? Ana. Gained $1516.24.
121. A sum of money is divided among four persons. The first ro- ceived 1*0 of the amount, the second J, the third J, and the fourth the remainder which is $5000. What is the amount each received?
Ans. $6000; $5000; $4000; $5000.
122. If it takes 4 tons of coal to heat as much as 6 cords of wood, which is the cheaper if coal is $7J a ton and wood $5J a cord? How much more will one cost than the other to heat a house that requires 1 1 tons of coal a year? Ans. Coal, $lf.
123. Which is the cheaper and how much, to have a 17J cent an hour boy take 13 J hours to do a certain piece of work, or have a 42 § cent an hour man do it in 4$ hours? Ans. The man 38} cents cheaper.
124. A piece of work when forged weighed 214$ Ib. After being turned down it weighed 156f Ib. The forging cost 16i cents a pound and the metal turned off sold at 3} cents a pound. Find the net cost of the metal in the finished piece. Ans. $33.51^4.
126. A machinist drills 6 holes through a piece that is 2J in. thick. The drill is 1J in. in diameter and makes 154 R. P. M. with a feed of B'O in. How many minutes does it take if 3 minutes are used in setting for each hole? Ans. 23fi.
126. A hole 6f in. deep is drilled with a 1 J-m. drill making 126 R. P. M. What feed is required to drill the hole in 3} minutes? Ans. fa in.
127. To change from Centigrade thermometer reading to Fahrenheit, the following formula is used: F = gC+32°, where C is the Centigrade reading and F the Fahrenheit. (1) Find F when C=22i°. (2) Find F when C is 720°. Ana. (1) 72i°5 (2) 1328°.
128. To change Fahrenheit thermometer reading to Centigrade, the following formula is used: C = 8(F — 32°), where C is the Centigrade reading and F the Fahrenheit. (1) Find C when F=8H°. (2) Find C when F = 1760°. Ans. (1) 27 J°; (2) 960°.
129. In inspecting steamboilers, the following formula is often used :
PRF
t — — .
TX%
where / = thickness of plate in inches,
P= steam pressure in pounds per square inch, R = radius of boiler in inches (i of diameter), F = factor of safety,
T = tensile strength of boiler plate in pounds, % = percentage of strength in joints.
COMMON FRACTIONS
37
Find the thickness of the boiler plate which should be used for a boiler 50 in. in diameter to carry 120 Ib. of pressure if the tensile strength is (>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 * A8
If
2
Width across flats F
i
2
3~2
i¥
&
1*'
n
i&
1 i4
1 16
29 T6
2f
Ql «*5
5 0 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, TVo is written 0.53; fWA = 0.3756; TJJ0 = 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, T98 =-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 $=0.2857+ Process. 7)2.0000
0.2857+
The sign, +, placed after the number indicates that there are still other figures if the division is carried further.
A common fraction in its lowest terms will reduce to an exact decimal only when its denominator contains no other prime factors than 2 and 5.
Thus, 88« reduces to an exact decimal for 64 is made up of 2X2X2X 2X2x2, while j7z cannot be reduced to an exact decimal for its de- nominator contains the factor 3.
35. Decimal fraction to common fraction. — To change a decimal fraction to a common fraction proceed as follows:
COMMON FRACTIONS 41
RULE. Replace the decimal point by a denominator having 1 and as many ciphers as there are decimal places in the original fraction. (See Art. 32.)
Thus, 2. 375 =f§?$, 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<r; dlft&for-
16. Tell without trial which of the following common fractions will reduce to exact decimals: f; ^; T75; J|; /g; ¥\; ^; ^ ; f|; f; J||.
17. Change the following decimals of an inch to the nearest 64ths of an inch: 0.394; 0.709; 1.416; 1.89.
36. Addition of decimals. — RULE. Write the numbers so that the decimal points are under each other. Add as in whole numbers, and place the decimal point in the sum under the other decimal points.
Example. Add 36.036, 7.004, 0.00236, 427, 723.0026.
36.036 7.004 0.00236 427.
723.0026 1193.04496 Ans.
37. Subtraction of decimals. — RULE. Write the numbers so that the decimal points are under each other; subtract as in
42 PRACTICAL MATHEMATICS
whole numbers, and place the decimal point of the remainder under the other decimal points.
Example. Subtract 46.8324 from 437.421.
437.421 46.8324
390.5886 Ans.
38. Multiplication of decimals. — RULE. Multiply as in whole numbers, and point off as many decimal places in the product as the sum of the numbers of the places in the factors.
Example 1. Example 2.
Multiply 7.32 by 0.032. Multiply 0.00264 by 0.000314. 7.32 0.00264
0.032 0.000314
1464 1056
2196 264
0.23424 Ans. 792
0.00000082896 Ans.
Multiplying a whole number or a decimal by 0.1 moves the decimal point one place to the left; by 0.01, two places; by 0.001, three places; etc. If it is necessary, zeros are prefixed to the multiplicand.
Thus, 32.4 X 0.0001 = 0.00324.
Multiplying by 10, 100, 1000, etc., moves the decimal point 1, 2, 3, etc., places to the right.
39. Division of decimals. — RULE. // the number of decimal places in the dividend is kss than the number in the divisor, annex ciphers to the dividend till there are as many or more decimal places as in the divisor. Divide as in whole numbers, and point off as many decimal places in the quotient as there are more decimal places in the dividend than in the divisor.
Example 1. Example 2.
Divide 0.4375 by 0.125. Divide 4365 by 0.005.
0.43750.125 0.005)4365.000 375
3.5 Ans. 873.000 Ans.
625 625
Dividing by 0.1, 0.01, 0.001, etc., moves the decimal point 1, 2, 3, etc., places to the right. Dividing by 10, 100, 1000, etc., moves the decimal point 1, 2, 3, etc., places to the left.
DECIMAL FRACTIONS 43
40. Accuracy of results. — Often we are asked to give a result correct to a certain number of decimal places. Thus, if in working a problem we have a result as 47.264735, and wish to write it correct to three places, it is 47.265 — . Correct to two places, it is 47.26+, correct to one place, 47.3 — , correct to five places 47.26474-.
The last place taken is written one larger when the next figure to the right is 5 or more.
The part to the right of the last place taken is thrown away when the first figure of it is less than 5.
In this way we call a half or more of the last unit taken, a whole one of those units, and throw away anything less than a half.
The sign, +, is used to show that the accurate result is larger than the one given, that is, that something has been thrown away; and the sign, — , is used to show that the accu- rate result is smaller than the one given, that is, that something has been added.
EXERCISES 12
Add up and test by adding down :
1. 864.2 2. 5.82 3. 49.235
43.276 .486 86.426
21.004 41.987 92.784
9824.246 987. 46.324
47.02 201.478 33.867
39.09 804.008 99.847
34.396
4. 18i + 17f +29^+14.672+34? = ? Ans. 114.791+.
5. 87.46 \ +93.27^+43.2906 +0.0047 + 17£ = ? Ans. 241.2027-.
6. 8.706 + 7.898+43|+89A+14§ =? Ans. 163.833 + . 7.1-0.640726 = ? ' Arts. 0.359274.
8. 2-1.798642 = ? Ans. 0.201358.
9. 4.8728-0.987 = ? Ans. 3.8858.
10. 75.0075-2.75903 = ? Ans. 72.24847.
11. 470.84-86.4396 = ? Ans. 390.4004.
12. From one thousand take five-thousandths. Ans. 999.995.
13. From three million and one-millionth take one-tenth.
Ans. 2999999.900001.
14. 78.896-53.5987 = ? Ans. 25.2973.
15. 81.35-11.678956 = ? Aws. 69.671044.
16. From nine hundred nine take nine hundred and nine thousandths.
Ans. 8.991.
44
PR A CTICAL MA THEM A TICS
17. From 37.75J take 4.43f. Ana. 33.32126.
18. One quart liquid measure has 57.75 cu. in., and 1 quart dry mca*- ure has 67.200025 cu. in. How many cubic inches larger is the dry
quart than the liquid quart?
19. 3.62X0.0037 = ?
20. 2.53X0.00635 = ?
21. 0.00076X0.0015 = ?
22. 7.789X4.924 = ?
23. 2.236X799 = ?
24. 2.967X2.967 = ?
25. 8.943X1§ = ?
Why would it be best not to reduce 1J to a decimal before multiplying?
The multiplication can be carried out readily as shown here, the process is as if the multiplicand were a whole number.
Ans. 9.460626.
Ans. 0.013394.
Ant. 0.0160655.
Ans. 0.00000114.
Ans. 38.353+.
Ans. 1786.564.
Ans. 8.803 +.
Ans. 14.905.
Process.
8.943
2981 5962 8 943
14.905
= J of 8943 = § of 8943 = 1 X 8943 = 1§X8.943
Ans. 23.218-.
Ans. 2.1657 + .
Ans. 1499.2176 -.
Ans. 0.000105+.
Ans. 0.9745 + .
Ans. 2.7781 +.
26. 2.43|X9.5? = ?
27. 0.0439X491 = ?
28. 1X3.1416X7.1X7.1X7.1 = ?
29. 9.3X0.0042X0.0027 = ?
30. 49367 X 0. 002 1 X 0. 0094 = ?
31. 0.81X91X0.3375 = ?
32. Multiply 3f thousandths by 3f hundredths.
Ans. 0. 0001406 +.
33. 1 kilogram equals 2.2046 pounds; how many pounds in 275.3 kilograms? .4ns. 606. 926+.
34. Multiply each of the following numbers by 0.1, 0.01, 0.001, 0.0001, 10, 100, 1000, 10,000: 94, 47.368, 0.023, 3.42.
36. 67.56785 -=-0.035 = ? Ans. 1930.51.
36. 0.567891-i-8.2 = ? Ans. 0.069255.
37. Divide 43.769 by 4.76 correct to four places.
Process. 43.7690004.70 4284
9.1951
Explanation. Since the quotient is to be correct to four places, the dividend must contain four more decimal places than the divisor. Three zeros are added to make this number. Since the fifth decimal figure in the quotient is not less than 5. the answer is 9.1952 — .
929
476 An*. 4530
1JM
2460
is 9.1 952-.
800 476 324
DECIMAL FRACTIONS
45
In the next 5 exercises, find the result correct to four decimal places.
38. 9.375+4.76 = ? Ans. 1.9695 + .
39. 89.7201 + 3.276 = ? Ans. 27.3871-.
40. 34.675+4.375 = ? Ans. 7.9257 + .
41. 43.45+3.1416 = ? Ans. 13.8305+.
42. 3.1416 + 6.67 = ? .4ns. 0.4710 + .
43. Divide 324.8 by 4000.
Explanation. Cancel the zeros in the divisor. Since this divides the divisor by 1000, the dividend must be divided by 1000, this is done by moving the decimal point three places to the left.
44. (a) 25 + 500 (b) 2+2000
(d) 1.44 + 12,000 (e) 3.075 + 5000
45. (a) 1+0.0001 (b) 0.66+0.011 (d) 0.00072+8 (e) 6600 + 0.0022
46. Divide 3.1416X1.25X50 by 0.8X2.75X3. Explanation. The cancelling may
be done as in whole numbers, paying no attention to the decimal point. When through, point off as many places in the result as the difference
between the sum of those above and 0.$X2./75X 3
the sum of those below the line. tf
Thus, in the example there are six
places above and three below the line; hence, the result has three decimal places. If there are more places below than above, add ciphers until there are as many above as below. Find the value of the following: 2.75X46.2X100
Process. 4p0p)0.3248 0.0812
(c) 9009 + 11,000 Cf) 5684 + 14,000 (c) 525+0.025 (/) 3.03+0.03
Process.
= 29.750.
47.
48.
49.
60.
61.
to units.
2.5X2.8 0.7854X5X5X8
1.25 6.4X0.84X9.6X1.44
8X9.2X1.28 0.7854X6X6X12.5X1728
231X31.5 8.5X9.25X3.66X1728
to three places.
to three places.
to two places.
Ans. 1815.
Ans. 125.664.
Ans. 0.789-.
Ans. 83.93 + .
to two places.
2150.42
52. What is the inside diameter of a pipe which is 7.34 in. outside diameter and is made of iron 0.743 in. thick? Ans. 5.854 in.
63. A hundred pounds of coke were found to contain 5.79 Ib. of ash and 0.597 Ib. of sulphur, the rest was carbon. How much carbon was there? Ans. 93.613 Ib.
54. A steam pump delivers 26.44 gallons per stroke. A gallon of water weighs 8.355 Ib. What weight of water will it deliver in 117 strokes? Ans. 25,846 Ib.
Ans. 231.24-.
0.743
FIG.
8.— Cross of pipe.
46 PRACTICAL MATHEMATICS
65. In 1 Ib. of phosphor bronze 0.925 is copper, 0.07 w tin, and 0.00.r> 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 (i±M> 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 0 173 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 739~ =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
0 398 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* 47«X333*
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 0°C. 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
— 0 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 =1I0 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 = TY<r = 40%; or I = | of 100% = 40%. Similarly I = f of 100% = 87$%.
A per cent expressed as |%, does not mean | but | of 1%, which is the same as f of 1^5- = ?|7 = -j^ = 0.004. In the same way f % = f of y|~¥ = ¥-|T = 0.00375 = 0.375%.
It should be carefully noticed that the sign % does the duty of two decimal places.
Thus, 0.05 = 5%, 0.0005 = 0.05%, 1.07 = 107%, and 4.33$ = 433$%.
58. Cases. — The problems of percentage usually occurring are of the following forms:
(1) What is 37£% of 720?
(2) 45 is what per cent of 450?
(3) 85 is 62$% of what number?
73
74 PRACTICAL MATHEMATICS
These three forms can be stated in general terms if the following definitions are given :
The number of which the per cent is taken is the base.
The number of per cent taken is called the rate.
The part of the base determined by the rate is the per- centage.
The sum of the base and percentage is the amount.
The base minus the percentage is the difference.
The three problems now become the cases:
Case I. Base and rate given to find percentage.
Case II. Base and percentage given to find rate.
Case III. Percentage and rate given to find base.
These three cases of percentage correspond to the throe cases in multiplication; when any two of the numbers, multi- plicand, multiplier, and product, are given, to find the third.
multiplicand corresponds to base
In multiplication,
multiplier corresponds to
> 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 T4<y°o is useless for doing work. This does not state the actual numerical amount of energy wasted; all it tells is that for every 100 units of work expended on the machine, 40 units disappear. Such per- centages enable comparisons of different machines to be made. If one machine has an efficiency of 60% and another of 70%, we know that the second is 10% more efficient than the first. If we know that | or 12|% of the 40% loss is in a certain part, this gives a percentage of a percentage. The solution of the problem is:
| = part of the energy lost,
f = 60% = efficiency of the machine,
I of $ = -sV = 5 % = loss in the particular part of the machine.
62. Averages and per cent of error. — The data for practical calculations are* in many cases either the result of measuring quantities, or of experimental observations, and in each case are liable to error. To obtain a result which can be relied upon, a number of measurements or observations are taken and the average or mean result calculated.
The average, or mean result, is obtained by adding all the measured results together and dividing the sum by the number of them. This average is accepted as the best approximation to the truth. The error of any particular observation is obtained by finding the difference between it and the average. This error can often be most conveniently expressed as a per cent, and is spoken of as the per cent of error. We always take the correct value, or in this case the average value, as the base.
Example. In measuring the diameter of a steel rod with a micrometer, the separate measurements are: 0.3562 in., 0.3569 in., 0.3567 in., 0.3570 in., and 0.3565 in. Find the average measurement, and the per cent of error in the largest and the smallest measurements.
78 PRACTICAL MATHEMATICS
Solution ami discussion.
0.3562 in. + 0.3569 in. -I- 0.3567 in. + 0.3570 in. + 0.3565 in. = 1.7833 in.
1.7833 in. -*- 5 = 0.35666 in. = average.
0.3570 in. - 0.35666 in. = 0.00034 in. = error in largest measurement.
Using the formula r = p -r- b gives
r = 0.00034 in. -*• 0.35666 in. = 0.00095 + = 0.095% = per cent of error in largest measurement.
0.35666 in. -0.3562 in. = 0.00046 in. = error in smallest measurement.
r = 0.00046 in. -5- 0.35666 in. = 0.0013 = 0.13% = percent of error in the smallest measurement.
It should be emphasized that the per cent of error in any measurement is always found by using the correct measure- ment as the base and the error as the percentage.
63. List prices and discounts. — The prices of machines and materials, printed in catalogs and price lists, are usually subject to discounts. Often the discount is so large that the list price gives no idea of the actual cost. In preparing an estimate, it is necessary to know what discounts are given from a price list.
Discounts are usually given thus: 60% and 10% off or simply 60 and 10 or perhaps "sixty and ten." This does not mean a discount of 70%, but that a discount of 60% is first made and then a discount of 10% on the remainder. Thus, if the list price is $3.50 with 60% and 10% off, we find 60% of $3.50, which is $2.10. Then deduct this from $3.50 leaving $1.40. Now get 10% of $1.40, which is $0.14, and deduct it from $1.40, leaving $1.26 as the actual cost.
Similarly we may have discounts of 40%, 10%, and 4%, or 40, 10, and 4 off. These are deducted in turn as with the two discounts.
EXERCISES 18
1. 62% of 2000 = ? Am. 1240.
2. What is |% of $28.80? An*. $0.108.
3. 37i% of 4000 = ? Ana. 1500.
4. 300 is 1\% of what number? Ans. 4000. 6. What per cent of $104 is $18.20? Ans. 17$%. 6. What per cent of 300 is 272? Ana. 90|%.
PERCENTAGE AND APPLICATIONS 79
7. The indicated horse-power of an engine is 10.6, the actual effective horse-power is 8.96. What per cent of the indicated horse-power is the actual? Ans. 84|$%.
8. A bankrupt has $5760, and with that sum can pay 40% of his debts; find his entire indebtedness. Ans. $14,400.
9. For collecting a bill an attorney received $2.52, which was \\% of the bill; find the amount of bill. Ans. $224.
10. A milkman sold milk at 7 cents a quart, which was 233|% of the cost; find cost per quart. Ans. 3 cents.
Suggestion. 233 \% =\. If 7 cents is I of the cost, what is the cost?
11. What is the net price per barrel of oil, the list price of which is $18.00, subject to a discount of 12^% and 4% off for cash?
Ans. $15.12.
12. A tradesman marks his goods at 25% above cost and deducts 12% of the amount of a customer's bill for cash. What per cent does he make?
Ans. 10%.
Suggestion. Suppose the cost is $20. The marked price is 25% above $20 or $25. A deduction of 12% on $25 is $3. Hence the selling price is $25— $3 =$22. The gain is $22 -$20 = $2. What per cent is $2 of $20? This gives the gain per cent.
13. If 1225 pounds of coal were fired to a boiler and 152 pounds were taken out of the ashpit as ash and waste, what per cent of the coal was taken from the ashpit. Ans. 12.4 + %.
14. The weight resting on the drivers of a locomotive is 158,700 pounds. If this is 68.72% of the total weight; find the weight of the locomotive.
15. A man who receives 42| cents an hour works a day of 8 hours, and 4 hours overtime at pay for time and a half. What does he receive in all? What per cent is the overtime pay of the total?
Ans. $5.95;426T%.
15. A firm increases the wages of its employees 12| %. Find the wages of a man who was getting $3.40. Of a boy who was getting $1.60 a day. A man now receives $6.30 a day. What did he receive before the increase? Ans. $3.82|; $1.80; $5.60.
17. A tank whose capacity is 168 gallons, discharges 72 gallons per hour, which equals 25% less than it receives. In what time will it be filled? Ans. 7 hours.
Suggestion. If 72 gal. is 25% or j less than it receives per hour, 72 gal. = I of what it receives per hour. Hence it receives 96 gal. per hour. Then the tank receives 24 gal. per hour more than it discharges.
18. One mill is gaged at 767 barrels of flour a day, which equals 18% more than the amount for another. What is the value of the daily out- put of the latter at $5 a barrel? Ans. $3250.
19. The usual allowance made for shrinkage when casting iron pipes is £ in. per foot. What per cent is this? Ans. 1.04 + %.
20. Find the cost of an article that is listed at 80 cents, 40 and 6 off.
Ans. 45.12 cents.
80 PRACTICAL MATHEMATICS
21. Find the cost of a machine quoted at $25.50, 40% and 10% off with a further discount of 4% for cash. Ana. $13.22.
22. When rock is crushed or broken into fragments of nearly uniform size it increases in hulk and has voids, or inter-spaces, of from 30 to 55 per cent of the whole volume. Find the number of cubic yards when crushed occupied by 1 cu. yd. of solid rock, if voids are (a) 30%; (b) 35%; (c) 40%; (d) 45%; (e) 55%.
Ans. (a) 1.43; (6) 1.54; (c) 1.67; (d) 1.82; (e) 2.22. Suggestion. The volume of the rock, 1 cu. yd., is 30% less than or 70% of bulk of crushed rock. Using formula
b = p •*• r gives 6 = 1 -f- 0.70 = 1.43 - . Ann.
23. A team of horses and a wagon cost, say $400. If money is worth 6%, depreciation in value of team and wagon is 25% per year, teamster's wages are $2.00 per day while working, and cost per month for keeping team is $18.50; find the amount that should be charged per day for man and team, counting 250 days actually worked per year. Ans. $3.384.
24. In the preceding problem what would be the gain per year from the team if $5.00 a day was charged for services, deduction being made for depreciation in value? Ans. $404.
26. In estimating the amount to charge per day for the use of a steam- roller, a contractor has the following data: first cost of steam-roller $3000; money worth 6%; days actually worked per year, 100; deprecia- tion in value of the machine, $200 per year. Find price to be charged per day for use of roller. Ans. $3.80.
26. The actual cost of removing a cubic yard of rock in excavating a certain canal is $1.10. What price should be put in the estimate, if 12% is to be allowed for superintending, and 10% on the cost, including superintending, is allowed for profit? Ans. $1.3552.
27. A house depreciates in value each year at the rate of 4% of its value at the beginning of each year, and its value at the end of two years is $6451.20. Find the original value. Ans. $7000.
Suggestion. $645 1. 20 -J- 96 = $6720. This is the value at beginning of second year.
28. A house valued at $4000 rents for $27.50 per month. The repairs on house each year amount to $40, and the taxes are $17.50. What interest does the property pay on the investment, no allowance being made for change in value of house? Ans. 6H%-
29. In making a certain machine, 750 Ib. of iron are used at an average cost of 8 cents per pound. There are used in the work on the machine, 20 hr. of time at 30 cents per hour, 7 hr. at 60 cents and 4 hr. at 16 cents. If 20% is allowed on cost as profit, what is the selling price of the machine? What will it be listed at if sold at 30 and 5 off?
Ans. $85 nearly, $127.82.
30. An article is listed at $225, and sells at 40 and 10 off. How will the 40% discount be changed to offset an increase of 15% in cost of production? Am. 31%, or better, 30%.
PERCENTAGE AND APPLICATIONS 81
Solution. It is desired to find a first discount so that the net price, that is, the price after the discounts are made from the list price will be 15% more than the net price when discounts of 40% and 10% are used.
A discount of 40% and 10% off from $225 leaves $121.50. This is the old net price.
$121.50 + 15% of $121.50 = $139.725=new net price.
$139.725-^0. 90 = $155.25 = price after new first discount is deducted from list price.
$225 -$155.25 = $69.75 = amount of first discount.
$69.75 -^ $225 =0.31 =31%= the first discount.
31. The composition of white metal is to be 4 parts by weight of cop- per, 9 antimony, and 97 tin. Express these as per cents, and find the weight of each material required to make 2376 Ib. of the alloy.
Ans. Copper 3rV%, 86.4 Ib.; antimony 81SI%, 194.4 Ib.; tin 88ft %, 2095.2 Ib.
32. For the two months ending Feb. 28, 1908, there were exported from the United States $27,531,617 worth of iron and steel, including machinery. During the same time in 1909 it was $21,276,547. Find the decrease per cent. Ans. 22.7 + %.
33. The output of Canadian pig iron for 1908 was 563.672 tons, a decrease of 3% from 1907. What was the output in 1907?
Ans. 581,105 tons.
34. Steel billets that were selling at $26 per ton dropped to $23 per ton. What is the per cent of reduction? Ans. Ili73%.
35. The water-power in use in the United States is 5,300,000 horse- power. The undeveloped is 8,100,000 horse-power. What per cent of the total water-power is developed ? Ans. 39.55 + %.
36. A ton of coal from the Rock Island field has 11.57% moisture, and 6.27% of the dry coal is ash. How many pounds of ash in a ton of the coal? Ans. 110.89+.
37. If 2.346 g. of an ore give 0.362 g. of copper, what per cent of copper does the ore contain? Ans. 15.43 + %.
38. 2.3656 Kg. of ore give 0.7 g. of gold and 2.5 g. of silver. Find the per cent of each. Ans. 0.0296 - % ; 0.1057 - %.
39. A merchant buys rubber door mats at $48.00 a dozen less discounts of 40%, 15%, and 5%. What should he sell them apiece in order that he may make 35%? Ans. $2.62-.
40. lj-in. basin plugs are listed by the jobber at $1.20 a dozen. The retailer gets discounts of 50 and 10 off, and sells them at 15 cents each. Find his gain per cent. Ans. 233$%.
41. If the author gets 10% of the selling price of a book, how many hooks, selling at 75 cents each, must be sold to pay the author $117.30?
Ans. 1564.
42. In a compound of two substances A and B, their weights are in the ratio of 1.3498 to 1. What is the per cent of each in the compound?
Ans. 57.44 + %; 42.56-%.
82 PRACTICAL MATHEMATICS
43. Two substances A and B form a compound and have a total weight of 3.267 g. If the compound has 24.725% of A and 75.275% of B, find the weight of each substance in the compound.
Ana. 0.8078- g. ; 2.4592 +g.
44. Find the cost of a steam boiler listed at $500 subject to discounts of 40%, 10%, and 7J%. Ana. $249.75.
45. Marshall Field and Co. quotes an article of silverware at $25 with discounts of 40, 10, 5, and 6% off in 10 days. Find net cost if paid in 10 days. Ana. $12.06-.
46. The recorded measurement of a city block is 528 ft. By chaining carefully the length is 527.75 ft. Find the per cent of error in the re- corded length. How wide is a man's lot recorded as 30 ft.?
Ana. 0.047 + %; 29. 986- ft.
47. A sample of nickel-steel contained 24.51% of nickel and 0.16% of carbon. How much of each nickel and carbon in 2240 Ib. of nickel-steel?
Ana. 549.024 Ib.; 3.584 Ib.
48. If a 3i% nickel-steel rail is used to maintain a curve in a street-car track, it lasts three times as long as carbon-steel. How much will be saved per ton when one nickel-steel rail is worn out, if nickel-steel costs $56 per long ton and carbon-steel $28? It costs $2.00 a ton for laying, and the old rails are worth $16.00 per ton, besides 20 cents a pound is realized on the nickel. Ana. $15.68.
49. In an experiment to show the loss of pressure for different kinds of valves in water pipes, a globe valve in a 3-in. pipe caused the pressure to fall from 80 Ib. to 41 Ib. per square inch; while a gate valve caused a loss of pressure of 4 Ib. per square inch. Find (a) the per cent of loss for globe valve, (b) for gate valve, (c) what per cent loss through gate valve is of loss through globe valve. Ana. 48J% ; 5% ; 10.26 - %.
60. In an analysis of the best quality of crucible cast steel, the follow- ing was found: carbon 1.2%, silicon 0.112%, phosphorus 0.018%, man- ganese 0.36%, sulphur 0.02%, iron 98.29%. Find the number of pounds of each substance if the total weight is 176.5 Ib.
Ana. 2.118; 0.1977- ; 0.0318- ; 0.6354; 0.0353; 173.4S18+.
61. Find the cost of the following at 83% discount:
350 ft., 8-in. sewer pipe at $0.50
4 elbows at $2.00
3 T branches at $2.25
4 traps at $6.60
Ana. $36.75.
62. The mean effective pressure on the piston of a steam engine, found from the indicator diagram, was 59.75 Ib. per square inch. The boiler pressure was 87 Ib. per square inch. What per cent of the boiler pressure was the mean effective pressure? Ana. 68.7 — %.
63. The grade of a railroad track is given in per cent. A grade of 1% is a rise of 1 ft. in 100 ft. If a railroad has a constant grade of 1J%, what is the rise in 3J miles? Ana. 231 ft.
PERCENTAGE AND APPLICATIONS 83
54. The total rise in a If % grade is 43.6 ft. Find the length of the track having this grade. Ans. 249 If ft.
55. A railroad rises 112.7 ft. in 3£ miles. Find the average grade.
Ans. 0.61-%.
56. The mechanical efficiency of a machine is the relation between the work put into the machine and the work gotten out of it. Mechanical efficiency is usually stated as a per cent. Thus, if 100 units of work are put into a machine and only 80 units gotten out the mechanical efficiency, or simply the efficiency, is 80%. What is the efficiency of the engine of exercise 7?
57. At what advance must a shopkeeper mark goods costing 90 cents that he may allow a 20% discount and yet gain 25%? Ans. 50f cents.
58. A man purchases ice at 50 cents per 100 Ib. At what rate must he sell it after it has lost 10% of its weight by melting to gain 20%.
64. Interest. — Interest is money that is paid for the use of money. It is usually reckoned at a certain rate per cent per year. The base on which the interest is reckoned is called the principal.
In percentage, the time did not enter, but in reckoning interest the time has to be taken into account. The interest on a sum of money for one year at a certain rate is the princi- pal multiplied by the rate; for two years it is twice as much; and for any period of time it is the interest for one year multi- plied by the time in years.
If p stands for principal, / for interest, r for rate per cent, and t for time in years, the interest is found by the formula
The amount, A, is the principal plus the interest.
Many short methods for reckoning interest can be given, but here it is not the intention to enter into them.
Example 1. Find the interest and amount of $350 for 5 years at 6%.
Z = pXrX£ = $350X0.06X5 = $105.00. Ans. A = p+I = $350+$105.00 = $455.00. Ans.
Example 2. Find the interest on $750 for 2 yr. 7 mo. at 8%.
Here the time is f £ years, since in getting the time in years we use 12 months for a year, 30 days for a month, and 360 days for a year.
/ . / = $750 X 0.08 X H = $1 55.00. A ns.
84 PRACTICAL MATHEMATICS
It is usually best to use cancellation. 750X8X31
Example 3. Find the interest on $375 for 2 yr. 5 mo. 15 da at 5%.
Here the time is f| ft years.
/. 7 = $375X0.05X|f! = $46.09. Ans.
EXERCISES 19
Find the interest and amount of each of the following:
1. $700 for 3 yr. at 8%. Ans. $168.00; $868.00.
2. $14.30 for 2 yr. 9 mo. at 8%. Ans. $3.15; $17.45.
3. $245.60 for 2 yr. 7 mo. 21 da. at 8%. Ans. $51.90; $297.50.
4. $436.75 for 1 yr. 2 mo. 15 da. at 5%. Ans. $26.39; $463.14. 6. $325.25 for 2 yr. 9 mo. 12 da. at 6J%. Ans. $58.84; $384.09.
6. $87.50 lor 3 yr. 3 mo. at 7%. Ans. $19.91; $107.41.
7. $480 for 6 yr. 3 mo. at 15%. Ans. $450; $930.
8. $18.20 for 9 yr. 9 mo. 9 da. at 5J%. Ans. $10.23; $28.43.
9. A note for $225 at 6% runs for 9 mo. What is the amount of the- note when due? Ans. $235.13.
10. A note for $390.00 at 7% runs for 3 yr. 6 mo. What is the amount due? Ana. $485.55.
CHAPTER VII RATIO AND PROPORTION
65. Ratio. — There are several ways of stating the relation of one quantity to another. If the size or magnitude of the quantities are thought of, a very convenient way of com- paring them is to state the ratio of one to the other.
The ratio of one number to another is the quotient of the first number divided by the second.
fB»/»
Thus, the ratio of $6 to $2 is 3, and may be stated in the form -^ or
V™
$6 : $2. In either case it is read "the ratio of $6 to $2."
From the idea of a ratio it is evident that we can state a ratio between two magnitudes only when the magnitudes are alike. That is, a ratio cannot be stated between such quantities as dollars and bushels.
The two numbers used in a ratio are called the terms of the ratio. The first one is named the antecedent and is the dividend; the second is named the consequent and is the divisor.
The ratio 2 : 3 is the inverse of the ratio 3 : 2.
Since a ratio in the form 4:3 is an indicated division or a fraction, the principles applying in division or to a fraction likewise apply to a ratio.
The expressions "in the same ratio as," "in the same pro- portion," "proportionally," and "pro rata" all have practically the same meaning.
When it is said that $20 is divided between two men in the ratio of
2 to 3, it is meant that one gets $2 as often as the other gets $3. That is, of each $5, one gets $2 and the other $3. Hence one gets | of $20 or $8, and the other gets | of $20 or $12.
EXERCISES 20
1. Find the value of the following ratios: 8:2; 9:4; 17:2J; 44 hours:
3 hours; 7bu.:2 bu.; 4^:3£; 9*: 16.
2. A room is 16 ft. by 12 ft. What is the ratio of its length to its width?
85
80 PRACTICAL MATHEMATICS
3. Two gear wheels have 80 teeth and 30 teeth respectively. What is the ratio of the numbers of teeth?
4. One city has a population of 8000 and a second a population of 20,000. What is the ratio of their populations? What part is the first of the second? What per cent? How many times as large as the first is the second? What difference is there in the ideas involved in the questions?
6. W'rite the inverse ratios to the following: 7:2; 9:2J; 10 ft. :90 ft,; 23J:2t.
6. Divide $50 between A and B in the ratio of 3:7.
7. A man rode 250 miles partly by rail and partly by boat. What distance did he travel by each if their ratio is as 3 to 2?
,4ns. 150 mi. ; 100 mi.
8. Fifty-one students entered a class and 33 of them finished the work. What per cent finished? What is the ratio of the number that finished to the whole number?
9. A worm wheel makes 6 turns per minute and the worm 180 turns per minute. What is the ratio of the reduction of speed ? Ans. 30tol.
66. Proportion. — A proportion is a statement of equality between two ratios.
Thus, 2:3=4:6 and 4 men: 8 men =$6: $12, are proportions.
The first and last terms of a proportion are called the extremes. The second and third terms are called the means.
In the first proportion above, 2 and 6 are the extremes and 3 and 4 the means.
By inspecting several proportions the following principles will be evident:
(1) The product of the means of any proportion is equal to the product of the extremes.
(2) The product of the two means divided by either extreme gives the other extreme.
(3) The product of the two extremes divided by either mean gives the other mean.
Example 1. Find the value of h from the proportion 25 : 100 = 7 :h.
100X7
Solution. Applying principle (2), h = - -~= — = 28. Ans.
£\t
Example 2. If 15 tons of coal cost $63 what will 27 tons cost at the same rate per ton ?
Solution. Since the same relation holds between the cost
RATIO AND PROPORTION
87
prices as between the amounts of coal, the ratio of 15 tons to 27 tons must equal the ratio $63 to the cost of 27 tons. Let x stand for the number of dollars 27 tons cost, and we can state the proportion,
15 : 27 = 63 : x. 27X63
15
= 113.40.
.'. 27 tons cost $113.40. Ans.
Example 3. If 25 men can do a piece of work in 30 days, in how many days can 35 men do the same work?
Solution. It is evident that 35 men can do the work in less time than 25 men, hence the ratio of the number of days is equal to the inverse ratio of the number of men. Using x for the number of days required,
35 :25 = 30 25X30 •
x.
X —
35
.'. 35 men can do the work in 21-f days. Ans. Example 4. An inclined plane as shown in the figure rises 38 ft. in 100 ft., find the height h it will rise in 28 ft.
Solution. Here the proportion is
100 : 28 = 38 : h.
, 28X38 ..h=~m- = 10.64.
.'. the rise in 28 ft. is 10.64 ft. Ans.
The proportion could as well be stated 100 : 38 = 28 : h.
Definition. If the rise of a road bed is h ft. in 100 ft., the
grade of the road is r^' or the ratio of the rise to the horizontal
88 PRACTICAL MATHEMATICS
distance. Thus, if a road rises 3 ft. in 100 ft. the grade is
JL _ QC7 100 "
Example 5. What is the grade of a road bed that rises 1.2 ft. in a horizontal distance of 40 ft.? Solution. Let h stand for the number of feet rise in 100 ft.
It is evident that the ratio T ~ = the ratio —• But -
100 40 40
0.03. .*. the grade is 0.03 or 3%, Ans.
Example 6. If a bell metal is 25 parts copper to 12 parts tin, what is the weight of each in a bell weighing 1850 lb.?
Solution. The ratio of the number of parts of each metal to the whole number of parts equals the ratio of the weight of each metal to the whole weight. Use c to stand for the number of pounds of copper, and t for the tin. Then we have
25 : 37 = c : 1850, and 12 : 37 = t : 1850.
25X1850 19_n . . c = — = 1250,
12X1850
and t = 5= = 600.
o7
.'. weight of copper is 1250 lb. and tin is 600 lb. Ans.
EXERCISES 21
Find the value of the letter in the exercises 1 to 6.
1. 17:45 = 14:*. Ans. z=37iV.
2. 3|:9j=6:x. Ans. z = 15H-
3. 16|:29J=50&:z. Ans. z = 88*.
4. 3:z = 5:25. Ans. x-15. 6. 75:85 =x: 170. Ans. a: = 150.
6. r: 11 = 17: 121. Ans. r = l^.
7. If a train travels 378 miles in 11 hours, how far will it travel in 17 hours? Ans. 684ft miles.
8. If 10 men can do a piece of work in 20 days, how long will it take 25 men to do it? Ans. 8 days.
9. If a ship sails 256 miles in 11' hours, how far will it sail a the same rate in 179 hours? Ans. 3984lj miles.
10. The roof of a house rises 2 ft. in a run of 3 ft., how far will it rise in a run of 20 ft. ? Ans. 13 ft. 4 in.
11. A road bed rises 2\ ft. in 200 ft., what is the grade? In how many feet will it rise 1 ft. ? Ans. 1 J % ; 80.
RATIO AND PROPORTION 89
12. If 16f tons of hay cost $61.875, find the cost of 28 tons at the same rate. Ans. $105.
13. The mixture for a casting has 4 parts of copper, 3 parts lead, and 2 parts tin. How many pounds of each in a casting weighing 96 Ib. ?
Ans. 42f ; 32; 21J. Work the following 4 exercises by proportion.
14. What per cent is 59.1 of 51.3?
Solution. 100% stands for the base, then, using x for the number of per cent required, the proportion is
51.3 : 51.9 = 100 :z. . 59.1X100
51.3 .'.59.1 is 115.2% of 51.3.
15. 46 is what per cent of 79? Ans. 58.23 - %.
16. 146 is 17% of what number? Ans. 858.82+.
17. 3% of a number is 426, what is the number? Ans. 14,200.
67. Measuring heights. — There are several methods for determining the height of a standing tree. One of the simplest is to measure the shadow of the tree and the shadow of a straight pole of known length set upright in the ground. Then if H stands for the height of the tree, h for the height of the pole, S for the length of the shadow of the tree, s for that of the pole, we have the proportion
s:S=h:H.
EXERCISES 22
1. Find the height of a tree that casts a shadow 115 ft. long when a pole 8 ft. high casts a shadow of 5 ft. Solution. 5:8 = 115: H.
= =
5
/. height of tree is 184 ft. Ans.
2. Find the height of a church steeple that casts a shadow 84 ft. long when a pole 1 1 ft. long casts a shadow of 7 ft. 9 in. Ans. 1 19 ft. nearly.
The two following methods with the figures are given in Bulletin 36 of the Bureau of Forestry, U. S. Department of Agriculture.
(1) A method used when the sun is not shining is to set two poles in a line with the tree as shown in Fig. 16. From a point S on one pole sight across the second pole to the base and to the top of the tree. Let an assistant note the points a and b where the lines of vision cross the second pole and measure the distance between these points, ab, also measure the distance from the sighting point on the first pole to the base
90 PRACTICAL MATHEMATICS
of the tree, SB, and to the lowest point on the second pole, 56. Then the following proportion is true:
56 : SB = 06 : AB.
3. Find the height of a tree when 56 = 6 ft., 5fi=40 ft., and 06 = 9 ft.
Ans. 60 ft.
(2) Another method sometimes used is as follows: The observer walks on level ground to a point A at a convenient distance AD from the foot of the tree. He then lies on his back as shown in Fig. 17. An assistant
RATIO AND PROPORTION
91
notes on an upright staff erected at his feet the exact point C where his line of vision to the top of the tree E crosses the staff. The height of the staff BC is measured, and his own height AB from his feet to his eyes, then the following proportion is true:
AB:BC = AD:DE.
4. Find the height of the tree DE if AB=5± ft., BC = 8 ft., and AD = 90 ft. Ans. 131 ft, nearly.
68. The lever. — A stiff bar or rod supported at some pivotal point, about which it can move freely, is called a lever. The pivotal point is called the fulcrum. The lever enters in one form or another into many mechanical devices.
-D-
(a)
In Fig. 18, F stands for fulcrum, W for the weight lifted, P for the force that does the lifting, D for the distance from the fulcrum to the point of application of the force, and d for the distance from the fulcrum to the point where the weight is attached. In all possible relations of the fulcrum, weight, and force the following proportion holds:
P:W = d:D.
That is, the applied force is to the weight inversely as their distances from the fulcrum. This means that a small force will balance a larger weight only if the weight is nearer the fulcrum than the force.
'.'•J
PR A CT1CAL MA Til K\tA TICS
EXERCISES 23
1. Given P - 150 Ib., D-12* ft., d-lj ft, find W. Ans. 1250 Ib.
2. Given P -200 Ib., D-9J ft, TF-775 Ib., find d.
Ans. 2ft. 6.2- in.
3. Given P = 160 Ib., TF=900 Ib., rf-ljft., find D.
Ana. 7 ft f in.
4. Given IF = 160 Ib., D = 3$ ft, rf = 8J ft, find P.
Ans. 381 Ib. nearly. 6. In a wire cutter the wire is placed i in. from the fulcrum and the pressure of the hand is 7 in. from the fulcrum. Find the re- sistance of the wire if the hand exerts a force of 40 Ib.
Ans. 560 Ib.
6. In pulling a nail from a board with a hammer as shown in Fig. 19, find the resistance of the nail
at the start if P = 50 Ib., D = 10 in., and d = H in. Ans. 333 J Ib.
7. In the ordinary steel-yard, Fig. 20, what must be the weight P to
FIG. 20.
balance a weight W of 17 J Ib. if it is 1$ in. from fulcrum to application of W and 8J in. from application of P to fulcrum? Ana. 2 Ib. 9.2— oz.
8. If the steel-yard is turned over so that the distance from the ful- crum to W is i in., what weight W will 1J Ib. at P balance when P is 20 J in from the fulcrum? Ans. 62 i Ib.
RATIO AND PROPORTION
93
69. Hydraulic machines. — A principle known as Pascal's Law states that pressure exerted on a liquid in a closed vessel is transmitted equally and undiminished in all directions.
In Fig. 21, if the area of a is 1 sq. in. then a pressure of 1 Ib. at a gives a pressure of 1 Ib. on each square inch of the surface of C. If the area of the top of C is 100 sq. in. then a pressure of 1 Ib. at a will lift a weight of 100 Ib. at A.
If a, A, p, and P are the areas and pressures respectively then we have the proportion
a : A=p :P.
EXERCISES 24
1. A pressure of 5 Ib. on the cork of a jug filled with water gives how many pounds pressure tending to force out the bottom of the jug? The area of the
cork is 1| sq. in. and the area of the bottom is 245.6 sq. in.
Ans. 818| Ib.
2. A hydraulic lifter used to raise heavy weights has the pressure ap- plied to a piston having an area of 5 sq. in. by a lever. From the ful- crum to the point attached to the small piston is 4 in., and to the point where a force of 100 Ib. is applied is 22 in. Find the weight that can be raised on a piston having an area of 75.6 sq. in.
Solution. Let x = pressure in pounds applied on small piston. Then 1 00 : x = 4 : 22. From which x = 550. And 0.5: 75.6 = 550: P.
FIG. 21.
550 X 75.6 0.5
= 82,160.
Hence a weight of 82,160 Ib. can be raised.
3. A supply pipe for a 14-in. plunger hydraulic elevator piston is 1J sq. in. in area, and the pressure in the supply pipe is pumped up to 150 Ib. per square inch. What is the total pressure on the 14-in. plunger if it has an area of 153.94 sq. in.? Ans. 23,091 Ib.
CHAPTER VIII DENSITY AND SPECIFIC GRAVITY
70. Density. — Experience tells us that some bodies are heavier than others; that is, of two bodies of the same size, one weighs more than the other. Take a cubic foot of metal and one of wood; suppose the metal weighs 500 Ib. and the wood 50 Ib., then the metal is ten times as heavy as the wood, or the ratio of their densities is as 10 to 1 . We also say that the density of the metal is 500 Ib. per cubic foot.
Water has a density of about 62.5 Ib. per cubic foot. In the metric system the density of water under standard con- ditions is one gram per cubic centimeter.
The density of a body is its mass per unit volume. For our purpose the mass is the same as the weight. Strictly speaking, the weight of a body near the earth is the force with which the earth attracts the mass of the body.
71. Specific gravity. — The term specific gravity is used for the ratio of the densities of two bodies. Thus, the specific gravity of the metal with reference to the wood is 10, which means that the metal is ten times as heavy as the wood. It should be carefully noticed that the specific gravity of a substance is an abstract number, that is, a number with no name attached.
72. Standards. — For convenience the standard to which other substances are referred, in stating specific gravities, is water for solids and liquids.
RULE. The specific gravity of a substance is obtained by finding the weight of a certain volume of it and dividing thin weight by the weight of the same volume of the standard.
Thus, to find the specific gravity of a stone it is necessary to find its weight, and the weight of an equal volume of water. The weight of the stone divided by the weight of the water gives the specific gravity of the stone.
The specific gravity of any other body could be found in the same manner. Some difficulty might be found in doing
94
DENSITY AND SPECIFIC GRAVITY 95
the weighing, but a little ingenuity will devise a plan. Vari- ous methods for doing the weighing are discussed in physics.
Water is taken as the standard because of its abundance. All substances can be referred to it, but gases are usually compared with air or hydrogen gas.
If w stands for the weight of the body whose specific gravity is to be found, s the weight of the same volume of the stand- ard, and g for the specific gravity, the rule may be stated as a formula :
w -T- s = g.
73. Use. — Tables of the specific gravities of the various substances are given for use in making computations. In Table VIII are given the specific gravities of a few of the more common substances.
If it is required to find the weight of a block of iron 2 ft. by 3 ft. by 1 ft. we could find the number of cubic feet in the block which is 6. This times the weight of a cubic foot of water gives the weight of an equal volume of water, or 62.5X6 = 375 Ib. The weight of the water multiplied by the specific gravity of iron gives the weight of the iron, or 375 Ib. X7.2 = 2700 Ib.
In terms of the letters already used, since w + s = g,
Example 1. Find the specific gravity of a rock if 1 cu. ft. of it weighs 182 Ib.
Solution. Since water weighs 62.5 Ib. per cubic foot the specific gravity of the rock is found thus: 182 lb.^62.5 lb. = 2.912. .'. specific gravity of the rock is 2.912.
Example 2. How many cubic inches are there in 1 Ib. of cork, if its specific gravity is 0.24?
Solution. Since 1728 cu in, of water weigh 62.5 Ib.
1728 cu. in. of cork weigh 0.24X62.5 Ib.
.'. 1 cu. in. of cork weighs —— 7-700-" - Ib.
A nrl 1 -• _"••"•* ^"*-" 1728 _ 1 i K o
1728 ~ 0.24X6275"
.'.there are 115.2 cu. in. in 1 Ib. of cork.
96 PRACTICAL MATHEMATICS
EXERCISES 26
1. Find the weight of 176 cu. in. of copper. An*. 56+ Ib.
2. Find the weight of 37 cu. ft. of cast iron. Ans. 16,650 Ib.
3. A stone weighs 3 Ib. in air and 1.75 Ib. in water. Find its specific gravity. Ans. 2.4.
4. What is the specific gravity of a substance 40 cu. in. of which weighs 6 Ib.? Ans. 4.1 47 + .
6. Two cubic feet of cast iron immersed in water weigh how much? Solution. From Table VIII, 2 cu. ft. of cast iron weigh in air 2X450 Ib. =9001b.
2 cu. ft. of water weigh 2X62.5 Ib. = 125 Ib. Weight of iron in water = 900 Ib. -125 Ib. =775 Ib.
6. A piece of metal weighing 243 Ib. floats in mercury (s. g. 13.6) with 1*7 of its volume immersed. Determine the volume and the specific gravity of the metal. Ans. s. g. =7.2; Vol. =933.1 in.*
7. The specific gravity of ice is 0.92, of sea water 1.025. What part of an iceberg is below the surface of the water when floating?
Ans. 0.8975.
8. A balloon containing 10,200 cu. ft. will lift how great a weight if filled with hydrogen gas? Ans. Less than 756.5 Ib.
9. An irregular shaped mass of iron (s. g. 7.22) weighed in air 126 Ib. Find its volume. What would be its weight if immersed in water?
Ans. 482.5- in.»; 108.55- Ib.
10. A pond f acre in area is frozen over. Find the weight in tons of the ice if it is 3J in. thick and the specific gravity of ice is 0.92.
Ans. 273.95 tons.
11. Find the weight of a cubic meter of iron (s. g. 7.22) in kilograms. What is the weight in pounds? Ans. 7220 Kg. ; 15935.6 Ib.
12. Find the number of liters in a vat 2 m. X75 cm. X50 cm. Also find the weight in Kg. of the sulphuric acid (s.g. 1.84) required to fill it.
Ans. 7501.; 1380 Kg.
13. Find the value of 17 1. of sulphuric acid at 5 cents per Kg.
Ans. $1.56.
14. Mercury weighs 13.596 times as much as water at its greatest density. What is the pressure per square centimeter of a column of mercury 76 cm. high? Ans. 1033.296 g.
16. A column of mercury how high would cause a pressure per square inch equal to 14.7 Ib.? Ans. 29.89+ in.
16. A tank 1.85 m. long, 1.35 m. wide, and 85 cm. deep is filled with sea water (s. g. 1.025). What is the weight of the water?
Ans. 2175.95- Kg.
17. Sandstone of specific gravity 2.5 is crushed. Find the weight of 1 cu. yd. of the crushed stone if the voids are 35%. (See Ex. 22, p. 80).
Solution. If 35% are voids, 65% is rock. Weight = 0.65 X27 X62.5 X2.5 =-2742 + Ib.
DENSITY AND SPECIFIC GRAVITY 97
18. Granite of specific gravity 2.8 is crushed. Find the weight of 1 cu. yd. of the crushed rock if voids are 40%. Ans. 2835 Ib.
19. A casting of iron when immersed in water displaces 2 quarts; find the weight of the casting. Ans. 30 Ib.
20. An irregular shaped steel forging was found to displace 6.75 quarts of water; find the weight of the forging. (Use s. g. of steel = 7.85).
Ans. Ill Ib. nearly.
21. A wooden pattern for a casting weighs 2f pounds. An aluminum casting is to be made. Find the weight of the casting if the specific gravity of the wood is 0.52 and that of the aluminum is 2.6.
Ans. 13f Ib.
CHAPTER IX POWERS AND ROOTS
74. Powers. — When we have several numbers multiplied together, as 3X4X6 = 72, we call the numbers 3, 4, and 6, factors and 72 the product. If now we make all the factors alike, as 3X3X3X3=81, we call the product by the special name power. We say 81 is a power of 3, and 3 is the base of the power.
A power is a product obtained by using a base a certain number of times as a factor.
If the base is used twice as a factor the power is called the second power; three times as a factor, the third power; and so on for any number of times.
75. Exponent of a power. — Instead of 3X3X3X3, we may write 34. The small figure, placed at the right and above the base, shows how many times the base is to be used as a factor, and is called an exponent.
The exponent of a power is a number placed to the right and above a base to show how many times the base is used as a factor.
It should be noted that the use of the exponent gives us a short concise way of writing a continued product where the factors are all alike.
76. Squares, cubes, involution. — The second power of a number is called the square of the number, as 32.
The third power of a number is called the cube of the number, as 5s.
The higher powers have no special names. 34 is called the fourth power of 3, 57 the seventh power of 5, etc.
Involution is the process of finding the powers of numlx?rs.
98
POWERS AND ROOTS 99
EXERCISES 26
1. Find the square of 7, of 27, of 92, of 736. Find the square of the square of 3, of 7, of 10.
2. Find the cube of 7, of 8. Find the square of the cube of 3.
Arcs. 343; 512; 729.
3. Find the fourth power of 5. What is the difference between the fourth power of a number and the square of the square of the same number?
4. Find values of the following: (a) 7922, (6) 353, (c) 34, (d) 216.
Ans. (a) 627,264, (6) 42,875, (c) 81, (d) 65,536.
77. Roots. — If we take 9 and separate it into the two equal factors 3 and 3, that is, 9 = 3X3, then one of these factors, 3, is called the square root of 9. The process is just the inverse of that by which the power is found. Similarly 64=4X4X4, and we say 4 is the cube root of 64.
The square root of a number is one of the two equal factors into which a number is divided.
The cube root is one of the three equal factors into which a number is divided; the fourth root is one of the four equal factors; and so on for the higher roots.
78. Radical sign and index of root. — To indicate a root, we use the sign \/, which is called the radical sign. A small figure, called the index of the root, is placed in the opening of the radical sign to show what root is to be taken. Thus, -y/64 indicates the cube root of 64. The small 3 is the index of the root.
Since the square root is the most frequently written root, the index 2 is omitted. Thus, the square root of 625 is written \/625 and not -v/625. Higher roots are indicated as A/243, \/l28.
Evolution is the process of finding a root of a given number.
79. Square root— The numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, which are the squares of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, respectively, should be carefully remembered. It will be noticed that these are the only whole numbers less than 100 of which we can find the square roots. Such numbers as these are called perfect squares. As we pass to numbers above 100, the perfect squares become still more scarce.
The square root of 49 is 7, but the square root of 56 cannot
327 284
KM) PRACTICAL MATHEMATICS
be expressed as a whole number, nor can it be expressed as a decimal exactly. We can find it to any desired number of decimal places, and so as accurately as we wish. It remains to devise a method by which this may be done.
The practical man who wishes to find the square root of a number does not care greatly why he goes through a certain process; but it is very important to him that he shall be able to find the root quickly and accurately. In what follows then the attempt is made to tell in as simple a manner as passible how to find the root.
80. Process for the square root of a perfect square.—
Example 1. Find \/522729.
Explanation. Process.
First, separate the number 52'27'29 (723 Ans.
into periods of two figures 49
each, beginning at the right, 142 and placing a mark between them. The number of periods 1443 thus formed is equal to the number of figures in the root.
Find the largest perfect square which is equal to, or less than, the left-hand period, 52. This perfect square is 49. Write it under 52; and put its square root, 7, to the right as the first figure of the root. Now subtract 49 from 52 and. bring down the next period, 27, and unite with the remainder 3, thus obtaining 327.
Take twice 7, the first figure of the root, and write it to the left of 327. Find how many times this, 14, is contained in 32, which is 2, for the second figure of the root. Place this figure 2 in the root, and also to the right of 14, making 142. Now multiply 142 by 2, and write the product, 284, under 327. Subtract 284 from 327 and bring down and unite the next period, 29, with the remainder, 43, thus obtaining 4329.
In the above work 327 is called the first remainder; 14, the trial divisor; 142, the true divisor; and 4329, the second remainder.
Next multiply 72 by 2, and write it at the left of 4329 as the second trial divisor. Find how many times 144 is contained
4ML*) 4329
20816 20816
POWERS AND ROOTS 101
in 432, which is 3, for the third figure of the root. Place this figure, 3, in the root and also at the right of 144, making 1443, the second true divisor. Multiply 1443 by 3 and write the product under 4329. This gives no remainder. There- fore, 723 is the exact square root of 522,729, that is, 723X723 -522,729. Example 2. Find \/67808T6.
Explanation. Process.
First, separate the 6'78'08'16 (2604 Ans.
number into periods of 4
two figures each as in 45 1273 example 1. As before, i276
we find the greatest 5204 square, 4, in the left- hand period, write it
under 6 and put the square root, 2, of this square for the first figure of the root. Subtract the square, 4, from 6, and bring down the next period, 78, and unite it with the 2, making the first remainder, 278.
Take twice 2 for a trial divisor. . Find how many times it is contained in the first remainder, excepting the right-hand figure ; that is, find how many times 4 is contained in 27. The number is 6, which write as the second figure of the root, and also at the right of the trial divisor. This makes 46 the true divisor. Multiply the true divisor by 6, and subtract the product, 276, from the first remainder. Bring down and unite the next period to the difference, making the second remainder, 208.
Multiply the root already found by 2, and get the second trial divisor, 52. Find how many times this is contained in 20, which gives 0 for the next figure of the root. Place this 0 in the root and also to the right of 52, making 520, the second true divisor. Now, since the 0 written in the root is the multiplier, nothing is gained by multiplying the true divisor by it, and subtracting from 208. This part of the process is omitted, and the next period, 16, is united with 208, making 20,816, the third remainder.
The third trial divisor is twice the root, 260, which gives
102 PRACTICAL MATHEMATICS
520. This is contained in 2081, 4 times. Place the 4 as the next figure of the root, and also to the right of 520, making 5204, the third true divisor. Multiply this by 4 and subtract from the third remainder. As the remainder is zero, 2604 is the exact square root of 6,780,816.
81. Square root of a number containing a decimal.-
Example. Find V 665. 1241. Process.
Here the division into periods 6'65.'12'41 (25.79 Ans.
is made by beginning at the decimal point and going in both directions. The rest of the work is the same as in examples
265 225
507
4012 3549
1 and 2, Art. 80. 5149T634T
46341
The student should note that the second trial divisor, 50, is contained 8 times in the first three figures of the second remainder, 4012. However, if 8 were used as the root, it would give a number larger than 4012 when the true divisor was multiplied by it. The relations noted here should help to make clear why we give to the trial divisor its name.
The decimal point in the root is so placed that there are as many whole number figures in the root as there are whole number periods in the number of which the root is extracted. The position of the decimal point can also be determined so that there will be as many decimal places in the root as there are decimal periods in the number of which the root is being extracted.
If the decimal part of the number consists of an odd number of figures a cipher is annexed to make a full period at the right .
Thus, in pointing off 53.76542 into periods it is 53'.76'54'20.
82. Roots not exact. — Most numbers are not perfect squares, but the roots may be found to any desired number of decimal places. When extracting the root of a number not a perfect square, one must determine how many decimal places he wishes in the answer, and then annex ciphers to the right of the number till there are as many decimal periods as there are to be decimal places in the root. The root is then extracted
POWERS AND ROOTS 103
in the usual manner. We stop when the desired number of figures is found in the root.
Example. Find \/27 to three decimal places in the root.
Explanation. Process.
Since three decimal places 27.'00'00'00(5.196 Ans.
are required in the root, annex
three periods of ciphers to the right of 27. These are the deci-
1029
9900 9261
mal periods. Extract the root
as before. Place the decimal ]Q386 53900
point in the root as in example 62316
of Art. 81. It will be noticed L584~
that there is a remainder;
this is disregarded as it affects the next figures only, that is,
the fourth and following figures in the decimal part of the root.
83. Root of a common fraction. — If the numerator and the denominator of the fraction are each a perfect square, find the square root of each separately.
Example 1. Find
The V 144 = 12, and \/625 = 25.
Hence V'iHfr£ = H> 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. TV
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) =62:42=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|)3:33=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*37rd.; (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.
JTIG 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. 0 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. $222.30.
27. Find the area in acres of a farm which is represented on paper as a rectangle 3f in. by 102 in. on a scale of •?$ in- 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 AC2 = AB2+BC2, or in general, if c stands for the hypotenuse, 6 for base, and a for altitude, then c2 = a2+62. 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 = \/a2+&2.
= Vl62+142=\/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=\/92+92=Vl62.
D=\/162+92=\/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, a1 : 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-T87=lTin. 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'^XS1 = 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+V2 = 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 0 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)2X 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 0 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«)2X 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 = 600, 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-Vrf-(|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 = 7rr2 = 3.1416 Xr2.
[25] A= i7rd2 = 0.7854 Xd2.
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 Ar for the area of the ring, then
[28] Ar=A-a=rR2-7rr2=7r(R2-r2)=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,
Ar = 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 6c%X7rr2. If the radius is 10 in., the area of the sector is
B A=iX3.1416X102 = 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 = |h2^-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 0 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(a2+b2).
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(92+32) =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 =0.7854X62 in.2 Area of 1-in. pipe =0.7854 XI2 in.2 Area of 6-in. pipe 0.7854X62 62
1-1 — .— . — __ — — __ -IK 4 99£
Area of 1-in. pipe 0.7854 XI2 I2
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 =62X0.7854 in.2
Areaof 8-in. circle =82X 0.7854 in.2
Sum of areas =(62+82)X 0.7854 in.2 But 62+82 = 102, 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 102X0.7854 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 X62 = 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 <l(DE + }AB) as the circumference? Ans. 0.65 — % too small.
48. If statements in numbers 46 and 47 gave the exact length of the circumference, what would be the value of v in each case?
Ans. In No. 46, ir=\/T6 = 3.16228- ; in No. 47, r=3.1213 + .
49. Make a construction as shown in Fig. 103, and A B is approxi- mately the quadrant of the circle. Find the per cent it differs from the correct value. Ans. 0.4 + % too large.
FIG. 104.
Suggestion. By the quadrant of the circle is meant the length of the arc BN.
60. Justify the following rule used by sheet-metal workers, or show the per cent of error if it is not correct: Divide the radius AO, Fig. 104, into four equal parts; place one of these parts from A to C and another from B to D, the ends of two perpendicular diameters. Connect C and D, which gives the side of the square of the same area as the circle.
Ans. 0.5 + % too small.
CIRCLES 163
51. The stem of a 4-in. safety valve, Fig. 105, is 2j in. from the ful- crum. Supposing the valve will blow when the gage reads 7 Ib. without any weight on the lever (i.e.,
that 7 Ib. per square inch on the valve overcomes weight of valve and lever), at what pressure would it blow with a weight of 75 Ib. 32 in. from the fulcrum?
Ans. 76.4+ Ib. per square inch.
Solution. If P represents the total number of pounds pressure on the valve necessary to lift the weight, then
2.75: 32 = 75: P. From which FIG. 105.
P= 872.73.
Area of valve =0.7854 X42 = 12.5664 sq. in.
Pressure to raise weight = 872.73 -^ 12.5664 =69.4 Ib. per sq. in.
Total pressure =69.4 Ib. +7 Ib. =76.4 Ib. per sq. in.
52. What weight of ball would be required to allow the valve in exer- cise 51 to blow off at 80 Ib.? Ans. 78.8+ Ib.
53. If the original weight of 75 Ib. is used, at what distance from the fulcrum should it be placed to allow the valve to blow off at 80 Ib.?
Ans. 33.6+ in.
54. The following is an approximate formula for determining the number of inscribed tangent circles in a larger circle:
N = 0.907 [-j- 0.94 +3.7,
\ Of j
where N is the number, D the diameter of the enclosing circle, and d the diameter of the inscribed circles. This formula can be used to find the number of wires that can be put in a casing of given size.
Apply the above formula and find how many wires | in. in diameter can be placed inside a 5-in. pipe. Ans. 78.
65. How many steel balls 0.4 in. in diameter can rest at the bottom of a closed pipe 2| in. in diameter? Ans. 29.
56. In a circle of radius 5 ft. there is a chord 6 ft. 6 in. in length. Find the height of the segment. Ans. 1.200+ ft.
57. A segment of a circle cut off by a chord 4 ft. 6 in. in length has a height of 1 ft. 10 in. Find the radius of the circle. Ans. 27.57 — in.
58. Find the area of a sector in a circle whose radius is 28 cm., if the sector contains an angle of 50° 36'. Ans. 346.19— cm.2
59. Find the area of a sector whose arc is 99.58 m. long and radius 86.34m. Ans. 4298.87- m.2
60. Test formula [30 (a)] by taking the segment as half the circle 60 in. in diameter.
Ans. By [30(a)j A = 1425 in.2 As f a circle A =1414— in.2
61. Find the radius of a circle in which a chord of 10 ft. has a middle ordinate of 3 in, Ans. 50 ft. 1.5 in.
Kit
PRACTICAL MATHEMATICS
62. The Gothic Arch is formed by two arcs each i of a circle. The center of each circle is at the extremity of the width of the arch, that is, the radius equals the width. Find the area of such an arch of radius 6ft. (See Fig. 106.) Ann. 22.11 ft.*
Fia. 106.
63. In Fig. 107, W, W is a wall with a round corner, of dimensions as given from A to B, on which a molding, gutter, or cornice is to be placed ; find the radius of the circle of which arc ANB is a part. Ans. 3 ft. J in.
64. Each of four steam engines is supplied by a 6-in. steam pipe.
These open off from a single steam pipe. Find the diameter of the larger pipe that it may have the same capacity as the four 6-in. pipes.
Ans. 12 in.
65. A milling cutter 4J in. in diameter is cutting soft steel at the rate of 45 ft. per minute. Find the number of revolutions per minute.
Ans. 38.2-.
66. How many turns per second must a drill \ in. in diameter make so that the outer edge of the lip will have a cutting speed of 35 ft. per minuteV Ans. 4.46 — .
67. The distance between the center of the crank pin C, Fig. 109, and the center of the flywheel at D is 20 in. What is the length of the stroke
FIG. 108. — Milling cutter.
Fio. 109.
of the piston? If the flywheel makes 144 R. P. M., find the average speed of the piston in feet per minute. Ans. 40 in.; 960 ft. per min. 68. The "piston speed " in a Corliss engine should be 600 ft. per minute.
CIRCLES
165
FIG. 110.
How many revolutions per minute should be made by an engine having a 20-in. stroke? By an engine having a 36-in. stroke?
Ana. 180; 100.
69. Which would occupy the greater portion of the square shown in Fig. 110, the four small circles or the large circle? Ans. Both the same.
70. The drivers on a locomotive are making 210 R. P. M., and are 76 in. in diameter. Find the speed of the loco- motive in miles per hour if 2% is allowed for
slipping. Ans. 46. 53 mi. per hr.
71. In a Corliss engine the high-pressure cylin- der is 22 in. in diameter. What must be the diameter of the low-pressure cylinder in order that it may have double the area of the high-pressure cylinder? Ans. 31.1+ in.
72. If the total pressure on the piston of a brake cylinder is 8100 lb., what is the diameter of the
cylinder if the pressure is 60 lb. per square inch? Ans. 13.1+ in.
73. A 10-in pipe is to be branched off into two equal pipes. What must be the diameter of each of these pipes if the two pipes shall equal the area of the 10-in. pipe? Ans. 7.07+ in.
74. A machine screw f in. in diameter has 12 sharp V-threads to the inch. Find the root diameter. Find the tensile strength at 50,000 lb. per square inch. Give answer to the nearest 100 lb. Ans. 9100 lb.
75. What should be the area of the opening of a cold-air box for a hot- air furnace in order to supply 7 hot-air pipes 9 in. in diameter and one pipe 14 in. in diameter, if the area of the cold-air opening is -| the area of the hot-air pipes? Ans. 449.4+ sq. in.
76. A cylinder of a double-acting engine is 26 in. in diameter and the length of the stroke is 30 in. Compute the pressure on each side of the piston if the piston rod is 3| in. in diameter and the steam pressure in the cylinder is 150 lb. per square inch. (Use formula [28].)
FIG. 111.
77. In practice piston rings for a steam engine are turned so that they are \\% larger in diameter than the diameter of the cylinder barrel. They then have a piece cut out and are sprung into place. Find the diameter of the ring for a cylinder 24 in. in diameter. Find the length of the piece to be cut out if when sprung into place it has a clearance of i^ in. between ends. Give dimensions to the nearest 32nd of an inch.
Ans. 24 -f in.; 1^ in.
160 PRACTICAL MATHEMATICS
78. Find the speed of a belt running over a pulley having a diameter of 22 in. and making 320 R. P. M., if \% is allowed for slipping.
Ana. 1834— ft. per minute.
79. If the greatest and least diameters of an elliptical manhole arc 2 ft. 7 in. and 2 ft. 3 in. respectively, find its area. Find its perimeter, using formulas [32 (b)] and [32 (c)].
Ans. 4.565 ft.1; 91.2+ in. and 91.3 in.
80. Find the length in feet of the arc of contact of a belt with a pulley, if the pulley is 3 ft. 6 in. in diameter and the arc of contact is 210°.
Ans. 6 ft. 5— in.
81. As in the preceding, find the length of the arc of contact if it is 120° and the diameter of the pulley is 16 in. Ans. 16 J in.
REGULAR POLYGONS AND CIRCLES
127. It is often necessary to determine the dimensions of a regular polygon inscribed in or circumscribed about a given circle, or to determine the size of a circle that can be inscribed in or circumscribed about a given polygon.
Such problems are readily solved by trigonometry, and some of them may be solved by geometry. In either case, though, the computation may be long and tedious. For this reason handbooks give rules by which the computations can be readily made. In the table on page 167 are classified certain facts about the regular polygons named. These facts can readily be applied to polygons of any size.
(1) To find the area of a polygon when the length of one side is given. — Multiply the square of the side by a number given in column (3).
This rule is an application of the principle that similar areas are in the same ratio as the squares of their like dimensions. Show that this is the case.
Example. Find the area of a regular pentagon having sides of 7 in.
Area = 72X 1.7204774 in.2 = 84.30339 in.2
(2) To find the side of a polygon when its area is given. — Diinde the area of the polygon by a number from column (3). The square root of the quotient is the required side of the polygon.
Example. The area of an octagon is 4376 ft.2; find a side of the polygon.
Side = v/4376-*- 4.828 = 30.1 (Hi ft.
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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 0_ 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-rrd2, 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 Ad2
for the areas of the circles are in the same ratio as the squares of the diameters. This proportion gives
A =d2. 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 4602 =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 0.7854d2, 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 0 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. — T