ornia tal y A TEXTBOOK ON STEAM ENGINEERING INTERNATIONAL CORRESPONDENCE SCHOOLS SCRANTON, PA. ARITHMETIC MENSURATION AND USE OF LETTERS IN FORMULAS PRINCIPLES OF MECHANICS MACHINE ELEMENTS MECHANICS OF FLUIDS STRENGTH OF MATERIALS 1048! SCRANTON INTERNATIONAL TEXTBOOK COMPANY A-2 Copyright, 1897, by THE COLLIERY ENGINEER COMPANY. Copyright, 1902, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. Arithmetic : Copyright, 1893, 1894, 1895, 1896, 1897, 1898, 1899, by THE COLLIERY EN- GINEER COMPANY. Copyright, 1901, by INTERNATIONAL TEXTBOOK COMPANY. Mensuration and Use of Letters in Formulas : Copyright, 1894, 1895, 1897, 1898, by THE COLLIERY ENGINEER COMPANY. Principles of Mechanics: Copyright, 1901, by INTERNATIONAL TEXTBOOK COM- PANY. Entered at Stationers' Hall, London. Machine Elements: Copyright, 1901, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. Mechanics of Fluids : Copyright, 1901, by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall. London. Strength of Materials : Copyright, 1901. by INTERNATIONAL TEXTBOOK COMPANY. Entered at Stationers' Hall, London. Arithmetic, Key : Copyright, 1893, 1894. 1896, 1897, 1898, by THE COLLIERY ENGI- NEER COMPANY. Mensuration and Use of Letters in Formulas, Key : Copyright, 1894, 1895, by THE COLLIERY ENGINEER COMPANY. All rights reserved. Cl BURR PRINTING HOUSE, FRANKFORT AND JACOB STREETS, NEW YORK. 715 PREFACE. In preparing this set of textbooks, which replaces those formerly sent to students of our Stationary Engineering Course under title of "The Elements of Steam Engineer- ing," we have aimed to produce a practical treatise on the care, operation, and management of all kinds of steam plants, with the result that these volumes form the most thorough, practical, and comprehensive treatise on the sub- ject that has yet been published. While these textbooks contain much that is of direct value to the designer, they have been prepared more particularly to meet the wants of those engaged in actual work in steam plants and of those desiring to prepare themselves to pass examinations for license. The subject of steam boilers and the various appliances found in boiler rooms has been very thoroughly covered. Enough of the theory has been given to enable the fireman or engineer to surmount the usual difficulties encountered and to select boilers and appliances best suited to particular conditions; the best modern practice of installing, managing, and testing boiler plants is also given. Steam engines and their appliances, and condensers, are treated in the same thorough manner as steam boilers. The general principles underlying the operation of all makes of steam engines, engine appliances, and condensers are so treated that the engineer will be enabled not only to satis- factorily and economically run a particular engine, but will also be able to rapidly trace out the peculiarities of other iii iv PREFACE. engines, make adjustments of all kinds of governors and valve gears, make engine tests, take and read indicator dia- grams, and select engines and appliances to suit existing conditions. The sections relating to pumps are especially valuable to pump runners and engineers in charge of water-works plants. Our thanks are due to the different pump manu- facturers, notably to Henry R. Worthington, New York, for information furnished and courtesies extended. On account of the rapid and growing increase in the use of elevators in large buildings, we have devoted a large part of one volume to this subject. The majority of elevators in use are looked after by the engineers in charge of the buildings in which they are located, and a knowledge of their construction, care, and management is an invaluable aid to their successful operation. The present is the first attempt to treat all existing types of elevators, and we record with pleasure that in preparing this portion of the Course we have had the hearty cooperation of the following firms: Otis Elevator Company of New York; Otis Elevator Company of Chicago (formerly Crane Elevator Company) ; Morse, Williams & Company, Philadelphia ; A. B. See Manufacturing Company, Brooklyn; The Elektron Manu- facturing Company, Springfield, Mass. ; Burdett-Rowntree Manufacturing Company, Chicago; Automatic Switch Com- pany, Baltimore, Md. ; The Winslow Brothers Company, Chicago; The Mason Regulator Company, Boston, Mass.; The Cutler-Hammer Manufacturing Company, Milwaukee, Wis. ; Sprague Elevator Company, New York; The Plunger Elevator Company, Worcester, Mass. ; Elevator Supply and Repair Company, New York and Chicago. Special thanks are due to The Electrical World and Engineer for infor- mation gleaned from its files, and to Mr. R. C. Smith and Mr. E. W. Marshall of the Otis Elevator Company for assistance rendered in the preparation of this treatise. By courtesy of the Otis Elevator Company we have incorpo- rated a few valuable articles on elevator management appear- ing in their pamphlet " Information for Engineers." PREFACE. v The paper on steam heating will be valuable to those engineers who look after the steam-heating systems in large buildings and also to those who desire to use an existing boiler plant to heat buildings with steam. On account of the inconvenience resulting from the large size of the former drawing volume, it has been especially reset for this edition and the pages made uniform with the other volumes. On account of its thinness, it has been combined with one of the other volumes, thus making five volumes in the set instead of six, as formerly. The paper on Mechanical Drawing has been thoroughly revised and much new matter added. Among those who assisted in the preparation of these volumes, in addition to members of our own staff, were Charles J. Mason, Chief Engineer in charge of the State Street power station of the Brooklyn Rapid Transit Company, Brooklyn, N. Y. , and L. Hollingsworth, Jr., formerly Chief Draftsman of the Dickson Manufacturing Company, Scranton, Pa., and now Chief Draftsman of the Pennsylvania Iron Works Company, Philadelphia, Pa. The whole Course was under the editorial supervision of John A. Grening. The method of numbering the pages, cuts, articles, etc., is such that each paper and part is complete in itself; hence, in order to make the indexes intelligible it was necessary to give each paper and part a number. This number is placed at the top of each page, on the headline, opposite the page number; to distinguish it from the page number it is pre- ceded by the printer's section mark (). Consequently, a reference such as 7, page 28, is readily found as follows: Look along the inside edges of the headlines until 7 is found, and then through 7 until page 28 is found. The Examination Questions and their answers are grouped together at the ends of the volumes containing the papers to which they refer, and are given the same section numbers. INTERNATIONAL CORRESPONDENCE SCHOOLS. CONTENTS. ARITHMETIC. S> Definitions action. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 Page. I I 4 9 11 16 20 23 25 28 31 33 37 39 41 42 44 52 1 4 9 10 13 16 18 20 21 Notation and Numeration Addition Subtraction Multiplication Division ... Cancelation Fractions Reduction of Fractions Addition of Fractions Multiplication of Fractions Division of Fractions Decimals Addition of Decimals Subtraction of Decimals . ... Multiplication of Decimals Division of Decimals . Signs of Aggregation Percentage Calculations Involving Percentage . . Denominate Numbers Measures Reduction of Denominate Numbers . Addition of Denominate Numbers Subtraction of Denominate Numbers Multiplication of Denominate Numbers Division of Denominate Numbers XV. . vii viii CONTENTS. ARITHMETIC Continued. Section. Page. Involution 2 25 Evolution 2 27 Square Root . . . 2 28 Cube Root 2 35 Ratio 2 40 Proportion 2 43 Unit Method 2 48 MENSURATION AND USE OF LETTERS IN FORMULAS. Formulas 3 1 Mensuration ..." 3 13 Quadrilaterals 3 17 The Triangle 3 20 Polygons 3 24 The Prism and Cylinder 3 32 The Pyramid and Cone 3 36 The Frustum of a Pyramid or Cone . . 3 37 The Sphere and Cylindrical Ring ... 3 39 PRINCIPLES OF MECHANICS. Matter and Its Properties 4 1 Motion and Velocity 4 5 Force 4 10 Gravitation and Weight ...... 4 15 Work, Power, and Energy 4 20 Composition and Resolution of Forces . 4 26 Friction ........... 4 33 Center of Gravity . . 4 39 Centrifugal Force ........ 4 43 Equilibrium 4 45 MACHINE ELEMENTS. The Lever, Wheel, and Axle .... 5 1 Pulleys . . . 5 7 Belts . 5 12 Wheel Work 5 25 Gear Calculations . 5 34 CONTENTS. x MACHINE ELEMENTS Continued. Section. Page. Fixed and Movable Pulleys ..... 5 42 The Inclined Plane ....... 5 47 The Screw .......... 5 53 Velocity Ratio and Efficiency .... 5 59 MECHANICS OF FLUIDS. Hydrostatics ........ , . . 6 1 Specific Gravity ........ 6 15 Buoyant Effects of Water ..... 6 19 Hydrokinetics ......... 6 22 Pneumatics .......... 6 27 Pneumatic Machines ....... 6 40 Pumps ............ 6 47 STRENGTH OF MATERIALS. General Principles ........ 7 1 Tensile Strength of Materials .... 7 2 Crushing Strength of Materials ... 7 15 Transverse Strength of Materials 7 23 Shearing, or Cutting, Strength of Mate- rials ............ 7 29 Torsion ........... 7 32 EXAMINATION QUESTIONS. Section. Arithmetic, Part 1 ....... ,. 1 Arithmetic, Part 2 ........ 1 Arithmetic, Part 3 ........ 1 Arithmetic, Part 4 . ....... 2 Arithmetic, Part 5 ........ 2 Mensuration and Use of Letters in For- mulas ........... 3 Principles of Mechanics ...... 4 Machine Elements ........ 5 Mechanics of Fluids . ..... 6 Strength of Materials ....... 7 ANSWERS TO EXAMINATION QUESTIONS. Section. Arithmetic, Part 1 ........ 1 Arithmetic, Part 2 . 1 x CONTENTS. ANSWERS TO EXAMINATION QUESTIONS Con- tinued. Section. Arithmetic, Part 3 1 Arithmetic, Part 4 . ' 2 Arithmetic, Part 5 . . ".'.. . * . . 2 Mensuration and Use of Letters in For- mulas . . . . . . . v ... 3 Principles of Mechanics 4 Machine Elements ........ 5 Mechanics of Fluids G Strength of Materials . 7 ARITHMETIC. (PART 1.) DEFINITIONS. 1. Arithmetic is the art of reckoning, or the study of numbers. 2. A unit is one, or a single thing, as one, one bolt, one pulley, one dozen. 3. A number is a unit, or a collection of units, as one^ three engines, five boilers. 4. The unit of a number is one of the collection of units which constitutes the number. Thus, the unit of twelve is one, of twenty dollars is one dollar, of one hundred bolts is one bolt. 5. A concrete number is a number applied to some particular kind of object or quantity, as three grate bars, five dollars, ten. pounds. 6. An abstract number is a number that is not applied to any object or quantity, as three, five, ten. 7. Iiike numbers are numbers which express units of the same kind, as six days and ten days, two feet and five feet. 8. Unlike numbers are numbers which express units of different kinds, as ten months and eight miles, seven wrenches and five bolts. NOTATION AND NUMERATION. 9. Numbers are expressed in three ways: (1) by words; (2) by figures; (3) by letters. 10. Notation is the art of expressing numbers by figures or letters. 11. Numeration is the art of reading the numbers which have been expressed by figures or letters. 1 For notice of copyright, see page immediately following the title page. 2 ARITHMETIC. 1 12. The Arabic notation is the method of expressing numbers by figures. This method employs ten different figures to represent numbers, viz. : Figures 0123456789 naught, one two three four five six seven eight nine Names cipher, or zero. The first character (0) is called naught, cipher, or zero, and when standing alone has no value. The other nine figures are called digits, and each has a value of its own. Any whole number is called an Integer. 13. As there are only ten figures used in expressing numbers, each figure must have a different value at different times. 14. The value of a figure depends upon its position in relation to other figures. 15. Figures have simple values, and local, or relative, values. 16. The simple value of a figure is the value it expresses when standing alone. 17. The local, or relative, value of a figure is the increased value it expresses by having other figures placed on its right. For instance, if we see the figure 6 standing alone, thus 6 we consider it as six units, or simply six. Place another 6 to the left of it ; thus 66 The original figure is still six units, but the second figure is ten times 6, or 6 tens. If a third 6 be now placed still one place further to the left, it is increased in value ten times more, thus making it 6 hundreds 666 A fourth 6 would be 6 thousands 6666 A fifth 6 would be 6 tens of thousands, or sixty thousand 66666 A sixth 6 would be 6 hundreds of thousands . . 666666 A seventh 6 would be 6 millions 6666666 1 ARITHMETIC. The entire line of seven figures is read six millions six hundred sixty-six thousands six hundred sixty-six. 18. The increased value of each of these figures is its local, or relative, value. Each figure is ten times greater in value than the one immediately on its right. 19. The cipher (0) has no value in itself, but it is useful in determining the place of other figures. To represent the number four hundred five ^ two digits only are necessary, one to represent four hundred and the other to represent five units; but if these two digits are placed together, as 45, the 4 (being in the second place) will mean 4 tens. To mean 4 hun- dreds, the 4 should have two figures on its right, and a cipher is therefore inserted in the place usually given to tens, to show that the number is composed of hundreds and units only, and that there are no tens. Four hundred five is therefore ex- pressed as 405. If the number were four thousand and five, two ciphers would be inserted ; thus, 4005. If it were four hundred fifty, it would have the cipher at the right-hand side to show that there were no units, and only hundreds and tens; thus, 450. Four thousand and fifty would be expressed 4050. 20. In reading figures, it is usual to point offihe number into groups of three figures each, beginning at the right- hand, or units, column, a comma (,) being used to point off these groups. Billions. Millions. Thousands. Units. S' en 1 _o 3 1 3 % & O a en | i 1 1 i S 2 "8 3 "o ^ Q o r* rn "8 '3 Hundreds Tens of Bi Billions. Hundreds Tens of M Millions. Hundreds Tens of Tl Thousands Hundreds Tens of U OT '3 4 3 2,1 9 8,7 G 5,4 3 2 In pointing off these figures, begin at the right-hand figure and count units, tens of units, hundreds of units; the next 4 ARITHMETIC. 1 group of three figures is thousands; therefore, we insert a comma (,) before beginning with them. Beginning at the figure 5, we say thousands, tens of thousands, hundreds of thousands, and insert another comma. We next read millions, tens of millions, hundreds of millions (insert another comma). billions, tens of billions, hundreds of billions. The entire line of figures would be read: Four hundred thirty-two billions one htindred ninety-eight millions seven hundred sixty- five thousands four hundred thirty-two. When we thus reads, line of figures, it is called numeration, and if the numeration be changed back to figures, it is called notation. For instance, the writing of the following figures, 72,584,623, would be the notation, and the numeration would be seventy- two millions Jive hundred eighty-four thousands six hundred twenty-three. 21. NOTE. It is customary to leave the "s" off the words millions, thousands, etc. in cases like the above, both in speaking and writing; hence, the above would usually be expressed seventy-two million five hundred eighty-four thousand six hundred twenty-three. 22. The four fundamental processes of arithmetic are addition, subtraction, multiplication, and division. They are called fundamental processes because all operations in arith- metic are based upon them. ADDITION. 23. Addition is the process of finding the sum of two or more numbers. The sign of addition is -f. It is read plus, and means more. Thus, 5 + 6 is read 5 plus 6, and means that 5 and 6 are to be added. 24. The sign of equality is =. It is read equals or is equal to. Thus, 5 -f 6 = 11 may be read 5 plus 6 equals 11, or 5 plus 6 is equal to 11. 25. Like numbers can be added; unlike numbers cannot be added. 6 dollars can be added to 7 dollars, and the sum will be 13 dollars; but 6 dollars cannot be added to 7 feet. 1 ARITHMETIC. 26. The following table gives the sum of any two numbers from 1 to 12 ; it should be carefully committed to memory : 1 and 1 is 2 2 and 1 is 3 3 and 1 is 4 4 and 1 is 5 1 and 2 is 3 2 and 2 is 4 3 and 2 is 5 4 and 2 is 6 1 and 3 is 4 2 and 3 is 5 3 and 3 is 6 4 and 3 is 7 1 and 4 is 5 2 and 4 is 6 Sand 4 is 7 4 and 4 is 8 1 and 5 is 6 2 and 5 is 7 3 and 5 is 8 4 and 5 is 9 1 and 6 is 7 2 and 6 is 8 3 and 6 is 9 4 and 6 is 10 1 and 7 is 8 2 and 7 is 9 3 and 7 is 10 4 and 7 is 11 1 and 8 is 9 2 and 8 is 10 3 and 8 is 11 4 and 8 is 12 1 and 9 is 10 2 and 9 is 11 3 and 9 is 12 4 and 9 is 13 1 and 10 is 11 2 and 10 is 12 3 and 10 is 13 4 and 10 is 14 1 and 11 is 12 2 and 11 is 13 3 and 11 is 14 4 and 11 is 15 1 and 12 is 13 2 and 12 is 14 3 and 12 is 15 4 and 12 is 16 5 and 1 is 6 6 and 1 is 7 7 and 1 is 8 8 and 1 is 9 5 and 2 is 7 6 and 2 is 8 7 and 2 is 9 8 and 2 is 10 5 and 3 is 8 6 and 3 is 9 7 and 3 is 10 8 and 3 is 11 5 and 4 is 9 6 and 4 is 10 7 and 4 is 11 8 and 4 is 12 5 and 5 is 10 6 and 5 is 11 7 and 5 is 12 8 and 5 is 13 5 and 6 is 11 6 and 6 is 12 7 and 6 is 13 8 and 6 is 14 5 and 7 is 12 6 and 7 is 13 7 and 7 is 14 8 and 7 is 15 5 and 8 is 13 6 and 8 is 14 7 and 8 is 15 8 and 8 is 16 5 and 9 is 14 6 and 9 is 15 7 and 9 is 16 8 and 9 is 17 5 and 10 is 15 6 and 10 is 16 7 and 10 is 17 8 and 10 is 18 5 and 11 is 16 6 and 11 is 17 7 and 11 is 18 8 and 11 is 19 5 and 12 is 17 6 and 12 is 18 7 and 12 is 19 8 and 12 is 20 9 and 1 is 10 10 and 1 is 11 11 and 1 is 12 12 and 1 is 13 9 and 2 is 11 10 and 2 is 12 11 and 2 is 13 12 and 2 is 14 9 and 3 is 12 10 and 3 is 13 11 and 3 is 14 12 and 3 is 15 9 and 4 is 13 10 and 4 is 14 11 and 4 is 15 12 and 4 is 16 9 and 5 is 14 10 and 5 is 15 11 and 5 is 16 12 and 5 is 17 9 and 6 is 15 10 and 6 s 16 11 and 6 is 17 12 and 6 is 18 9 and 7 is 16 10 and 7 s 17 11 and 7 is 18 12 and 7 is 19 9 and 8 is 17 10 and 8 s 18 11 and 8 is 19 12 and 8 is 20 9 and 9 is 18 10 and 9 s 19 11 and 9 is 20 12 and 9 is 21 9 and 10 is 19 10 and 10 s 20 11 and 10 is 21 12 and 10 is 22 9 and 11 is 20 10 and 11 s 21 11 and 11 is 22 12 and 11 is 23 9 and 12 is 21 10 and 12 is 22 11 and 12 is 23 12 and 12 is 24 27. For addition, place the numbers to be added directly under one another, taking care to place units under units, tens under tens, Jiundreds under hundreds ^ and so on. When the numbers are thus written, the right-hand figure of one number is placed directly under the right-hand figure of the one above it, thus bringing- units under units, tens under tens, etc. Proceed as in the following examples: 6 ARITHMETIC. 1 28. EXAMPLE. What is the sum of 131, 222, 21, 2, and 413 ? SOLUTION. 131 222 21 2 413 sum 789 Ans. EXPLANATION. After placing the numbers in proper order, begin at the bottom of the right-hand, or units, col- umn and add; thus, 3 + 2 + 1 + 2 + 1 = 9, the sum of the numbers in units column. Place the 9 directly beneath as the first, or units, figure in the sum. The sum of the numbers in the next, or tens, column equals 8 tens, which is the second, or tens, figure in the sum. The sum of the numbers in the next, or hundreds, col- umn equals 7 hundreds, which is the third, or hundreds, figure in the sum. The sum, or answer, is 789. 29. EXAMPLE. What is the sum of 425, 36, 9,215, 4, and 907 ? SOLUTION. 425 sum 10587 Ans. EXPLANATION. The sum of the numbers in the first, or units, column is 27 units; i. e., 2 tens and 7 units. Write 27 as shown. The sum of the numbers in the second, or tens, column is 6 tens, or 60. Write 60 underneath 27 as shown. The sum of the numbers in the third, or hundreds, column is 15 hundreds, or 1,500. Write 1,500 under the two pre- ceding results as shown. There is only one number in the fourth, or thousands, column, 9, which represents 9,000. 1 ARITHMETIC. 7 Write 9,000 under the three preceding results. Adding these four results, the sum is 10,587, which is the sum of 425, 36, 9,215, 4, and 907. 3O. The addition may also be performed as follows: 425 36 9215 4 907 sum 10587 Ans. EXPLANATION. The sum of the numbers in units col- umn = 27 units, or 2 tens and 7 units. Write the 7 units as the first, or right-hand, figure in the sum. Reserve the 2 tens and add them to the figures in tens column. The sum of the figures in the tens column plus the 2 tens reserved and carried from the units column is 8, which is written down as the second figure in the sum. There is nothing to carry to the next column, because 8 is less than 10. The sum of the numbers in the next column is 15 hundreds, or 1 thousand and 5 hundreds. Write down the 5 as the third, or hundreds, figure in the sum and carry the 1 to the next column. 1 -j- 9 = 10, which is written down at the left of the other figures. The second method saves space and figures, but the first is to be preferred when adding a long column. 31. EXAMPLE. Add the numbers in the column below : SOLUTION. 890 82 90 393 281 80 770 83 492 80 383 84 191 sum 3899 Ans. H. S. I.2 8 ARITHMETIC. 1 EXPLANATION. The sum of the digits in the first column equals 19 units, or 1 ten and 9 units. Write the 9 and carry 1 to the next column. The sum of the digits in the second column + the 1 carried from the first column is 109 tens, or 10 hundreds and 9 tens. Write down the 9 and carry the 10 to the next column. The sum of the digits in this column plus the 10 carried from the second column is 38. The entire sum is 3,899. 32. Rule. I. Begin at the right, add each column separately, and write the sum, if it be only one figure, under the column added. II. If the sum of any column consists of two or more fig- ures, put the right-hand figure of the sum under that column and add the remaining figure or figures to the next column. 33. Proof. To prove addition, add each column from top to bottom. If you obtain the same result as by adding from bottom to top, the work is probably correct. EXAMPLES FOR PRACTICE. 34. Find the sum of: (a) 104 + 203 + 613 + 214. () 1,875 + 3,143 + 5,826 + 10,832. (. o SOLUTION. Sxl^^'' = : f = t- Ans. o Or, 2 X f = I 9 = f. Ans. O -T- 6 32 ARITHMETIC. 1 41. EXAMPLE. What is the product of T \ and | ? 4- v 7 SOLUTION.- A X f =^ * g- = is = A- Ans - 4v7 7 Or, by cancelation, #T^~8 = ?^8 = &' AnS< 4 42. EXAMPLE. What is | of | of if ? SOLUTION.- 8 X ^ X ^ = 8^2 = *' Ans ' 2 43. EXAMPLE. What is the product of 9f and 5f ? SOLUTION. 9| = \ 9 - ; 5| = - 4 /-. X = ^*f- = 4f* = 54B. Ans. 44. EXAMPLE. Multiply 15 J by 3. SOLUTION. 15| 15| 3 or 3 475 45 + _2i_ _ 45 + 2| = 47|. Ans. 45. Rule. I. Divide the product of the numerators by the product of tJie denominators. All factors common to tJie numerators and denominators should first be cast out by can- celation. II. To multiply one mixed number by another, reduce them both to improper fractions. III. To multiply a mixed number by a ^cvhole number, first multiply the fractional part by the multiplier, and if the product is an improper fraction, reduce it to a mixed num- ber and add the whole number part to the product of the mul- tiplier and the whole number. EXAMPLES FOR PRACTICE. 46. Find the product of: () 7 X T V f(a) 1 T V (*> . 14 X A. (6) 4|. (0 11 X A- (O ** (*) tf X 4. (d) 2tf. W H X 7. Ans - \ (e} 7A- (/) l?if X 7. I ( /) 125. () m X 32. ( g ) 15. (*) tf X 14. (A) 7i. 1 ARITHMETIC. 33 1. A single belt can transmit 107f horsepower, but as it is desired to use more power, a double belt of the same width is substituted for it. Supposing the double belt to be capable of transmitting ^P- as much power as the single belt, how many horsepower can be used after the change ? Ans. 153f H. P. 2. What is the weight of 2f miles of copper wire weighing 5| pounds per 100 feet ? There are 5,280 feet in a mile. Ans. 796 JJ Ib. 3. The grate of a steam boiler contains 20 square feet. If the boiler burns 8 fa pounds of coal an hour per square foot of grate area and can evaporate 7-J pounds of water an hour per pound of coal burned, how many pounds of water are evaporated by the boiler in 1 hour ? Ans. 1,276^ Ib. DIVISION OF FRACTIONS. 47. In division of fractions it is not necessary to reduce the fractions to fractions having a common denominator. 48. Dividing the numerator or multiplying the denomi- nator divides the fraction. EXAMPLE 1. Divide f by 3. SOLUTION. When dividing the numerator, we have |-3 = |^ 3 = f = i. Ans. When multiplying the denominator, we have |H-3 = ^ x3 = ^ r = i. Ans. EXAMPLE 2. Divide & by 2. O SOLUTION. ^^ 2 = iQ X 2 = A< AnS * EXAMPLE 3. Divide -J-f by 7. 14 7 SOLUTION. $| -*- 7 = ^ ' =^ = ^. Ans. 49. To invert a fraction is to turn it upside down; that is, make the numerator and denominator change places. Invert f and it becomes -f . 50. EXAMPLE. Divide T 9 ^ by T \. SOLUTION. 1. The fraction T S 6 is contained in T 9 ff , 3 times, for the denominators are the same, and one numerator is contained in the other 3 times. 2. If we now invert the divisor -fa and mul- tiply, the solution is This brings the same quotient as in the first case. 34 ARITHMETIC. 1 51. EXAMPLE. Divide f by J. SOLUTION. We cannot divide $ by , as in the first case above, for the denominators are not the same ; therefore, we must solve as in the second case. 2 52. EXAMPLE 1. Divide 5 by |. SOLUTION. j inverted becomes ^f. EXAMPLE 2. How many times is 3f contained in 7^ ? SOLUTION. 3f = Y ; V 4 53. Rule. Invert the divisor and proceed as in multipli- cation. 54. We have learned that a line placed between two numbers indicates that the number above the line is to be divided by the number below it. Thus, - 1 -/- shows that 18 is to be divided by 3. This is also true if a fraction or a frac- tional expression be placed above or below a line. 9 3 y 7 - means that 9 is to be divided by f ; means that 16 3 X 7 is to be divided by the value of . . | is the same as -H f . It will be noticed that there is a heavy line between the 9 and the f . This is necessary, since otherwise there would be nothing to show as to whether 9 was to be divided by f or f was to be divided by 8. Whenever a heavy line is used, as shown here, it indicates that all above the line is to be divided by all below it. 1 ARITHMETIC. 35 55. Whenever an expression like one of the three fol- lowing ones is obtained, it may always be simplified by transposing the denominator from above to below the line, or from beloiv to above, as the case may be, taking care, how- ever, to indicate that the denominator when so transferred is a multiplier. so 1- | = g-^-j = A = TIT J for , regarding the fraction above the heavy line as the numerator of a fraction whose denominator is 9, ^ = -, as before. y x 4 9x4 * = 12. The proof is the same as in the first = . ; for, regarding f as the numerator 5 X 9 5 of a fraction whose denominator is , |- = ; and if X 9 o X 9 5X45X4 4 = ^TTTT = f 4> as above. X9 X4 4 This principle may be used to great advantage in cases like i X 310 X 14 X 72 _ L - 7? . ^ . Reducing the mixed numbers to frac- 4:1) X 4^- X o-g- . , i X 310 X H X 72 __ tions, the expression becomes = -=-j^ . Now trans- ferring the denominators of the fractions and canceling, J0 3 3 1 x 310 x 27 X 72 x 2 x 6 _ 1 X m X ?T X n X % X 40 X 9 X 31 X 4 X 12 ^0 x ^ X n X X ft I 2 .= = 13*. ? O Greater exactness in results can usually be obtained by using this principle than can be obtained by reducing the fractions to decimals. The principle, however, should not be employed if a sign of addition or subtraction occurs either above or below the dividing line. ARITHMETIC. EXAMPLES FOB PRACTICE. 56. Divide: (a) 15by6f (a) () SObyf (r) 172 by f W) ttbyl* ( DUt is expressed by placing a period (.), which is called a decimal point, to the left of the figures of the numerator, to indicate that the figures on the right form the numerator of a fraction whose denominator is ten, one Jnmdred, one thousand, etc. 3. The reading of a decimal number depends upon the number of decimal places in it, i. e.,the number of figures to the right of the decimal point. One decimal place expresses tenths. Two decimal places express hundredths. Three decimal places express thousandths. Four decimal places express ten-thousandths. Five decimal places express hundred-thousandths. Six decimal places express millionths. 1 For notice of copyright, see page immediately following the title page. 38 ARITHMETIC. 1 Thus: .3 = ^ =3 tenths. .03 = yfo =3 hundredths. .003 = T^O- = 3 thousandths. .0003 = T olnro 3 ten-thousandths. .00003 = ToifVoir 3 hundred-thousandths. .000003 = nnrioTro = 3 millionths. We see in the above that the number of decimal places in a decimal equals the number of ciphers to the right of the figure 1 in the denominator of its equivalent fraction. This fact kept in mind will be of much assistance in reading and writing decimals. Whatever may be written to the left of a decimal point is a whole number. The decimal point affects only the figures to its right. When a whole number and decimal are written together, the expression is a mixed number. Thus, 8.12 and 17.25 are mixed numbers. The relation of decimals and whole numbers to each other is clearly shown by the following table : a o 45 BJ | I # *j rt 4-5 ifi m CO o millions. hundreds of O os 1 thousands. hundreds. to R units. c *o OH E 'o 'O 09 V hundredths. thousandths. ten-thousand hundred-thoi (A c .2 ' 1 I hundred-mill 7 6 5 4 3 2 1 a 3 4 5 6 f 8 9 The figures to the /*// of the decimal point represent whole numbers ; those to the right are decimals. In both decimals and whole numbers, the units place is made the starting point of notation and numeration. The decimals decrease on the scale of ten to the right, and the whole numbers increase on the scale of ten to the left. The first figure to the left of units is tens, and the _/&*/ figure to the right of units is tenths. The jrawdT figure tQ the / or T 9 * f a foot. This reduced to a decimal by the above rule shows what decimal part of a foot 9 inches is. 1 2) 9.00 (.7 5 of afoot. Ans. 84 ~60 60 35. Rule. I. To reduce inches to a decimal part of a foot, divide the number of inches by 12. 50 ARITHMETIC. 1 II. Should the resulting decimal be an unending one, and it is desired to terminate the division at some point, say tJic fourth decimal place, carry the division one place farther, and if the fifth figure is 5 or greater, increase the fourtJi figure by 1, omitting the signs -\- and . EXAMPLES FOR PRACTICE. 36. Reduce to the decimal part of a foot: (a) 3 in. C (a) .25 ft. (d) 44 in. () .375 ft. (c) 5 in. Ans. \ (c) .4167 ft. (d) 6| in. (d) .5521 ft. (e) 11 in. [(*) .9167ft. 1. The lengths of belting required to connect three countershafts with the main line shaft were found with the tape measure to be 27 feet 4 inches, 23 feet 8 inches, and 38 feet 6 inches. How many feet of belting were necessary ? Ans. 89.5ft. 2. The stroke of an engine is 14 inches. What is the length of the crank in feet measured from center of shaft to center of crankpin, knowing that length of crank is one-half the stroke ? Ans. .5833+ ft. 3. A steam pipe fitted with an expansion joint was found to expand 1.668 inches when steam was admitted into it. How much was its expansion in decimal parts of a foot ? Ans. .139 ft. TO REDUCE A DECIMAL TO A FRACTION. 37. EXAMPLE 1. Reduce .125 to a fraction. SOLUTION. .125 = ^flfr = ^ = $. Ans. EXAMPLE 2. Reduce .875 to a fraction. SOLUTION. .875 = T 8 ^ ff = _ |. Ans. 38. Rule. Under the figures of the decimal, place 1 with as many ciphers at its right as there are decimal places in the decimal, and reduce the resulting fraction to its lowest terms by dividing both numerator and denominator by the same number. 1 ARITHMETIC. 51 EXAMPLES FOR PRACTICE. 30. Reduce the following to common fractions: (a) .375. (6) .625. (c) .3125. (d) .04. (e) .06. &) .875. Ans. () f- (6) f. (0 A- (<0 A- W A- (/) t- 40. To express a decimal approximately as a frac- tion having a given denominator : EXAMPLE 1. Express .5827 in 64ths. 37 2928 SOLUTION. .5827 X fj = ^ , say |f. Hence, .5827 = f J, nearly. Ans. EXAMPLE 2. Express .3917 in 12ths. SOLUTION.- .3917 x H = ^^- Sa 7 A- Hence, .3917 = &, nearly. Ans. 41. Rule. Reduce 1 to a fraction having the given denominator. Multiply the given decimal by the fraction so obtained, and the result will be the fraction required. EXAMPLES FOR PRACTICE. 4:2. Express: () .625 in 8ths. (a) f. (*) .3125 in 16ths. (b) ^g. w .15625 in 32ds. Ans (?) A- (d} .77 in 64ths. ( -oir> and T 4 A it is usual to express them as .06, .25, and .43. 6. The following table will show how any per cent, can be expressed either as a decimal or as a fraction : 1$ .01 tfcr w .0025 ior^ 2$ .02 dhr r A w .005 | l 5$ .05 dhr r A iw .015 ^ r * 10$ .10 TTTO or A 6W .06i ^or^ 25$ .25 TO or i 8^ .08^ ^ r A 50$ .50 T S A or i 12** .125 i^or^ 75$ .75 yVo or f 18** 16f lS r - 100$ 1.00 IHorl 33 W .33| li r * 125$ 1.25 W or li 37i!< .37i lS^ 150$ 1.50 m or H 62^ .625 Ib1 0r * 500$ 5.00 |^ or 5 ^ .875 lS r ^ 2 ARITHMETIC. 3 7. The names of the different terms used in percentage are : the base, the rate or rate per cent. , the percentage, the amount, and the difference. 8. The base is the number or quantity which is supposed to be divided into 100 equal parts. 9. The rate per cent, is that number of the 100 equal parts into which the base is supposed to be divided which is taken or considered. The rate is the number of hun- dredths of the base that is taken or considered. The distinction between the rate per cent, and the rate is this: The rate per cent, is always 100 times the rate. Thus, 7$ of 125 and .07 of 125 amount in the end to the same thing; the former, 7, is the rate per cent. the number of hundredths of 125 intended; the latter, .07, is the rate, the part of 125 that is to be found; 7$ is used in speech, .07 is the form used in computation. So, also, 12$ = . 125, \% .005, lf$ = .0175. In the table just given, the num- bers in the first column are rates per cent. ; those in the second column are rates. 10. The percentage is the result obtained by multiply- ing the base by the rate. Thus, 7$ of 125 125 X .07 = 8.75, the percentage. 11. The amount is the sum of the base and the per- centage. 12. The difference is the remainder obtained when the percentage is subtracted from the base. 13. The terms amount and difference are ordinarily used when there is an increase or a decrease in the base. For example, suppose the population of a village is 1,500 and it increases 25 per cent. This means that for every 100 of the original 1,500 there is an increase of 25, or a total increase of 15 X 25 = 375. This increase added to the original population gives the amount, or the population after the increase. If the population had decreased 375, 4 ARITHMETIC. 2 the final population would have been 1,500 375= 1,125, and this would be the difference. The original popula- tion, 1,500, is the base on which the percentage is com- puted; the 25 is the rate per cent., and the increase or decrease, 375, is the percentage. If the base increases, the final value is the amount ; and if it decreases, its final value is the difference. CALCULATIONS INVOLVING PERCENTAGE. 14. From the foregoing it is evident that to find the percentage, the base must be multiplied by the rate. Hence, the following Rule. To find the percentage, multiply the base by the rate. EXAMPLE. Out of a lot of 300 boiler tubes 76$ was used in a boiler. How many tubes were used ? SOLUTION. The rate is .76; the base is 300; hence, the number of tubes used, or the percentage, is by the above rule 300 x- 76 = 228 tubes. Ans. Expressing the rule as a Formula, percentage = base X rate. 15. When the percentage and rate are given, the base may be found by dividing the percentage by the rate. For, suppose that 12 is 6$, or yf^, of some number; then 1$, or -j-J-g-, of the number, is 12 -r- 6, or 2. Consequently, if 2 = 1$, or T fg-, 100$, or {$$ = 2 X 100 = 200. But as the same result may be arrived at by dividing 12 by .06, since 12 -H .06 = 200, it follows that: Rule. When the percentage and rate are given, to find the base, divide the percentage by the rate. Formula, base = percentage -=- rate. 2 ARITHMETIC. 5 EXAMPLE. 76$ of a lot of boiler tubes was used in the construction of a boiler. If the number of tubes used was 228, how many tubes were in the lot ? SOLUTION. Here 228 is the percentage and .76 is the rate; hence, applying the rule, 228 -j- . 76 = 300 tubes. Ans. 16. When the base and percentage are given, to find the rate, the rate may be found by dividing the percentage by the base. For, suppose that it is desired to find what per cent. 12 is of 200, \% of 200 is 200 X .01 = 2. Now, if 1# is 2, 12 is evidently as many per cent, as 2 is contained times in 12, or 12 -4- 2 = 6#. But the same result may be obtained by dividing 12, the percentage, by 200, the base, since 12 -j- 200 = .06 = 6#. Hence, Rule. When the percentage and base are given, to find the rate, divide the percentage by the base, and the result will be the rate. Formula, rate = percentage -r- base. EXAMPLE 1. Out of a lot of 300 boiler tubes 228 were used. What per cent, of the total number was used ? SOLUTION. Here 300 is the base and 228 is the percentage ; hence, applying rule, Rate = 228 -f- 300 = . 76 = 76*. Ans. EXAMPLE 2. What per cent, of 875 is 25 ? SOLUTION. Here 875 is the base and 25 is the percentage; hence, applying rule, Rate = 25 H- 875 = .02$ = 2f*. Ans. PROOF. 875 X .024 = 25. EXAMPLES FOB PRACTICE. 17. What per cent of: (a) 360 is 90 ? \b) 900 is 360 ? (c) 125 is 25 ? (OP) -y 1 O that 12 earn = , or $75, since either extreme equals the product of the means divided by the other extreme. (c) 12 men : 4 men = the amount 12 men earn : 25 ; or the amount that 12 men earn = ' * , or $75, since either mean equals the product of the extremes divided by the other mean. 61. If the proportion is an inverse one, first form it as though it were a direct proportion and then invert one of the couplets. 62. EXAMPLES FOR PRACTICE. Find the value of x in each of the following: (a) $16 : $64 = x : $4. f (a) = $1. () x : 85 = 10 : 17. (b) - 50. (c) 24 : x = 15 : 40. (c} = 64. (d) 18 : 94 = 2 : x. Ans. J (d) = lOf. (?) $75 : $100 = x : 100. (e) 75. (/) 15 pwt. : x = 21 : 10. (/) = 7| pwt. (g~) x : 75 yd. = 15 : $5. [ (g) = 225 yd. 1. If 75 pounds of lead cost $2.10, what would 125 pounds cost at the same rate? Ans. $3.50. 2. If A does a piece of work in 4 days and B does it in 7 days, how long will it take A to do what B does in 63 days 1 Ans. 36 days. 3. The circumferences of any two circles are to each other as their diameters. If the circumference of a circle 7 inches in diameter is 22 inches, what will be the circumference of a circle 31 inches in diameter? Ans. 97f in. INVERSE PROPORTION. 63. In Art. 53, an inverse proportion was defined as one which required one of the couplets to be expressed as an inverse ratio. Sometimes the word inverse occurs in the statement of the example; in such cases the proportion can be written directly, merely inverting one of the couplets. 2 ARITHMETIC. 47 But it frequently happens that only by carefully studying the conditions of the example can it be ascertained whether the proportion is direct or inverse. When in doubt, the student can always satisfy himself as to whether the propor- tion is direct or inverse by first ascertaining what is required, and stating the proportion as a direct proportion. Then, in order that the proportion may be true, if the first term is smaller than the second term, the third term must be smaller than the fourth ; or if the first term is larger than the second term, the tJiird term must be larger than the fourth term. Keeping this in mind, the student can always tell whether the required term will be larger or smaller than the other term of the couplet to which the required term belongs. Having determined this, the student then refers to the example and ascertains from its conditions whether the required term is to be larger or smaller than the other term of the same kind. If the two determinations agree, the proportion is direct; otherwise, it is inverse, and one of the couplets must be inverted. 64. EXAMPLE. A's rate of doing work is to B's as 5:7; if A does a piece of work in 42 days, in what time will B do it ? SOLUTION. The required term' is the number of days it will take B to do the work. Hence, stating as a direct proportion, 5 : 7 = 42 : x. Now, since 7 is greater than 5, x will be greater than 42. But, refer- ring to the statement of the example, it is easy to see that B works faster than A ; hence it will take B a less number of days to do the work than A. Therefore, the proportion is an inverse one, and should be stated 5 : 7 = x : 42, from which x = 5 * 42 = 30 days. Ans. Had the example been stated thus: The time that A requires to do a piece of work is to the time that B requires, as 5 : 7 ; A can do it in 42 days, in what time can B do it ? it is evident that it would take B a longer time to do the work than it would A; hence, x would be greater than 42, and the proportion would be direct, the value of x being *4*? = 68.8 days. 48 ARITHMETIC. 2 EXAMPLES FOR PRACTICE. 65. Solve the following: 1. If a pump which discharges 4 gal. of water per min. can fill a tank in 20 hr., how long will it take a pump discharging 12 gal. per min. to fill it ? Ans. 6 hr. 2. The circular seam of a boiler requires 50 rivets when the pitch is 2-J- in. ; how many would be required if the pitch were 3 in. ? Ans. 40. 3. The spring hangers on a certain locomotive are 2-J in. wide and f in. thick ; those on another engine are of same sectional area, but are 3 in. wide; how thick are they ? Ans. f in. 4. A locomotive with driving wheels 16 ft. in circumference runs a certain distance in 5,000 revolutions; how many revolutions would it make in going the same distance if the wheels were 22 ft. in circum- ference (no allowance for slip being made in either case) ? Ans. 3,636^- rev. UNIT METHOD. 66. In the older books on arithmetic, a large number of problems were solved by proportion ; but these problems can be solved much more easily by the unit method, which we now proceed to explain by means of examples. EXAMPLE 1. If a pump discharging 4 gallons of water per minute can fill a tank in 20 hours, how long will it take a pump discharging 12 gallons per minute to fill the tank ? SOLUTION. A pump discharging 4 gallons per minute fills the tank in 20 hours. Therefore, a pump discharging 1 gallon per minute fills it in 4 X 20 hours. Hence, a pump discharging 12 gallons per minute fills it in 4x2 ? h urs = 20h UrS = 6| hr. Ans. 1* o EXAMPLE 2. If 4 men earn $65.80 in 7 days, how much can 14 men, paid at the same rate, earn in 12 days ? SOLUTION. 4 men in 7 days earn $65.80. Therefore, 1 man in 7 days earns ^ . Therefore, 1 man in 1 day earns ' . 4 X Therefore, 1 man in 12 days earns $65 . 8 * 12 . 4 X ' Therefore, 14 men in 12 days earn S 65 - 80 ^ *2 X 14 Canceling, 14 men in 12 days earn $65.80 X 3 X 2 = $65.80 X 6 - $.394.80. Ans, 2 ARITHMETIC. 49 67. The student will notice that in the solution of these examples, in the successive steps, the operations of mul- tiplication and division were merely indicated, and no multiplication or division was performed until the very last, and then the answer was 'obtained easily by cancelation. In arithmetical calculations, the student should make it an invariable habit to indicate the multiplications and divisions that occur in the successive steps of a solution, and not to perform these operations until the very last. Then, he will probably be able to use the principle of cancelation. EXAMPLE. If a block of granite 8 feet long, 5 feet wide, and 3 feet thick weighs 7,200 pounds, what is the weight of a block of granite 12 feet long, 8 feet wide, and 5 feet thick ? SOLUTION. If a block 8 feet long, 5 feet wide, and 3 feet thick weighs 7,200 pounds, a block 1 foot long, 5 feet wide, and 3 feet thick weighs -^ pounds ; a block 1 foot long, 1 foot wide, and 3 feet o 7 900 thick weighs 0^5; and a block 1 foot long, 1 foot wide, and 1 foot 7 200 thick weighs 5 5 pounds. Therefore, by the same reasoning, a " X 5 X o block 12 feet long, 8 feet wide, and 5 feet thick weighs 7.0 g x 12 X 8 X ^ = T.0 X ff X > X , = ^ , b An , EXAMPLES FOB PRACTICE. 1. If a pump discharges 90,000 gallons of water in 20 hours, in what time will it discharge 144,000 gallons ? Ans. 32 hr. 2. When the barometer stands at 30 inches, the pressure of the atmosphere is 14.7 pounds per square inch. What is the atmospheric pressure per square inch when the barometer stands at 29.5 inches? Give answer correct to three figures. Ans. 14.5. //. -Sf. /. 8 MENSURATION AND USE OF LETTERS IN FORMULAS. FORMULAS. 1. The term formula, as used in mathematics and in technical books, may be defined as a rule in which symbols are used instead of words ; in fact, a formula may be regarded as a shorthand method of expressing a rule. Any formula can be expressed in words, and when so expressed it becomes a rule. Formulas are much more convenient than rules; they show at a glance all the operations that are to be performed ; they do not require to be read three or four times, as is the case with most rules, to enable one to understand their meaning; they take up much less space, both in the printed book and in one's note book, than rules; in short, whenever a rule can be expressed as a formula, the formula is to be preferred. As the term " quantity " is a very convenient one to use, we will define it. In mathematics, the word quantity is applied to anything that it is desired to subject to the ordi- nary operations of addition, subtraction, multiplication, etc., when we do not wish to be more specific and state exactly what the thing rs. Thus, we can say "two or more num- bers," or "two or more quantities"; the word quantity is more general in its meaning than the word number. 2. The signs used in formulas are the ordinary signs indicative of operations and the signs of aggregation. All 3 For notice of copyright, see page immediately following the title page. 2 MENSURATION AND 3 these signs are explained in arithmetic, but some of them will here be explained in order to refresh the student's memory. 3. The signs indicative of operations are six in number, viz.: +, -, X, -*-, | , V- Division is indicated by the sign -i- , or by placing a straight line between the two quantities. Thus, 25 | 17, 25 / 17, and f4 all indicate that 25 is to be divided by 17. When both quantities are placed on the same horizontal line, the straight line indicates that the quantity on the left is to be divided by that on the right. When one quantity is below the other, the straight line between indicates that the quan- tity above the line is to be divided by the one below it. The sign ( \/} indicates that some root of the quantity on the right is to be taken ; it is called the radical sign. To indicate what root is to be taken, a small figure, called the index, is placed within the sign, this being always omitted when the square root is to be indicated. Thus, |/25 indi- cates that the square root of 25 is to be taken; |/25 indicates that the cube root of 25 is to be taken ; etc. 4. The signs of aggregation are four in number; viz., 1 ( )> [ ]> and I } respectively called the vinculum, the parenthesis, the brackets, and the brace ; they are used when it is desired to indicate that all the quantities included by them are to be subjected to the same operation. Thus, if we desire to indicate that the sum of 5 and 8 is to be mul- tiplied by 7, and we do not wish to actually add 5 and 8 before indicating the multiplication, we may employ any one of the four signs of aggregation as here shown : 5 + 8 X 7, (5 + 8) X 7, [5 + 8] X 7, {5 + 8} X 7. The vinculum is placed above those quantities which are to be treated as one quantity and subjected to the same operations. 5. While any one of the four signs may be used as shown above, custom has restricted their use somewhat. The vin- culum is rarely used except in connection with the radical sign. Thus, instead of writing y (5 + 8), ^ [5 + 8], or y 1 5 -j- 8 1 for the cube root of 5 plus 8, all of which would be correct, the vinculum is nearly always used, 4/5 + 8. 3 USE OF LETTERS IN FORMULAS. 3 In cases where but one sign of aggregation is needed (ex- cept, of course, when a root is to be indicated), the paren- thesis is always used. Hence, (5 -f- 8) X 7 would be the usual way of expressing the product of 5 plus 8, and 7. If two signs of aggregation are needed, the brackets and parenthesis are used, so as to avoid having a parenthesis within a parenthesis, the brackets being placed outside. For example, [(20 5) -r- 3] X 9 means that the difference between 20 and 5 is to be divided by 3, and this result mul- tiplied by 9. If three signs of aggregation are required, the brace, brackets, and parenthesis are used, the brace being placed outside, the brackets next, and the parenthesis inside. For example, { [(20 5) -^ 3] X 9 21} -h 8 means that the quotient obtained by dividing the difference between 20 and 5 by 3 is to be multiplied by 9, and that after 21 has been subtracted from the product thus obtained, the result is to be divided by 8. Should it be necessary to use all four of the signs of aggre- gation, the brace would be put outside, the brackets next, the parenthesis next, and the vinculum inside. For example, {[(20-5 -T- 3) X 9 - 21] -^ 8} X 12. 6. As stated in Arithmetic, when several quantities are connected by the various signs indicating addition, subtrac- tion, multiplication, and division, the operation indicated by the sign of multiplication must always be performed first. Thus, 2 + 3x4 equals 14, 3 being multiplied by 4 before adding to 2. Similarly, 10 -4- 2 X 5 equals 1, since 2x5 equals 10, and 10 -f- 10 equals 1. Hence, in the above case, if the brace were omitted, the result would be , whereas, by inserting the brace, the result is 36. Following the sign of multiplication comes the sign^of division in order of importance. For example, 5 9 -f- 3 equals 2, 9 being divided by 3 before subtracting from 5. The signs of addition and subtraction are of equal value; that is, if several quantities are connected by plus and minus signs, the indicated operations may be performed in the order in which the quantities are placed. 4 MENSURATION AND 3 7. There is one other sign used, which is neither a sign of aggregation nor a sign indicative of an operation to be performed; it is (=), and is called the sign of equality; it means that all on one side of it is exactly equal to all on the other side. For example, 2 2, 5 3 = 2, 5 X (14 9) = 25. 8. Having called particular attention to certain signs used in formulas, the formulas themselves will now be explained. First, consider the well-known rule for finding the horsepower of a steam engine, which may be stated as follows : Divide the continued product of the mean effective pressure in pounds per square inch, the length of the stroke in feet, the area of tJie piston in square inches, and the number of strokes Per minute, by 33,000 ; the result will be the horsepower. This is a very simple rule, and very little, if anything, will be saved by expressing it as a formula, so far as clear- ness is concerned. The formula, however, will occupy a great deal less space, as we shall show. An examination of the rule will show that four quantities (viz., the mean effective pressure, the length of the stroke, the area of the piston, and the number of strokes) are mul- tiplied together, and the result is divided by 33, 000. Hence, the rule might be expressed as follows: mean effective pressure stroke Horsepower = . . X ,. , (in pounds per square inch) (in feet) area of piston number of strokes X (in square inches) X (per minute) This expression could be shortened by representing each quantity by a single letter; thus, representing horsepower by the letter "//," the mean effective pressure in pounds per square inch by "/*," the length of stroke in feet by " L" the area of the piston in square inches by "A," the number of strokes per minute -by "TV, "and substituting these letters for the quantities that they represent, the above expression would reduce to Px L X A X N H = 33,000 3 USE OF LETTERS IN FORMULAS. 5 a much simpler and shorter expression. This last expres- sion is called a formula. 9. The formula just given 'shows, as we stated in the beginning, that a formula is really a shorthand method of expressing a rule. It is customary, however, to omit the sign of multiplication between two or more quantities when they are to be multiplied together, or between a number and a letter representing a quantity, it being always understood that, when two letters are adjacent with no sign between them, the quantities represented by these letters are to be multiplied. Bearing this fact in mind, the formula just given can be further simplified to PLAN 33,000 ' 10. The sign of multiplication, evidently, cannot be omitted between two or more numbers, as it would then be impossible to distinguish the numbers. A near approach to this, however, may be attained by placing a dot between the numbers which are to be multiplied together, and this is frequently done in works on mathematics when it is desired to economize space. In such cases it is usual to put the dot higher than the position occupied by the decimal point. Thus, 2-3 means the same as 2 X 3; 542-749-1,006 indicates that the numbers 542, 749, and 1,006 are to be multiplied together. It is also customary to omit the sign of multiplication in expressions similar to the following: a X tfb-\-c, 3 X (b + <0, (b + c) X a, etc., writing them a \/b + c, 3 (b + c), (b + c) a, etc. The sign is not omitted when several quantities are included by a vinculum and it is desired to indicate that the quantities so included are to be multiplied by another quantity. For example, 3 X ^ + , b + c X a, \/b + c y. a, etc. are always written as here printed. 11. Before proceeding further, we will explain one other device that is used by formula makers and which is apt to puzzle one who encounters it for the first time it is the use of what mathematicians call primes and subs., and 6 MENSURATION AND 3 what printers call superior and inferior characters. As a rule, formula makers designate quantities by the initial letters of the names of the' quantities. For example, they represent volume by v, pressure by /, height by //, etc. This practice is to be commended, as the letter itself serves in many cases to identify the quantity which it represents. Some authors carry the practice a little further and repre- sent all quantities of the same nature by the same letter throughout the book, always having the same letter repre- sent the same thing. Now, this practice necessitates the use of the primes and subs, above mentioned when two quantities have the same name but represent different things. Thus, consider the word pressure as applied to steam at dif- ferent stages between the boiler and the condenser. First, there is absolute pressure, which is equal to the gauge pres- sure in pounds per square inch plus the pressure indicated by the barometer reading (usually assumed in practice to be 14.7 pounds per square inch, when a barometer is not at hand). If this be represented by/, how shall we represent the gauge pressure ? Since the absolute pressure is always greater than the gauge pressure, suppose we decide to repre- sent it by a capital letter and the gauge pressure by a small (lower-case) letter. Doing so, P represents absolute pres- sure and /, gauge pressure. Further, there is usually a "drop " in pressure between the boiler and the engine, so that the initial pressure, or pressure at the beginning of the stroke, is less than the pressure at the boiler. How shall we represent the initial pressure ? We may do this in one of three ways and still retain the letter /or Pto represent the word pressure: First, by the use of the prime mark; thus, /' or P' (read / prime and/ 3 major prime] may be con- sidered to represent the initial gauge pressure or the initial absolute pressure. Second, by the use of sub. figures ; thus, /, or P l (read/ sub. one and P major sub. one). Third, by the use of sub. letters ; thus, /,- or P t (read / sub. i and P major sub. i). In the same manner /" (read / second}, / 2 , or / r might be used to represent the gauge pressure at release, etc. The sub. letters have the advantage of still further identifying the quantity represented: in many instances, 3 USE OF LETTERS IN FORMULAS. 7 however, it is not convenient to use them, in which case primes and subs, are used instead. The prime notation may be continued as follows: /'", p iv , p v , etc.; it is inad- visable to use superior figures, for example, /', /", p*, /, etc., as they are liable to be mistaken for exponents. 12. The main thing to be remembered by the student is that when a formula is given in which the same letters occur several times, all like letters having the same primes or subs. represent the same quantities, while those which differ in any respect represent different quantities. Thus, in the formula . t _. w,, w v and w 3 represent the weights of three different bodies; s lt s y , and s s , their specific heats; and / 15 / 2 , and / s , their temperatures; while t represents the final temperature after the bodies have been mixed together. It should be noted that those letters having the same subs, refer to the same bodies. Thus, w,, s^, and t^ all refer to one of the three bodies; w 2 , s^ t^ to another body, etc. It is very easy to apply the above formula when the values of the quantities represented by the different letters are known. All that is required is to substitute the numer- ical values of the letters and then perform the indicated operations. Thus, suppose that the values of ; s lt and ^ are, respectively, 2 pounds, .0951, and 80; of ; s t , and t s , 7.8 pounds, 1, and 80; and of zu 3 , s 3 , and A,, 3 pounds, .1138, and 780; then, the final temperature t is, substi- tuting these values for their respective letters in -the formula, 2 X .0951 X 80 + 7.8 X 1 X 80 + 3j X .1138 X 780 2 X .0951 + 7.8 X 1 -f 3 X .1138 _ 15.216 + 624 + 288.483 _ 927. 699 . 1902 + 7. 8 + . 36985 ~ 8. 36005 In substituting the numerical values, the signs of multi- plication are, of course, written in their proper places; all the multiplications are performed before adding, according to the rule previously given. 8 MENSURATION AND 3 13. The student should now be able to apply any for- mula involving only algebraic expressions that he may meet with, and which does not require the use of logarithms for its solution. We will, however, call his attention to one or two other facts that he may have forgotten. Expressions similar to -7-7- sometimes occur, the heavy line 25 indicating that 160 is to be divided by the quotient obtained by dividing 660 by 25. If both lines were light, it would be impossible to tell whether 160 was to be divided by , or whether - was to be divided by 25. If this latter 160 result were desired, the expression would be written. In every case the heavy line indicates that all above it is to be divided by all below it. In an expression like the following, -, the heavy line is not necessary, since it is impossible to mistake the operation that is required to be performed. But, since , 660 175 + 600 ., 175 -f 660 , , 660 7 + -25-= -35 , ^ we substitute^ for 7 + , the heavy line becomes necessary in order to make the resulting expression clear. Thus, 160 160 _ 160 660 ~~ 175 + 660 835" + 25 25 25 14. Fractional exponents are sometimes used instead of the radical sign. That is, instead of indicating the square, cube, fourth root, etc. of some quantity, as 37, by 4/37, 1/37^4/37^ etc., these roots are indicated by 37', 37 , 37 1 , etc. Should the numerator of the fractional exponent be some quantity other than 1, this quantity, whatever it may be, indicates that the quantity affected by the exponent is to be raised to the power indicated by the numerator; the 3 USE OF LETTERS IN FORMULAS. 9 denominator is always the index of the root. Hence, instead of writing |/37" for the cube root of the square of 37, it may be written 37 ? , the denominator being the index of the root ; in other words, |/37 2 = 37 s . Likewise, |/(1 -(- a* b} % may also be written (1 -f- a' 1 /)*, a much simpler expression. 15. We will now give several examples showing how to apply some of the more difficult formulas that the student may encounter. 1. The area of any segment. of a circle that is less than (or equal to) a semicircle is expressed by the formula TT r 1 E c , , . ^ = "sir -s <'-*> in which A = area of segment; TT = 3.1416; r = radius; E = angle obtained by drawing lines from the cen- ter to the extremities of arc of segment ; c = chord of segment ; 1i = height of segment. EXAMPLE. What is the area of a segment whose chord is 10 inches long, angle subtended by chord is 83.46, radius is 7.5 inches, and height of segment is 1.91 inches? SOLUTION. Applying the formula just given, = 40.968 - 27.95 .- 13.018 sq. in., nearly. Ans. 2. The area of any triangle may be found by means of the following formula, in which A = the area, and a, b, and c represent the lengths of the sides: EXAMPLE. What is the area of a triangle whose sides are 21 feet, 46 feet, and 50 feet long ? SOLUTION. In order to apply the formula, suppose we let a repre- sent the side that is 21 feet long; b, the side that is 50 feet long; andr, the side that is 46 feet long. Then, substituting in the formula 10 MENSURATION AND = 25 |/441 - 8.25* = 25 4/441 - 68.0625 = 25 |/372.9375 = 25 X 19.312 = 482.8 sq. ft., nearly. Ans. The operations in the above examples have been extended much farther than was necessary ; it was done in order to show the student every step of the process. The last for- mula is perfectly general, and the same answer would have been obtained had the 50-foot side been represented by a, the 46-foot side by , and the 21-foot side by c. 3.. The Rankine-Gordon formula for determining the least load in pounds that will cause a long column to break is D _ SA r> in which P load (pressure) in pounds ; 5 ultimate strength (in pounds per square inch) of the material composing the column ; A = area of cross-section of column in square inches ; q = a factor (multiplier) whose value depends upon the shape of the ends of the column and on the material composing the column ; / = length of column in inches; G == least radius of gyration of cross-section of column. The values of 5, g, and G* are given in printed tables in books in which this formula occurs. EXAMPLE. What is the least load that will break a hollow steel column whose outside diameter is 14 inches; inside diameter, 11 inches; length, 20 feet, and whose ends are flat ? SOLUTION. For steel, 5 = 150,000, and q = --. for flat-ended steel columns; A, the area of the cross-section, = .7854 (K '( 2 + 25) T ~~25 * 25 Reducing the fraction to a decimal before extracting the cube root, E = ^6.0606 = 1.823. Ans. (&) Substituting, 10-1/4 16. In the preceding pages, the unknown quantity has always been represented by the single letter at the left of the sign of equality, while the letters at the right have represented known values from which the required values could be found. It is possible, however, to find the value of the quantity represented by any letter in a formula, if the values represented by all the others are known. For example, let it be required to find how many strokes per minute an 12 MENSURATION AND 3 engine having a piston area of 78.54 square inches must make in order to develop 60 horsepower, if the mean effective pressure is 40 pounds per square inch and the length of stroke is 1 ft. By substituting the given values in the for- . PLAN 40 X Ij-X 78.54 X N 33,000 in which N, the number of strokes, is to be found. But it is evident that the expression on the right of the sign of equality is equal to - ^ - ' X N, a fraction oo,000 whose numerator is composed of three factors. Reducing the numerator to a single number by performing the indi- cated multiplications, we obtain, after canceling, If 60 equals .119 N, then TV equals 60 divided by .119; hence, N = = 504.2 strokes per minute. . 119 The method of procedure is essentially the same when the unknown quantity occurs in the denominator of a formula. 111 7 2 Thus, in the formula/= - , suppose that/= 375, /;/ = 1.25, and v = 60. Then, substituting, 1.25 X 60' 4,500 o75 = - = - . r r But, if 375 equals 4,500 divided by r, then 375 X r = 4,500; hence, r must equal 4,500 divided by 375, or r = -^ = 12. o75 EXAMPLES FOR PRACTICE. Find the numerical values of x in the following formulas, when A 9, B = 8, d = 10, e = 3, and c = 2 : Ans. x = 1 J. 3 USE OF LETTERS IN FORMULAS. 13 Ans. x = 29. Ans.* = 8. Ans. jr=12&. Ans. JT= .396+. MENSURATION. 17. Mensuration treats of the measurement of lines, angles, surfaces, and solids. LINES AND ANGLES. 18. A straight line is one that does not change its direction throughout its whole length. To distinguish one straight line from another, two of its points are designated by letters. The j B line shown in Fig. 1 would be called the FIG. i. line A B. 19. A curved line changes its di- rection at every point. Curved lines are designated by three or more letters, as the curved line ABC, Fig. 2. 20. Parallel lines (Fig. 3) are those which are equally distant from each other at all points. 21. A line, is perpendicular to another (see Fig. 4) when it meets that line so as not to incline towards it on either side. 22. A vertical line is one that points towards the center of the earth, j and is also known as a plumb-line. B 23. A horizontal line (see Fig. 5) is one that makes a right angle with I - .. Horizontal. any vertical line. FIG. 5. FIG. 3. FIG. 4. 14 MENSURATION AND FIG. 6. 24. An angle is the opening between two lines which intersect or meet ; the point of meeting is called the vertex of the angle. Angles are distinguished by naming the vertex and a point on each line. Thus, in Fig. 6, the angle formed by the lines A B and C B is called the angle ABC, or the angle C B A ; the letter at the vertex is always placed in the middle. When an angle stands alone so that it cannot be mistaken for any other angle, only the vertex letter need be used. Thus, the angle referred to might be designated simply as the angle B. 25. If one straight line meets an- J other straight line at a point between its ends, as in Fig. 7, two angles, ABC and A B D, are formed, which are c called adjacent angles. FIG. 26. When these adjacent angles, ABC and A B D, are equal, as in Fig. 8, they are called right angles. B FIG. 8. 27. An acute angle is less than a right angle. A C, Fig. 9, is an acute angle. FIG. 9. 28. An obtuse angle is greater than a right angle. A B D (Fig. 10) is an obtuse angle. 29. A circle (see Fig. 11) is a figure bounded by a curved line, called the circum- ference, every point of which is equally dis- tant from a point within, called the center. 3 USE OF LETTERS IN FORMULAS. 15 FIG. 12. 30. An arc of a circle is any part of its circumference ; thus a e b, Fig. 12, is an arc of the circle. 31. The circumference of every circle is considered to be divided into 360 equal parts, or arcs, called degrees; every degree is subdivided into GO equal parts, called min- utes, and every minute is again divided into 60 equal parts, called seconds. Since 1 degree is 7 ^ of any circumference, it follows that the length of a degree will be different in circles of different sizes, but the proportion of the length of an arc of 1 degree to the whole circumference will always be the same, viz., ^J-g- of the circumference. Degrees, minutes, and seconds are denoted by the symbols , ', '. Thus, the expression 37 14' 44* is read 37 degrees 14 minutes 44 seconds. 32. The arcs of circles are used to measure angles. An angle having its vertex at the center of a circle is measured by the arc in- cluded between its sides ; thus, in Fig. 13, the arc F B measures the angle FOB. If the arc FB contains 20, or ^/ v of the circumference, the angle FOB would be an angle of 20; if it contained 20 14' 18", it would be an FIG. 13. angle of 20 14' 18", etc. In the figure, if the line C D be drawn perpendicular to A B, the adjacent angles will be equal, and the circle will be divided into four equal angles, each of which will be a right angle. A right angle, therefore, is an angle of , or 90 ; two right angles are measured by 180, or half the circumference, and four right angles by the whole circum- ference, or 360. One-half of a right angle, as E O B, is an angle of 45. An acute angle may now be defined as an angle of less than 90, and an obtuse angle as one of more 16 MENSURATION AND than 90. These values are important and should be remembered. 33. From the foregoing it will be evident that if a number of straight lines on the same side of a given straight line meet at the same point, the sum of all the angles formed is equal to two right angles, or 180. Thus, in Fig. 14, angles COB' +DOC+EOD+FOE+AOF o FIG. 14. = 2 right angles, or 180. 34. Also, if through a given point any number of straight lines be drawn, the sum of all the angles formed about the points of intersection equals four right angles, or 360. Thus, in Fig. 15, angles/fOF+FOC+COA + AOG +GOE+EOD+DOB+BOH = 4 right angles, or 360. EXAMPLE. In a flywheel with 12 arms, how many degrees are there in the angle included between the center lines of any two arms, the arms being spaced equally ? SOLUTION. Since there are 12 arms, there are 12 angles, which together equal 360. Hence, one angle equals ^ of 360, or ^-r- = 30. Ans. EXAMPLES FOR PRACTICE. 1. How many seconds are in 32 14' 6" ? Ans. 116,046 sec. 2. How many degrees, minutes, and seconds do 38,582 seconds amount to? Ans. 10 43' 2". 3. How many right angles are there in an angle of 170 ? Ans. If right angles. 4. In a pulley with five arms, what part of a right angle is included between the center lines of any two arms ? Ans. | of a right angle. 5. If one straight line meets another so as to form an angle of 20 r ' 10', what does its adjacent angle equal ? Ans. 159 50'. 6. If a number of straight lines meet a given straight line at a given point, all being on the same side of the given line, so as to form six equal angles, how many degrees are there in each angle ? Ans. 30. 3 USE OF LETTERS IN FORMULAS. 1? QUADRILATERALS. 35. A plane figure is any part of a plane or flat sur- face bounded by straight or curved lines. 36. A quadrilateral is a plane figure bounded by four straight lines. 37. A parallelogram is a quadrilateral whose opposite sides are parallel. There are four kinds of parallelograms : the square, the rectangle, the rhombus, and the rhomboid. 38. A rectangle (Fig. 16) is a parallelo- gram whose angles are all right angles. FIG. 16. 39. A square (Fig. 17) is a rectangle whose sides are all of the same length. FIG. 17. 4O. A rhomboid (Fig. 18) is a parallelogram whose opposite sides are equal and parallel, and whose angles are not right angles. 41. A rhombus (Fig. 19) is a parallelogram having equal sides, and whose angles are not right angles. FIG. 19. 42. A trapezoid (Fig. 20) is a quadrilateral which has only two of its sides parallel. FIG. 20. 43. The altitude of a parallelogram or a trapezoid is the perpendicular distance between the parallel lines, as shown by the vertical lines in Figs. 18, 19, and 20. 44. The base of any plane figure is the side on which it is supposed to stand. 18 MENSURATION AND 3 45. The area of a surface is expressed by the number of unit squares it will contain. 46. A unit square is the square having a unit for its side. For example, if the unit is 1 inch, the unit square is the square each of whose sides measures 1 inch in length, and the area of a surface would be expressed by the number of square inches it would contain. If the unit were 1 foot, the unit square would measure 1 foot on each side, and the area of the given surface would be the number of square feet it would contain, etc. The square that measures 1 inch on a side is called a square inch, and the one that measures 1 foot on a side is called a square foot. Square inch and square foot are abbreviated to sq. in. and sq. ft. 47. To find the area of any parallelogram: Rule 1. Multiply the base by the altitude. NOTE. Before multiplying, the base and altitude must be reduced to the same kind of units ; that is, if the base should be given in feet and the altitude in inches, they could not be multiplied together until either the altitude had been reduced to feet or the base to inches. This principle holds throughout the subject of mensuration. EXAMPLE 1. The sides of a square piece of sheet iron are each 10^ inches long. How many square inches does it contain ? SOLUTION. 10 inches 10.25 inches when reduced to a decimal. The base and altitude are each 10.25 inches. Multiplying them together, 10.25 X 10.25 = 105.06+ sq. in. Ans. EXAMPLE 2. What is the area in square rods of a piece of land in the shape of a rhomboid, one side of which is 8 rods long and whose length, measured on a line perpendicular to this side, is 200 feet ? SOLUTION. The base is 8 rods and the altitude 200 feet. As the answer is to be in rods, the 200 feet should be reduced to rods. Reducing 200 H- 16$ = 200 H- y = 12.12 rods. Hence, area = 8 X 12.12 = 96.96sq. rd. Ans. 48. To find the area of a trapezoid: Rule 2. Multiply one-Jialf the sum of the parallel sides by the altitude. EXAMPLE. A board 14 feet long is 20 inches wide at one end and 16 inches wide at the other. If the ends are parallel, how many square feet does the board contain ? 3 USE OF LETTERS IN FORMULAS. 19 SOLUTION. One-half the sum of the parallel sides = ^ = 18 inches = l feet. The length of the board corresponds to the altitude of a trapezoid. Hence, 14 X H = 21 sq. ft. Ans. 49. Having given the area of a parallelogram and one dimension, to find the other dimension: Rule 3. Divide the area by the given dimension. EXAMPLE. What is the width of a parallelogram whose area is 212 square feet and whose length is 26^ feet ? SOLUTION. 212 -=- 26^ = 212 -s- ^- = 8 ft. Ans. The following examples illustrate a few special cases: EXAMPLE 1. An engine room is 22 feet by 32 feet. The engine bed occupies a space of 3 feet by 12 feet; the flywheel pit, a space of 2 feet by 6 feet, and the outer bearing a space of 2 feet by 4 feet. How many square feet of flooring will be required for the room ? SOLUTION. Area of engine bed = 3 X 12 = 36 sq. ft. Area of flywheel pit = 2 X 6 = 12 sq. ft. Area of outer bearing = 2 X 4 = 8 sq. ft. Total, 56 sq. ft. Area of engine room = 22 X 32 = 704 sq. ft. 704 56 = 648 sq. ft. of flooring required. Ans. EXAMPLE 2. How many square yards of plaster will it take to cover the sides and ceiling of a room 16 X 20 feet and 11 feet high, having four windows, each 7x4 feet, and three doors, each 9x4 feet over all, the baseboard coming 6 inches above the floor ? SOLUTION. Area of ceiling = 16 X 20 = 320 sq. ft. Area of end walls = 2(16 X 11) = 352 sq. ft. Area of side walls = 2(20 X 11) = 440 sq. ft. Total area = 1,112 sq. ft. From the above must be deducted: Windows = 4(7 X 4) = 112 sq. ft. Doors = 3(9 X 4) = 108 sq. ft. Baseboard less the width of three doors = (72 12) X ^ = 30 sq. ft. Total number of feet to be deducted = 112 + 108 + 30 = 250 sq. ft. Hence, number of square feet to be plastered = 1,112 250 = 862 sq. ft., or 95J sq. yd. Ans. EXAMPLE 3. How many acres are contained in a rectangular tract of land 800 rods long and 520 rods wide ? SOLUTION. 800 X 520 = 416,000 sq. rd. Since there are 160 square rods in 1 acre, the number of acres = 416,000 H- 160 = 2,600 acres. Ans. MENSURATION AND EXAMPLES FOR PRACTICE. 1. What is the area in square feet of a rhombus whose base is 84 inches and whose altitude is 3 feet ? Ans. 21 sq. ft. 2. A flat roof, 46 feet by 80 feet in size, is covered by tin roofing weighing one-half pound per square foot ; what is the total weight of the roofing ? Ans. 1,840 Ib. 3. One side of a room measures 16 feet. If the floor contains 240 square feet, what is the length of the other side ? Ans. 15 ft. 4. How many square feet in a board 12 feet long, 18 inches wide at one end, and 12 inches wide at the other end ? Ans. 15 sq. ft. 5. How much would it cost to lay a sidewalk a mile long and 8 feet 6 inches wide, at the rate of 20 cents per square foot ? How much at the rate of $1.80 per square yard ? Ans. $8,976 in each case. 6. How many square yards of plastering will be required for the ceiling and walls of a room 10 ft. X 15 ft. and 9 feet high ; the room con- tains one door 3 ft. X 7 ft., three windows 3 ft. X 6 ft., and a baseboard 8 inches high ? Ans. 53.5 sq. yd. THE TRIANGLE. 5O. A triangle is a plane figure having three sides. 51. An isosceles triangle is one having two of its sides equal ; see Fig. 21. 52. An equilateral triangle (Fig. 22) is one having all of its sides of the same length. FlG- 33. 53. A scalene triangle (Fig. 23) is one having no two of its sides equal. FIG. 23. 54. A right-angled triangle (Fig. 24) is any triangle having one right angle. The side opposite the right angle is called the hypotenuse. A right-angled triangle is now usually called a right triangle. 55. In any triangle the sum of the three angles equals two right angles, or 180. Thus, in Fig. 25, the sum of the angles A, B, and ^equals two right angles, or 180. Hence, if any two angles of a tri- C angle are given and it is required to find the third angle: FIG. 25. 3 USE OF LETTERS IN FORMULAS. 21 Rule 4. Add the two given angles and subtract their sum from 180 ; the remainder will be the third angle. EXAMPLE. If two angles of a triangle are 48 16' and 47 50', what does the third angle equal ? SOLUTION. First reduce 48 16' and 47 50' to minutes, for conve- nience in adding and subtracting the angles. 48 = 48 X 60'= 2,880'; 2,880' + 16' = 2,896' ; hence, 48 16' = 2,896'. In like manner, 47 50' = 47 X CO' + 50' = 2,820'+ 50' = 2,870'. Adding the two angles and sub- tracting the sum from 180 reduced to minutes, 2,896' + 2,870' = 5,766'; 180 = 180 X 60'= 10,800' ; 10,800 - 5,766 = 5,034'. Reducing this last number to degrees and minutes, ' = 83f = 83 54'. Hence, the third angle in the triangle = 83 54'. Ans. 56. In any right triangle there can be but one right angle, and since the sum of all the tt angles is two right angles, it is evident that the sum of the two acute angles must equal one right angle, or 90. Therefore, if in any right triangle one acute angle is known, to find the other acute angle: FIG. 26. Rule 5. Subtract the known acute angle from 90; the result will be the other acute angle. EXAMPLE. If one acute angle, as A, of the right triangle ABC, Fig. 26, equals 30, what does the angle B equal? SOLUTION. 90 30 = 60'. Ans. 57. If a straight line be drawn through two sides of a triangle, parallel to the third side, a second triangle will be formed whose sides will be propor- tional to the corresponding sides of the first triangle. Thus, in the triangle ABC, Fig. 27, if the line D E be drawn parallel to the side B C, the triangle FIG. 27. v A D E will be formed and we shall have (1) vSide A D : side D E = side A B : side B C ; and, (2) Side A E : side D E = side A C : side B C ; also, (3) Side A D : side A E = side A B : side A C. 22 MENSURATION AND 3 FIG - EXAMPLE In Fig. 27, if A B = 24, B C = 18, and D E = 8, what does A D equal ? SOLUTION. Writing these values for the sides in (1), 24- v 8 A D : 8 = 24 : 18 ; whence, A D = * = lOf . Ans. io 58. In any right triangle, the square described on the hypotenuse is equal to the sum H of the squares described upon the other two sides - If A B C> Fig. 28, is a right triangle right-angled at .5, then the square described upon the hypotenuse A C is equal to the sum f the s q uares de - scribed upon the sides A B an( j ^ ^7 Hence, having given the two sides forming the right angle in a right triangle, to find the hypotenuse : Rule 6. Square each of the sides forming the right angle ; add the squares together and take the square root of the sum. EXAMPLE. If A B 3 inches and D C 4 inches, what is the length of the hypotenuse A C 1 SOLUTION. Squaring each of the given sides, 3 2 = 9 and 4 2 = 16. Taking the square root of the sum of 9 and 16, the hypotenuse = 1/9 + 16 = V'25 = 5 in. Ans. 59. If the hypotenuse and one side are given, the other side can be found as follows: Rule 7. Subtract the square of the given side from the square of the hypotemise, and extract the square root of the remainder. EXAMPLE 1. The side given is 3 inches, the hypotenuse is 5 inches; what is the length of the other side ? SOLUTION. 3 2 = ; 5* = 25. 25 - 9 = 16, and 4/16 = 4 in. Ans. 3 USE OF LETTERS IN FORMULAS. 23 150 EXAMPLE 2. If from a church steeple which is 150 feet high a rope is to be attached to the top and to a stake in the ground, which is 85 feet from the center of the base (the ground being supposed to be level), what must be the length of the rope ? SOLUTION. In Fig. 29, A B represents the stee- ple, 150 feet high ; C a stake 85 feet from the foot of the steeple, and A C the rope. Here we have a right triangle right-angled at B, and A C is the hypotenuse. The square of A B ISO 2 =22,500; of C />', 85* = 7,225. 22,509 + 7,225 = 29,725 ; 4/29,725 = 172.4 ft., nearly. Ans. B 6O. The altitude of any triangle is a line, as B D, drawn from the vertex B of the -angle opposite the base A C, perpendicular to the base, as in Fig. 30, or to the base extended, as in Fig. 31. 61. If in any parallelogram a straight line, called the diagonal, be drawn, connecting two opposite corners, it will divide the parallelogram into two equal triangles, as A D B and D B C in Fig. 32. The area of each triangle will equal one-half the area of the parallelogram, i. e., one-half the product of the base and the altitude. Hence, to find the area of any triangle : Rule 8. Multiply the base by the altitude and divide the product by 2. EXAMPLE. What is the area in square feet of. a triangle whose base is 18 feet and whose altitude is 7 feet 9 inches ? SOLUTION. 7 ft. 9 in. = 7f ft. = -j- ft, 18 X -r = 139|, and one- half of 139 = 69f sq. ft. Ans. To find the altitude or base of a triangle, having given the area and the base or altitude: Rule 9.. Multiply the area by 2 and divide by the given dimension 24 MENSURATION AND 3 EXAMPLE. What must be the height of a triangular piece of sheet metal to contain 100 square inches, if the base is 10 inches long ? SOLUTION. 100 X 2 = 200 ; 200 -*- 10 = 20 in. Ans. EXAMPLES FOR PRACTICE. 1. What is the area of a triangle whose base is 18 feet long and whose altitude is 10 feet 6 inches ? Ans. 94.5 sq. ft. 2. Two angles of a scalene triangle together equal 100 4'. What is the size of the third angle ? Ans. 79 56'. 3. One angle of a right triangle equals 20 10' 5". What is the size of the other acute angle ? Ans. 69 49' 55". 4. A ladder 65 feet long reaches to the top of a wall when its foot is 25 feet from the wall. How high is the wall ? Ans. 60 ft. 5. Draw a triangle, and through two of its sides draw a line parallel to the base. Letter the different lines, and then, without referring to the text, write out the proportions existing between the sides of the two triangles. 6. A triangular piece of sheet metal weighs 24 pounds. If the base of the triangle is 4 feet and its height 6 feet, how much does the metal weigh per square foot ? Ans. 2 Ib. 7. The area of a triangle is 16 square inches. If the altitude is 4 inches, what does the base measure ? Ans. 8 in. 8. Two sides of a right triangle are 92 feet and 69 feet long. How long is the hypotenuse ? Ans. 115 ft. POLYGONS. 62. A polygon is a .plane figure bounded by straight lines. The term is usually applied to a figure having more than four sides. The bounding lines are called the sides, and the sum of the lengths of all the sides is called the perimeter of the polygon. 63. A regular polygon is one in which all the sides and all the angles are equal. 64. A polygon of five sides is called a pentagon ; one of six sides, a hexagon ; one of seven sides, a heptagon, Pentagon. Hexagon. Heptagon. Octagon. Decagon. Dodecagon. FIG. 33. etc. Regular polygons having from five to twelve sides are 3 USE OF LETTERS IN FORMULAS. 25 shown in Fig. 33. interior angles, as In any polygon, the sum of all the A-\-B+C-\-D-\-E, Fig. 34, equals 180 multiplied by a number which is two less than the number of sides in the poly- gon. Hence, to find the size of any one of the interior angles of a regular polygon: Rule 1O. Multiply 180 by the num- ber of sides less two and divide the result by the number of sides ; the quotient will be the number of degrees in each interior angle. EXAMPLE 1. If Fig. 34 is a regular pentagon, how many degrees are there in each interior angle ? SOLUTION. In a pentagon there are five sides; hence, 5 2 = 3 and 180 X 3 = 540 ; 540 -v- 5 = 108 in each angle. Ans. FIG. si. EXAMPLE 2. It is desired to make a miter-box in which to cut a strip of molding to fit around a column having the shape of a regular hexa- gon. At what angle should the saw run across the miter-box ? SOLUTION. In Fig. 35, let A B, B C, CD, etc. represent the pieces of molding as they will fit around the column. First find the size of. one of the equal angles of the polygon by FIG. 35. the a bove rule. Number of sides = 6; 6-2 = 4; hence, 180x4 = 720, and 720^-6 = 120 in each angle. Now, let M N represent the miter-box and O S the direction in which the saw should run ; then, A B O is the angle made by the saw with the side of the miter-box ; but as the polygon is a regular one, this angle is one-half the interior angle ABC, which we have found to be 120. 120 Hence, the saw should run at an angle of -^- = 60 with the side of the miter-box. Ans. 65. The area of any regular polygon may be found by drawing lines from the center to each angle and computing the area of each triangle thus formed. Hence, to find the area of any regular polygon : Rule 11. Multiply the' length of a side by half the distance from the side to the FIG. se. center, and that product by the number of sides. The last product will be the area of the figure. 26 MENSURATION AND 3 EXAMPLE. In Fig. 36 the side B C of the regular hexagon is 12 inches and the distance A O is 10.4 inches; required the area of the polygon. SOLUTION. 10.4 -*- 2 = 5.2; 12 X 5.2 X 6 = 374.4 sq. in. Aris. 66. To obtain the area of any irregular polygon, draw diag- onals dividing the polygon into triangles and quadrilaterals, and compute the areas of these sepa- rately ; their sum will be the area of the figure. EXAMPLE. It is required to find the area of the polygon A B CD EF, Fig. 37. SOLUTION. Draw the diagonals B F and CF and the line FG perpendicular to D E, dividing the figure into the triangles A B F, B C F, and FG E and the rectangle FC D G. Let it be supposed that the altitudes of the figures and the lengths of the sides A B, D G, and G E are as indicated in the polygon above. Then, Area A B F = 16 7 = 56 sq. in. Area FC D G = 14 X 10 = 140 sq. in. Area FG E = 9 * 10 = 45 sq. in. Total area = 56 + 35 + 140 + 45 = 276 sq. in. Ans. EXAMPLES FOR PRACTICE. 1. How many degrees are there in one of the angles of a regular octagon? Ans. 135. 2. Find the area of the polygon A B CD EF (see Fig. 37), suppo- sing each of the given dimensions to be increased to 1^ times the length given in the figure. Ans. 621 sq. in. 3. What is the area of a regular heptagon whose sides are 4 inches long, the distance from one side to the center being 4.15 inches? Ans. 58.1 sq. in. 4. At what angle should the saw run in a miter-box to cut strips to fit around the edge of a table top made in the shape of a regular pentagon? Ans. 54. USE OF LETTERS IN FORMULAS. THE CIRCLE. 67. A circle (Fig. 38) is a figure bounded by a curved line, called the circumference, every point of which is equally distant from a point within, called the center. 11- -J* FIG. 39. 68. The diameter of a circle is a straight line passing through the center and terminated at both ends by the cir- cumference; thus, A B (Fig. 39) is a diam- eter of the circle. 69. The radius of a circle, A O (Fig. 40), is a straight line drawn from the center O to the circumference. It is equal in length to one-half the diameter. ;6 The plural of radius is radii, and all radii of a circle are equal. FIG. 40. 7O. An arc of a circle (see a e b, Fig. 41) is any part of its circumference. 71. A chord is a straight line joining any two points in a circumference ; or it is a straight line joining the extremities of an arc ; thus, the straight line A J5, Fig. 42, is a chord of the circle whose corresponding arc is A E B. 72. An inscribed angle is one whose vertex lies on the circumference of a circle and whose sides are chords. It is measured by one-half the intercepted arc. Thus, in Fig. 43, A B C is an in- scribed angle, and it is measured by one- half the arc ADC. 28 MENSURATION AND 3 EXAMPLE. If in Fig. 43, the arc A D C = f of the circumference, what is the measurement of the inscribed angle A B C ? SOLUTION. Since the angle is an inscribed angle, it is measured by one-half the intercepted arc, or f X 1 = \ of the circumference. The whole circumference = 360 ; hence, 360 X \ = 72 ; therefore, angle A B C is an angle of 72. 73. If a circle is divided into halves, each half is called a semicircle, and each half circumference is called a semi-circumference. Any angle inscribed in a semicircle is a right angle, since it is measured by } c one-half a semi-circumference, or 180 -=- 2 = 90. Thus, the angles ADC and A B C, Fig. 44, are right angles, since they are inscribed in a semicircle. 74. An inscribed polygon is one whose vertexes lie on the circumference of a circle and whose sides are chords, as A B C D E, Fig. 45. The sides of an inscribed regular hex- agon have the same length as the radius of the circle. If, in any circle, a radius be drawn perpendicular to any chord, it bisects (cuts in halves) the chord. Thus, if the FIG. 45. radius O C, Fig. 46, is perpendicular to the chord A J3, A D = DB. EXAMPLE. If a regular pentagon is inscribed in a circle and a radius is drawn perpendicular to one of the sides, what are the lengths of the two parts of the side, the perimeter of the pen- tagon being 27 inches ? SOLUTION. A pentagon has five sides, and" since it is a regular pentagon, all the sides are of equal lengths ; the perimeter of the pentagon, which equals the distance around it, or equals the sum of all the sides, is 27 inches. There- fore, the length of one side = 27 -f- 5 = 5| inches. Since the penta- gon is an inscribed pentagon, its sides are chords, and as a radius perpendicular to a chord bisects it, we have 5| -r- 2 = 2 T 7 ff inches, which equals the length of each of the parts of the side cut by a radius per- pendicular to it. Ans. 3 USE OF LETTERS IN FORMULAS. 29 75. If, from any point on the circumference of a circle, a perpendicular is let fall upon a diameter, it will divide the diameter into two parts, one of ^^ which will be in the same ratio to the perpendicular as the perpendicular is to the other part. That is, the perpendic- A\ ular will be a mean proportional between the two parts of the diameter. If A B, Fig. 47, is the given diameter and C any point on the circumference, FlG - 4 ~- then AD: CD=CD:DB,CD being a mean proportional between A D and D B. EXAMPLE. If H K= 30 feet and IB 8 feet, what is the diameter of the circle, H K being perpendicular to A B ? SOLUTION. 30 feet -f- 2 feet = 15 feet = I H. And B I : I H = SH : I A, or 8: 15 = 15: I A. Therefore, I A = ~ = ~ = 28 feet and IA + IB = 28 + 8 feet = A B, the diameter of the circle. Ans. 76. When the diameter of a circle and the lengths of the two parts into which it is divided are given, the length of the perpendicular may be found by multiplying the lengths of the two parts together and extracting the square root of the product. EXAMPLE. In Fig. 47, the diameter of the circle A B is 36 feet and the distance B /is 8 feet; what is the length of the line H K 1 SOLUTION. As the diameter of the circle is 36 feet and as B I is 8 feet, I A is equal to 36| 8 = 28 feet. The two parts, therefore, are 8 and 28 feet, and their product = 8 X 28 = 8 X ^ = 225 ; the square root of their product = 4/225 = 15 feet, and as H K = IH+ IK, or 2 Iff, /fAT= 15X2 = 80 ft. Ans. 77. To find the circumference of a circle, the diameter being given: Rule 12. Multiply the diameter by 3.1416. EXAMPLE. What is the circumference of a circle whose diameter is 15 inches ? SOLUTION. 15 X 3.1416 = 47.124 in. Ans. 78. To find the diameter of a circle, the circumference being given: 30 MENSURATION AND 3 Rule 13. Divide the circumference by 3.1416. EXAMPLE. What is the diameter of a circle whose circumference is 65. 973 inches? SOLUTION. 65.973 -*- 3.1416 = 21 in. Ans. 79. To find the length of an arc of a circle : Rule 14. Mtilttply the length of the circumference of the circle of wJiich the arc is a part by the number of degrees in the arc and divide by 360. EXAMPLE. What is the length of an arc of 24 U , the radius of the arc being 18 inches ? SOLUTION. 18 X 2 = 36 in. = the diameter of the circle. 36 X 3.1416 = 113.1 in., the circumference of the circle of which the arc is a part. 24 113.1 X ogn = 7 - 54 in -> or the length of the arc. Ans. 80. To find the area of a circle : Rule 15. Square the diameter and multiply by .7854- EXAMPLE. What is the area of a circle whose diameter is 15 inches ? SOLUTION. 15* = 225; and 225 X .7854 = 176.72 sq. in. Ans. 81. Given the area of a circle, to find its diameter: Rule 16. Divide the area by . 7854 and extract the square root of the quotient. EXAMPLE 1. The area of a circle = 17,671.5 square inches. What is its diameter in feet ? SOLUTION. y ^ = 150 inches. rp-j- = 12^ feet, or the diameter. Ans. EXAMPLE 2. What is the area of a flat circular ring, Fig. 48, whose outside diameter is 10 inches and inside diameter is 4 inches ? SOLUTION. The area of the large circle = 10 s X- 7854 = 78. 54 sq. in.; the area of the small circle = 4 2 X -7854 = 12.57 sq. in. The area of the ring is the difference between these areas, or FIG. 48. 78.54 12.57 = 65.97 sq. in. Ans. 82. To find the area of a sector (a sector of a circle is the area included between two radii and the circumference, as, for example, the area B A CO, Fig. 36): 3 USE OF LETTERS IN FORMULAS. Rule 17. Divide tJic number of degrees in the arc of the sector by 360. Multiply the result by the area of the circle of i^Jiich the sector is a part. EXAMPLE. The number of degrees in the angle formed by drawing radii from the center of a circle to the extremities of the arc of the circle is 75. The diameter of the circle is 12 inches; what is the area of the sector ? SOLUTION. ^ = A ; and 12 2 X .7854 = 113.1 sq. in. 113.1 X 52 = 23.56 sq. in., the area. Ans. 83. To find the area of a segment of a circle (a seg- ment of a circle is the area included between a chord and its arc; for example, the area ABC, Fig. 49) when its chord and height are given. There is no exact method, except by applying principles of trigonometry. The follow- ing rule gives results that are exact enough for practical purposes. Rule 18. Divide the diameter by the height of the seg- ment ; subtract .608 from the quotient and extract the square root of the remainder. This result multiplied by 4 times the square of the height of the segment and then divided by 3 will give the area, very nearly. The rule, expressed as a formula, is as follows, where D = the diameter of the circle and h = the height of the segment (see Fig. 49) : Area =- * - . 608. EXAMPLE. What is the area of the segment of a circle whose diameter is 54 inches, the height of the segment being 20 inches ? SOLUTION. Substituting in the formula, ~ 4x400 FIG. 49. Ans. NOTE. Had the chord A C, Fig. 49, been given instead of the diameter, the diameter would have been found as explained in Art. 75. //. X. J.1Q 32 MENSURATION AND 3 EXAMPLES FOR PRACTICE. 1. An angle inscribed in a circle intercepts one-third of the circum- ference. How many degrees are there in the angle ? Ans. 60. 2. Suppose tjiat in Fig. 47, the diameter A B = 15 feet and the distance B I 3 f eet. What is the length of the line H K ? Ans. 12 ft. 3. The diameter of a flywheel is 18 feet. What is the distance around it to the nearest 16th of an inch ? Ans. 56 ft. 6^ in. 4. A carriage wheel was observed to make 71f turns while going 300 yards. What was its diameter ? Ans. 4 ft. , nearly. 5. What is the length of an arc of 04, the radius of the arc being 30 inches? Ans. 33.51 in. 6. Find the area of a circle 2 feet 3 inches in diameter. Ans. 3.976 sq. ft. 7. What must be the diameter of a circle to contain 100 square inches? Ans. 11.28 in. 8. Compute the area of a segment whose height is 11 inches and the radius of whose arc is 21 inches. Ans. 289.04 sq. in. 9. Find the area of a flat circular ring whose outside diameter is 12 inches and whose inside diameter is 6 inches. Ans. 84.82 sq. in. THE PRISM AND CYLINDER. 84. A solid, or body, has three dimensions: length, breadth, and thickness. The sides which enclose it are called the faces, and their lines of intersection are called the edges. 85. A prism is a solid whose ends are equal and par- allel polygons and whose sides are parallelograms. Prisms take their names from the form of their bases. Thus, a tri- angular prism is one having a triangle for its base; a hex- agonal prism is one having a hexagon for its base, etc. 86. A cylinder is a body of' uniform diameter whose ends are equal parallel circles. 87. A parallelopipedoii (Fig. 50) is a prism whose bases (ends) are parallelograms. 88. A cube (Fig. 51) is a prism whose faces and ends are squares. All the faces of a cube are equal. FlG - 51 - In the case of plane figures, we are concerned with perimeters and areas. In the case of solids, we are 3 USE OF LETTERS IN FORMULAS. 33 concerned with the areas of their outside surfaces and with their contents or volumes. 89. The entire surface of any solid is the area of the whole outside of the solid, including the ends. The convex surface of a solid is the same as the entire surface, except that the areas of the ends are not included. 90. A unit of volume is a cube each of whose edges is equal in length to the unit. The volume is expressed by the number of times it will contain a unit of volume. Thus, if the unit of length is 1 inch, the unit of volume will be the cube whose edges each measure 1 inch, this cube being 1 cubic inch ; and the number of cubic inches the solid contains will be its volume. If the unit of length is 1 foot, the unit of volume will be 1 cubic foot, etc. Cubic inch, cubic foot, and cubic yard are abbreviated to cu. in., cu. ft., and cu. yd., respectively. Instead of the word volume, the expression cubical con- tents is sometimes used. 91. To find the area of the convex surface of a prism or cylinder: Rule 19. Multiply the perimeter of the base by the altitude. EXAMPLE 1. A block of marble is 24 inches long and its ends are 9 inches square. What is the area of its convex surface ? SOLUTION. 9 X 4 = 36 = the perimeter of the base; 36 X 34 = 864 sq. in., the convex area. Ans. To find the entire area of the outside surface, add the areas of the two ends to the convex area. Thus, the area of the two ends = 9 X 9 X 2 = 162 sq. in. ; 864 + 162 = 1,026 sq. in. Ans. EXAMPLE 2. How many square feet of sheet iron will be required for a pipe H feet in diameter and 10 feet long, neglecting the amount necessary for lapping ? SOLUTION. The problem is to find the convex surface of a cylinder 1 feet in diameter and 10 feet long. The perimeter, or circumference, of the base = !$ X 3.1416 = 1.5 X 3.1416 = 4.712 ft. The convex sur- face = 4.712 X 10 = 47.12 sq. ft. of metal. Ans. 92. To find the volume of a prism or a cylinder: Rule 2O. Multiply the area of the base by the altitude. EXAMPLE 1. What is the weight of a length of wrought-iron shaft- ing 16 feet long and 2 inches in diameter ? Wrought iron weighs .28 pound per cubic inch. MENSURATION AND SOLUTION. The shaft is a cylinder 16 ft. long. The area of one end, or the base, = 2* X .7854 = 3.1416 sq. in. Since the weight of the iron is given per cubic inch, the contents of the shaft must be found in cubic inches. The length, 16 ft., reduced to inches = 16 X 12 = 192 in.; 3.1416x192 = 603.19 cu. in. = the volume. The weight = 603. 19 X .28 = 168.89 Ib. Ans. EXAMPLE 2. Find the cubical contents of a hexagonal prism, Fig. 52, 12 inches long, each edge of the base being 1 inch long. SOLUTION. In order to obtain the area of one end, the distance CD from the center Cto one side must be found. In the right triangle C D A, side A D = $ A B, or | inch, and since the polygon is a hexagon, side C A = distance A B, or 1 inch (Art. 74). Hence, C A being the hypotenuse, the length of side C D = ^l*~ (*) 2 = 4/1* - .5* = 4//T5, or 1 X FIG. 52. .866 inch. Area of triangle A C B = = .433 sq. in. ; area of the whole polygon = .433 X 6 = 2.598 sq. in. Hence, the contents of the prism = 2.598 X 12 = 31.176 cu. in. Ans. EXAMPLE 3. It is required to find the number of cubic feet of steam space in the boiler shown in Fig. 53. The boiler is 16 feet long between FIG. 53. heads, 54 inches in diameter, and the mean water-line M ' N is at a distance of 16 inches from the top of the boiler. The volume of the steam outlet casting may be neglected. SOLUTION. The volume of the steam space, which is that space within the boiler above the surface M NO P of the water, is found by the rule for finding the volume of a prism or cylinder, the area M N S being the base and the length NO the altitude. First obtain the area of the segment M N S, whose height // is 16 inches, in square feet; then multiply the result by 16, the length of the boiler. By the formula given in Art. 83, the area of the segment = 3 USE OF LETTERS IN FORMULAS. 35 27767 = 1.663. Hence, the area = 341.33 X 1-663 = 567.63 sq. in. This reduced to sq. ft. = 567.63 -5- 144 = 3.942 sq. ft., and the volume therefore = 3.942 X 16 = 63.07 cu. ft. Ans. In the above solution, the space occupied by the stays is not considered, for sake of simplicity. They are not shown in the figure. EXAMPLE 4. In the above boiler there are 60 tubes, 3J inches outside diameter. How many gallons of water will it take to fill the boiler up to the mean water level, there being 231 cubic inches in a gallon ? SOLUTION. Find the volume in cubic inches of that part of the boiler below the surface of the water M NO P, since the contents of a gallon is given in cubic inches, and from it subtract the volume of the tubes in cubic inches. This may be done by first finding the total area of one end of the boiler in square inches, from it subtracting the area of the seg- ment M N S, and the areas of the ends of the tubes in square inches, and then by multiplying the result by the length of the boiler in inches. Total area of one end = 54* X -7854 = 2,290.23 sq. in. Area of segment M N S, as found in last example, = 567.63 sq. in. Area of the end of one tube = 8.25* X -7854 = 8.2958 sq. in. Area of the ends of the 60 tubes = 8.2958 X 60 = 497.75 sq. in. Hence, the area to be subtracted = 567.63 + 497.75 = 1,065.38 sq. in. Subtracting, 2,290.23 - 1,065.38 = 1,224.85 sq. in. = net area. The cubical contents = 1,224.85 X 16 X 12 = 235,171.2 cu. in. This divided by 231 will give the number of gallons; whence, 235,171.2 -T- 231 = 1,018.06 gal. of water. Ans. EXAMPLES FOR PRACTICE. 1. Find the area in square inches of the convex surface of a bar of iron 4J inches in diameter and 8 feet 5 inches long. Ans. 1,348.53 sq. in. 2. Find the area of the entire surface of the above bar. Ans. 1,376.9 sq. in. 3. What is the area of the entire surface of the hexagonal prism whose base is shown in Fig. 52 ? Ans. 77.196 sq.,in. 4. A multitubular boiler has the following dimensions: diameter, 50 inches ; length between heads, 15 feet ; number of tubes, 56 ; outside diameter of tubes, 3 inches ; distance of mean water-line from top of boiler, 16 inches, (a) Compute the steam space in cubic feet, (b) Find the number of gallons of water required to fill the boiler up to the mean water-line. . ( (a) 56.4 cu. ft. \(b) 800 gal. 36 MENSURATION AND THE PYRAMID AND CONE. 93. A pyramid (Fig. 54) is a solid whose base is a polygon and whose sides are triangles uniting at a common point, called the vertex. 94. A cone (Fig. 55) is a solid whose base is a circle and whose convex surface tapers uniformly to a point called FIG. 55. FIG. 54. the vertex. 95. The altitude of a pyramid or cone is the perpen- dicular distance from the vertex to the base. 96. The slant height of a pyramid is a line drawn from the vertex perpendicular to one of the sides of the base. The slant height of a cone is any straight line drawn from the vertex to the circumference of the base. 97. To find the convex area of a pyramid or cone : Rule 21. Multiply the perimeter of tJie base by one-Jialf the slant height. EXAMPLE 1. What is the convex area of a pentagonal pyramid if one side of the base measures 6 inches and the slant height is 14 inches ? SOLUTION. The base of a pentagonal pyramid is a pentagon, and, consequently, has fives sides. 6 X 5 = 30 inches, or the perimeter of the base. 30 X - = 210 sq. in., or the convex area. Ans. EXAMPLE 2. What is the entire area of a right cone whose slant height is 17 inches and whose base is 8 inches in diameter ? SOLUTION. The perimeter of the base = 8 X 3.1416 = 25.1328 in. 25.1328 X = 213.63 sq. in. Convex area Area of base = 8 s X .7854 = 50.27 sq. in. Entire area = 263.90 sq. in. Ans. 98. To find the volume of a pyramid or cone: Rule 22. Multiply the area of the base by one-third of the altitude. EXAMPLE 1. What is the volume of a triangular pyramid, each edge of whose base measures 6 inches and whose altitude is 8 inches ? USE OF LETTERS IN FORMULAS. 37 SOLUTION. Draw the base as shown in Fig. 56; it will be an equilateral triangle, all of whose sides are 6 inches long. Draw a perpendicular B D from the vertex to the base; it will divide the base into two equal parts, since an equilateral triangle is also isosceles, and will be the altitude of the triangle. In order to obtain the area of the base, this altitude must be determined. In the right triangle B D A, the hypotenuse B A = 6 inches and side A D = 3 inches, to find the other side, B D = |/6* 2 3 2 = 5.2 in., nearly. 6X5.2 Area of the base, or B A C, = Q ume = 15.6 X o = 41.6 cu. in. Ans. 15.6 sq. in. Hence, the vol- Ex AMPLE 2. What is the volume of a cone whose altitude is 18 inches and whose base is 14 inches in diameter ? SOLUTION. Area of the base = 14 2 X .7854 = 153.94 sq. in. Hence, 1 8 the volume = 153.94 X -5- = 923.64 cu. in. o Ans. EXAMPLES EOB PRACTICE. 1. Find the convex surface of a square pyramid whose slant height is 28 inches and one edge of whose base is 7 inches long. Ans. 420 sq. in. 2. What is the volume of a triangular pyramid, one edge of whose base measures 3 inches and whose altitude is 4 inches ? Ans. 5.2cu.in. 3. Find the volume of a cone whose altitude is 12 inches and the circumference of whose base is 31.416 inches. Ans. 314.16 cu. in. NOTE. Find the diameter of the base and then its area. THE FRUSTUM OF A PYRAMID OB CONE. 99. If a pyramid be cut by a plane, parallel to the base, so as to form two parts, as in Fig. 57, the lower part is called the frustum of the pyramid. If a cone be cut in a similar man- ner, as in Fig. 58, the lower part is called the frustum, of the cone. FIG. 57. FIG. 58. 38 MENSURATION AND 3 100. The upper end of the frustum of a pyramid or cone is called the tipper base, and the lower end the lower "base. The altitude of a frustum is the perpendicular dis- tance between the bases. 101. To find the convex surface of a frustum of a pyramid or cone: Rule 23. Multiply one-half the sum of the perimeters of the t^vo bases by the slant height of tlie frustum. EXAMPLE 1. Given, the frustum of a triangular pyramid, in which one side of the lower base measures 10 inches, one side of the upper ' base measures 6 inches, and whose slant height is 9 inches ; find the area of the convex surface. SOLUTION. 10 in. X 3 = 30 in., the perimeter of the lower base. 6 in. X 3 = 18 in., the perimeter of the upper base. OA I -J Q = 24 in., or one-half the sum of the perimeters of the two bases. 24 X 9 = 216 sq. in., the convex area. Ans. EXAMPLE 2. If the diameters of the two bases of a frustum of a cone are 12 inches and 8 inches, respectively, and the slant height is 12 inches, what is the entire area of the frustum ? SOLUTION.- - xia = 876j99 ^ , the area of the convex surface. Area of the upper base = 8 s X -7854 = 50.27 sq. in. Area of the lower base = 12* X .7854 = 113.1 sq. in. The entire area of the frustum = 376.99 -f- 50.27 + 113.1 = 540.36 sq. in. Ans. 1O2. To find the volume of the frustum of a pyramid or cone : Rule 24. Add togetJier the areas of the upper and lower bases and the square root of the product of the two areas ; multiply the sum by one-third of the altitude. EXAMPLE 1. Given, a frustum of a square pyramid (one whose base is a square); each edge of the lower base is 12 inches, each edge of the upper base is 5 inches, and its altitude is 16 inches ; what is its volume? SOLUTION. Area of upper base = 5 X 5 = 25 sq. in. ; area of lower base = 12 X 12 = 144 sq. in. ; the square root of the product of the two areas = |/25 X 144 = 60. Adding these three results, and multiplying by one-third the altitude, 25 +144 + 60 = 229; 229 X ^ = 1.221$ cu. in, = the volume. Ans. 3 USE OF LETTERS IN FORMULAS. 39 EXAMPLE 2. How many gallons of water will a round tank hold which is 4 feet in diameter at the top, 5 feet in diameter at the bottom, and 8 feet deep ? SOLUTION. There are 231 cubic inches in a gallon, and the volume of the tank should be found in cubic inches. The tank is in the shape of the frustum of a cone. The upper diameter = 4 X 12 = 48 inches ; the lower diameter = 5 X 12 = 60 inches, and the depth = 8 X 12 96 inches. Area of upper base = 48 s X .7854= 1,809.56 sq. in. ; area of lower base = 60' 2 X .7854 = 2,827.44 sq. in. ; -/1, 809. 56^X^7827^4 = 2,261.95. Qfi Whence, 1,809.56 + 2,827.44 + 2,261.95 = 6,898.95; 6,898.95 X y 220,766.4 cu. in. = contents. Now, since there are 231 cu. in. in 1 gallon, the tank will hold 220,766.4 -=- 231 = 955.7 gal., nearly. Ans. EXAMPLES FOR PRACTICE. 1. Find the convex surface of the frustum of a square pyramid, one edge of whose low.er base is 15 inches long, one edge of whose upper base is 14 inches long, and whose slant height is 1 inch. Ans. 58 sq. in. 2. Find the volume of the above frustum, supposing its altitude to be 3 inches. Ans. 631 cu. in. 3. Find the volume of the frustum of a cone whose altitude is 12 feet and the diameters of whose upper and lower bases are 8 and 10 feet, respectively. Ans. 766.55 cu. ft. 4. If a tank had the dimensions of example 3, how many gallons would it hold? Ans. 5,734.2 gal., nearly. THE SPHERE AND CYLINDRICAL RING. 103. A sphere (Fig. 59) is a solid bounded by a uniformly curved surface, every point of which is equally distant from a point within, called the center. The word ball, or globe, is generally used instead of sphere. 104. To find the area of the surface of a FIG 59 sphere : Rule 25. Square the diameter and multiply the result by 3. 1416. EXAMPLE. What is the area of the surface of a sphere whose diam- eter is 14 inches ? SOLUTION. Diameter squared X 3.1416 = 14 J X 3.1416 = 14 X 14 X 3.1416 = 615.75 sq. in. Ans. 40 MENSURATION AND 3 From this it will be seen that the surface of a sphere equals the circumference of a great circle multiplied by the diameter, a rule often used ; a great circle of a sphere is the intersection of its surface with a plane passing through its center ; for instance, the great circle of a sphere 6 inches diameter is a circle of 6 inches diameter. Any number of great circles could be described on a given sphere. 1O5. To find the volume of a sphere : Rule 26. Cube the diameter and multiply the result by EXAMPLE. What is the weight of a lead ball 12 inches in diameter, a cubic inch of lead weighing .41 pound ? SOLUTION. Diameter cubed X .5236 = 12 X 12 X 12 X .5236 = 904.78 cu. in., or the volume of the ball. The weight, therefore, =904.78 X .41=370.961b. Ans. 1O6. To find the convex area of a cylindrical ring: A cylindrical ring (Fig. 60) is a cyl- inder bent to a circle. The altitude of the cylinder before bending is the same \ B as the length of the dotted center line D. The "base will correspond to a cross- section on the line A B drawn from the center O. Hence, to find the convex FIG. eo. area : Rule 27. Multiply the circumference of an imaginary cross-section on the line A B by the length of the center line D. EXAMPLE. If the outside diameter of the ring is 12 inches and the inside diameter is 8 inches, what is its convex area ? SOLUTION. The diameter of the center circle equals one-half the sum -| e) . Q of the inside and outside diameters = 5 = 10, and 10x3.1416 = 31.416 in., the length of the center line. The radius of the inner circle is 4 inches; of the outside circle, 6 inches ; therefore, the diameter of the cross-section on the line A B is 2 inches. Then, 2 X 3.1416 = 6.2832 in., and 6.2832 X 31.416 = 197.4 sq. in., the convex area. Ans. 3 USE OF LETTERS IN FORMULAS. 41 1O7. To find the volume of a cylindrical ring: Rule 28. TJic volume will be tlie same as that of a cylin- der whose altitude equals the length of the dotted center line D (Fig. 61) and whose base is the same as a cross-section of the ring on the line A B drawn from the center O. Hence, to find the volume of a cylindrical ring, multiply the area of an imaginary cross-section on the line A B by the length of the center line D. FlG - 01 - EXAMPLE. What is the volume of a cylindrical ring whose outside diameter is 12 inches and whose inside diameter is 8 inches? SOLUTION. The diameter of the center circle equals one-half the sum of the inside and outside diameters = -^ = 10. 10 X 3.1416 = 31.416 inches, the length of the center line. The radius of the outside circle = 6 inches; of the inside circle = 4 inches ; therefore, the diameter of the cross-section on the line A B = 2 inches. Then, 2 2 X .7854 = 3.1416 sq. in., the area of the imaginary cross- section. And 3.1416 X 31.416 = 98.7 cu. in., the volume. Ans. EXAMPLES FOR PRACTICE. 1. What is the volume of a sphere 30 inches in diameter ? Ans. 14, 137. 2 cu. in 2. How many square inches in the surface of the above sphere ? Ans. 2,827.44 sq. ia 3. Required the area of the convex surface of a circular ring, the outside diameter of the ring being 10 inches and the inside diameter 7fc inches. Ans. 107.95 sq. in. 4. Find the cubical contents of the ring in the last example. Ans. 33.73 cu. in. 5. The surface of a sphere contains 314.16 square inches. What K the volume of the sphere ? Ans. 523.6 cu. in. PRINCIPLES OF MECHANICS. MATTER AOT) ITS PROPERTIES. DEFINITION OF MECHANICS. 1. Mechanics is that science which treats of the action of forces on bodies and the effects that they produce ; it treats of the laws that govern the movement and equilibrium of bodies and shows how they may be applied. MATTER. 2. Matter is anything that occupies space. It is the substance of which all bodies consist. Matter is composed of molecules and atoms. 3. A molecule is the smallest portion of matter that can exist without changing its nature. 4. An atom is an indivisible portion of matter. Atoms unite to form molecules, and a collection of mole- cules forms a mass or body. A drop of water may be divided and subdivided until each particle is so small that it can only be seen by the moat powerful microscope, but each particle will still be water. Now, imagine the division to be carried on still further, until a limit is reached beyond which it is impossible to go without changing the nature of the particle. The particle of water is now so small that, if it be divided again, it will For notice of the copyright, see page immediately following the title page. 2 PRINCIPLES OF MECHANICS. 4 cease to be water, and will be something else; we call this particle a molecule. If a molecule of water be divided, it will yield 2 atoms of hydrogen gas and 1 of oxygen gas. If a molecule of sul- phuric acid be divided, it will yield 2 atoms of hydrogen, 1 of sulphur, and 4 of oxygen. 5. Bodies are composed of collections of molecules. Matter exists in three conditions or forms : solid, liquid, and gaseous. 6. A solid "body is one whose molecules change their relative positions with great difficulty; as iron, wood, stone, etc. 7. A liquid "body is one whose molecules tend to change their relative positions easily. Liquids readily adapt them- selves to the shape of vessels that contain them, and their upper surface always tends to become perfectly level. Water, mercury, molasses, etc., are liquids. 8. A gaseous body, or gas, is one whose molecules tend to separate from one another; as air, oxygen, hydrogen, etc. Gaseous bodies are sometimes called aeriform, (air-like) bodies. They are divided into two classes: the so-called "permanent" gases and vapors. 9. A permanent gas is one that remains a gas at ordi- nary temperatures and pressures. 10. A vapor is a body that at ordinary temperatures is a liquid or solid, but when heat is applied, becomes a gas, as steam. One body may, under different conditions, exist in all three states; as, for example, mercury, which at ordinary temperatures is a liquid, becomes a solid (freezes) at 40 below zero, and a vapor (gas) at 600 above zero. By means of great cold, all gases, even hydrogen, have been liquefied, and many solidified. By means of heat, all solids have been liquefied, and a great many vaporized. It is probable that, if we had the means of producing sufficiently great extremes of heat and 4 PRINCIPLES OF MECHANICS. 3 cold, all solids might be converted into gases, and all gases into solids. 1 1. Every portion of matter possesses certain qualities called properties. Properties of matter are divided into two classes : general and special. 12. General properties of matter are those that are common to all bodies. They are as follows: Extension, impenetrability, weight, indestructibility, inertia, mobility, divisibility, porosity, compressibility, expansibility, and elasticity, 13. Extension is the property of occupying space. Since all bodies must occupy space, it follows that extension is a general property. 14. By Impenetrability we mean that no two bodies can occupy exactly the same space at the same time. 15. Weight is the measure of the earth's attraction upon a body. All bodies have weight. In former times it was supposed that gases had no weight, since, if unconfmed, they tend to move away from the earth, but, nevertheless, they will finally reach a point beyond which they cannot go, being held in suspension by the earth's attraction. Weight is measured by comparison with a standard. The standard is a bar of platinum weighing 1 pound, owned and kept by the Government. 16. Inertia means that a body cannot put itself in motion nor bring itself to rest. To do either it must be acted upon by some force. 17. Mobility means that a body can be changed in position by some force acting upon it. 18. Divisibility is that property of matter by virtue^ of which a body may be separated into parts. 19. Porosity is the term used to denote the fact that there is space between the molecules of a body. The mole- cules of a body are supposed to be spherical, and, hence, there is space between them, as there would be between 4 PRINCIPLES OF MECHANICS. 4 peaches in a basket. The molecules of water are larger than those of salt ; so that when salt is dissolved in water, its molecules wedge themselves between the molecules of the water, and, unless too much salt is added, the water will occupy no more space than it did before. This does not prove that water is penetrable, for the molecules of salt occupy the space that the molecules of water did not. Water has been forced through iron by pressure, thus proving that iron is porous. 20. Compressibility is a natural consequence of poros- ity. Since there is space between the molecules, it is evi- dent that by means of force (pressure) they can be brought closer together, and thus the body be made to occupy a smaller space. 21. Expansibility is the term used to denote the fact that the molecules of a body will, under certain conditions (when heated, for example), move farther apart, and so cause the body to expand, or occupy a greater space. 22. Elasticity is that property of matter which enables a body when distorted within certain limits to resume its original form when the distorting force is removed. Glass, ivory, and steel are very elastic, clay and putty in their natural state being very slightly so. 23. Indestructibility is the term used to denote the fact that we cannot destroy matter. A body may undergo thousands of changes, be resolved into its molecules, and its molecules into atoms, which may unite with other atoms to form other molecules and bodies entirely different in appearance and properties from the original body, but the same number of atoms remain. The whole number of atoms in the universe is exactly the same now as it was millions of years ago, and will always be the same. Matter is indestructible. 24. Special properties are those that are not pos- sessed by all bodies. Some of the most important are as 4 PRINCIPLES OF MECHANICS. 5 follows: hardness, tenacity, brittleness, malleability, and ductility. 25. Hardness. A piece of copper will scratch a piece of wood, steel will scratch copper, and tempered steel will scratch steel in its ordinary state. We express all this by saying that steel is harder than copper, and so on. Emery and corundum are extremely hard, and the diamond is the hardest of all known substances. It can only be polished with its own powder. 26. Tenacity is the term applied to the power with which some bodies resist a force tending to pull them apart. Steel is very tenacious. 27. Brittleness. Some bodies possess considerable power to resist either a pull or a pressure, but they are easily broken when subjected to shocks or jars; for exam- ple, good glass will bear a greater compressive force than most woods, but may be easily broken when dropped upon a hard floor; this property is called brittleness. 28. Malleability is that property which permits of some bodies being hammered or rolled into sheets. Gold is the most malleable of all substances. 29. Ductility is that property which enables some bodies to be drawn into wire. Platinum is the most ductile of all substances. MOTION AOT) VELOCITY. DEFINITIONS. 3O. Motion is the opposite of rest and indicates a chang- ing of position in relation to some object which is for that purpose regarded as being fixe'd. If a large stone is rolled down hill, it is in motion in relation to the hill. If a person is on a railway train and walks in the oppo- site direction from that in which the train is moving, and with the same speed, he will be in motion as regards the H. 8. I. 11 6 PRINCIPLES OF MECHANICS. 4 train, but at rest with respect to the earth, since, until he gets to the end of the train, he will be directly over the spot at which he was when he started to walk. 31. The path, of a body In motion is the line described by a certain point in the body called its center of gravity. No matter how irregular the shape of the body may be, nor how many turns and twists it may make, the line that indi- cates the direction of this point for every instant that it is in motion is the path of the body. 32. Velocity is rate of motion. It is measured by a unit of space passed over in a unit of time. When equal spaces are passed over in equal times, the velocity is said to be uniform. If the flywheel of an engine keeps up a constant speed of a certain number of revolutions per minute, the velocity of any point is uniform. A railway train having a constant speed of 40 miles per hour moves 40 miles every hour, or -- = ! mile every minute, and since equal spaces are passed over in equal times, the velocity is uniform. 33. Variable Velocity. When a body moves in such a way that the spaces passed over in equal periods of time are unequal, its velocity is said to be variable. 34. The rate of motion of a body with a variable veloc- ity may increase or decrease at a uniform rate. When the velocity varies in either of these ways, the body is said to have a uniformly varying velocity. The most familiar example of uniformly varying velocity is a falling weight. Suppose a stone is dropped from a high bridge. It starts from a state of rest with no velocity, but its velocity constantly increases until it strikes. Its increase in velocity during any equal units of time is nearly the same. Thus, at the end of the first second its velocity will have increased from to a rate of 32.16 feet per second, nearly; during the next second the velocity will have increased by the same amount, making the velocity at the end of the second second 64. 32 feet. At the end of the third 4 PRINCIPLES OF MECHANICS. 7 second a like increase will have taken place, and the veloc- ity will then be 3 X 32.16 = 96.48 feet per second. 35. The change in the velocity of a body during a period of time is called its acceleration for that period. Thus, in the case of the falling weight just considered, the change of 32.16 feet per second that takes place in its velocity during each second is its acceleration in feet per second for each second considered. If the period of time considered had been 2 seconds, the acceleration would have been the increase in velocity during this time, that is, 64.32 feet per second for each 2 seconds considered. 36. The change in velocity may be from a higher to a lower rate. Thus, when a stone is thrown upwards, it leaves the hand with a given velocity; its upward motion is con- stantly resisted by the force of gravity and the resistance of the air, and in consequence of these resistances, it moves slower and slower until it finally stops and begins to return to the earth. A change in velocity of this kind is some- times called retardation. 37. The mean or average velocity of a body moving with a variable velocity can only be given for a stated period of time, and is numerically equal to the uniform velocity that will take the body over the same distance in the same time. RULES FOR VELOCITY PROBLEMS. 38. Uniform and Average Velocity. Let d= distance; / = time; v = velocity. Riile 1. To find the uniform or the average velocity that a body must have to pass over a certain distance or space in a given time, divide the distance by the time. Or, v= d 7 . 8 PRINCIPLES OF MECHANICS. 4 EXAMPLE 1. The piston of a steam engine travels 3,000 feet in 5 minutes ; what is its average velocity in feet per minute ? SOLUTION. Here 3,000 feet is the distance, and 5 minutes is the time. Applying the rule, 3,000 H- 5 = 600 ft. per min. Ans. CAUTION. Before applying the above or any of the succeeding rules, care must be taken to reduce the values given to the denomi- nations required in the answer. Thus, in the above example, if the velocity is required in feet per second instead of in feet per minute, the 5 minutes must be reduced to seconds before dividing. The oper- ation will then be: 5 minutes = 5 X 60 = 300 seconds. Applying the rule, 3,000 -H 300 = 10 ft. per sec. Ans. If the velocity is required in inches per second, it is necessary to reduce the 3,000 feet to inches and the 5 minutes to seconds, before dividing. Thus, 3,000 feet X 12 = 36,000 inches. 5 minutes X 60 = 300 seconds. Now applying the rule, 36,000 -f- 300 = 120 in. per Sfec. Ans. EXAMPLE 2. A railroad train travels 50 miles in l hours ; what is its average velocity in feet per second ? SOLUTION. Reducing the miles to feet and the hours to seconds, 50 miles X 5,280 = 264,000 feet. H hours X 60 X 60 = 5,400 seconds. Applying the rule, 264,000 -*- 5,400 = 48| ft. per sec. Ans. 39. If the uniform velocity (or the average velocity) and the time are given, and it is required to find the distance that a body having the given velocity will travel in the given time : Rule 2. Multiply the velocity by the time. Or, d=vt. EXAMPLE 1. The velocity of sound in still air is 1,092 feet per second; how many miles will it travel in 16 seconds ? SOLUTION. Reducing the 1,092 feet to miles', 1,092 H- 5,280 = \\\\. Applying the rule, X 16 = 3.31 mi., nearly. Ans. EXAMPLE 2. The piston speed of an engine is 11 feet per second, how many miles does the piston travel in 1 hour and 15 minutes ? SOLUTION. 1 hour and 15 minutes reduced to seconds = 4,500 seconds = the time. 11 feet reduced to railes = ^STF mile = velocity in miles per second. Applying the rule, ^H^ X 4,500 = 9.375 mi. Ans. 40. If the distance through which a body moves is given, and also its average or uniform velocity, and it is desired to know how long it takes the body to move through the given distance: 4 PRINCIPLES OF MECHANICS. 9 Rule 3. Divide the distance, or space passed over, by the velocity. '.- EXAMPLE 1. Suppose that the radius of the crank of a steam engine is 15 inches and that the shaft makes 120 revolutions per minute; how long will it take the crankpin to travel 18,849.6 feet ? SOLUTION. Since the radius, or distance from the center of the shaft to the center of the crankpin, is 15 inches, the diameter of the circle it moves in is 15 inches X 2 = 30 inches = 2.5 feet. The circum- ference of this circle is 2.5 X 3.1416 = 7.854 feet. 7.854 X 120 = 942.48 feet = distance that the crankpin travels in 1 minute = velocity in feet per minute. Applying the rule, 18,849.6 -H 942.48 = 20 min. Ans. EXAMPLE 2. A point on the rim of an engine flywheel travels at the rate of 150 feet per second ; how long will it take it to travel 45,000 feet ? SOLUTION. Applying the rule, 45,000 -f- 150 = 300 sec. = 5 min. Ans. EXAMPLES FOR PRACTICE. 1. A locomotive has drivers 80 inches in diameter. If they make 293 revolutions per minute, what is the velocity of the train in (a) feet- per second ? (V) miles per hour ? j () 102.277 ft. per sec. IS '1() 69. 734 mi. per hr. 2. Assuming the velocity of steam as it enters the cylinder to be 900 feet per second, how far can it travel, if unobstructed, during the time the flywheel of an engine revolves 7 times, if the number of revolutions per minute are 120 ? Ans. 3,150 ft. 3. The average speed of the piston of an engine being 528 feet per minute, how long will it take the piston to travel 4 miles ? Ans. 40 min. 4. A speed of 40 miles per hour equals how many feet per second ? Ans. 58f ft. 5. The earth turns around once in 24 hours. If the diameter be taken as 8,000 miles, what is the velocity of a point on the equator, in miles per minute ? Ans. 17.45 mi. per min. 6. The stroke of an engine is 28 inches. If the engine makes 11,400 strokes per hour, (a) what is its speed in feet per minute ? () how far will this piston travel in 11 minutes ? Ans -i (a} 448 * ft ' per min ' \ (b) 4,876 ft. 8 in. 10 PRINCIPLES OF MECHANICS. 4 FORCE. GENERAL PRINCIPLES. 41. Force is known only by the effects it produces on matter. Forces cannot, therefore, be compared in the same way that quantities of matter or distances are compared ; the only way in which forces can be examined in reference to one another is by noting their relative effects in the production of motion or a change of state in matter. The most familiar conception of force is that of a push or pull that tends to produce or destroy motion. If the force is great enough, its effect is seen in a change in the state of motion of the body on which it acts; that is, it either produces motion in the body or destroys some or all of the motion already existing. If, however, the body acted on by a force is so situated that the force applied to it is opposed by a resisting force of equal magnitude, no change in motion is produced. In this case the force is not perceiv- able, unless some other force is introduced whose effects will reveal the existence of the first force. 42. Forces are called by various names according to the ways in which they manifest themselves. Manifestations of force are: attraction, repulsion, coJiesion, adhesion, accelera- tion, retardation, resistance, etc., and the forces producing these manifestations are called attractive, repulsive, cohesive, adhesive, accelerating, retarding, resisting, etc. forces. 43. Comparison of Forces. In considering the effects of a force on a body, some standard of comparison must be used. The standard most commonly adopted in English- speaking countries is the pound, which is the force required to raise a standard mass of matter from the ground under certain specified conditions. In practice, force is always regarded as a pressure; that is, a force is considered the equivalent of the pressure exerted by a weight. For example, the effect of a force of 4 PRINCIPLES OF MECHANICS. 11 20 pounds acting upon a body is the same as the pressure of 20 pounds exerted by a weight of 20 pounds. 44. In order that we may compare the effect of a force on a body with that of another force on another body, it is necessary that the following three conditions be fulfilled in regard to both bodies: 1. The point of application, or point at which the force acts upon the body, must be known. 2. The direction of the force, or, what is the same thing, the straight line along which the force tends to move the point of application, must be known. 3. The magnitude or value of the force in comparison with the given standard must be known. 45. Reduction of Forces. When a number of forces act upon a body and produce a certain effect, it is often necessary to find a single force that, when substituted for the given forces, will produce the same effect. In order to find the point of application, the direction, and the magni- tude of this single force, it is necessary to know the above three conditions of every one of the given forces. NEWTON'S L.AWS OF MOTION. 46. The fundamental principles of the relations between force and motion were first stated by Sir Isaac Newton, and are called Newton s Three Laws of Motion. They are as follows: 1. All bodies continue in a state of rest, or of uniform motion in a straight /me, unless acted upon by some external force that compels a change. 2. Every motion or change of motion is proportional to the acting force, and takes place in the direction of the straight line along which the force acts. 3. To every action there is always opposed an equal reaction. These laws in their accepted form, as just given, have been more or less indirectly derived from experiment, but 12 PRINCIPLES OF MECHANICS. 4 they are so comprehensive as to defy complete experimental verification. 47. Exemplification of the First Law. In the first law of motion, it is stated that a body once set in motion by any force, no matter how small, will move forever in a straight line, and always with the same velocity, unless acted upon by some other force that compels a change. It is not pos- sible to actually verify this law, on account of the earth's attraction for all bodies, but, from astronomical observa- tions, we are certain that the law is true. This law is often called the law of inertia. 48. The word inertia is so abused that a full under- standing of its meaning is important. Inertia is not a force, although it is often so called. If a force acts upon a body and puts it in motion, the effect of the force is stored in the body, and a second body, in stopping the first, will receive a blow equal in every respect to the original force, assuming that there has been no resistance of any kind to the motion of the first body. It is dangerous for a person to jump from a fast-moving train, for the reason that, since his body has the same veloc- ity as the train, it has the same force stored in it that would cause a body of the same weight to take the same velocity as the train, and the effect of a sudden stoppage is the same as the effect of a blow necessary to give the person that velocity. By " bracing " himself and jumping in the same direction that the train is moving, and running, he brings himself gradually to rest, and thus reduces the danger. If a body is at rest, it must be acted upon by a force in order to be put in motion, and, no matter how great the force may be, it cannot be instantly put in motion. The resistance thus offered to being put in motion is com- monly, but erroneously, called the resistance of inertia. It should be called the resistance due to inertia. 49. Exemplification of the Second !Law. From the second law, we see that if two or more forces act upon a PRINCIPLES OF MECHANICS. 18 body, their final effect on the body will be in proportion to their magnitudes and to the directions in which they act. Thus, if the wind is blowing due west with a velocity of 50 miles per hour, and a ball is thrown due north with the same velocity, or 50 miles per hour, the wind will car- ry the ball west while the force of the throw is carrying it north, and the combined effect will be to cause it to move northwest. The amount of depar- ture from due north will be proportional to the force of the wind and independent of the ve- locity due to the force of the throw. 5O. In Fig. 1 a ball e is supported in a cup, the bottom of which is attached to the lever o in such a manner that o will swing the bot- tom horizontally and allow the ball to drop. Another ball b rests in a horizontal groove that is provided with a slit in the bottom. A swing- ing arm is actuated by the spring d in such a manner that, when drawn back, as shown, and then released, it will strike the lever o and the ball b at the same time. This gives b 14 PRINCIPLES OF MECHANICS. 4 an impulse in a horizontal direction, and swings o so as to allow e to fall. On trying the experiment, it is found that b follows a path shown by the curved dotted line, and reaches the floor at the same instant as e, which drops vertically. This shows that the force that gave the first ball its horizontal movement had no effect on the vertical force that com- pelled both balls to fall to the floor; the vertical force pro- duces the same effect as if the horizontal force had not acted. The second law may also be stated as follows: A force has the same effect in producing motion, whether it acts upon a body at rest or in motion, and whether it acts alone or with other forces. 51. Exemplification of the Third Law. The third law states that action and reaction are equal. A man can- not lift himself by his boot straps for the reason that he presses downwards with the same force that he pulls upwards; the downward reaction equals the upward action, and is opposite to it. In springing from a boat, we must exercise caution or the reaction will drive the boat from the shore. When we jump from the ground, we tend to push the earth from us, while the earth reacts and pushes us from it. EXAMPLE. Two men pull on a rope in opposite directions, each exerting a force of 100 pounds; what is the force that the rope resists ? SOLUTION. Imagine the rope to be fastened to a tree, and that one man pulls with a force of 100 pounds. The rope evidently resists 100 pounds. According to Newton's third law, the reaction of the tree is also 100 pounds. Now, suppose the rope to be slacked, but that one end is still fastened to the tree ; the second man then takes hold of the rope near the tree, and pulls with a force of 100 pounds, the first man pulling as before. The resistance of the rope is 100 pounds, as before, since the second man merely takes the place of the tree. He is obliged to exert a force of 100 pounds to keep the rope from slipping through his fingers. If the rope is passed around the tree, and each man pulls an end with a force of 100 pounds in the same and parallel directions, the stress in the rope is 100 pounds, as before, but the tree must resist the pull of both men, or 200 pounds. 4 PRINCIPLES OF MECHANICS. 15 52. Dynamics, also called kinetics, is that branch of mechanics that treats of forces and their effects when they produce a change in motion in the bodies on which they act. 53. Statics is that branch of mechanics that treats of forces and their effects when they do not produce a change in motion in the bodies on which they act. GRAVITATION AND WEIGHT. 54. Every body in the universe exerts a certain attract- ive force on every other body, which tends to draw the two together. To scientists, this attractive force is known as gravitation. If a body is held in the hand, a downward pull is felt, and if the hold is loosened, the body will fall to the ground. This pull, which we commonly call weight, is the attrac- tion between the earth and the body. 55. The attraction between the earth and bodies at or near its surface is denoted by the term force of gravity. This attraction is generally considered as acting along the line joining the center of gravity of the body and the center of the earth. By center of gravity is meant that point of a body at which its whole weight may be assumed to be concentrated. 56. The weight of a body is directly proportional to the force of gravity. From this it follows that the weight of a body can only be uniform everywhere if the force of gravity is uniform. As a matter of fact, the force of gravity varies in different locations; consequently, the weight of the body is not the same at all points on the surface of the earth. This has been conclusively shown by sensitive spring balances. ACCELERATING AND RETARDING FORCES. 57. According to the first law of motion, if a body is set in motion by a force and the force then ceases to act, the body will continue to move at the rate it had at the 16 PRINCIPLES OF MECHANICS. 4 instant the action of the force was discontinued, unless acted upon by some other force. If, however, a force acts upon a body for a given period of time, say 1 second, and imparts to it a certain amount of motion, and then, instead of ceasing to act, acts with the same intensity during the next second, it will impart to the body an increase in velocity equal to the velocity imparted during the first second. Then, the velocity at the end of the second second will be twice that at the end of the first. During the third second a like increase in velocity will be produced, making the velocity at the end of the third sec- ond three times as great as at the end of the first. This uniform increase in velocity will continue as long as the constant force continues to act on the body. A constant force when producing a constant acceleration is called a constant accelerating force. 58. If a constant force is applied to a body in motion in such a manner that it opposes the motion, its effect will be to reduce the motion by a certain amount, which will be the same for each second during which it acts. In this case, the force is called a constant retarding force. We thus see that the effect of a constant force acting upon a body in motion is to produce a uniform acceleration or retardation in the velocity of motion of the body, it being assumed that the motion of the body is not opposed by varying resisting forces. 59. Acceleration Due to the Force of Gravity. If a body falls freely under the action of the force of gravity, its velocity will increase at a uniform rate; in other words, it will be accelerated. Since the force of gravity varies in different localities, it follows that the acceleration produced by it is not everywhere the same. The greatest range in the acceleration due to the force of gravity in the United States is from a minimum of about 32.089 feet per second up to a maximum of about 32.186 feet per second for each second. In the latitude of Scranton, Pa., and at the level of the sea, the acceleration is nearly 32.16 feet per second; 4 PRINCIPLES OF MECHANICS. 17 this value will be used in all calculations in this Course that involve the use of acceleration due to the force of gravity. In accordance with the practice of most scientific writers, we will denote the acceleration due to the force of gravity by the letter g. 60. Mass. If the weight of a body at any place, as determined by a spring balance, is divided by the accelera- tion due to the force of gravity at that place, a numerical value will be obtained that, for the same body, will be the same wherever it may be weighed. This quotient is called the mass of the body, and is generally designated by the letter m. Rule 4. To find the mass of a body, divide its weight by the acceleration due to the force of gravity. Let W= the weight of a body; g = acceleration due to gravity; m = mass of the body. Then, m . g EXAMPLE. What is the mass of a body weighing 96.48 pounds ? q/> AQ SOLUTION. Applying the rule, m = ^TQ = 3. Ans. 61. Law of Gravitation. The attractive force by which one body tends to draw another body toward it is directly proportional to its mass and inversely proportional to the square of the distance between their centers of gravity. 62. Laws of Weight. 1. Bodies weigh most at the surface of the earth. Below the surface, the weight decreases directly as the distance to the center of the earth decreases. * 2. Above the surface, the weight decreases inversely as the square of the distance. 63. Change in Motion of a Body. A change in the motion of a body cannot take place without the action of an accelerating or retarding force. The force required to 18 PRINCIPLES OF MECHANICS. 4 produce a given acceleration or retardation in a body is given by the following rule, where . f = force in pounds ; a = acceleration or retardation in feet per second. Rule 5. Multiply the mass of the body by the accelera- tion, or retardation, in feet per second. Or, f= ma. W W Since m = (see rule 4), this may also be written/" = a. EXAMPLE. What force will be required to give a body weighing 90 pounds an acceleration of 5 feet per second ? SOLUTION. Applying rule 5, we get QA 5 = 13.99+ lb., say 14 lb. Ans. 64. According to the first law of motion, a body in motion not acted upon by any external force will continue its motion without any further application of a force. In practice, however, the motion of a body is always opposed by some resisting force or forces. According to the third law of motion, the force required to overcome the resistance is equal to the resistance. The opposing forces are usually constant, or nearly so. Taking the opposing forces into account, the actual force required to accelerate a body meeting with resistance will be the sum of the accelerating force and the opposing forces. ILLUSTRATION. Imagine a weight of 321.6 pounds to be lying on a smooth plane surface. Assume that it has been determined experi- mentally that a force of 100 pounds is required to be exerted con- tinually to overcome the friction between the weight and the surface. What force will be required to produce an acceleration of 2 feet per second ? By rule 5, the accelerating force is ^-f^ X 2 = 20 pounds. As a force of 100 pounds must be exerted continually to overcome the resist- ance due to friction, a force of 100 + 20 = 120 pounds will be required to produce the required acceleration. 65. The question is often asked, what force is required to start a flywheel and keep it going at a stated number of 4 PRINCIPLES OF MECHANICS. 19 revolutions per minute ? This question, or similar questions, cannot be answered without knowing the time in which the flywheel is to attain the given speed. An accelerating force depending on the mass of the wheel, its diameter, the distribution of the material of which it is made, the time of acceleration, and the constant resistances, is required to bring the wheel up to its speed. When this speed has been attained, the force required to keep it going will be that required to overcome the frictional and air resistances. MOMENTUM. 66. Experience teaches us that the same force acting upon bodies of different weights produces different effects. For example, if a given force imparts a velocity of 10 feet per second in a certain time to a body weighing 1 pound, we know from experience and observation that it cannot impart the same velocity in the same time when acting upon a body weighing 1,000 pounds. Scientists have shown that the velocity imparted to a body in a given time by a force varies directly as the force and inversely as the mass of the body. Hence, forces may be compared with one another by comparing their effects in imparting velocities to bodies whose masses are known. 67. The product obtained by multiplying the mass of a body by its velocity in feet per second is called the momen- tum of the body ; it represents the magnitude of the force that will produce the given velocity in the body in 1 second. Hence, we may call momentum the time effect of a force. 68. According to the third law of motion, action and reaction are equal to each other. Consequently, if the force required to produce a stated momentum in a given time is known, it is likewise known what force is required to destroy this momentum, or to bring the body to rest, in an equal period of time. When a liquid body is flowing in a stream, as from a nozzle, the weight to be considered in problems involving 20 PRINCIPLES OF MECHANICS. 4 momentum is the weight of the liquid discharged in 1 second. For example, let it be required to estimate the force with which a man must hold the nozzle of a fire hose to prevent its slipping through his hands when a stream of water issues from it with a velocity of 20 feet per second, the area of the opening in the nozzle being 1 square inch and the weight of a cubic inch of water .0361 pound. The volume of water discharged in 1 second is 1 X 12 X 20 = 240 cubic inches, and since the weight of 1 cubic inch is .0361 pound, the weight discharged per second is 240 X .0361 = 8.664 pounds. The momentum of the stream is -^ ^ X 20 = 5.38 pounds. o*.lo , This is the constant force required to give a body of water weighing 8.664 pounds a velocity of 20 feet per second, in 1 second ; it also represents the magnitude of the reaction on the nozzle, and the man must hold the nozzle with a force equal to the reaction, or 5.38 pounds, in order to prevent its slipping through his hands. WORK, POWER, AOT> WORK. 69. Work is the overcoming of resistance continually occurring along the path of motion. Motion in itself is not work ; a force must overcome a resistance in order that work may be done. 70. Unit of Work. The unit by which the work done by a force is measured is the work done in overcoming a resistance of 1 pound through a space of 1 foot; this unit is called a foot-pound. According to the definition, it may be considered as the force required to raise 1 pound 1 foot vertically. All work is measured by this standard. A horse going up a hill does an amount of work equal to its own weight, plus the weight of the wagon and its con- tents, plus the frictional resistances reduced to an equivalent 4 PRINCIPLES OF MECHANICS. 21 weight, multiplied by the vertical height of the hill. Thus, if the horse weighs 1,200 pounds, the wagon and contents 1,200 pounds, and the frictional resistances equal 400 pounds, then if the vertical height of the hill is 100 feet, the work done is equal to (1,200 + 1,200 + 400) X 100 = 280,000 foot- pounds. Rule 6. In all cases the force (or resistance] multiplied by the distance through wJncJi it acts equals the work. If a weight is raised, the weight multiplied by the vertical height of the lift equals the work. 71. The total amount of work done in overcoming a given resistance through a given distance is independent of time; that is, it. is immaterial whether it takes 1 minute or 1 year in which to do it ; but in order to compare the rate at which work is done by different machines with a common standard, time must be considered. If one machine does a certain amount of work in 10 minutes and another machine does exactly the same amount of work in 5 minutes, the second machine can do twice as much work as the first in an equal period of time. POWER. 72. Power is a term used to denote the rate at which work is done. 73. The common unit used for expressing the rate at which work is done is the horsepower. One horsepower is 33,000 foot-pounds of work per minute; in other words, it is 33,000 pounds raised vertically 1 foot in 1 minute, or 1 pound raised vertically 33,000 feet in 1 min- ute, or any combination that will, when multiplied together, give 33,000 foot-pounds in 1 minute. Thus, the work done in raising 110 pounds vertically 5 feet in 1 second is a horsepower; for, since in 1 minute there are 60 seconds, 110 X 5 X 60 = 33,000 foot-pounds in 1 minute. EXAMPLE. -In a steam engine the force impelling the piston for- wards and backwards is 10,000 pounds. This force overcomes the H. 8. /. 12 22 PRINCIPLES OF MECHANICS. 4 resistance due to the load at the rate of 600 feet per minute ; that is, it moves the piston back and forth at that rate. What is the horsepower of the engine ? SOLUTION. According to rule 6, the work done is 10,000 X 600 = 6,000,000 foot-pounds per minute. Then, as 33,000 foot-pounds per minute represent a horsepower, the horsepower of the engine is 6,000,000 H- 33,000 = 181.818. Ans. ENERGY. 74. Energy is a term used to express the ability of an agent to do work. 75. Kinetic Energy. If we have a body at rest, a cer- tain amount of force must be exerted and a certain amount of work must be done to set it in motion. A part of this force will be required to overcome those resistances outside of the body, such as friction and the resistance of the air, that oppose the motion of all bodies with which we have to do ; another part acts to overcome the inertia of the body, to start it from its state of rest, and give it motion (see Newton's first law). The force that overcomes the resist- ance due to the inertia of the body does work, and the work so performed is stored in the body; in being brought to rest, the body is capable of overcoming a resistance and of doing an amount of work exactly equal to the work expended in giving it motion. The ability that the moving body has to do work while being brought to rest is called the kinetic energy of the body. 76. Rule for the Energy of a Moving Body. Let w weight of body in pounds; v = its velocity in feet per second; E = kinetic energy in foot-pounds. Then, the kinetic energy of a moving body may be found as follows: Rule 7. Multiply the weight of the body by the square of its velocity and divide the product by twice the acceleration due to the force of gravity. 4 PRINCIPLES OF MECHANICS. 23 wv* Or, E = 64.32' Thus, if a weight is raised a certain height, an amount, of work is done equal to the product of the weight and the vertical height. If a weight is suspended at a certain height and allowed to fall, it will do the same amount of work in foot-pounds that was required to raise the weight to the height through which it fell. EXAMPLE 1. If a body weighing 25 pounds falls from a height of 100 feet, how much work can it do ? SOLUTION. Work = w h = 25 x 100 = 2,500 ft.-lb. Ans. It requires the same amount of work or energy to stop a body in motion within a certain time as it does to give it that velocity in the same length of time. EXAMPLE 2. A body weighing 50 pounds has a velocity of 100 feet per second; what is its kinetic energy ? SOLUTION. Kinetic energy = -^ = ^ = 7.773.63 ft.-lb. o4.o^ O4.OA . Ans. EXAMPLE 3. In the last example, how many horsepower will be required to give the body this amount of kinetic energy in 3 seconds ? SOLUTION. 1 horsepower = 33,000 pounds raised 1 foot in 1 minute. If 7,773.63 foot-pounds of work are done in 3 seconds, in 1 second there will be done ^ = 2,591.21 foot-pounds of work. 1 horse- power = 33,000 foot-pounds per minute = 33,000 -f- 60 = 550 foot-pounds per second. The number of horsepower required will be = 4.7113 H. P. An, 77. Potential energy is latent energy; it is the energy that a body at rest is capable of giving out under certain conditions. If a stone is suspended by a string from a high tower, it has potential energy. If the string is cut, the stone will fall to the ground, and during its fall its potential energy will change into kinetic energy, so that at the instant it strikes the ground its potential energy is wholly changed into kinetic energy 24 PRINCIPLES OF MECHANICS. 4 At a point equal to one-half the height of the fall, the potential and kinetic energies are equal. At the end of the first quarter the potential energy is three-fourths and the kinetic energy one-fourth; at the end of the third quarter the potential energy is one-fourth and the kinetic energy three-fourths. A pound of coal has a certain amount of potential energy. When the coal is burned, the potential energy is liberated and changed into kinetic energy in the form of heat. The kinetic energy of the heat changes water into steam, which thus has a certain amount of potential energy. The steam acting on the piston of an engine causes it to move through a certain space, thus overcoming a resistance, changing the potential energy of the steam into kinetic energy, and thus doing work. Potential energy, then, is the energy stored within a body that may be liberated and produce motion, thus generating kinetic energy and enabling work to be done. 78. The principle of conservation of energy teaches that energy, like matter, can never be destroyed. If a clock is put in motion, the potential energy of the spring is changed into kinetic energy of motion, which turns the wheels, thus producing friction. The friction produces heat, which dissipates into the sur- rounding air, but still the energy is not destroyed it merely exists in another form. 79. Work of Acceleration and Retardation. The theoretical amount of work that must be done in order to start a body from a state of rest and accelerate it until it reaches a given velocity is equal to the kinetic energy of the body at the given velocity. Likewise, the theoretical amount of work that must be done on a moving body to retard it and finally bring it to rest is equal to the kinetic energy the moving body possessed at the moment retarda- tion began. The work that must be done in changing the velocity of a body is equal to the difference in the kinetic energies at the initial and final velocities. Since the motion 4 PRINCIPLES OF MECHANICS. 25 of all bodies is opposed by some resisting force or forces, the actual amount of work required to give a body the given velocity will be the sum of the work of acceleration and the work required to overcome the outside resisting forces. EXAMPLE 1. A body weighing 1,000 pounds is started from rest and is to attain a velocity of 88 feet per second in 2 minutes, passing over a distance of 5,280 feet in that time. If a constant force of 120 pounds must be exerted to overcome the frictional resistances, what work must be done ? SOLUTION. According to this article, the work required to accelerate 1 000 v 88'^ the body is ^ = 120,398 foot-pounds. As a constant force of 120 pounds must act through a distance of 5,280 feet to overcome the frictional resistances, the work done in overcoming friction is 5,280 X 120 = 633,600 foot-pounds. Then, the total amount of work done is 120,398 + 633,600 = 753,998 ft.-lb. Ans. EXAMPLE 2. What horsepower will be required to do the work cal- culated in the last example ? SOLUTION. As the work is done in 2 minutes, the horsepower is 753,998 2 X 33,000 = 11.424 H. P., nearly. Ans. FORCE OF A BL.OW. 8O. The questions are frequently asked, with what force will a falling hammer strike, or with what force will a pro- jectile fired from a gun strike an object ? These questions catinot be answered directly, as they are based on a miscon- ception. A moving body possesses kinetic energy, or ability to do work, which ability can only be expressed in foot- pounds, but not in pounds of force, since the work done by the hammer or projectile in coming to rest is not a manifes- tation of force, but of energy. Work is the product of force into distance ; hence, if the amount of work a body has done or is capable of doing 'is known, the force can be determined for each case if, by some means, it is possible to determine exactly the distance in which the work is done. This distance depends on vari- ous resistances, such as that due to moving the object struck, the resistance to penetration, friction, the resistance 26 PRINCIPLES OF MECHANICS. 4 to shearing or deformation of the body, etc. The distance through which these resisting forces act is generally inde- terminate, and since the average of the resisting forces varies generally with the distance, this average resisting force is also indeterminate; hence, the force that, acting through a distance, will absorb all the kinetic energy of the hammer or projectile cannot be determined for the practical reasons given. COMPOSITION A1STD RESOLUTION OF FORCES. 81. According to Art. 44, in order that forces may be compared with one another, three conditions must be ful- filled. These conditions may all be repre- sented by a line ; hence, we may represent FIG - 2 - forces by lines. Thus, in Fig. 2, let A be the point of application of the force, let the length of the line A B represent its magnitude, and let the arrowhead indicate the direction in which the force acts; then, the line A B fulfils the three conditions and the force is fully represented. COMPOSITION OF FORCES. 82. When two or more forces act upon a body at the same time along lines that meet in a common point, their combined effect on the body may be obtained by an applica- tion of the principle of the triangle of forces. In Fig. 3 (a), let A and B be two forces having the mag- nitudes and directions represented by the two lines. To find .. A the effect due to the combined action of these two forces, draw in any convenient location a line parallel to either of the 4 PRINCIPLES OF MECHANICS. 27 lines representing the two forces, making it equal in length, to some scale, to the magnitude of the force. Mark upon it an arrowhead pointing in the same direction as the arrow- head on the line representing the force. Then, from one extremity of the line just drawn draw a line parallel to the line representing the second force and equal in length, to the same scale, to the magnitude of the second force, and mark the arrowhead upon it. It is essential that the second line be so drawn that when passing over the two lines with a pencil, commencing at the beginning of either force, the arrowheads will both point in the same direction in reference to the direction of motion of the pencil. The second line may be drawn from either end of the first line, but its direction must be made to fulfil the above abso- lutely essential condition. Thus, in Fig. 3 (6), the line B' has been drawn from the right-hand extremity of A'. Start- ing at a with a pencil and moving toward b, the pencil will move in the direction in which the arrowhead on A' points. Passing over B' from b to c, the pencil will move in the direction in which the arrowhead on B' points; we thus see that the lines are drawn correctly in reference to each other. In Fig. 3 (c), the line H'has been drawn from the left-hand extremity of A'. Starting at c and following up the lines, it is seen that in this case the arrowheads both point in the same direction relative to the direction of motion of the pencil, thus showing the lines to be located correctly. The two lines having been drawn, complete the triangle by drawing the line C. This line, called the resultant, represents the combined effect of the two forces; it gives the direction along which the two forces will act when com- bined. The magnitude of their combined effect is found by measuring this line by the scale with which A' and B' were laid off. The resultant will always have a direction opposite in sense to that of the forces; that is, if we pass a pencil around the triangle in the direction in which the arrowheads on the lines A' and B' point, the arrowhead on C, represent- ing the direction of action of the resultant, must point in a direction opposite to that in which the pencil moves. 28 PRINCIPLES OP MECHANICS. 4 83. In practice it is often desired to find not only the magnitude and direction, but also the actual location of the resultant, the magnitudes and lines of action of the two forces being known. (By location of the forces and result- ant is meant the location of the lines along which the forces actually act.) This can readily be done by producing the lines giving the location of the forces until they meet and then drawing the triangle of forces with their point of intersection as the starting point. ILLUSTRATION. In Fig. 4 is shown a head-frame erected at the mouth of a deep, vertical shaft. The hoisting rope that leads to the Scale 1=1000 Jfc FIG. 4. hoisting engine passes over a sheave at the top of the head-frame. A weight of 1,500 pounds hangs at the end of the rope. Neglecting the weight of the rope and sheave, what is the total pressure on the bear- ings of the sheave ? According to Art. 51, the stress in both parts of the hoisting rope is 1,500 pounds. The part A B of the rope supports the weight; 4 PRINCIPLES OF MECHANICS. 29 consequently, the force acting along A B is downwards. The force acting along CD is the pull exerted by the engine and is toward the engine. To find the resultant of these two forces, draw the triangle of forces. Choosing a scale of 1 inch per 1,000 pounds, draw the line E F parallel to CD, making it |{j{$ = 1.5 inches long. From F draw FG parallel to A B and jgg = 1.5 inches long. Join E and G; then EG will be the resultant whose magnitude is measured by the same scale. Measuring EG, it is found to be 2.75 inches long. As each inch repre- sents 1,000 pounds, the magnitude of the resultant is 2.75 X 1,000 = 2,750 pounds. To find the actual location of the resultant in reference to A B and CD, produce both lines, as shown in dotted lines, until they intersect at E'. Starting the triangle of forces at E', lay off E' F' 1.5 inches. From F' draw F' G' 1.5 inches and parallel to A B. Join E' G'. Upon measurement it is found to be 2,750 pounds. As inspection shows that the resultant passes through the center of the bearings, the total pressure on the bearings is 2,750 pounds. Ans. RESOLUTION OF FORCES. 84. Since two forces can be combined to form a single resultant force, we may also treat a single force as if it were the resultant of two forces whose action upon a body will be the same as that of a single force. Thus, in Fig. 5, the force OA may be resolved into two forces OandA. If the force O A acts upon a body moving or at rest upon a FIG - 5 - horizontal plane, and the resolved force O B is vertical and B A horizontal, OB, measured to the same scale as O A, is the magnitude of that part of O A that pushes the body downwards, while B A is the magnitude of that part of the force O A that is exerted in pushing the body in a horizontal direction. O B and B A are called the components of the force O A, and when these components are vertical and horizontal, as in the present case, they are called the verti- cal component and the horizontal component of the force O A. 30 PRINCIPLES OP MECHANICS. 4 85. It frequently happens that the position, magnitude, and direction of a certain force is known, and that it is desired to know the effect of the force in some direction other than that in which it acts. Thus, in Fig. 5, suppose that O A represents, to some scale, the magnitude, direction, and line of action of a force acting upon a body at A, and that it is desired to know what effect O A produces in the direction B A. Now B A, instead of being horizontal, as in the figure, may have any direction. To find the value of the component of O A which acts in the direction B A, we use the following rule: Rule 8. From one extremity of the line representing the given force, draw a line parallel to the direction in which it is desired that the component shall act / from the other extremity of the given force, draw a line perpendicular to the component first drawn, and intersecting it. The length of the component, measured from the point of intersection to the intersection of the component with the given force, will be the magnitude of the effect produced by the given force in the required direction. Thus, suppose O A, Fig. 5, represents a force acting upon a body resting upon a horizontal plane, and it is desired to know what vertical pressure O A produces on the body. Here the desired direction is vertical; hence, from one extremity, as O, draw OB parallel to the desired direc- tion (vertical in this case), and from the other extremity draw A B perpendicular to OB and intersecting OB at B. Then OB, when measured to the same scale as O A, will be the value of the vertical pressure produced by O A. 86. Tangential Pressure. One of the most familiar applications of the principle of the resolution of forces occurring in steam engineering is the case of the connecting- rod and crank. When the piston is at the end of its stroke and the crankpin is in a line drawn through the center of the cylinder and crank-shaft, a position that is expressed by saying the engine is "on the center," the pressure of the connecting-rod on the crankpin acts directly against the PRINCIPLES OF MECHANICS. 31 bearings of the shaft and there is no turning effect on the pin. After the pin leaves the center, the pressure exerted on it by the connecting-rod may be resolved into two components: One of these components acts in the direction of the center line PO of the crank (see Fig. 6) and merely Scale 1*200O lb. exerts a pressure on the bearings of the shaft; the other component acts along the line A B at right angles to the center line PO and tangent to the circle described by the crankpin. This is the force that tends to turn the crank and is called the tangential pressure on the crankpin. When the crank is on the center, there is no tangential pressure on the pin and no tendency to turn the crank. The tangential pressure gradually increases as the pin leaves the center and becomes greatest at the point where the con- necting-rod and crank are at right angles to each other; it then decreases until the next center is reached. At 'the position where the connecting-rod and crank are at right angles to each other, the tangential pressure on the pin is equal to the total pressure exerted on it by the connecting- rod, and there is no component in the direction of the center line PO of the crank. 32 PRINCIPLES OF MECHANICS. 4 EXAMPLE. If a force of 3,000 pounds acts along the connecting-rod a in the direction of the arrow (see Fig. 6), what is the tangential pres- sure on the crankpin ? SOLUTION. As the tangential pressure is the pressure perpendicu- lar 'to the crank, draw A B through the crankpin center at right angles to the center line of the crank ; A B then represents the line along which the tangential pressure acts. Then, in any convenient location, draw CD parallel to the connecting-rod. Choosing a scale of 2,000 pounds = 1 inch, make the line CD 3,000 -H 2,000 = 1| inches long. From D draw an indefinite line D E parallel to A B, and draw C E perpendicular to D E. Now ED, measured to the scale adopted, will be the magnitude of the tangential pressure for the posi- tion of the crank and connecting-rod shown in the figure. Upon measurement, it is found to be 1.3 inches long. Then 1.3 X 2,000 = 2,600 lb., the tangential pressure. Ans. 87. When the total pressure on the piston rod of a steam engine is known, the force acting along the connecting-rod and the force acting upon the guides can be determined by the following application of the principle of the resolution of force: Draw a line, as A B, Fig. 7, parallel to the line of motion of the crosshead a. Make its length, to some scale, equal in magnitude to the force impelling the piston. From one extremity of A B draw a line B C parallel to the center line of the connecting-rod b. From the other extremity of A B draw a line at a right angle to A B, producing it until it intersects the line B C at D. Then, B D represents the force acting along the connecting-rod and A D represents the force acting upon the guides. 4 PRINCIPLES OF MECHANICS. 33 FRICTION. 88. Friction is the resistance that a body meets with from the surface upon which it moves. 89. The ratio between the resistance to the motion of a body due to friction and the perpendicular pressure between the surfaces is called the coefficient of friction. If a weight W, as in Fig. 8, rests upon a horizontal plane, and has a cord fastened to it passing over a pulley , from 100 Ib. W 10 Ib. which a weight Pis suspended, then, if P is just sufficient p to start W, the ratio of P to W, or yr>, is the coefficient of friction between Wand the surface it slides upon. The weight W is the perpendicular pressure, and P is the force necessary to overcome the resistance to the motion of W due to friction. If W = WO pounds and P= 10 pounds, the p coefficient of friction for this particular case would be - - 100 9O. Laws of Friction. 1. Friction is directly proportional to the perpendicular pressure between the two surfaces in contact. 2. Friction is independent of the extent of the surfaces in contact when the total perpendicular pressure remains the same. 3. Friction increases with the roughness of the surfaces. 4. Friction is greater between surfaces of the same mate- rial than between those of different materials. 34 PRINCIPLES OF MECHANICS. 4 5. Friction is greatest at the beginning of motion. 6. Friction is greater between soft bodies than between hard ones. 7. Rolling friction is less than sliding friction. 8. Friction is diminished by polishing' or hibricating the surfaces. 91. Law 1 shows why the friction is so much greater on journals after they begin to heat than before. The heat causes the journal to expand, thus increasing the pressure between the journal and its bearing, and, consequently, increasing the friction. Law 2 states that, no matter how small may be the sur- face that presses against another, if the perpendicular pres- sure is the same the friction will be the same. Therefore, large surfaces are used where possible, not to reduce the friction, but to reduce the wear and diminish the liability of heating. For instance, if the perpendicular pressure between a journal and its bearing is 10,000 pounds, and the coefficient of friction is .2, the amount of friction is 10,000 X .2 = 2,000 pounds. Suppose that the area receiving the pres- sure is 80 square inches, then the amount of friction for each square inch is 2,000 -f- 80 = 25 pounds. If the area receiving the pressure had been 160 square inches, the friction would have been the same, that is, 2,000 pounds; but the friction per square inch would have been 2,000 -=- 160 = 12 pounds, just one-half as much as before, and the wear and liability to heat would be one-half as great also. 92. The values of the coefficient of friction given in the following tables are average values determined by General Morin, many years ago. Under certain conditions the coefficient may be considerably less than is given in the tables; it varies greatly, but the variation depends on so many conditions and the numerous experiments that have been made have given such contradictory results that no definite rules have yet been derived for determining the 4 PRINCIPLES OF MECHANICS. 35 exact values under any condition. The student is, there- fore, advised to use the values given in the tables, except where careful experiments have been made that give relia- ble values for the particular case under consideration. To find the force, in pounds, necessary to overcome friction, the coefficient taken from the table is multiplied by the perpendicular pressure, in pounds, on the surface consid- ered. If the force acts at an angle to the surface, the perpendicular force can be found by resolving the given force into two components, one perpendicular and the other parallel to the surface. TABLE I. COEFFICIENTS OF FRICTION FOR PLANE SURFACES. (Reduced from M. Morin's Data.) Description of Surfaces in Contact. State of the Surfaces. Coefficient of Friction. Wrought iron on cast iron. Slightly greasy .18 Wrought iron on bronze.. Slightly greasy .18 Cast iron on cast iron. . . . Slightly greasy .15 Cast iron on bronze Slightly greasy .15 Bronze on bronze Drv .20 Bronze on cast iron J Dry .22 Bronze on wrought iron. . Slightly greasy .16 Cast iron, wrought iron, 1 steel, and bronze sli- Ordinary lubrication ding on one another \ with lard, tallow, and .07-. 08 or sliding on them- oil selves. Cast iron, wrought iron, ' \r steel, and bronze sli- ding on one another , Continuous lubrication .05 or sliding on them- selves. PRINCIPLES OF MECHANICS. TABLE II. COEFFICIENTS OF FRICTION FOB JOURNAL FRICTION. (Reduced from M. Morin's Data.) Journals. Bearings. Lubricant. Coefficient of Friction. Ordinary Lubrication. Continuous Lubrication. Cast iron Cast iron Lard, olive oil, tallow .07-. 08 .03-.054 Cast iron Bronze Lard, olive oil, tallow .07-.08 .03-.054 Wrought iron Cast iron Lard, olive oil, tallow .07-. 08 .03. -054 Wrought iron Bronze Lard, olive oil, tallow .07-. 08 .03-. 054 Wrought iron Lignum vita? Lard, olive oil .11 Bronze Bronze Oil .10 Bronze Bronze Lard .09 93. In the case of a weight sliding along a horizontal plane surface, the pressure is equal to the weight. When the surface is inclined, 'the weight acts vertically down- wards, and the pressure perpendicular to the surface can be found by the principle of the resolution of forces. In many cases the pressure on the surfaces is due to the com- bined action of several forces that must be combined into one common resultant force. 94. The work that must be done in overcoming the resistance of friction depends on the distance through which the resistance is overcome. It may be calculated by the following rule: Rule 9. Multiply the total pressure in pounds by the dis- tance in feet and by the coefficient of friction. 4 PRINCIPLES OF MECHANICS. 37 Let W = work in foot-pounds; f coefficient of friction; / = total pressure in pounds ; d= distance in feet. Then, W = pdf. EXAMPLE. The average perpendicular pressure on the guide of a steam engine due to the force impelling the piston is 2,500 pounds. The pressure due to the weight of the crosshead and connecting-rod is 400 pounds. The crosshead moves at the rate of 500 feet per minute ; what horsepower is required to overcome the friction on the guides ? SOLUTION. The total perpendicular pressure is 2,500 + 400 = 2,900 pounds. Since the lubrication is usually intermittent, the coefficient of friction, for a brass slipper working on a cast-iron guide, may be taken as .08. The resistance being overcome through a distance of 500 feet each minute, the work done in overcoming friction is 2,900 X 500 X .08 = 116,000 foot-pounds per minute. Then, the horse- power is 116,000 -=- 33,000 = 3.52 H. P., nearly. Ans. 95. In the case of a shaft rotating in a bearing, the dis- tance through which friction is overcome each minute is found by multiplying the circumference of the journal by the number of revolutions per minute. For a shaft, or any other body rotating in a bearing, the force required to overcome friction, as calculated by multiplying the pres- sure by the coefficient of friction, is the force that must be applied at the surface of the journal. 96. Allowable Pressures. It has been found by expe- rience that when the pressure per unit of area exceeds a certain amount, the lubricant will be forced out and the bodies rubbing on each other will heat and, finally, seize. The pressures that can safely be allowed on the bearings of steam engines, on guides, thrust bearings, crankpins, crosshead pins, etc. vary considerably, being dependent to a large extent on the character of the workmanship, the degree of finish, the variation of the pressure, the character of the lubrication, and the quality of the lubricant. With fair workmanship, the following pressures per square inch represent average practice in steam-engine work: 11. 8. I. IS PRINCIPLES OF MECHANICS. TABLE III. PRESSURES PER SQUARE INCH ALLOWABLE IN STEAM-ENGINE \VORK. Engine Bearing. Slow-Speed Stationary Engines. Pounds. High-Speed Stationary and Marine Engines. Pounds. Crankpins iron .... 800) Crankpins steel 1 200 f 400 to 600 WVistpin 1 200 000 to 800 Main bearings Guides 200 100 200 100 Thrust bearings ... . 60 97. For crankpins, wristpins, and guides, the allowable pressures given represent the pressures corresponding to the maximum load, which in the case of a wristpin and crankpin occurs when the crank, connecting-rod, and piston rod are in a straight line, and in the case of guides, when the con- necting-rod and crank are at right angles to each other. In the case of pins and journals, the area to be considered in calculating the pressure on the bearing is the projected area, which is found by multiplying the length of the journal by its diameter. EXAMPLES FOR PRACTICE. 1. A body weighs 90 pounds ; what is its mass ? Ans. 2.799, nearly. 2. What force will be required to accelerate a body at the rate of 2 feet per second, the body weighing 450 pounds, and the frictional resistances being equal to 10 per cent, of the weight of the body ? Ans. 72.99 lb., nearly. 8. What work is done in raising 950 pounds 17 feet ? Ans. 16,150 ft.-lb. 4. If an engine does 205,000 foot-pounds of work per minute, what is its horsepower ? Ans. 6.21 H. P., nearly. PRINCIPLES OF MECHANICS. 39 5. What is the kinetic energy of a shell fired from a cannon with a velocity of 1,800 feet per second, the shell weighing 1,000 pounds? Ans. 50,373,135 ft.-lb., nearly. 6. Taking the coefficient of friction at .15, what horsepower will be required to pull 100 pounds at a uniform speed of 5 feet per second along a level surface ? Ans. .136 H. P. CENTER OF GRAVITY. 98. The center of gravity of a body is tJiat point at which the body may be balanced, or it is the point at which the whole weight of a body may be considered as concentrated. This point is not always in the body; in the case of a horseshoe or a ring it lies outside of the substance of, but within the space enclosed by, the body. In a moving body, the line described by its center of gravity is always taken as the path of the body. In finding the distance that a body has moved, the distance that the center of gravity has moved is taken. The definition of the center of gravity of a body may be applied to a system of bodies if they are considered as being connected at their centers of FIG. 9. gravity. If w and W, Fig. 9, are two bodies of known weight, their center of gravity will be at C. This point C may be readily determined as follows: Rule 1O. The distance of tlie common center of gravity- front the center of gravity of the large weight is equal to the weight of the smaller body multiplied by the distance between the centers of gravity of t/ie two bodies, and this product divided by the sum of the weights of the two bodies. EXAMPLE. In Fig. 9, w = 10 pounds, IV 30 pounds, and the dis- tance between their centers of gravity is 36 inches; where is the center of gravity of both bodies situated ? 40 PRINCIPLES OF MECHANICS. SOLUTION. Applying the rule, 10 X 36 = 360. 10 + 30 = 40. 360 -s- 40 = 9 in., distance of center of gravity from center of large weight. Ans. 99. It is now very easy to extend this principle to find the center of gravity of any number of bodies, when their weights and the distances apart of their centers of gravity are known, by the following rule : Rule 11. Find the center of gravity of two of the bodies, as W l and W < in Fig. 10. As- sume that the weight of both bodies is concen- trated at Cj, and find the center of gravity of this combined weight at C t and the weight of W^. Let it be at C^; then find the center of gravity of the combined weights of W^, W 4 , W^ (concentrated at C^), and W 3 . Let it be at C ; then C will be the center of gravity of the four bodies. 100. To find the center of gravity of any parallel- ogram: Rule 12. Draw the two diagonals, Fig. 11, and their point of intersection C will be the center of gravity. 1O1. To find the center of gravity of a triangle, as ABC, Fig. 12: 4 PRINCIPLES OF MECHANICS. 41 Rule 13. From any vertex, as A, draw a line to the mid- dle point D of the opposite side B C. From one of the other vertexes, as C, draw a line to F, the middle point of the opposite side A B ; the point of intersection O of these two lines is the center of gravity. It is also true that the distance D O \ D A and that F O = \ F C\ the center of gravity could have been found by drawing from any vertex a line to the middle point of the opposite side and measuring back from that side of the length of the line. The center of gravity of any regular plane figure is the same as the center of the inscribed or circumscribed circle. 1O2. To find the center of gravity of any Irregular plane figure, but of uniform thickness throughout, divide one of the parallel surfaces into triangles, parallelograms, circles, ellipses, etc., according to the shape of the figure; find the area and center of gravity of each part separately, and combine the centers of gravity thus found in the same 42 PRINCIPLES OF MECHANICS. 4 manner as in rule 11, in this case, however, dealing with the area of each part instead of its weight. See Fig. 13. 1O3. Center of Gravity of a Solid. In a body free to move, the center of gravity will lie in a vertical plumb- line drawn through the point of support. Therefore, to FIG. 14. find the position of the center of gravity of an irregular solid, as the crank, Fig. 14, suspend it at some point, as B, so that it will move freely. Drop a plumb-line from the point of suspension, and mark its direction. Suspend the body at another point, as A, and repeat the process. The intersection C of the two lines will be directly over the center of gravity. It is often desired to find the horizontal distance of the center of gravity from a given point of the body. In many cases this can readily be done by balancing the body on a knife edge, and then measuring, horizontally, the distance between the knife edge and the given point. Since the position of the center of gravity depends wholly on the shape and weight of a body, it may be without the body, as in the case of a circular ring, whose center of gravity is the same as the center of the circumference of the ring. By considering the symmetry of form, the position of the 4 PRINCIPLES OF MECHANICS. 43 center of gravity of homogeneous solids may often be deter- mined without analysis, or it may be limited to a certain plane, line, or point. Thus, the center of gravity of a sphere, or any other regular body, is situated at its center; of a cylinder, in the middle of its axis; of a thin plate having the form of a circle or regular polygon, in the center of the figure; of a straight wire of uniform cross-section, in the middle of its length. EXAMPLES FOR PRACTICE. 1. A spherical shell has a wrought-iron handle attached to it. The shell is 10 inches in diameter and weighs 20 pounds. The handle is 1| inches in diameter, and the distance from the center of the shell to the end of the handle is 4 feet. Where is the center of gravity ? Take the weight of a cubic inch of wrought iron as .278 pound. Ans. 13.612 in. from center of shell. 2. The distance between the centers of two bodies is 51 inches. The weights of the bodies being 20 and 73 pounds, where is the center of gravity ? Ans. 10.968 in. from the center of the larger weight. 3. Weights of 5, 9, and 12 pounds lie in one straight line in the order named. Distance from the 5-pound weight to the 9-pound weight is 22 inches, and from the 9-pound weight to the 12-pound weight is 18 inches. Where is the center of gravity ? Ans. 13.923 in. from 12-lb. weight. CF/NTRIFTJGAI, FORCE. 1O4. If a body is fastened to a string and whirled, so as to give it a circular motion, there will be a pull on the string that will be greater or less, according as the velocity increases or decreases. The cause of this pull on the string will now be explained. Suppose that the body is revolved horizontally, so that the action of gravity upon it will always be the same. According to the first law of motion, a body put in motion tends to move in a straight line unless acted upon by some other force, causing a change in the direction. When the body moves in a circle, the force that causes it to move F IG . 15. 44 PRINCIPLES OF MECHANICS. 4 in a circle instead of in a straight line is exactly equal to the tension of the string. If the string were cut, the pulling force that drew it away from the straight line would be removed, and the body would then "fly off at a tangent"; that is, it would move in a straight line tangent to the circle, as shown in Fig. 15. Since, according to the third law of motion, every action has an equal and opposite reaction, we call the force that acts as an equal and opposite force to the pull of the string the centrifugal force, and it acts away from the center of motion. 105. The other force, or tension, of the string is called the centripetal force, and it acts toward the center of motion. It is evident that these two forces, acting in oppo- site directions, tend to pull the string apart, and, if the velocity be increased sufficiently, the string will break. It is also evident that no body can revolve without generating centrifugal force. 106. To Find the Centrifugal Force of Any Re- volving Body. The value of the centrifugal force, expressed in pounds, of any revolving body is calculated by the following rule: Rule 14. The centrifugal force equals the continued product of .00034, the weight of the body in pounds, the radius in feet (taken as the distance between the center of gravity of the body and the center about which it revolves), and the square of the number of revolutions per minute. Let F ' = centrifugal force in pounds; W= weight of revolving body in pounds; R = radius in feet of circle described by center of gravity of revolving body; A r = revolutions per minute of revolving body. Then, F= .00034 WRN*. EXAMPLE. What is the tension in the string of Fig. 15, if the ball weighs 2 pounds and is revolved around at the rate of 500 revolutions per minute ? The string is 2 feet long. 4 PRINCIPLES OF MECHANICS. 45 SOLUTION. Applying the rule just given, we get F= .00034 X 2 x 2 X 500- = 340 Ib. Ans. 107. In flywheels, belt wheels, and pulleys the centrif- ugal force tends to tear the rim asunder; this tendency is resisted by the tenacity of the material of which the wheel is composed. Since the centrifugal force increases as the square of the number of revolutions, it will be seen that an apparently slight increase in the number of revolutions per minute may be sufficient to burst the wheel. 108. For solid cast-iron wheels and for built-up wheels of cast iron where the strength of the joint is equal to the strength of the rim, the greatest number of revolutions per minute that practice has indicated to be safe may be found by the following rule : Rule 15. Divide 1,930 by the diameter of the wheel. Or. where d = diameter of the wheel in feet; N= number of revolutions per minute. EXAMPLE. What is the maximum number of revolutions allowable for a cast-iron flywheel 27 feet 6 inches in diameter ? SOLUTION. Applying rule 15, we get N = ' = 70 rev. per min. Ans. EQUILIBRIUM. 1O9. When a body is at rest, all the forces that act upon it balance one another and are said to be in equilibrium. The most important force to be considered is the attraction of the earth, which acts upon every particle of a body. There are three states of equilibrium : stable, unstable, and neutral. HO. A body is in stable equilibrium when, if slightly rotated about its support in such manner as to change the 46 PRINCIPLES OF MECHANICS. 4 elevation of its center of gravity, it tends to return to its position of rest. Examples of bodies in a state of stable equilibrium are a cube resting on one of its sides, a cone resting on its base, a pendulum, etc. A body can only be in stable equilibrium when a rotation about its support raises the center of gravity. 111. A body is in unstable equilibrium when a rota- tion about its support, so as to change the elevation of its center of gravity, tends to make it fall farther from its posi- tion of rest. Examples of bodies in unstable equilibrium are a cube balanced on one of its edges, a cone standing upon its point, an egg balanced upon its end, etc. When a body is in unstable equilibrium, any rotation, no matter how slight, tends to lower its center of gravity. A body is in neutral equilibrium when a rota- tion about its support does not change the elevation of its center of gravity. Examples of bodies in neutral equilib- rium are a sphere of uniform density and a cone resting on its side. 113. A vertical line drawn through the center of grav- ity of a body is called the line of 'direction. So long as the line of direction falls within the base, the body will stand; when the line of direction falls outside of the base, the body will fall. B Let A CB, Fig. 16, be a cylinder whose base is oblique to the center line B O D, and let O be the center of gravity of 4 PRINCIPLES OF MECHANICS. 4? the cylinder. So long as the vertical line drawn through O falls between A and C, the cylinder will stand, but the instant it falls without the base, the cylinder will fall. 114. The center of gravity of a body has a tendency to always seek the lowest possible position. MACHINE ELEMENTS. THE LEVER, WHEEL, AND AXLE. FUNDAMENTAL PRINCIPLES. 1. A lever is a bar capable of being turned about a pivot, or point, as in Figs. 1, 2, and 3. r The object IV to be lifted is called the weight; the force is represented by P* ; and the point, or pivot F, is called the fulcrum. * The force applied to a lever, screw, wheel and axle, or any similar machine element in order to raise a given weight, was formerly called the power, and the arm of a lever to which the force is applied was called the power arm ; at present, however, the term power is uni- versally used to represent the rate at which work is done, and, hence, its application to those cases where a simple force is meant often leads to a serious confusion of ideas regarding the relation between force, work, and power. To prevent this confusion, we will use the word power only in accordance with the definition given. 5 For notice of the copyright, see page immediately following the title page. 2 MACHINE ELEMENTS. 5 2. That part of the lever F b between the fulcrum and the weight is called the weight arm, and the part F c between the fulcrum and the force is called the force arm. In order that the lever will be in equilibrium (balance), the force multiplied by the force arm must equal tlie weight multiplied by the weight arm ; that is, P X F c must equal W X F b. 3. If F is taken as the center of a circle, and arcs are described through b and , it will be seen that if the weight arm is moved through a certain angle, the force arm will move through the same angle. Since in the same or equal angles the lengths of the arcs are proportional to the radii with which they were described, it is seen that the force arm is proportional to the distance through which the force acts, and the weight arm is proportional to the distance through which the weight moves. Hence, instead of wri- ting Px F c IV X F b, we might have written P X (dis- tance through which P acts) = W X (distance through which Amoves). This is the general law of all machines, and can be applied to any mechanism from the simple lever up to the most complicated arrangement. When stated in the form of a rule it is as follows: Rule 1. The force multiplied by the distance through which it acts equals the weight multiplied by the distance through which it moves. 4. In the above rule, it will be noticed that there are four requirements necessary for a complete knowledge of the lever, viz. : the force, the weight, the force arm, or distance through which the force acts, and the weight arm, or distance through which the weight moves. If any three are given, the fourth may be found by letting x represent the requirement that is to be found, and multiplying the force by the force arm and the weight by the weight arm ; then, dividing the product of the two known numbers by the number by which x is multiplied, the result will be the requirement that is to be found. 5 MACHINE ELEMENTS. 3 EXAMPLE. If the weight arm of a lever is 6 inches long and the force arm is 4 feet long, how great a weight can be raised by a force of 20 pounds at the end of the force arm ? SOLUTION. In this example, the weight is unknown; hence, repre- senting it by x, we have, after reducing the 4 feet to inches, 20 X 48 = 960 = force multiplied by the force arm, and x X 6 = weight mul- tiplied by the weight arm. Dividing the 960 by 6, the result is 160 pounds, the weight. Ans. 5. If the distance through which the force acted or the weight had moved had been given instead of the force arm or weight arm, and it were required to find the force or weight, the process would have been exactly the same, using the given distance instead of the force arm or weight arm. EXAMPLE. If, in the above example, the weight had moved 2^ inches, through what distance would the force have acted ? SOLUTION. In this example, the distance through which the force acts is required. Let x represent the distance. Then, 20 X x = dis- tance multiplied by force, and 2 X 160 = 400 = distance multiplied by the weight. Hence, x = 4^- = 20 inches = distance through which the force arm moves. Ans. The ratio between the weight and the force is 160 H- 20 = 8. The ratio between the distance through which the weight moves and the distance through which the force acts is 2 -=- 20 = \. This shows that while a force of 1 pound can raise a weight of 8 pounds, the 1-pound weight must move through 8 times the distance that the 8-pound weight does. It will also be noticed that the ratio of the lengths of the two arms of the lever is also 8, since 48 -*- 6 = 8. 6. The law that governs the straight lever also governs the bent lever, but care must be taken to determine the true lengths of the lever arms, which are, in every case, the per- pendicular distances from the fulcrum to the line of direction of the weight or force. Thus, in Figs. 4, 5, 6, and 7, F c in each case represents the force arm and F b the weight arm. 7. A compound lever is a series of single levers arranged in such a manner that when a force is applied to the first it is communicated to the second, and from that to the third, and so on. 4 MACHINE ELEMENTS. 5 Fig. 8 shows a compound lever. It will be seen that when a force is applied to the first lever at P it will be communi- cated to the second lever at /*, from this to the third lever at P, and thus raise the weight W. FIG. 5. The weight that the force applied to the first lever could raise acts as the force of the second, and the weight that this force could raise through the second lever acts as the force FIG. 0. FIG. 7. of the third lever, and so on, no matter how many single levers make up the compound lever. In this case, as in every other, the force multiplied by the distance through which it acts equals the weight multiplied by the distance through which it moves. 5 MACHINE ELEMENTS. 5 Hence, if we move the P end of the lever, say 4 inches, and the end carrying the weight W moves | inch, we know that the ratio between P and W is the same as the ratio between \ and 4; that is, 1 to 20, and, hence, that 10 pounds at /'will balance 200 pounds at W, without measuring the lengths of the different lever arms. If the lengths of the lever arms are known, the ratio between P and J^may be readily obtained from the following rule: Rule 2. The continued product of the force and each force arm equals the continued product of the weight and each weight arm. EXAMPLE. If, in Fig. 8, P F= 24 inches, 18 inches, and 30 inches, respectively, and W F=. inches, 6 inches, and 18 inches, respectively, how great a force at P will it require to raise 1,000 pounds at W? What is the ratio between W and P ? SOLUTION. Let x represent the force ; then, x X 24 X 18 X 30 = 12,960 X x = continued product of the force and each force arm. 1,000 X 6 X 6 X 18 = 648,000 = continued product of the weight and each weight arm; and, since 12,960 X x = 648,000, 648,000 = 50 Ib. = the force. Ans. 1,000 -H 50 = 20 = ratio between fFand P. Ans. 8. The wheel and axle consists of two cylinders of different diameters rigidly connected, so that they turn FIG. 9. //. s. LU 6 MACHINE ELEMENTS. 5 together about a common axis, as in Fig. 9. Then, as before, P X distance through which it acts = W X distance through which it moves ; and, since these distances are pro- portional to the radii of the force cylinder and weight cylin- der, P X Fc = W X Fb. It is not necessary that an entire wheel be used ; an arm, projection, radius, or anything P \\ that the force causes to revolve in a circle, may be considered as the wheel. Consequently, if it is desired to hoist a weight with a windlass, Fig. 10, the force is ap- plied to the handle of the crank, and the distance between the center line of the crank-handle and the axis of the drum corresponds to the radius of the wheel. EXAMPLE. If the distance between the center line of the handle and the axis of the drum in Fig. 10 is 18 inches and the diameter of the drum is 6 inches, what force will be required at P to raise a load of 300 pounds ? SOLUTION. P x (18 X 2) = 300 x 6, or P = 50. Ans. EXAMPLES FOR PRACTICE. 1. The lever of a safety valve is of the form shown in Fig. 1, where the force is applied at a point between the fulcrum and the weight lifted. If the distance from the fulcrum to the valve is 5 inches and from the fulcrum to the weight is 42 inches, what total force is neces- sary to raise the valve, the weight being 78 pounds and the weight of valve and lever being neglected ? Ans. 595.64 Ib. 2. If, in Fig. 8, P F 10 inches, 12 inches, 14 inches, and 16 inches, respectively, and W F= 2 inches, 3 inches, 4 inches, and 5 inches, respectively, (a) how great a weight can a force of 20 pounds raise ? (b) What is the ratio between W and P ? (c) If P moves 4 inches, how far will W move ? i (a) 4,480 Ib. Ans. \ (6) 224. (') A in- 5 MACHINE ELEMENTS. 7 3. A windlass is used to hoist a weight. If the diameter of the drum on which the rope winds is 4 inches, and the distance from the center of the handle to the axis of the drum is 14 inches, how great a weight can a force of 32 pounds applied to the handle raise ? Ans. 234 Ib. PTJI.LEYS. 9. Pulleys for the transmission of power by belts may be divided into two principal classes: (1) The solid pulley shown in Fig. 11, in which the hub, arms, and rim are one entire casting. (2) The split pulley shown in Fig. 12, which is cast in halves. The latter style of pulley is more readily placed on and removed from the shaft than the solid pulley. Pulleys are generally cast in halves or parts when they are more than 6 feet in diameter; this is done on account of the shrink- age strain in large pulley castings, which renders them liable to crack as a result of the unequal cooling of the metal. 8 MACHINE ELEMENTS. 10. Of late years, wooden pulleys have come into exten- sive use. They are built of segments securely glued together, maple being the wood used. Wooden split pulleys can be procured that are fitted with removable bushings, thus allow- ing the same pulley to be readily adapted to various diam- eters of shafting. They are somewhat lighter than cast-iron pulleys. 11. Crowning Pulley Faces. In Fig. 13, suppose the shafts a and b to be parallel. Let the pulley on the shaft a be cone-shaped, as shown. The right-hand side of the belt will be pulled ahead more rapidly than the left-hand side, because of the greater diameter and, consequently, greater speed of the right-hand end of the pulley. It has been observed that in this case the belt will leave its normal position, which is indi- cated by the dotted lines, and climb toward the part of the pulley that has the largest diameter, as shown in the illustration. This tendency of the belt to climb toward the high side is taken advantage of to make the belt stay on a pulley. Suppose the pulley on the shaft a is replaced with one formed of two equal cones, with the large diameter in the KIG> 13> center. Then, each side of the belt will tend to climb toward the highest point, and the consequence is that the belt will stay in the center of the pulley. A pulley with its surface formed in this way is said to be crowned. The surface need not necessarily represent the frustums of two cones; it may simply be curved, as shown in Fig. 14. It is only required that the pulley be larger in diameter in the center. As to the proper amount of crowning necessary, the practice of the makers FIO.M. of pulleys differs considerably ; usually, though, it is from $ to inch per foot of width of the pulley face. MACHINE ELEMENTS. 9 12. Balancing Pulleys. All pulleys that rotate at high speeds should be balanced. If they are not, the centrifugal force that is gener- ated by the rota- tion of the pulley, is greater on one side than on the other and will cause the pulley shaft to vibrate and shake. Pulleys are usually balanced in the manner shown in FIG. is. Fig. 15. A closely fitting arbor, which is simply a truly cylindrical piece of iron or steel, is driven into the bore of the pulley, which is then placed on the so-called " balancing ways." These are two planed iron or steel bars, preferably planed to a knife edge. These bars are placed on conve- nient supports far enough apart to allow the pulley to go between them. The bars should be carefully leveled with a spirit level, so that both bars are in the same horizontal plane. When the arbor rests on the ways and the pulley is slightly rotated, the latter will quickly come to rest with the heavy part downwards. Now, either some metal must be removed from the heavy part or some weight added to the light part. For small pulleys, a convenient substance to use for finding the proper location and weight of the counterweight is ordinary glazier's putty. By repeated trials the proper weight of the counterweight is found, and then a mass of metal of convenient shape equal in weight to the putty is fastened to the light part of the pulley in what- ever way is safe and convenient. The proper weight and location of the counterweight will have been obtained when the pulley will be at rest on the balancing ways in any posi- tion in which it is put. In other words, the pulley is balanced when it is in neutral equilibrium. This balance is called a standing balance. The method just explained answers very well for pulleys that are narrow in comparison 10 MACHINE ELEMENTS. 5 to their diameter. When pulleys with a wide face are run at a high speed, it is often found that serious vibrations are set up even when they are in perfect standing balance, thus showing that they do not possess the so-called running balance. No method has yet been found that will insure a running balance at all speeds, nor has any method become known by which it can be discovered directly where to apply the counterweight. The usual remedy for pulleys not pos- sessing a running balance is to turn carefully the inside of the rim to run true with the outside of the pulley. Absence of running balance is, in all cases, due to an unequal distri- bution of metal in reference to the axis of the shaft, this unequal distribution being due to lack of homogeneity of the metal, to poor foundry work, or to poor lathe work. 13. Pulleys should run true in order that the strain, or tension, of the belt will be equal at all parts of the revolu- tion, thus making the transmitting power equal. The smoother the surface of a pulley, the greater is its driving power. The transmitting power of a pulley can be increased by covering its face with a leather or rubber band; this in- creases the driving power about one-quarter. 14. Relation Between Speeds of Drivers and Driven Pulleys. The pulley that imparts motion to the belt is called the driver ; that which receives the motion is called the driven. The revolutions of any two pulleys over which a belt is run vary in an inverse proportion to their diameters; conse- quently, if a pulley 20 inches in diameter is driven by one 10 inches in diameter, the 20-inch pulley will make 1 revolu- tion while the 10-inch pulley makes 2 revolutions, or they are in the ratio of 2 to 1. From this fact, the following formulas have been deduced : Let D = diameter of the driver; d = diameter of the driven ; N= number of revolutions of the driver; n = number of revolutions of the driven. 5 MACHINE ELEMENTS. 11 NOTE. The words revolutions per minute are frequently abbre- viated to R. P. M. 15. To find the diameter of the driving pulley when the diameter of the driven pulley and the number of revolutions per minute of each is given: Rule 3. The diameter of the driving pulley equals the product of the diameter and the number of revolutions of the driven pulley divided by the number of revolutions of the driving pulley. 0, , . *= . | EXAMPLE. The driving pulley makes 100 revolutions per minute, the driven pulley makes 75 revolutions per minute and is 18 inches in diameter; what is the diameter of the driving pulley ? SOLUTION. Applying the rule just given and substituting, we have D = ^- = 134 in. Ans. 16. The diameter and number of revolutions per minute of the driving pulley being given, to find the diameter of the driven pulley, which must make a given number of revolu- tions per minute : Rule 4. The diameter of the driven pulley equals the prod- uct of the diameter and the number of revolutions of the driving pulley divided by the number of revolutions of the driven pulley. , DN Or, d = - . EXAMPLE. The diameter of the driving pulley is 13| inches and it makes 100 revolutions per minute; what must be the diameter of the driven pulley to make 75 revolutions per minute ? SOLUTION. Applying rule 4 we have d = 2_2l - = jg in. Ans. 17. To find the number of revolutions per minute of the driven pulley, its diameter and the diameter and the number of revolutions per minute of the driving pulley being given: 12 MACHINE ELEMENTS. 5 Rule 5. The number of revolutions of the driven pulley equals the product of the diameter and the number of revolu- tions of the driver divided by the diameter of the driven pulley. DN Or, n = - r . EXAMPLE. The driving pulley is 13| inches in diameter and makes 100 revolutions per minute; how many revolutions will the driven pulley make in 1 minute if it is 18 inches in diameter ? D N SOLUTION. Formula, n y-. a Substituting, we have n = 13 ^ * 10 _ 75 R. p. M. Ans. lo 18. To find the number of revolutions per minute df the driving pulley, its diameter and the diameter and the number of revolutions per minute of the driven pulley being given : Rule 6. The number of revolutions of the driving pulley equals the product of the diameter and the number of revolu- tions of the driven pulley divided by the diameter of the driving p u I ley. "*-% EXAMPLE. The driven pulley is 18 inches in diameter and makes 75 revolutions per minute; how many revolutions will the driving pulley make in 1 minute if it is 13 inches in diameter ? SOLUTION. Formula, TV- -^. Substituting, we have N= 18 * 75 = 100 R. P. M. Ans. BELTS. 19. A belt is a flexible connecting band that drives a pulley by its frictional resistance to slipping at the surface of the pulley. Belts are most commonly made of leather, cotton, or rubber, and are united in long lengths by cement- ing, riveting, or lacing. 5 MACHINE ELEMENTS. 13 20. Leather belts are made single and double. A single belt is one composed of a single thickness of leather; a double belt is one composed of two thicknesses of leather cemented and riveted together the whole length of the belt. 21. Cotton belts are in use to some extent, as are also belts made of a number of layers of duck sewed together and impregnated with a preparation rendering them waterproof. These belts, in accordance with the number of layers or "plys, " are called two-ply, three-ply, etc. Four-ply cotton and duck belting is about equal to single leather belting, and eight-ply to double leather belting. 22. Rubber belts are especially adapted for use in damp or wet places ; they will endure a great degree of heat or cold without injury, are quite durable, and are claimed to be less liable to slip than leather belts. CALCULATIONS FOR BELTS. 23. To Find the Length of a Belt. In practice, the necessary length for a belt to pass around pulleys that are already in their position on a shaft is usually obtained by passing a tape line around the pulleys, the stretch of the tape line being allowed as that necessary for the belt. The lengths of open-running belts for pulleys not in position can be obtained approximately as follows : Rule 7. The length of a belt for open-running pulleys equals 3^ times one-half the sum of the diameters of the pulleys plus 2 times the distance between the centers of the shaft. Let D = diameter of one pulley in inches; d = diameter of the other pulley in inches; L = distance between the centers of the shafts in inches; B length of the belt in inches. Then, B = \ 14 MACHINE ELEMENTS. 5 EXAMPLE. The distance between the centers of two shafts is 9 feet 7 inches ; the diameter of the large pulley is 36 inches and the diameter of the small one is 14 inches ; what is the necessary length of the belt ? SOLUTION. Applying the rule just given and substituting the values given, we have, since 9 feet 7 inches = 115 inches, B - 8j ( 36 + 14 ) + 2 X 115 = Slli in., or 25 ft. llj in. Ans. The length of crossed belts cannot be determined by any simple calculation, it being a rather difficult mathematical problem. 24. To Find the Width of Belts. A belt should be wide enough to bear safely and for a reasonable length of time the greatest tension that will be put upon it. This will be the tension of the driving side. The safe tension for single belts may be taken as 60 pounds per inch of width; single belts average T 3 inch in thickness. The tension on the driving side, however, does not represent the force tend- ing to turn the pulley. The force tending to turn the pulley, or the effective pull, is the difference in tension between the driving side and the slack side of the belt. The tension on the driving side depends on three factors: the effective pull of the belt, the coefficient of friction between the belt and pulley, and the size of the arc of contact of the belt on the smaller pulley. 25. The effective pull that may be allowed per inch of width for single leather belts with different arcs of contact (the arc in which the belt touches the smaller pulley), is given in Table I. 26. To Find the Arc of Contact. The arc of con- tact in degrees, or as a fraction of the circumference, can be determined, practically, as follows: Stretch a string over the two pulleys to represent the belt. Then, take another string, wrap it around the small pulley and cut it off so that the ends meet. This represents the circumference of the small pulley. Now take a third string, hold one end at the MACHINE ELEMENTS. TABLE I. 18 ALLOWABLE BELT PTJI/LS. Arc Covered by Belt. Allowable Effective Pull Degrees. Fraction of Cir- cumference. Per Inch of Width in Pounds. 90 .250 23.0 112 .312 27.4 120 .333 28.8 135 .375 31.3 150 .417 33.8 157 .437 34.9 180 or over .500 38.1 beginning of the arc of contact, as shown by the string stretched around both pulleys, wrap it around the smaller pulley, and cut it off at the end of the arc of contact. The length of this last string represents the length of the arc of contact. We now have the proportion: the length of the string representing the circumference : the length of the string representing the arc of contact :: 360 (the number of degrees in a circle) : the number of degrees in the arc of con- tact. Whence, the number of degrees in the arc of contact equals the quotient obtained by dividing the product of the length of the arc of contact and 360 by the circumference of the pulley. To obtain the fraction of the circumference, divide the length of the string representing the arc of contact by the circumference. 27. To use the table for finding the width of a single leather belt for transmitting a given number of horsepower, we have the following rule, 16 MACHINE ELEMENTS. 5 where C = allowable effective pull, from table ; H = horsepower to be transmitted; W= width of single belt in inches; V = velocity of belt in feet per minute. Rule 8. Multiply the horsepower to be transmitted by S3, 000, and divide this product by the product of the velocity of the belt and the allowable effective pull, as taken from the table. The quotient will be the width of the belt. 33, 000 # Or, vc EXAMPLE. What width of single belt is needed to transmit 20 horse- power, the arc of contact on the small pulley being 135 and the speed of the belt 1,500 feet per minute ? SOLUTION. According to Table I, the allowable effective pull for 135 is 31.3 pounds. Then, applying rule 8, we have 33,000 X 20 28. To Find the Horsepower of a Belt. The horse- power that a single belt will transmit is given by the follow- ing rule: Rule 9. Multiply together the effective pull taken from the table, the widtli of the belt in inches, and the speed of the belt in feet per minute. Divide the product by 33,000. CWV Or, 33,000' 29. Speed of Belts. By applying rule 8 to the same belt running at different velocities, it will be seen that the higher the velocity, the greater is the horsepower that the same belt can transmit, and from rule 9 it will be seen that the higher the speed of the belt, the less may be its width to transmit a given horsepower. From this it follows that a belt should be run at as high a velocity as conditions will permit, the maximum velocity allowable for a laced belt being about 3,500 feet per minute for ordinary single leather and double leather belts. For belts spliced by 5 MACHINE ELEMENTS. 17 cementing, where the splice is practically as strong as the belt itself, the velocity may be as high as 5,000 feet per minute. Cases are on record where wide main belts have been run at as high a velocity as 6,000 feet per minute. 3O. In choosing a proper belt speed, due regard must be paid to commercial conditions. While a high speed of the belt means a narrow and, consequently, a cheaper belt, the increased cost of the larger pulleys that may be required may offset the gain due to the high belt speed, at least as far as first cost is concerned. For illustration, let the problem be to transmit 10 horse- power from one shaft to another. Let the revolutions of both the driven and the driving shaft be equal, and let the shafts make 200 revolutions per minute. Choosing a belt speed of 2,000 feet per minute, the width of a single belt to transmit the given horsepower will be, by rule 8, say, 4 inches. The diameter of the pulley that at 200 revolu- tions per minute will give a belt speed of 2,000 feet per C\ AAA x/ "I O minute is - = 38J inches, nearly. Taking the &\J\J /\ O. - price of a 38-inch cast-iron pulley, 5-inch face, as $15, we have the price of two pulleys as $30. Let the distance between the pulleys be 20 feet. Then, the length of belt, according to rule 7, is 50 feet 4 inches. Taking 51 feet as the length of the belt, and the price of a single leather belt 4 inches wide at 50 cents per foot, the price of the belt will be $25.50. Then, the cost of belt and pulleys, not count- ing freight or express charges, etc., will be $25.50 -f- $30 $55.50. Choosing a belt speed of 3,500 feet per minute, the width of belt will be 2 inches, nearly. The proper diameter of Q f)f\0 v 1 9 the pulley is ^ * ,g- = 66 inches, say, 67 inches. The ZOO X o.l41o length of the belt will be 59 feet, about. The price of the belt at 30 cents per foot is $17.70. The price of two 67-inch pulleys 3^-inch face is, say, $80. Then, the total first cost is $80 + $17.70 = $97.70, showing that in this 18 MACHINE ELEMENTS. 5 particular case the use of a low belt speed reduces the first cost by $97.70 $55.50 = $42.20. The above illustration is not intended, and must not "be construed, to be an argument against high belt speed; it simply shows the advisability of considering the commercial features in each and every case. In many cases it will be found that the narrow, high-speed belt is by far the more economical one to use. . , 31. Double belts are made of two single belts cemented and, usually, riveted together their whole length, and are used where much power is to be transmitted. As the effective pull for single belts, as given in Table I, is based primarily on the strength through the lace holes, a double belt, which is twice as thick, should be able to transmit twice as much power as a single belt, and in fact more than this, where, as is quite common, the ends of the belt are cemented instead of laced. Where double belts are used on small pulleys, however, the contact with the pulley face is less perfect than it would be if a single belt were used, owing to the greater rigidity of the former. More work is, also, required to bend the belt as it runs over the pulley than in the case of the more pliable single belt, and the centrifugal force tending to throw the belt from the pulley also increases with the thickness. Moreover, in practice, it is seldom that a double belt is put on with twice the tension of a single belt. For these reasons, the width of a double belt required to transmit a given horsepower is generally assumed to be seven-tenths the width of a single belt to transmit the same power. On this basis, rules 8 and 9 become for double belts, by multi- plying rule 8 by T 7 T and dividing rule 9 by T \, as follows: Rule 1O. To find the widtJi of a double belt, multiply the horsepower to be transmitted by 23, 100. Divide this product by the product of the velocity of the belt and the allowable effective pull, as taken from the table. u/ 23, 100 H Or, W= .^ . 5 MACHINE ELEMENTS. 19 Rule 11. To find the horsepower that a double belt can transmit, multiply together the effective pull taken from the table, the width of the belt, and its velocity. Divide the product by 23, 100. Or 23,100' EXAMPLE 1. What width of double belt is required to transmit 20 horsepower, the arc of contact on the smaller pulley being 135 and the speed of the belt 1,500 feet per minute ? SOLUTION. According to Table I, the effective pull is 31.3 pounds. Then, by rule 1O, 23,100 x 20 W = r500 X 31.3 = 10 m " nearly " AnS " EXAMPLE 2. What horsepower can be transmitted by a 6-inch double belt running at 400 feet per minute with an arc of contact of 180 ? SOLUTION. According to Table I, the effective pull is 38.1 pounds. Applying rule 11, we have THE CARE AND USE OP BELTS. 32. Leather Belts. It is a much disputed question as to which side of the belt should be run next to the pulley. The more common practice, it is believed, is to run the belt with the hair or grain side nearest the pulley. This side is harder and more liable to crack than the flesh side. By running it on the inside the tendency is to cramp or com- press it as it passes over the pulley, while if it ran on the outside, the tendency would be for it to stretch and crack. The flesh side is the tougher side, but for the reason given above the life of the belt will be longer if the wear comes upon the grain side. The lower side of the belt should be the driving side, the slack side running from the top of the driving pulley. The sag of the belt will then cause it to encompass a greater length of the circumference. Long belts, running in any other direction than vertical, work 20 MACHINE ELEMENTS. 5 better than short ones, as their weight holds them more firmly to their work. It is bad practice to use rosin to prevent slipping. It gums the belt, causes it to crack, and prevents slipping for only a short time. If a belt, in good condition, persists in slipping, a wider belt should be used. Sometimes larger pulleys on the driving and driven shafts are of advantage, as they increase the belt speed and reduce the stress on the belt. Belts may be kept soft and pliable by oiling them once a month with castor oil or neatsfoot oil. 33. The Flapping of Belts. One of the most annoy- ing troubles experienced with belting of all kinds is the violent flapping of the slack side. Flapping may be due to any one of several causes, or to a combination of them. The most usual cause is that one or both of the pulleys run out of true. The belt is then alternately stretched and released, and while this may not cause flapping at one speed, it will usually do so at a higher speed. If the belt is rather slack, tightening it somewhat may cure or alleviate the flap- ping. The most obvious and best remedy, but the most expensive, is to turn the pulleys to run true. Pulleys being out of line with each other are another pro- lific source of flapping, especially when one or both run out of true. First, bring the pulleys in line; if this fails, tighten the belt if it is rather loose. If no improvement is noticed and it is not possible to turn the pulleys, try to lower the belt speed a little, either by the substitution of smaller pulleys or by changing the speed of the driven shaft, accord- ing to circumstances. With belts running at speeds above 4,000 feet per minute, flapping may occur when the pulleys are perfectly true and in line with each other, even when the belt has the proper tension. This is believed to be due to air becoming entrapped between the face of the pulley and the belt. At any rate, it has been observed that perforating the belt with a series of small holes will cure this trouble. Perforated belts may now be bought in the market. 5 MACHINE ELEMENTS. 21 Lack of steadiness in running, due either to sudden varia- tions in the speed of the engine, or sudden changes in the load of the machines driven by the belt, will produce a flap- ping that it is almost impossible to cure. The only known cure is to take such steps as will insure steady running, as for instance, increasing the weight of the flywheel on the engine, or placing a flywheel instead of a pulley on the driven machine. The belt not being joined square will also cause flapping, especially when the belt is running at a rather high speed. The remedy is to unlace or unfasten the joint and make it square. Too great a distance between the pulleys may also cause flapping. In general, the distance between the pulleys should not exceed 15 feet for belts up to 4 inches in width ; 20 feet for belts above 4 and below 12 inches; 25 feet for belts above 12 inches and below 18 inches ; and 30 feet for larger belts. The distances here given are occasionally exceeded considerably, but as no experiments have ever been made public that would enable a fairly correct formula to be deduced for the distance between pulleys when the belt speed, width of belt, and effective pull are known, they must be taken as representing average practice. A horizontal belt is, by many, considered to have the proper tension when it has about 1 inch of sag, while in motion, for every 8 feet between the pulleys. For belts other than horizontal, this should be less, there being no sag at all for vertical belts. JOINING THE ENDS OF BELTS. 34. Lacing. The ends of a belt may be joined by lacing, sewing, riveting, or cementing. Many ways of lacing belts are used. A very satisfactory method for belts up to 3 inches in width is' shown in Fig. 16. Cut the ends of the belt square, using a sharp knife and a try square. Punch a row of holes according to the width of the belt, punching 22 MACHINE ELEMENTS. f -i corresponding holes exactly opposite each other in each end of the belt, using 3 holes in belts up to 2 inches wide, and 5 holes in belts between 2 and 3 inches wide. The number of holes in the row should always be uneven for the style of lacing shown. In the figure, A is the outside of the belt and B the side running nearest the pulley. The lacing should be drawn half- way through one of the middle holes from the under side, as at 1- before going any further, it is well to see to it that the belt FIG - 16 - has no twists in it, or, in the case of a crossed belt, that it has not been given a wrong twist. The same side of the belt should run over both pulleys, which will be the case with a crossed belt if it has been twisted correctly. Having made sure that the belt is fair, pass the end of the lace on the upper side of the belt through 2 under the belt and up through 3, back again through 2 and 3, through 4 an d up through 5, where an incision is made in one side of the lacing, which forms a barb that will prevent the end from pulling through. Lace the right-hand side in the same manner. The lacing may advantageously be carried on at once to the right and left alternately. 35. For belts wider than 3 inches, the lacing shown in Fig. 17 is a good one. As will be observed, there are two rows of holes. The number of holes in the row nearest the joint should exceed by one the number of holes in the second row. For belts up to 4 inches wide, use 3 holes in the row nearest the joint and 2 holes in the second row. For belts up to 6 inches wide, use 4 and 3 holes, respect- ively. For larger belts, make the total number of holes in each end either one or two more than the number of inches of width, with the object of getting an odd total number of holes. For example, for a 10-inch belt, the total number of MACHINE ELEMENTS. 23 holes should be 10 + 1 11. For a 13-inch belt, it should be 13 + 2 = 15 holes. The outside holes of the first row should not be nearer the edges of the belt than f inch, nor should the first row be nearer the joint than inch. The second row should be at least If inches from the end. In the figure, A is the outside and B the side nearest the pulley. Begin at one of the center holes in the outside row, as 1, and continue through #, S,4,S, 6, 7, 4,3, 8,3, etc. Another method is to begin the lacing on one side instead of in the middle. This method will give the rows of lacing on the under side of the belt the same thickness all the way across. The lacing should not be crossed on the side of the belt that runs next to the pulley. Lacing affords convenient means of shortening belts when they stretch and of increasing their tension. If the belts are at all large, the ends of the belt should be drawn together by belt clamps. 36. Cementing. Cementing makes probably the best kind of a joint ever devised. It has the serious disadvantage, however, that the stretch of the belt cannot be readily taken up, and, hence, tightening pulleys must be used when the center-to-center distance of the pulleys is not adjustable. In dynamo driving, where endless belts are used to the exclusion of all others, the dynamo is usually mounted on a slide, so that the tension of the belt can be adjusted. For a cemented joint, the ends of the belt should be pared down with a very sharp knife to the form shown in Fig. 18, which shows a form that is recommended by belt manufacturers. 24 MACHINE ELEMENTS. 5 Warm the belt ends near a fire, apply the belt cement while hot, and press the joint together by two boards, one on the top and one on the bottom. Belt cement can be obtained of any dealer in engineer's supplies and full directions for using it will always be found on the can. These directions should be implicitly followed. If belt cement cannot be obtained, a good cement may be made by melting together over a slow fire 16 parts of gutta percha, 4 parts of india rubber, 2 parts of pitch, 1 part of shellac, and 2 parts of linseed oil, by weight. Cut all ingre- dients very small, mix well, and use while hot. 37. Stretch of Belts. New leather belts will stretch from one-fourth to one-half of an inch per foot of length, and hence must be taken up until the limit of stretch has been reached. Rubber belts are said to stretch continuously. Cotton and duck belts are said not to stretch with use. 38. Precautions to l>e Observed When Using Rubber Belts. When rubber belts are used, animal oils or animal grease should never be used on them. If the belt should slip, it should be lightly moistened on the side nearest the pulley with boiled linseed oil. EXAMPLES FOR PRACTICE. 1. The main line shaft is driven at 90 revolutions per minute and is to drive another shaft at 120 revolutions per minute; the latter shaft carries a pulley 20 inches in diameter. How large a pulley should be used on the main line shaft ? Ans. 26J in. 2. The belt wheel of an engine is 10 feet in diameter and makes 65 revolutions per minute ; the line shaft is to run at 150 revolutions per minute. What size pulley should be used ? Ans. 52 in. 3. The driving pulley is 48 inches in diameter and makes 90 revolu- tions per minute. The driven pulley being 20 inches in diameter, how many revolutions per minute will the driven shaft make ? Ans. 216 R. P. M. 4. The belt wheel of an engine being 5 feet in diameter, the driven pulley 45 inches, and the driven shaft to make 90 revolutions per minute, what should be the number of revolutions of the belt wheel ? Ans. 67* R. P. M. 5 MACHINE ELEMENTS. 25 5. Two pulleys, 48 and 32 inches in diameter, respectively, are 10 feet 6 inches from center to center. What is the length of an open belt for these pulleys ? Ans. 31 ft. 10 in. 6. A belt speed of 2,400 feet per minute having been chosen, what width of single belt will be needed to transmit 30 horsepower ? The pulleys over which the belt runs are equal in size. Ans. 11 in., nearly. 7. What horsepower can be transmitted by a single belt 11 inches wide traveling at the rate of 1,000 feet per minute over a small pulley on which its arc of contact is 150 ? Ans. 11 H. P., nearly. 8. With a belt speed of 1,300 feet per minute, what width of double belt will be required to transmit 5 horsepower over pulleys equal in diameter? Ans. 2 in., nearly. WHEEL WORK. DRIVER AND FOLLOWER. 39. A combination of wheels and axles, as in Fig. 19, is called a train. The wheel in a train to which motion is imparted from a wheel on another shaft, by such means as a belt or gearing, is called the driven wheel or follower; the wheel that imparts the motion is called the driver. It will be seen that the wheel and axle bears the same relation to the train that a simple lever does to the com- pound lever. Letting D v , D^ D 3 , etc. represent the diam- eters of the driven wheels, and d^ d^ d^ etc. the diameters of the different drivers, we have the following rule : Rule 13. The continued product of the force and the diameters of the driven wheels equals the continued product of the weiglit, tJie diameter of the drum that moves the weight, and the diameters of the drivers. Or, Px D l X A X D v etc. = IVx < X < X =^ = 10in. Ans. 59. To find the diameter over all, that is, the diameter of the blank from which the gear-wheel is cut, the number of teeth and the diametral pitch being given : Rule 17. Add 2 to the number of teeth and divide by the dia metral pitch. A EXAMPLE. In the last example, what is the diameter over all the blank? SOLUTION. Applying the rule just given, we get = 10* in. Ans. 60. The number of teeth and the outside diameter of the gear-wheel being known, to find the diametral pitch: Rule 18. Add 2 to the number of the teeth and divide by the outside diameter. p. EXAMPLE. A gear-wheel has 60 teeth and is 6^ inches in diameter over all ; what is the diametral pitch ? SOLUTION. By applying rule 18, we get Pa = M* = W , Ans. 61. To find the diametral pitch, the number of teeth and the pitch diameter being known: Rule 19. Divide the number of teeth by the pitch diam- eter. **% EXAMPLE. A wheel has 90 teeth and its pitch diameter is 30 inches; what is the diametral pitch ? 5 MACHINE ELEMENTS. 37 SOLUTION. By rule 19, we get on P d = g = 3. Ans. 62. The pitch diameter and the diametral pitch being given, to obtain the number of teeth : Rule 2O. Multiply the pitch diameter by the diametral pitch. Or, N=DP d . EXAMPLE. How many teeth are there in a gear-wheel having a pitch diameter of 36 inches and a diametral pitch of 5 ? SOLUTION. Applying the rule just given, we get N= 36 X 5 = 180 teeth. Ans. 63. The diameter over all and the diametral pitch being given, to find the number of teeth : Rule 21. Subtract 2 from the product of the outside diameter and the diametral pitch. Or, N = D P d - 2. EXAMPLE. How many 10-pitch teeth has a gear-wheel having an outside diameter of 8^ inches ? SOLUTION.. By rule 21, we get ^=81x10-2 = 80166^1. Ans. 64. The standard diametral pitches used for cut gears are as follows: 2, 2, 2|, 2f, 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 36, 40, 48. Gears hav- ing a diametral pitch differing from those given here are, usually, either very large or very small. 65. Diameters and Distances Between Centers. The distance between the centers of two gear-wheels being known and the ratio of their speeds, the diameter of the pitch circle of the smaller wheel is given by the following rule, where A = center-to-center distance; R = revolutions per minute of large gear; r = revolutions per minute of small gear; , d = pitch diameter of small gear. //. s. /. 16 38 MACHINE ELEMENTS. 5 Rule 22. Multiply twice the center-to-center distance by the number of revolutions of the large gear, and divide the product by the sum of the revolutions of the large and small gear. , %AR Or, d =R + ~f EXAMPLE. Given, the distance between centers = 5 inches; the small gear is to make 24 revolutions for every 8 revolutions of the large gear. What is the diameter of each gear ? SOLUTION. By rule 22, Tfi/rr/7<: nr siffftctisp . downwards, upwards, or sideivise. 7. Downward Pressure. In Fig. 3 the pressure on the bottom of the vessel a is, of course, equal to the weight of the water it contains. If the area of the bottom of the vessel b and the depth of the liquid contained in it are the same as in the vessel a, the pressure on the bottom of b will be the same as on the bottom of a. Suppose the bottoms of the vessels a and b are 6 inches square, and that the part c d in the vessel b is 2 inches square, and that they are filled with water. Then, since 1 cubic foot of water weighs 62.5 pounds, the weight of 1 cubic inch of water is ' pound .03617 pound. The number of cubic inches l,72b MECHANICS OF FLUIDS. inrt = 6x6x24 = 864 cubic inches; and the weight of the water is 864 X .03617 = 31.25 pounds. Hence, the total pressure on the bottom of the vessel a is 31.25 pounds, 31 25 or g-^-g = -868 pound per square inch. The pressure in b due to the weight contained in the part 7< can result ; but, if the stop-cock is opened, there will be no resistance to the pressure acting on that part, and the water will flow out ; at the same time, the reaction of the jet issuing from the stop-cock causes the vessel to move backwards through the water in a direction opposite to that of the issuing jet. 16. Liquids Influenced by Gravity. Since the pres- sure on the bottom of a vessel due to the weight of the liquid is dependent only on the height of the liquid and not on the shape of the vessel, it follows that if a vessel has a number of radiating tubes (see Fig. 8) the water in each 6 MECHANICS OF FLUIDS. 9 tube will be on the same level, no matter what may be the shape of the tubes. For, if the water were higher in one tube than in the others, the downward pressure on the bottom due to the height of the water in this tube would be greater than that due to the height of the water in the other tubes. Consequently, the upward pressure would also be greater; the equilibrium would be destroyed and the water would flow from this tube into the vessel and rise in the other tubes until it was at the same level in all, when it would be in equilibrium. This principle is expressed in the familiar saying, water seeks its level. An application of this principle is the glass water gauge used for showing the level of the water in a steam boiler or tank. 17. The above principle explains why city water reser- voirs are located on high elevations and why water on leaving the hose nozzle spouts so high. If there were no resistance by friction and air, the water would spout to a height equal to the level of the water in the reservoirs. If a long, vertical pipe, whose length was equal to the vertical distance between the nozzle and the level of the water in the reservoir, were attached to the nozzle, the 10 MECHANICS OF FLUIDS. 6 water would just reach the end of the pipe. If the pipe were lowered slightly, the water would trickle out. Fountains, canal locks, and artesian wells are examples of the application of this principle. EXAMPLE. The water level in a city reservoir is 150 feet above the level of the street ; what is the pressure of the water per square inch on the hydrant ? SOLUTION. Applying rule 1, 1 X 150 X 12 X .03617 = 65.106 Ib. per sq. in. Ans. NOTE. In measuring the height of the water to find the pressure that it produces, the vertical height, or distance, between the level of the water and the point considered is always taken. This vertical height is called the head. The weight of a column of water 1 inch square and 1 foot high is 62.5 -f- 144 = .434 pound, nearly. Hence, if the depth (head) be given, the pressure per square inch may be found by multiplying the depth in feet by .434. The constant .434 is the one ordinarily used in practical calculations. 18. In Fig. 9, let the area of the piston a be 1 square inch; of b, 40 square inches. According to Pascal's law, 1 pound placed on a will balance 40 pounds placed on b. Suppose that a moves downwards 10 inches, then 10 cubic inches of water will be forced into the tube b. This will be distributed over the entire area of the tube , in the form of a cylinder whose cubical contents must be 10 cubic inches, whose base has an area of 40 square inches, and whose altitude must be = \ inch; that is, a movement of 10 inches of the piston a will cause a movement of inch in the piston b. This is another illustration of the well- known principle of machines: The force multiplied by the distance tJirough FIG - 9 - whicli it acts equals the weight multi- plied by the distance through which it moves, since, if 1 pound on the piston a represents the force /*, the equiva- lent weight W on b may be obtained from the equation IV X i = / 3 X 10, whence W= 40/ } = 40 pounds. Another familiar fact is also recognized, for the velocity MECHANICS OF FLUIDS. 11 ratio of P to W is 10 : \, or 40; and since in any machine the weight equals the force multiplied by the velocity ratio, WPx^, and when P= 1, W = 40. An interesting application of this principle is the hydraulic jack, by the aid of which one man can lift a very large load. 19. A Watson -Stillman hydraulic jack is shown in section in Fig. 10 (a). In this illustration, a is a lever that is depressed by the operator when he desires to raise the load. This lever freely fits a rectangular hole or socket in a shaft /; that extends clear through the ram head. The shaft b carries the crank c to which the piston d is hinged. A small valve is placed in the center of the piston and serves to open or close communication between the water space above and below the piston. The lower end of the ram is fitted with a small check valve, which normally is held closed by the small spring shown beneath the valve. The inside of the jack is filled with a mixture of alcohol and water, about 2 parts of alcohol to 3 parts of water, which prevents the freezing of the liquid when the jack is used in cold weather. The operation is as follows : The jack being clear down, as shown in Fig. 10 (a), it is placed under the load and the operator alternately depresses and raises the lever through its full range, the motion being limited by a projection on the under side of the lever. The slightest depression of the lever closes the valve in the pis- ton. The downward movement of the piston continuing, the water between the bottom of the piston and the end of the ram is subjected to a pressure that opens the valve in the end of the ram. This allows the water under pressure to pass into the space beneath the ram. The pressure so trans- mitted to the lower end of the ram forces it upwards. The lever having been depressed to its limit is again raised. This raises the piston, causes the piston valve to open, and allows the water above the piston to flow beneath it. The pressure of the water below the ram promptly closes the ram valve as soon as the piston commences to move upwards. The operation may now be repeated as often as required. 12 MECHANICS OF FLUIDS. As will readily be seen, the ram cannot descend under the weight of the load, as the only water passage leading from FIG. 10. the space below the ram is closed by the ram valve, which is kept tightly closed by the pressure of the water beneath 6 MECHANICS OF FLUIDS. 13 it and also by the spring. Hence, in order to lower the ram, this ram valve must be opened by some means. As previously mentioned, the lever has a projection on its under side to limit its travel and the travel of the piston. This projection is made of such a length that when the lever is down as far as it will go, the piston is still a short distance from the upper end of the ram valve stem. In order to lower the ram, the lever is withdrawn from its socket and inserted again with the projection upwards. It can now be depressed far enough to cause the end of the piston to strike and open the ram-head valve, as shown in Fig. 10 (d), where the parts of the jack are shown in the relative positions occupied while lowering. But before the water in the bottom of the ram can escape to the top, the valve in the piston must also be opened. This is accom- plished by a heavy spring that forces the sleeve e downwards. The lower end of the sleeve has a cotter working in a slot in the piston rod; the cotter bearing against the top of the valve forces the latter downwards. The sleeve is connected to the crank c by the so-called lowering wire f, in such a manner that the cotter will not strike the piston valve while the jack is raising the load. The ram will descend as long as the lever is kept hard down ; it can be stopped instantly by raising the lever slightly. 2O. If the area of the piston is 1 square inch and the area of the ram is 10 square inches, the velocity ratio will be 10. If the length of the lever between the hand of the oper- ator and the fulcrum is 10 times the length between the fulcrum and the piston, the velocity ratio of the lever will be 10, and the total velocity ratio of the hand to the piston will be 10 X 10 = 100. Taking the weight of the operator at 150 pounds, his whole weight to be thrown on the lever, the weight that can be raised is 150 X 100 = 15,000 pounds, or 7 tons. But if the average movement of the hand is 4 inches per stroke, it will require J-^ = 25 strokes to raise the load on the jack a distance of 1 inch, and it is again seen that what is gained in force is lost in speed. 14 MECHANICS OF FLUIDS. 6 21. Other applications of this principle are seen in vari- ous hydraulic machines used in boiler shops. A familiar example is the hand test pump used in testing boilers under hydraulic pressure. EXAMPLE 1. A vertical cylinder is tested for the tightness of its heads by filling it with water. A pipe whose inside diameter is ^ inch and whose length is 20 feet is screwed into a hole in the upper head and is then filled with water. What is the pressure per square inch on each head if the cylinder is 40 inches in diameter and 60 inches long ? SOLUTION. 40 2 X .7854 = 1,256.64 square inches = area of head. 1 X 60 X .03617 = 2.17 pounds pressure per square inch on the bottom head due to the weight of the water in the cylinder. (i) 2 X -7854 = .04909 square inch, the area of the pipe. .04909 X 20 X 12 X .03617 = .426 pound = the weight of water in the pipe = the pressure on a surface area of .04909 square inch. The pressure per square inch due to the water in the pipe is X -426 = 8.68 Ib. per sq. in. upon the upper head. Ans. The total pressure per square inch on the lower head is 8.68 + 2.17 = 10.85 Ib. Ans. EXAMPLE 2. In the other example, if the pipe be fitted with a piston weighing pound and a 5-pound weight be laid on it what is the pressure per square inch on the upper head ? SOLUTION. In addition to the pressure of .426 pound on the area of .04909 square inch, there is now an additional pressure on this area of 5 -h i = 5.25 pounds, and the total pressure on this area is .426 + 5.25 = 5.676 pounds. X 5.676 = 115.6 Ib. = the pressure per square . 047oy inch. Ans. 22. When calculating the weight that can be raised with a hydraulic jack, no allowance was made for the power lost in overcoming the friction between the cup leathers of the piston and the ram, and the cup leather of the ram and the cylinder; this varies according to the condition of the leathers, and, of course, the smoothness of the ram and cylinder; when the leathers are in good condition, the loss is about 5 per cent, of the total pressure on the ram ; when the leathers are old, stiff, and dirty, the loss may amount to 15 per cent, or more. MECHANICS OF FLUIDS. 15 SPECIFIC GRAVITY. 23. The specific gravity of a body is the ratio between its weight and the weight of a like volume of water. 24. Since gases are so much lighter than water, it is usual to take the specific gravity of a gas as the ratio between the weight of a certain volume of the gas and the weight of the same volume of air. EXAMPLE. A cubic foot of cast iron weighs 450 pounds; what is its specific gravity, a cubic foot of water weighing 62.42 pounds ? SOLUTION. According to the definition, 450 62.42 = 7.21. Ans. TABLE I. SPECIFIC GRAVITY AND WEIGHT PER CUBIC FOOT OF VARIOUS METALS. Substance. Specific Gravity. Weight per Cubic Foot. Pounds. Platinum 21.50 1 343 8 Gold 19.50 1 218.8 Mercury 13 60 850 Lead (ca'st) 11.35 709.4 Silver 10.50 656.3 Copper (cast) 8 79 549 4 Brass 8.38 523 8 Wrought Iron 7.68 480.0 Cast Iron 7 21 450 Steel 7.84 490 Tin (cast) 7.29 455.6 Zinc (cast) 6.86 428 8 Antimony 6 71 419 4 Aluminum 2.50 156*. 3 10 MECHANICS OF FLUIDS. TABLE n. SPECIFIC GRAVITY AND WEIGHT PER CUBIC FOOT OF VARIOUS WOODS. Substance. Specific Gravity. Weight per Cubic Foot. Pounds. Ash 845 52 80 Beech 852 53 . 25 Cedar 561 35 06 Cork 240 15 00 Ebony (American) 1 331 83.19 Lignum- vitas 1 333 83 30 Maple 750 46.88 Oak (old) 1 170 73 10 Spruce 500 31 25 Pine (yellow) ... .060 41.20 Pine (white) .554 34.60 Walnut 671 41 90 TABLE III. SPECIFIC GRAVITY AND WEIGHT PER CUBIC FOOT OF VARIOUS LIQUIDS. Substance. Specific Gravity. Weight per Cubic Foot. Pounds. Acetic acid 1 062 66 4 Nitric acid 1 217 76 1 Sulphuric acid Muriatic acid 1.841 1 200 115.1 75 Alcohol .800 50.0 Turpentine .870 54.4 Sea water (ordinary) 1 026 64 1 Milk 1.032 64.5 MECHANICS OF. FLUIDS. 17 25. The specific gravities of different bodies are given in printed tables; hence, if it is desired to know the weight of a body that cannot be conveniently weighed, calculate its cubical contents and multiply the specific gravity of the body by the weight of a like volume of water, remembering that a cubic foot of water weighs 62.^2 pounds. EXAMPLE 1. How much will 3,214 cubic inches of cast iron weigh ? Take its specific gravity as 7.21. SOLUTION. Since 1 cubic foot of water weighs 62.42 pounds, 3,214 cubic inches weigh 3,214 X 62.42 = 116.098 pounds. Then, 1,728 116.098 X 7.21 = 837.067 Ib. Ans. TABLE IV. SPECIFIC GRAVITY AND WEIGHT PER CTTBIC FOOT OF VARIOUS GASES AT 3S8f F. AND UNDER A PRESSURE OF 1 ATMOSPHERE. Substance. Specific Gravity. Weight per Cubic Foot. Pounds. Atmospheric air 1.0000 08073 Carbonic acid 1 5290 12344 Carbonic oxide 9674 07810 Chlorine ' 2 . 4400 19700 Oxygen 1.1056 .08925 Nitrogen 9736 07860 Smoke (bituminous coal) .1020 00815 Smoke (wood) .0900 .00727 *Steam at 212 F 4700 03790 Hydrogen .0692 .00559 * The specific gravity of steam at any temperature and pressure compared with air at the same temperature and pressure is .622. 18 MECHANICS OF FLUIDS. 6 EXAMPLE 2. What is the weight of a cubic inch of cast iron ? SOLUTION. ^|| X 7.21 = .26044 Ib. Ans. One cubic foot of pure distilled water at a temperature of 39.2 Fahrenheit weighs 62.425 pounds, but t lie value usually taken in making calculations is 62.5 pounds. EXAMPLE 3. What is the weight in pounds of 7 cubic feet of oxygen ? SOLUTION. One cubic foot of air weighs .08073 pound, and the specific gravity of oxygen is 1.1056, compared with air; hence, .08073 X 1.1056 X 7 = .62479 Ib., nearly. Ans. TABLE V. SPECIFIC GRAVITY AND WEIGHT PER CUBIC FOOT OF VARIOUS SUBSTANCES. Substance. Specific Gravity. Weight per Cubic Foot. Pounds. Emery . . . . 4.00 250 Glass (average) Chalk 2.80 2 78 175 174 Granite 2.65 166 Marble 2.70 169 Stone (common) 2.52 158 Salt (common) 2.13 133 Soil (common) 1.98 124 Clay 1.93 121 Brick .... 1.90 118 Plaster of Paris (average) 2.00 125 Sand . . 1.80 113 MECHANICS OF FLUIDS. BUOYANT EFFECTS OF WATER. 26. In Fig. 11 is shown a 6-inch cube entirely submerged in water. The lateral pressures are equal and in opposite directions. The upward pressure acting on the lower surface of the cube is 6 X 6 X 21 X .03617 ; the downward pressure acting on the top of the cube is 6 X 6 X 15 X .03617; and the difference is 6 X 6 X 6 X .03617, which equals the volume of the cube in cubic inches times the weight of 1 cubic inch of water. That is, the upward pres- sure exceeds the downward pressure by the weight of a volume of water equal to the volume of the body. This excess of upward pressure over the downward pressure acts against gravity ; that is, the water presses the body upwards with a greater force than it presses it downwards ; consequently, if a body is immersed in a fluid, it ^vill lose in weight an amount equal to the weight of the fluid it displaces. This is called the principle of Archi- medes, because it was first stated by him. 27. This principle may be experimentally demonstrated with the beam scales, as shown in Fig. 12. From one scale pan suspend a hollow cylinder of metal t, and below that a solid cylin- der a of the same size as the hollow part of the upper cylinder. Put weights in the other scale pan until they ex- actly balance the two in water, the scale pan FIG. 12. cylinders. If a be immersed 20 MECHANICS OF FLUIDS. 6 containing the weights will descend, showing that a has lost some of its weight. Now fill t with water, and the volume of water that can be poured into t will equal that displaced by a. The scale pan that contains the weights will gradually rise until t is filled, when the scales again balance. If a body be lighter than the liquid in which it is immersed, the upward pressure will cause it to rise and project partly out of the liquid, until the weight of the body and the weight of the liquid displaced are equal. If the immersed body be heavier than the liquid, the downward pressure plus the weight of the body will be greater than the upward pressure, and the body will fall downwards until it touches bottom or meets an obstruction. If the weights of equal volumes of the liquid and the body are equal, the body will remain stationary and will be in equilibrium in any position or depth beneath the surface of the liquid. 28. An interesting experiment in confirmation of the above facts may be performed as follows : Drop an egg into a glass jar filled with fresh water. The mean density of the egg being a little greater than that of the water, it will fall to the bottom of the jar. Now, dissolve salt in the water, stirring it so as to mix the fresh and salt water. The salt water will presently become denser than the egg and the egg will rise. Now, if fresh water be poured in until the egg and water have the same density, the egg will remain stationary in any position that it may be placed below the surface of the water. 29. The principle of Archimedes gives a very easy and accurate method of finding the volume of an irregularly shaped body. Thus, subtract its weight in water from its weight in air and divide by .03617; the quotient will be the volume in cubic inches, or divide by 62.5 and the quo- tient will be the volume in cubic feet. If the specific gravity of the body is known, its cubical contents can be found by dividing its weight by its specific gravity, and then dividing again by either .03617 or 62.5. MECHANICS OF FLUIDS. EXAMPLE. A certain body has a specific gravity of 4.38 and weighs 76 pounds; how many cubic inches are there in the body ? 76 SOLUTION. 4. 38 X. 03617 = 479.72 cu. in. Ans. THE HYDROMETER. 30. Instruments called hydrometers are in general use for determining quickly and accurately the specific gravities of liquids and some forms of solids. They are of two kinds, viz. : (1) Hydrometers of constant weigJit, as Beaume's ; (2) hydrometers of constant volume, as Nicholson's. A hydrometer of constant weight is shown in Fig. 13. It consists of a glass tube, near the bottom of which are two bulbs. The lower and smaller bulb is loaded with mercury or shot, so as to cause the instrument to remain in a vertical position when placed in the liquid. The upper bulb is filled with air, and its volume is such that the whole instrument is lighter than an equal volume of water. The point to which the hydrometer sinks when placed in water is usually marked, the tube being graduated above and below in such a manner that the specific gravity of the liquid can be read directly. It is customary to have two instruments, one with the zero point near the top of the stem for use in liquids heavier than water, and the other with the zero point near the bulb for use in liquids lighter than water. These instruments are more commonly used for determin- ing the degree of concentration or dilution of certain liquids, as acids, alcohol, milk, solutions of sugar, etc., rather than their actual specific gravities. They are then known as acidometers, alcoholometers, lactometers, saccharometers, sali- nometers, etc. , according to the use to which they are put. 31. Hydrometers of constant volume are not in com- mon use and are rarely found outside of laboratories. 22 MECHANICS OF FLUIDS. HYDROKI^TETICS. FLOW OF WATER IX PIPES. 32. Experience has demonstrated that for satisfactory work, the flow of water in the suction pipes of boiler feed- pumps and other comparatively small pumps should not exceed 200 feet per minute, and it should not be more than 500 feet in the delivery pipe for a duplex pump, or 400 feet for a single-cylinder pump. Knowing the volume of water that is to flow through or to be discharged from a pipe in 1 minute, the area of the suction and delivery pipes can readily be determined. 33. The volume of water in cubic feet discharged from a pipe in 1 minute is equal to the velocity in feet per minute times the area of the pipe in square feet. Then, the area of , volume in cubic feet per minute . the pipe equals . : -. ; . As there velocity in feet per minute are 144 square inches in a square foot, the area of the pipe . 144 X volume in cubic feet per minute in square inches is . velocity in feet per minute 34. If the volume is expressed in gallons per minute, then, as there are 7.48 gallons in a cubic foot, the area of the 144 X volume in gallons* pipe in square inches will be = . 7.48 X velocity in feet per minute Hence, the following rule, where n = gallons per minute ; v = velocity in feet per minute ; A = area of pipe in square inches. Rule 4. To find the area of a pipe in square inches, divide 19.25 times the number of gallons per minute by the velocity in feet per minute. 19.25 Or, A = v * The gallon here referred to is the Winchester or wine gallon of 231 cubic inches capacity and in common use in the United States of America. 6 MECHANICS OF FLUIDS. 23 EXAMPLE. If a duplex pump is used, what area of feed-pipe is required for a boiler into which 25 gallons of water per minute is to be pumped ? SOLUTION. The allowable velocity being 500 feet, by applying rule 4, we get 19.25 X 25 = - = - 9625 s - ln - Ans - 35. The quantity of water, expressed in gallons, that will flow through a given pipe in 1 minute, its velocity being known, is given by the following rule: Rule 5. Multiply the area of the pipe in square inches by the velocity in feet per minute. Divide the product by 19. 25. Av EXAMPLE. How many gallons of water per minute will flow through a pipe having an area of 2 square inches, the velocity of flow being 450 feet per minute ? SOLUTION. Applying the rule just given, we get Ans - 36. The velocity with which water will flow through the delivery pipe of a pump when the area of the water cylinder, the area of the delivery pipe, and the piston speed of the pump are known, is given by the following rule, where v = velocity in feet per minute ; A = area of delivery pipe in square inches; a = area of water piston in square inches; S = piston speed in feet per minute. Rule 6. Multiply the area of the water piston by the piston speed, and divide this product by the area of the delivery pipe. aS Or, p.p-p EXAMPLE. If the water piston of a pump has an area of 12 square inches and moves at a speed of 100 feet per minute, what will be the H. S. I. 19 24 MECHANICS OF FLUIDS. 6 velocity of the water in the delivery pipe if the latter has an area of 2 square inches ? SOLUTION. Applying rule 6, we get v = 12xl = 600 ft. per min. Ans. STANDARD PIPE DIMENSIONS. 37. The great majority of the pipes used about steam plants are made of wrought iron and are almost invariably made in accordance with the Briggs standard. It will be noticed that the diameter of the pipe by which it is known to the trade is not the actual diameter; hence, in calcula- ting the amount of water that will flow through a pipe, the actual diameter or actual internal area must be taken from Table VI. 38. As wrought-iron pipes are not made in sizes differing from those given in the table, it will be apparent that only in rare instances can a pipe be selected that will have the area calculated by rule 4. In practice the nearest commer- cial size of pipe would be selected. EXAMPLE. What commercial size of delivery pipe should be used for a single-cylinder pump to deliver 90 gallons of water per minute ? SOLUTION. For a single-cylinder pump, the velocity of flow should not be more than 400 feet per minute. Then, applying rule 4, we get 19.25 X 90 = 400 = 4 - 33 sc l- in - According to Table VI, the commercial size of pipe having an area nearest this is 2 inches; therefore, a 2^-inch pipe should be used. Ans. Rule 4 will be found to agree quite closely with the prac- tice of the leading pump manufacturers. In case they should, however, recommend a larger size of pipe, it is advisable to follow their advice. Pipes made in accordance with the Briggs standard, when below 1 inch nominal size, are butt-welded and proved to 300 pounds per square inch by hydraulic pressure. Pipes above 1 inch are lap-welded and proved to 500 pounds. MECHANICS OF FLUIDS. TABLE VI. TABLE OF STANDARD DIMENSIONS OF WROUGHT-IRON WELDED PIPES. Nominal Diameter. Inches. Actual Internal Diameter. Inches. Actual Internal Area. Square Inches. Actual External Diameter. Inches. Number of Threads Per Inch. * .27 .057 .40 27 i .36 .104 .54 18 t .49 .192 .67 18 1 .62 .305 .84 14 t .82 .533 1.05 14 1 1.05 .863 1.31 Hi H 1.38 1.496 1.66 Hi ii 1.61 2 . 038 1.90 11* 2 2.07 3.355 2.37 Hi n 2.47 4.783 2.87 8 3 3.07 7.388 3.50 8 3i 3.55 9.887 4.00 8 4 4.07 12.730 4.50 8 4* 4.51 15.939 5.00 8 5 5.04 19 990 5.56 8 6 6.06 28.889 6.62 8 7 7.02 38.737 7.62 8 8 7.98 50 . 039 8.62 8 9 9.00 63 . 633 9.62 8 10 10 . 02 78.838 10.75 8 PIPE FITTINGS. 39. In piping a steam plant, it is rarely possible to run the piping in a straight line, it being usually necessary to introduce one or more elbows or similar fittings to reach the point desired. The effect of T and L fittings is to increase the resistance to the flow of the water through the pipes, 26 MECHANICS OF FLUIDS. 6 thus requiring the pump to do more work for the same quan- tity of water delivered. As pumps for boiler feeding are always built with a large steam cylinder, there is enough excess of power to allow the pump under ordinary conditions to force the required quan- tity of water through the pipe. On the suction side of the pump, however, the force impelling the water to flow into the pump is quite small, and a very slight resistance will be sufficient to interfere with the flow of the water into the pump. Hence, it is important that the suction pipe be as straight as possible ; if it is impossible to make it straight, larger sizes of pipe should be used, or easy bends with as large a radius as possible should be substituted for the elbows. This applies especially to pumps that must lift the water more than 10 feet. It is impossible to lay down any hard-and-fast rules as to what numbers of elbows should not be exceeded in a suction and delivery pipe ; judgment will have to be used. Generally speaking, they should be as few as possible. EXAMPLES FOR PRACTICE. 1. Suppose a cylinder to be filled with water and placed in an upright position. If the diameter of the cylinder is 19 inches and its total length inside 26 inches, what will be the total pressure on the bottom when a pipe inch in diameter and 12 feet long is screwed into the cylinder head and filled with water ? The pipe is vertical. Ans. 1,743.2 Ib. 2. In the last example, what is the total pressure against the upper head? Ans. 1,476.6 Ib. 3. In example 1, a piston is fitted to the upper end of the pipe and an additional force. of 10 pounds is applied to the water in the pipe. What is the total pressure (a) on the bottom of the cylinder ? (b) on the upper head ? j (a) 16,184 Ib. Lns ' 1 (b) 15,906 Ib. 4. In example 3, what is the pressure per square inch in the pipe 2 inches from the upper cylinder head ? Ans. 56.0656 Ib. per sq. in. 5. A water tower 80 feet high is filled with water. A pipe 4 inches in diameter is so connected to the side of the tower that its center is 3 feet from the bottom. If the pipe is closed by a flat cover, what is the total pressure against the cover ? Ans. 420 Ib. MECHANICS OF FLUIDS. 27 6. In the last example, what is the upward pressure per square inch 10 feet from the bottom of the tower ? Ans. 30.3828 Ib. per sq. in. 7. A cube of wood, one edge of which measures 3 feet, is sunk until the upper surface is 40 feet below the level of the water ; what is the total force that tends to move the cube upwards ? Ans. 1,687.5 Ib. 8. A body weighs 72 pounds when immersed in water and 321 pounds in air; what is its volume in cubic feet ? Ans. 3.984 cu. ft. 9. The specific gravity of a body being 9.5 and its weight 81 pounds, what is its volume in cubic inches ? Ans. 235.73 cu. in. 10. What commercial size of pipe should be used for the suction and delivery pipes of a single-cylinder pump to deliver 50 gallons of water per minute ? . ( Suction pipe, 2^ in. Ans. {,-..,. . ~ . ( Delivery pipe, 2 in. 11. How many gallons per hour, at a velocity of 400 feet per minute, will flow through a 1-inch pipe? Ans. 1,076 gal., nearly. 12. The water piston of a pump is 3 inches ; the piston speed is 54 feet per minute; and a l|-inch delivery pipe is used; what is the velocity in the delivery pipe ? Ans. 255 ft. per min., nearly. PNEUMATICS. PRESSURE OF GASES. 40. Pneumatics is that branch of mechanics that treats of the properties and pressures of gases. 41. The most striking feature of all gases is their great expansibility. If we inject a quantity of gas, ho^vever small, into a vessel, it will expand and fill that vessel. If a blad- der or football be partly filled with air and placed under a glass jar (called a receiver), from which the air has been exhausted, the bladder or foot- ball will immediately expand, as shown in Fig. 14. The force that a gas always exerts when confined in a limited 28 MECHANICS OF FLUIDS. 6 space is called tension. The word tension in this case means pressure, and is only used in this sense in reference to gases. 4:2. As water is the most common type of fluids, so air is the most common type of gases. It was supposed by the ancients that air had no weight, and it was not until about the year 1650 that the contrary was proved. A cubic inch of air, under ordinary conditions, weighs .31 grain, nearly. At a temperature of 32 F. and a pressure of 14.7 pounds per square inch, the ratio of the weight of air to water is about 1 : 774 ; that is, air is only -^^ as heavy as water. It has been shown that if a body were immersed in water and weighed less than the volume of water displaced, the body would rise and project partly out of the water. The same is true, to a certain extent, of air. If a vessel made of light material is filled with a gas lighter than air, so that the total weight of the vessel and gas is less than the air they displace, the vessel will rise. It is on this principle that balloons are made. PRESSURE OF THE ATMOSPHERE. 43. Since air has weight, it is evident that the enormous quantity of air that constitutes the atmosphere must exert a considerable pressure on the earth. This is easily proved by taking a long glass tube closed at one end and filling it with mercury. If the finger be placed over the open end so as to keep the mercury from running out and the tube inverted and placed in a cup of mercury, as shown in Fig. 15, the mercury will fall, then rise, and after a few oscillations will come to rest at a height above the top of the mercury in the cup equal to about 30 inches at sea level. This height will always be the same under the same atmospheric conditions. Now, if the atmosphere has weight, it must press upon the upper surface of the mercury in the cup with equal intensity upon every square unit, except upon that part of the surface occupied by the tube. In order that there may be equilibrium, the weight of the mercury in the tube must be equal to the pressure of the air MECHANICS OF FLUIDS. 29 upon a portion of the surface of the mercury in the cup equal in area to the inside of the tube. Suppose that the area of the inside of the tube is 1 square inch, then, since mercury is 13.6 times as heavy as water, the weight of the mercurial column is .03617 X 13.6 X 30 = 14.7574 pounds. The actual height of the mercury is a little less than 30 inches, and the actual weight of a cubic inch of distilled water is a little less than .03617 pound. When these considerations are taken into ac- count, the average weight of the mercurial column at the level of the sea, when the temperature is 60 F., is 14.69 pounds, or prac- tically 14.7 pounds. Since this weight, when exerted upon 1 square inch of the liquid in the glass, just produces equilibrium, it is plain that the pressure of the outside air is 14. 7 pounds upon every square inch of surface. 44. Vacuum. The space be- tween the upper end of the tube and the upper surface of the mer- cury is called a vacuum, meaning that it is an entirely empty space and does not contain any sub- stance solid, liquid, or gaseous. FIG - 15 - If there were a gas of some kind there, no matter how small the quantity might be, it would expand and fill the space, and its tension would cause the column of mercury to fall and become shorter, according to the amount of gas or air present. The condition then existing in the space would be called a partial vacuum. 45. The Measurement of Vacuum. The degree to which the air has been exhausted from a closed vessel in 30 MECHANICS OF FLUIDS. (i which there is a partial vacuum is measured by the height to which a mercurial column in a vertical tube, whose top is connected to the vessel, will rise under the pressure of the atmosphere. Thus, when we say that there is a vacuum of 29 inches in a vessel, we mean that enough air has been exhausted from the vessel to enable the surrounding air to support a mercurial column 29 inches high, as shown in Fig. 16, where A is the vessel in which there is a partial vacuum. In other words, the pressure in the vessel is less than the pressure of the atmosphere by a column of mer- cury 29 inches in height. When there is a partial vacuum in- side a closed vessel, the air remaining in the vessel is under a tension, or pres- sure, less than its original tension of 14.7 pounds per square inch. This ten- sion, in inches of mercury, is equal to the difference between the height at which the mercurial column connected to the vessel stands and that at which it would stand if the vacuum were per- fect. Consider that the mercury col- umn will be in equilibrium when the hydrostatic pressure on its base equals the atmospheric pressure. The hydro- static pressure on the base is the sum of two pressures: (1) The pressure due to the weight of the mercury column; (2) the pressure in the space above the mercury. From this it follows that if the atmospheric pressure and the pressure due to the weight of the mercurial column are given, the pressure above the column must be that due to their difference. As the atmosphere will force a mercurial column to a height of 30 inches when there is a perfect vac- uum above it, it follows that to find the pressure in a vessel FIG. 16. 6 MECHANICS OF FLUIDS. 31 in which there is a partial vacuum, the number of inches of height of the mercurial column is to be subtracted from 30. Thus, if the vacuum in a vessel is 26 inches, the pressure in the vessel is 30 26 = 4 inches of mercury. 46. In practice it is convenient to always use the same unit of pressure, which is 1 pound per square inch. We know that 14.7 pounds per square inch will support a mercu- rial column 30 inches high. Hence, 1 inch of height of a 14 7 mercurial column represents a pressure of ' = .49 pound oO per square inch. Then, to reduce inches of mercury to pounds per square inch, multiply their number by .49. EXAMPLE. A gauge shows a vacuum of 22 inches in a condenser. What is the absolute pressure in the condenser ? SOLUTION. The pressure, in inches of mercury, is 30 22 = 8 inches, or 8 X .49 = 3.92 Ib. per sq. in. Ans. 47. The Vacuum Gauge. For engineering work, the glass tube of Fig. 16 would be a rather inconvenient form of gauge for measuring the vacuum. Hence, special metallic gauges, known as vacuum gauges, have been designed. Their dial is graduated to show the degree of vacuum, in inches of mercury. In steam-engineering work, they are used chiefly in connection with condensers for steam engines. It should be borne in mind that a vacuum gauge does not indicate the pressure in the vessel to which it is attached, but instead shows, in inches of mercury, how much the pressure has been lowered below the atmospheric pressure. In this respect a vacuum gauge differs essentially from a pressure gauge, which shows how much the pressure has been increased either above the pressure of a perfect vac- uum, which is zero, as in case of a gauge registering the pressure of the atmosphere, or above the atmospheric pres- sure, as in case of the ordinary pressure gauge. 48. If the tube of Fig. 15 had been filled with a liquid lighter than mercury, the height of the column required to balance the atmospheric pressure would be greater. The MECHANICS OF FLUIDS. height of the column depends on the specific gravity of the liquid. Thus, if the liquid had a specific gravity of 1, as water, its height would be = 408 inches, or 34 feet. This means that if a tube be filled with water, inverted, and placed in a dish of water, in a manner similar to the experiment made with the mercury, the height of the column of water will be 34 feet. 49. The Barometer. As is well known, the atmosphere does not exert exactly the same pressure at all times; the pressure varies with conditions. As the height of the mercury in the glass tube of Fig. 15 depends chiefly on the atmospheric pressure, it is evident that an instrument constructed on this principle will indicate the varying pressure by the height of the column. Such an instrument is called a mercurial barometer. 50. A standard form of barometer is shown in Fig. 17. The barometer is simply a pressure gauge that registers the pressure of the air. In this case the cup and tube at the bottom are pro- tected by a brass or iron casing. At the top of the tube is a graduated scale. Attached to the casing is an accurate thermometer for determin- ing the temperature of the outside air at the time the barometric observation is taken. This is necessary, since mercury expands when the temperature is increased and contracts when the temperature falls ; for this reason a standard temperature is assumed, and all barometer read- ings are reduced to this temperature. This standard temperature is usually taken at 32 F., at which temperature the height of the mercurial column is 30 inches under normal conditions at sea level. Another 6 MECHANICS OF FLUIDS. 33 correction is made for the altitude of the place above sea level, and a third correction for the effects of capillary attraction. 51. A mercurial barometer is not only a very bulky instrument, but is also quite heavy and, hence, is not very well adapted for transportation. To overcome these draw- backs, a form of portable barometer known as an aneroid barometer, has been designed, which operates on a some- what different principle. Such a barometer is shown in FIG. 18. Fig. 18. The principal part of the aneroid barometer is a cylindrical, air-tight box of metal closed by a corrugated top of thin, elastic metal. The air is exhausted from the box, which is then sealed. Evidently, the pressure of the air on the outside of the cover will cause the cover to curve inwards, as the space inside of the cover is void of pressure, until the 34 MECHANICS OF FLUIDS. 6 resistance due to the elasticity of the cover, aided by the resistance of a spring beneath it, is equal to the force exerted by the air. Now, if the air pressure increases, the cover will be curved further inwards ; if the air pressure decreases, the elasticity of the cover, aided by the spring beneath, will cause a reduction of its inward curvature. These movements of the cover are transmitted and multiplied by a combination of delicate levers that act on the index hand shown in the figure and cause it to move to the right or left over a grad- uated scale. These barometers are compensated (self-correcting) for variations in temperature. The better grade of these instru- ments have two graduations; the inner graduation corre- sponds to the graduations of the mercurial barometer; that is, it reads to inches of mercury. The outer graduation represents elevations above and below sea level, the distance between each graduation line representing a difference in elevation of 10 feet. These instruments are made in various sizes, from the size of a watch up to 8 or 10 inches in diameter. They are very portable, occupying but a small space, are very light, and are quite delicate. 52. Both the mercurial and the aneroid barometers operate on the same general principle; viz., that if two opposite forces act on a body, it will be in equilibrium when the forces are equal and will be set in motion when they are unequal, provided the difference in the magnitude of the two forces is sufficient to overcome the resistance. The two styles of barometer differ from each other only in the method by which equilibrium is established. In the mercurial barometer, the weight of the mercurial column inside the tube equalizes the outside air pressure; in the aneroid barometer, the outside pressure is equalized by the resistance of the flexible cover and spring beneath it. 53. Variations of Pressure at Different Elevations. With air, as with water, the lower we get the greater is the pressure, and the higher we get the less is the pressure. At 6 MECHANICS OF FLUIDS. 35 the level of the sea, the height of the mercurial column is about 30 inches; at 5,000 feet above the sea, it is 24.7 inches; at 10,000 feet above the sea, it is 20.5 inches; at 15,000 feet, it is 16.9 inches; at 3 miles, it is 16.4 inches; and at 6 miles above the sea level, it is 8.9 inches. Air being an elastic fluid, the density or weight of the atmosphere also varies with the altitude.; that is, a cubic foot of air at an elevation of 5,000 feet above the sea level will not weigh as much as a cubic foot at sea level. This is proved conclusively by the fact that at a height of 3 miles the mercurial column measures but 15 inches, indicating that half the weight of the entire atmosphere is below that height. It is known that the height of the earth's atmos- phere is at least 50 miles; hence, the air just before reaching the limit must be in an exceedingly rarefied state. It is by means of barometers that great heights are measured. The aneroid barometer has the heights marked on the dial, so that they can be read directly. With the mercurial barom- eter, the heights must be calculated from the reading. 54. Pressure In Different Directions. The atmos- pheric pressure is everywhere present and presses all objects in all directions with equal force. If a book is laid upon the table, the air presses upon it in every direction with an equal average force of 14.7 pounds per square inch. It would seem as though it would take considerable force to raise a book from the table, since, if the size of the book were 8 inches by 5 inches, the pressure upon it would be 8 X 5 X 14.7 = 588 pounds; but there is an equal pressure beneath the book that counteracts the pressure on the top. It would now seem as though it would require a great force to open the book, since there are two pressures of 588 pounds each acting in opposite directions and tending to crush the book ; so it would, but for the fact that there is a layer of air beneath each leaf, which acts upwards and downwards with a pressure of 14. 7 pounds per square inch. If a piece of flat glass be laid upon a flat surface that has been previously moistened with water, it will require 36 MECHANICS OF FLUIDS. 6 considerable force to separate them; this is because the water helps to fill the pores in the flat surface and glass and thus creates a partial vacuum between the glass and the surface, thereby reducing the counter pressure beneath the glass. 55. Tension of Gases. In Fig. 15, the space above the column of mercury was said to be a vacuum, and it was also stated that if any gas or air were present, it would expand and its tension would force the column of mercury down- wards. If sufficient gas be admitted to cause the mercury to stand at 15 inches, the tension of the gas is evidently 14 7 =7.35 pounds per square inch, since the pressure of 2 the outside air, 14.7 pounds per square inch, now balances only 15 instead of 30 inches of mercury ; that is, it bal- ances only one-half as much as it would if there were no gas in the tube ; therefore, the pressure (tension) of the gas in the tube is 7.35 pounds. If more gas be forced into the tube until the top of the mercurial column is just level with the mercury in the cup, the gas in the tube will then have a tension equal to the outside pressure of the atmosphere. Suppose that the bottom of the tube is fitted with a piston, and that the total length of the inside of the tube is 36 inches. If the piston be shoved upwards so that the space occupied by the gas is 18 inches long instead of 36 inches, the temperature remaining the same as before, it will be found that the tension of the gas within the tube is 29.4 pounds. It will be noticed that the volume occupied by the gas is only half that in the tube before the piston was moved, while the pressure is twice as great, since 14.7 X 2 = 29.4 pounds. If the piston be shoved up, so that the space occupied by the gas is only 9 inches instead of 18 inches, the temperature still remaining the same, the pressure will be found to be 58.8 pounds per square inch. The volume has again been reduced one-half and the pres- sure increased twofold, since 29.4 X 2 58.8 pounds. The space now occupied by the gas is 9 inches long, whereas, 6 MECHANICS OF FLUIDS. 37 before the piston was moved, it was 36 inches long; as the tube is assumed to be of uniform diameter throughout its length, the volume is now -^ = ^ of its original volume, and its pressure is = 4 times its original pressure. More- over, if the temperature of the confined gas remains the same, the pressure and volume will always vary in a similar way. 56. Absolute and Gauge Pressures. From the above explanation, it will be apparent that the pressure in the tube is the pressure above that of a perfect vacuum. Pressures reckoned above vacuum are called absolute pressures. The only pressure gauges that indicate absolute pressures are the mercurial and aneroid barometers; ordinary pressure gauges, such as the common steam gauge, indicate pressures above the pressure of the atmosphere. Pressures above that of the atmosphere are commonly called gauge pressures. Gauge pressures are changed to absolute pressures by adding 14.7 pounds per square inch to their readings. Truly speak- ing, the pressure indicated by a barometer, reduced to pounds pressure, should be added to the gauge pressure, since the value 14. 7 pounds only represents the mean atmos- pheric pressure under normal conditions at sea level. 57. Mariotte's Law. The law that states the effect of compressing and expanding gases is called Mariotte's law and is as follows : The temperature remaining the same, the volume of a given quantity of gas varies inversely as the absolute pressure. The meaning of the law is this: If the volume of a gas be diminished to , , \, etc. of its former value, the ten- sion will be increased 2, 3, 5, etc. times, or if the outside pressure be increased 2, 3, 5, etc. times, the volume of the gas will be diminished to , , ^, etc. of its original volume, the temperature remaining constant. It also means that if a gas is under a certain pressure, and this pressure is dimin- ished to , , -j 1 ^, etc. of its original intensity, the volume of 38 MECHANICS OF FLUIDS. 6 the confined gas will be increased 2, 4, 10, etc. times its tension decreasing at the same rate. Suppose 3 cubic feet of air to be under a pressure of 60 pounds per square inch in a cylinder fitted with a mov- able piston ; then the product of the volume and pressure is 3 x 60 = 180. Let the volume be increased to 6 cubic feet; then' the pressure will be 30 pounds per square inch, and 30 x 6 = 180, as before. Let the volume be increased to 24 cubic feet ; it is then - 2 ^ 4 - = 8 times its original volume, and the pressure is of its original pressure, or 60 X = 7 pounds, and 24 X 7 = 180, as in the two preceding cases. It will now be noticed that if a gas be enclosed within a confined space and allowed to expand without losing any heat, the product of the pressure and the corresponding volume for any one position of the piston is the same as for any other position. If the piston were forced inwards so as to com- press the air, the same results would be obtained. 58. Application of Mariotte's Law. If the volume of the vessel and the pressure of the gas are known, and it is desired to know the pressure after the first volume has been changed : Rule 7. Divide the product of the first, or original, volume and pressure by the new volume; the result will be the new pressure. Or, let / = original absolute pressure ; /, = final absolute pressure; v = volume corresponding to the pressure / ; v l = volume corresponding to the pressure / r Then, /,=?. , EXAMPLE. At the point of cut-off in a steam engine, the amount of steam in the cylinder is 862 cubic inches. The pressure at this point is 120 pounds per square inch. What will be the pressure of the steam when the piston has reached the end of its stroke and the volume is 1,800 cubic inches ? 6 MECHANICS OF FLUIDS. 39 SOLUTION. Applying the rule, we get ' A = 86 i ^r! 20 = 57 - 47 lb - P er sq- in - absolute. Ans. l,oUU 59. If it is required to determine the volume after a change in the pressure: Rule 8. Divide the product of tJie original volume and pressure by the new pressure; the result will be the new volume. Or, using the same letters as before, - EXAMPLE 1. At the commencement of compression, the volume of the steam is 380 cubic inches and the pressure is 18 pounds per square inch. At the end of compression, the pressure is 112 pounds per square inch. What is the final volume ? SOLUTION. Applying the rule, we get 380 X 18 112 = 61.07 cu. in. Ans. EXAMPLE 2. A vessel contains 10 cubic feet of air at a pressure of 15 pounds per square inch and has 25 cubic feet of air of the same pressure forced into it ; what is the resulting pressure ? SOLUTION. The original volume = 10 + 25 = 35 cubic feet ; the original pressure is 15 pounds per square inch; the final volume is 10 cubic feet. Hence, applying rule 7, we get OK \/ 1 5 p\ = ^r = 52.5 lb. per sq. in. absolute. Ans. It must be remembered that in the preceding examples the temperature is supposed to remain constant. When the temperature changes during expansion and compression, the problem becomes a rather difficult one and cannot be solved by ordinary arithmetic. EXAMPLES FOR PRACTICE. 1. A vessel contains 25 cubic feet of gas at a pressure of 18 pounds per square inch ; if 125 cubic feet of gas having the same pressure are forced into the vessel, what will be the resulting pressure ? Ans. 108 lb. per sq. in. H. 8. L 20 40 MECHANICS OF FLUIDS. 6 2. The volume of steam in the cylinder of a steam engine at cut-off is 1.85 cubic feet and the pressure is 85 pounds per square inch; the pressure at the end of the stroke is 25 pounds per square inch. What is the new volume? Ans. 4.59 cu. ft. 3. A receiver contains 180 cubic feet of gas at a pressure of 20 pounds per square inch ; if a vessel holding 12 cubic feet be filled from the larger vessel until its pressure is 20 pounds per square inch, what will be the pressure in the larger vessel ? Ans. 18 Ib. per sq. in. 4. A spherical shell has a part of the air within it removed, forming a partial vacuum ; if the outside diameter of the shell is 18 inches and the pressure of the air within is 5 pounds per square inch, what is the total pressure tending to crush the shell ? Ans. 9,873.42 Ib. PNEU1VIATIC MACHINES. THE AIR PUMP. 6O. The air pump is an instrument for removing air from a given space. A section of the principal parts is FIG. 19. shown in Fig. 19 and the complete instrument in Fig. 20. The closed vessel R is called the receiver, and the space MECHANICS OF FLUIDS. 41 that it encloses is that from which it is desired to remove the air. It is usually made of glass, and the edges are ground so as to be perfectly air-tight. When made in the form shown, it is called a bell-jar receiver. The receiver rests upon a horizontal plate, in the center of which is an opening communi- cating with the pump cylinder C by means of the passage tt. The pump piston accu- rately fits the cylinder, and has a valve V opening upwards. Where the passage // joins the cylinder, another valve V is placed, which also opens upwards. When the piston is raised, the valve V closes, and, since no air can get into the cylinder from above, the piston leaves a vacuum behind it. The pressure upon V being now removed, the tension of the air in the receiver R causes V to rise; the air in the receiver and passage 1 1 then expands so as to occupy the additional space provided by the upward movement of the piston. The piston is now pushed down, the valve V closes, the valve V opens, and the air in C escapes. The lower valve V is sometimes supported, as shown in Fig. 19, by a metal rod passing through the piston and fitting it some- what tightly. When the piston is raised or lowered, this rod moves with it.. A button near the upper end of the rod confines its motion within very narrow limits, the pis- ton sliding upon the rod during the greater part of the journey. In the complete form of the instrument shown 42 MECHANICS OF FLUIDS. 6 in Fig. 20, communication between receiver and pump is made by means of the tube /. 61. Degrees and limits of Exhaustion. Suppose that the volume of R and t together is four times that of C, Fig. 19, and that there are, say, 200 grains of air in R and /, and 50 grains in C when the piston is at the top of the cylin- der. At the end of the first stroke, when the piston is again at the top, 50 grains of air in the cylinder C will have been removed, and the 200 grains in R and t will occupy the space R, t, and C. The ratio between the sum of the spaces R and t and the total space R -f t + C is ; hence, 200 X -| = 160 grains = the weight of air in R and t after the first stroke. After the second stroke, the weight of the air in R and t would be (200 X |) X f = 200 X (|) s = 200 X if = 128 grains. At the end of the third stroke, the weight would be [200 X (|) 2 ] X - = 200 X (f) s = 200 X T 6 / T = 102.4 grains. At the end of n strokes the weight would be 200 X ()". It is evident that it is impossible to remove all the air that is contained in R and t by this method. It requires an exceed- ingly good air pump to reduce the tension of the air in R to -5*5- inch of mercury. When the air has become so rarefied as this, the valve V will not lift, and, consequently, no more air can be exhausted. 63. The Dashpot. The pressure of the atmosphere is utilized in Corliss engines to close the steam valves rapidly. For this purpose a so-called dashpot is used. The dash- pot of a Bullock-Corliss engine is shown in two positions in Fig. 21. It consists of the base A, which is fastened to the floor, and a plunger B pivoted to the end of a crank-arm keyed to the stem of the steam valve. The base is accurately turned and bored to fit the plunger; packing rings are fitted to both the plunger and the base in order to make an air-tight joint. The annular space around the central stationary piston is in communication with the outside air by a small passage, the opening of which can be increased or decreased, at will, MECHANICS OF FLUIDS. 43 by means of the valve C. In operation, the plunger is first in the position shown in Fig. 21 (b}. It is then picked up by the valve gear of the engine and drawn up until it is in the position shown in Fig. 21 (a). The large piston on the lower end of the plunger is in communication with the atmosphere through the valve C, and hence the pres- sure of the atmos- phere on both sides of the piston is equal. The upper part of the plunger, however, has an equal pressure on both its sides only when it is down, as in Fig. 21 (b). When it is drawn up, the air in the space D expands and a partial vacuum is formed. The valve gear next releases the plunger, and as the atmos- pheric pressure is much greater than the pressure within D, the plunger rapidly descends. During its descent, the large pis- ton at the end of the plunger gradually compresses the air in the annular space beneath it, and is 44 MECHANICS OF FLUIDS. thus gradually brought to rest. The valve C serves to regulate the amount of compression and at the same time admits air during the up stroke of the plunger. THE SIPIIOX. 63. Theory of the Siphon. The action of the siphon illustrates the effect of atmospheric pressure. It is simply ^s^L.^ a bent tube having unequal Jf^ ^"^^V branches, open at both ends, if \ and is used to convey a liquid from a higher point to a lower over an intermediate point that is higher than either. In Fig. 22, A and B are two vessels, B being lower than A, and A C B is the bent tube, or siphon. Suppose this tube to be filled with water and placed in the vessels as shown, with the short branch A C in the vessel A. The water will flow from the vessel A into the vessel B as long as the level of the water FIG - 22 - in B is below the level of the water in A and the level of the water in A is above the lower end of the tube A C. The atmospheric pressure on the sur- faces of A and B tends to force the water up the tubes A C and B C. When the siphon is filled with water, each of these pressures is counteracted in part by the pressure of the water in that branch of the siphon that is immersed in the water on which the pressure is exerted. The atmospheric pressure opposed to the weight of the longer column of water will, therefore, be resisted more strongly than that opposed to the weight of the shorter column ; consequently, the pressure exerted on the shorter column will be greater than that on the longer column, and this excess of pressure will produce motion. G MECHANICS OF FLUIDS. 45 64. Let A the area of the tube ; h = D C = the vertical distance between the surface of the water in B and the highest point of the center line of the tube ; k l = E C the distance between the surface of the water in A and the highest point of the center line of the tube. The weight of the water in the short column is .03617 X A /;,, and the resultant atmospheric pressure tending to force the water up the short column is 14.7 X A .03617 X A h^. The weight of the water in the long column is .03617 A /z, and the resultant atmospheric pressure tending to force the water up the long column is 14. 7 A .03617 A h. The difference between these two is (14.7 A ,03617 A k^ - (14.7 A .03617 A h) .03617 A (k //,). But// //, = E D = the difference between the levels of the water in the two vessels. In the above, h and // 1 were taken in inches and A in square inches. It will be noticed that the short column must not be higher than 34 feet for water, or the siphon will not work, since the pressure of the atmosphere will not support a column of water that is higher than 34 feet. 65. Fig. 23 shows the actual construction of a siphon. It is desired to convey the water from D to E. The point of discharge must always be lower than the point from which the water is taken, otherwise a siphon cannot work. The siphon consists of ordinary iron pipe jointed in any convenient manner so as to be air-tight. It has three valves A, B, and C. The suction end is provided with a perforated strainer c to keep out large particles of foreign matter. In order to start the siphon, it is necessary to remove the air in the pipe. This is done by closing the valves A and B and opening the valve C, which should always be located at the highest point of the siphon. Water is then poured into the funnel above the valve C until the pipe is filled. When 46 MECHANICS OF FLUIDS. 6 no more water can be poured in, that is, when all the air has been driven out, the valve C is closed and the valves A and B are opened ; the siphon will now start. FIG. 23. In practice, it has been found that the distance s must not exceed 28 feet at sea level, and that the siphon will work more satisfactory if it is less. The greater the distance h between the two water levels, the better the siphon will work. 66. In order that a siphon will work properly, it is nec- essary that air should be kept out of the pipe, or, if it does get in, means should be provided for its escape. The joints of the pipe must be perfectly air-tight. But some air will always be carried in with the water, and this air will collect at the highest point. The bad effects of this can be mini- mized by having a straight horizontal pipe at the highest point instead of a sharp bend. 67. A device that will remove the air is shown in Fig. 24. Here A is an air-tight vessel connected with the siphon by two small pipes B and C. The pipe B extends near- ly to the top of A, while the pipe C barely enters the bot- tom. Each pipe has a valve D and E. A valve F and funnel G are placed on top of the vessel. When the siphon ceases to flow, which is an indication that air has collected (it is here MECHANICS OF FLUIDS. 47 assumed that the suction end has not become uncovered), the valves D and E are closed and the valve F is opened. The vessel is now filled with water, the valve F is closed, and D and E are opened. The water will now flow through C into the siphon and force out the air collected there, which passes through B to the top of A. When all air is out of the siphon, D and E are shut and F is opened. The vessel A is now filled with water, F is shut, D and E are opened and left open. Any air that enters the siphon will, instead of collecting at //, ascend the pipe B and force out a small amount of water through C. This will continue until A is filled with air, when the valves D and E should be closed and A refilled. This arrangement may also be used to fill the siphon for the purpose of setting it to work. The highest point of the water in A should not be more than 28 feet at sea level above the water level at the suction end. PUMPS. 68. A pump is a machine for conveying liquids from one level to a higher level or for performing work equiva- lent to such an operation. 48 MECHANICS OF FLUIDS. 69. Classification of Pumps. Pumps may be divided into three general divisions, according to the service they perform, viz., suction pumps, lifting pumps, and force pumps. They may also be divided into two general classes, single- acting and double-acting pumps, according as they take water on one side or on both sides of the water piston. According to the arrangement of the pump cylinders, they are classified as simple, duplex, or triplex pumps. As pumps displace the liquids in various ways, they may also be divided according to the method of displacement, into reciprocating, centrifugal, and rotary pumps. Reciprocating pumps only will be considered here. 70. The Suction Pump. A section of an ordinary suction pump is shown in diagrammatic form in Fig. 25. Suppose the piston, or bucket as it is commonly terned, to be at the bottom of the cylinder and to be just on the point of moving upwards in the direction of the arrow. As the piston rises it leaves a par- tial vacuum behind it, and the atmospheric pressure on the sur- face of the water in the well causes it to rise in the pipe P for the same reason that FIG - 25- the mercury rises in the barometer tube. The water rushes up the pipe and lifts the suction valve V, filling the empty space in the cylin- der B caused by the displacement of the piston. When the piston has reached the end of its stroke, the water entirely fills the space between the bottom of the piston and the 6 MECHANICS OF FLUIDS. 49 bottom of the cylinder, and also the pipe P. The instant the piston begins its down stroke, the water in the chamber B begins to flow back into the well, and its downward flow forces the valve V to its seat, thus preventing any further escape of the water. As the piston descends, the water must give way to it, and since the suction valve V is closed, the bucket valves u, u must open, and thus allow the water to pass through the piston, as shown in the right-hand figure. When the piston has reached the end of its down- ward stroke and commenced its upward movement again, the water flowing through the piston quickly closes the valves tt, u. All the water resting on the top of the piston is then lifted by the piston on its upward stroke and dis- charged through the spout A ; the valve V again opens and the water fills the space below the piston, as before. 71. It is evident that the distance between the piston when at the top of its stroke and the surface of the water in the well must not exceed 34 feet, the highest column of water that the pressure of the atmosphere will sustain, since, otherwise, the water in the pipe would not rise and fill the cylinder as the piston ascended. In practice, this distance should not exceed 28 feet. This is due to the fact that there is a little air left between the bottom of the piston and the bottom of the cylinder, a little air leaks through the valves, which are not perfectly air-tight, and a pressure is needed to raise the valve against its own weight, which, of course, acts downwards. There are many varieties of the suction pump, differing principally in the construction of the valves and piston, but the principle is the same in all. 72. The Lifting Pump. In some cases it is desired to raise water higher than it can be forced by the pressure of the atmosphere into the chamber of a simple suction pump, such as is shown in Fig. 25. To accomplish this the pump chamber with its bucket and valves are set at a distance above the supply not exceeding that to which the air will successfully force the water. A closed pipe P\ Fig. 26, 50 MECHANICS OF FLUIDS. called the delivery, or discharge, pipe, is then led from the upper part of the chamber to the point where it is desired to deliver the water. Such a pump is often called a lifting pump. In order to prevent the leakage of water around the pis- ton rod, a stuffmgbox 5 is provided. The lower end of the discharge pipe P' is sometimes fitted with a valve c to pre- vent the water flowing back into the pump chamber; this valve is not essential to the operation of the pump, how- ever, since the valve V prevents the water in the chamber and discharge pipe discharging during the downward motion of the piston. While it is customary to consider lifting pumps and suc- tion pumps as two different types of pumps, there is in reality no difference in their operation, as a careful study of Figs. 25 and 26 will show. The only difference is that the water is discharged by a suction pump at the level of the MECHANICS OF FLUIDS. 51 pump, while a lifting pump discharges the water above the level of the pump. 73. Force Pumps. The force pump differs from the lifting pump in one important particular, that is, in the fact that its piston is solid. A section of a force pump is shown in Fig. 27. As the piston ascends, as shown in the left- hand figure, the pressure of the atmosphere forces the water up the suction pipe P\ the water opens the suction valve V and flows into the pump cylinder. When the piston moves down, as shown in the right-hand figure, the suction valve is closed and the delivery valve V opened. The water in the pump cylinder is now forced up the delivery pipe P' . When the piston again begins to move upwards, the deliv- ery valve is closed by the water above it and the suction valve opened by the pressure of the atmosphere on the water below it. 74. Plunger Pumps. When force pumps are used to convey water to great heights or to force water against heavy pressures, the great pressure of the water makes it extremely difficult to keep the water from leaking past the piston, and the constant repairing and renewal of the piston packing be- comes a nuisance and involves serious expense. To obvi- ate this drawback, plunger pumps have been designed, one of which is shown diagrammatically in Fig. 28. The action FIG. 28. MECHANICS OF FLUIDS. does not differ in any way from that of the piston force pump. During the up stroke of the plunger, the suction valve is open and the delivery valve is closed; during the down stroke, the suction valve is closed and the delivery valve is open. 75. The force pumps shown so far are single-acting, that is, the water is forced into the delivery pipe only during the downstroke or forward stroke of the piston or plunger. Force pumps, either of the piston or plunger pattern, may be constructed so as to force water into the delivery pipe both during the forward and return stroke. They are then called double-acting. 76. A double-acting force pump of the piston pattern is shown in Fig. 29. Such a pump has two sets of suction valves and delivery valves, one set for each side of the piston. With the piston mov- ing in the direction of the arrow, the pressure of the atmosphere forces the water up the suction pipe P into the left-hand end of the pump cylinder, the left-hand suction valve opens and the left-hand delivery valve is closed. P The piston in moving FlG - - the water in the right- hand end of the pump cylinder; as a consequence the right-hand suction valve is closed and the right-hand deliv- ery valve opens. The water now flows up the delivery pipe P'. Imagine that the piston is at the end of its stroke and commences to move to the left. Its first movement 6 MECHANICS OF FLUIDS. 53 promptly closes the left-hand suction valve and opens the left-hand delivery valve. It also closes the right-hand delivery valve and opens the right-hand suction valve. It is thus seen that with the arrangement given, which shows the principle of operation of all double-acting pumps, the piston will discharge water both during the forward and the return stroke. While the pump shown is a horizontal pump, it may be vertical as well. STRENGTH OF MATERIALS. GENERAL PRINCIPLES. 1. When a force is applied to a body, it changes either its form or its volume. A force, when considered with ref- erence to the internal changes it- tends to produce in any solid, is called a stress. Thus, if we suspend a weight of 2 tons by a rod, the stress in the rod is 2 tons. This stress is accompanied by a length- ening of the rod, which increases until the internal stress or resistance is in equilibrium with the external weight. 2. .Classification of Stresses. Stresses may be classi- fied as follows : Tensile, or pulling stress ; transverse, or bending stress ; compressive, or pushing stress ; shearing, or cutting stress ; torsional, or twisting stress. 3. A unit stress is the amount of stress on a unit of area, and may be expressed either in pounds or tons per square inch or per square foot ; or, it is the load per square inch or per square foot on any bod)'-. Thus, if 10 tons are suspended by a wrought-iron bar that has an area of 5 square inches, the unit stress is 2 tons per square inch, because J / = 2 tons. 4. Strain is the deformation or change of shape of a body resulting from stress. For example, if a rod 100 feet long is pulled in the direc- tion of its length, and if it is lengthened 1 foot, it is strained -j-Jrj- its length, or 1 per cent. For notice of the copyright, see page immediately following the title page. H. 8. I.-21 2 STRENGTH OF MATERIALS. 7 5. Elasticity is the property by virtue of which a body regains its original form after the external force on it is withdrawn, provided the stress has not exceeded the elastic limit. It is a property possessed by all bodies. Consequently, we see from this that all material is length- ened or shortened when subjected to either tensile or com- pressive stress, and the change of the length within the elastic limit is directly proportional to the stress. For stresses within the elastic limits, materials are per- fectly elastic, and return to their original length on removal of the stresses; but when their elastic limits are exceeded, the changes of their lengths are no longer regular, and a permanent set takes place. The destruction of the material has then begun. 6. The measure of elasticity of any material is the change of length under stress within the elastic limit. 7. The elastic limit is that unit stress under which the permanent set becomes visible. 8. The elasticity of wrought iron and of all grades of steel is practically the same; that is, within the clastic limit each material will change an equal amount of length under the same stress. The elastic limit, however, is not the same for steel as for iron ; it is higher for soft steel than for wrought iron, and, in general, the harder and stronger the steel the higher will be its elastic limit. As a conse- quence, steel will lengthen or shorten more than wrought iron, and hard steel more than soft, before its elasticity or ability to return to its original dimensions is injured. TENSILE STRENGTH OF MATERIALS. 9. The tensile strength of any material is the resistance offered by its fibers to being pulled apart. 10. The tensile strength of any material is proportional to the area of its cross-section. STRENGTH OF MATERIALS. 3 Consequently, when it is required to find the safe tensile strength of any material, we have only to find the area at the minimum cross-section of the body, and multiply it by the load per square inch that it can safely carry, as given in the following table under the heading " Safe Loads." NOTE. The minimum cross-section referred to in the above para- graph is that section of the material which is pierced with holes, such as bolt or rivet holes in iron, or knots in wood, if there are any. TABLE I. TEXSIL.E STRENGTH OF MATERIALS. Material. Ultimate Tensile Strength. Pounds per Square Inch. Safe Loads. Pounds per Square Inch. Sudden. Gradual. Steady. Timber 10,000 16,000- 20,000 30,000- 35,000 45,000- 55,000 45,000- 55,000 52,000- 62,000 52,000- 62,000 60,000- 75,000 75,000- 90,000 90,000-115,000 100,000-125,000 125,000-180,000 14,000- 20,000 30,000- 36,000 70,000- 80,000 30,000- 40,000 30,000- 36,000 600 1,600 2,000 4,500 6,000 6,000 6,000 6,500 7,000 7,000 8,000 15,000 1,400 3,000 7,000 2,500 3,500 900 2,400 3,000 9,000 10,000 10,000 10,000 11,000 14,000 14,000 15,000 23,000 2,200 5,000 11,000 3,700 6,000 1,200 3,200 4,000 13,000 14,000 14,000 14,000 15,000 20,000 20,000 20,000 30,000 3,000 7,000 14,000 5,000 8,000 Cast iron Gun-metal cast iron Wrought iron Extra soft steel . . . Flange steel Firebox steel Machinery steel. . . Axle steel Hard steel (rail steel) Tire steel Crucible (tool) steel Brass, cast Bronze cast . ... Tobin bronze Hard-drawn brass wire Copper, rolled 4 STRENGTH OF MATERIALS. 7 11. In metals, a high tensile strength in itself is no indi- cation of the ability of the metal to stand repeated applica- tions of sudden stresses, as a high tensile strength usually involves a lesser degree of ductility than is obtained in metals of lower tensile strength. Steel of a tensile strength higher than 75,000 pounds is seldom used merely on account of its superior strength, but rather on account of its hard- ness, which enables it to better withstand abrasion. 12. The loads given in Table I are conservative safe loads deducted from experience and observation. They are given for material free from welds for such material as can be welded. While it is possible to make a weld as strong as the solid bar, the chances of the weld being imperfect are so great that it is unsafe to rely on such a degree of strength in welded bars. Furthermore, the value of the weld is an uncertain quantity that cannot be determined by an ocular inspection or any other inspection short of actually pulling the weld apart in a testing machine. Hence, it is advisable to reduce the safe loads given in the above table by 25 per cent, when a welded bar is subjected to a tensile stress. When judging the safe load of timber, due allowance must be made for knot holes and sappy spots. 13. It is often rather hard to determine whether a stress is steady, gradually varying, or gradually applied, or sudden. For example, considering the shell of a boiler, it would appear on first thought as though the tensile stress in the shell plates was a steady stress. But looking further into the problem, it will become apparent that the stress varies with the steam pressure, which gradually fluctuates within narrow or wider limits. Hence, most designers would con- sider the stress in a boiler shell as a gradually applied stress. In a piston rod or connecting-rod the load is applied almost instantly as soon as the crank passes over the center, and, hence, the stress would be considered to be a suddenly applied stress. When in doubt, it is good policy to err on the side of safety, that is, to choose a smaller safe load per square inch of section. 7 STRENGTH OF MATERIALS. 5 For special work, experience has indicated safe loads for different materials that fall below, or are above, those given in the table. Examples of this will be given later on. 14. The safe load per square inch of section is often called the working stress per square Inch, or the "work- ing load per square inch. Care should be taken not to confound these terms with working load, 'Corking stress, safe load, or safe tensile strength, which, when applied without the limitation as to the unit of area, refer to the safe load the whole bar can carry. RULES AND FORMULAS FOR TENSILE STRENGTH. 15. Let W safe load in pounds; A = area of minimum cross-section ; 5 = working stress in pounds per square inch, as given in Table I. Rule 1. The working load in pounds for any bar sub- jected to a tensile stress is equal to the minimum sectional area of the bar, multiplied by the working stress in pounds per square inch, as given in the table. Or, W=AS. EXAMPLE. A bar of good wrought iron that is 3 inches square is to be subjected to a steady tensile stress; what is the maximum load that it should carry ? SOLUTION. According to the table, a working stress of 13,000 pounds may be allowed. Applying the rule, we have W = 3 X 3 X 13,000 = 117,000 Ib. Ans. Rule 2. The minimum sectional area of any bar sub- jected to a tensile stress should be equal to the load in pounds, divided by the working stress in pounds per square inch, as given in the table. Or, A =- 6 STRENGTH OF MATERIALS. 7 EXAMPLE. What should be the area of a machinery-steel bar to carry a steady load of 108,000 pounds ? SOLUTION. According to the table, a safe load of 15,000 pounds per square inch of section may be allowed. Then, applying the rule, 108,000 A = = Rule 3. The working stress in pounds per square inch is equal to the load in pounds divided by the minimum sectional area of the bar. W Or, S f % EXAMPLE. A piston rod 3 inches in diameter carries a piston 20 inches in diameter. With a steam pressure of 100 pounds to the square inch, what is the stress per square inch of section of the rod ? SOLUTION. The load on the piston is 20 2 X .7854 X 100 = 31,416 pounds. The area of the rod is 3* X .7854 = 7.0686 square inches. Then, applying the rule just given, we have ~ 31,416 = 7Q686 = 4,444.4 Ib. Ans. CHAINS. 16. Chains made of the same size iron vary in strength, owing to the different kinds of links from which they are made. It is a good practice to anneal old chains that have become brittle by overstraining. This renders them less liable to snap from sudden jerks. It reduces their tensile strength, but increases their toughness and ductility, which are sometimes more important qualities. When annealing, care should be taken that a sufficient heat be applied, otherwise no benefit will be gained; the chains ought to be heated to a cherry red, say 1,300 F., at least. 17. The loads that can safely be lifted with chains are given in the following table. The safe load given is one- quarter of the steady load at which they may be expected to break. The table has been deduced from one published by the Lukens Iron and Steel Company. 7 STRENGTH OF MATERIALS. TABLE II. STRENGTH OF CHAINS. Diameter of Rod of Which Link Is Made. Safe Load. Pounds. Diameter of Rod of Which Link Is Made. Safe Load. Pounds. A 430 1 12,500 1 770 *t 15,000 yV 1,200 H 18,000 t 1,750 If 22,000 yV 2,350 if 26,000 1 3,100 if 31,000 TV 4,000 if 36,000 1 4,800 if 41,000 tt 5,800 2 47,000 t 6,900 ^ 56,000 it 8,000 ** 69,000 1 9,400 2f 84,000 if 11,000 3 100,000 STRENGTH OF ROPES. HEMP ROPES. 18. The strength of hemp ropes does not depend entirely on the quality of the material and the cross-section of the rope, but to a large extent on the method of manufacture and the amount of twisting. A hemp rope is made up of fibers twisted together to form a yarn, several of which are laid up left-handed into strands, which in turn are twisted up right-handed into what is known as plain-laid rope. Cable-laid or hawser-laid rope is left-handed rope of nine (9) strands. Shroud-laid rope is formed by adding another strand to a plain-laid rope. Since the four strands on application of 8 STRENGTH OF MATERIALS. strain would sink in and detract from the rope's strength by an unequal distribution of strain, they are laid up around a heart, which is a small rope, made soft and elastic, and about one-third the size of the strands. 19. Ordinary hemp rope is designated by its circumfer- ence; rope for transmission purposes, which is either iron, steel, or manila rope, is designated by its diameter. It is well to keep this in mind. The loads that can safely be hoisted with hemp ropes of different sizes is given in the following table prepared by Ensign F. R. Brainard, United States Navy. TABLE III. STRENGTH OF HEMP ROPES. Circumference. -Inches. Safe Load. Pounds. Circumference. Inches. Safe Load. Pounds. 1.00 65 5.75 2,680 1.25 125 6.00 2,975 1.50 185 6.50 3,470 1.75 255 7.00 3,965 2.00 320 7.50 4,570 2.25 410 8.00 5,175 2.50 500 8.50 5,910 2.75 610 9.00 6,640 3.00 715 9.50 '7,370 3.25 850 10.00 8,105 3.50 980 10.50 8,970 3.75 1,135 11.00 9,840 4.00 1,285 11.50 11,015 4.25 1,450 12.00 12,190 4.50 1,610 12.50 13,365 4.75 1,750 13.00 14,540 5.00 1,990 13.50 15,845 5.25 2,190 14.00 17,150 5.50 2,385 14.50 18,450 7 STRENGTH OF MATERIALS. 9 20. The above table applies to new ropes or ropes in first-class condition. If the rope is used for a block and tackle, the bending of the rope around the pulleys will cause a rapid deterioration of the inside, owing to the chafing and sliding of the strands and yarns upon one another. The outside appearance of a rope used for block and tackle is in itself no indication of its quality ; ropes are frequently found that appear perfectly sound when judged by their outside appearance and yet are entirely unsafe. To inspect the rope, take it in both hands and untwist it sufficiently to expose the inner surfaces that have been chafing against one another. If a considerable number of broken fibers are found, the safe load should be reduced by 50 per cent. If a rather large quantity of the fibers have been reduced to powder, the rope should be condemned and preferably destroyed immediately, in order to remove any temptation to use it. A rope in this condition is liable to give way suddenly under a very light load. Ropes used for slings and lashings, as a general rule, are ruined by the external chafing they receive and, hence, their safety can be determined by their outward appearance. 21. Slings are chiefly used for attaching tackles to machine parts during the erection and repair of machinery. They are made by taking a piece of rope a little more than twice the required length of the rope and joining the ends by splicing, the splice known as a short splice being usually employed. It should be distinctly understood that this splice is not intended nor adapted for ropes that are to pass over a pulley. 22. Splicing a Sling. To make a short splice in a sling, untwist the strands for some distance from each end of the rope. Take an end in each hand and lay the strands within one another, as shown in Fig. 1 (a). The strands 1', 2', and 3' may be tied to the left-hand end for convenience in handling. Now take the strand 1, pass it over strand 2' and under strand ./', as shown in Fig. 1 (b] ; then, draw it 10 STRENGTH OF MATERIALS. tight. Next, take strand 2, pass it over 3' and under 2', drawing it up tightly. Now take strand A -(- b L/CO /\ ~i j ^ 20 STRENGTH OF MATERIALS. 7 TABLE Till. CONSTANTS FOR WOODEN PILLARS. When One Cross-Section of Pillar. When Both Ends of the Pillar Are Flat or Fixed. End of the Pillar Is Flat or Fixed and the Other Round or When Both Ends of the Pillar Are Round or Movable. Movable. Round. 187.5 125.00 93.75 M<| |"- be used is W ^ . According to Table X, the value of 5 for steel for a gradually applied load is 7,500. By Table XI for an I beam, R = 5-55. Substituting values in this last formula, we get 7i = ' o.oo o.oo = 22.52, nearly. Substituting values in the formula for the safe work- ing load, we get 2 X 7,500 X 22.52 100 = 3,378 lb. Ans. TABLE XI. VALUES OF R. Section. K. B T T b h* - V /i' s AD AD 1 -ad* 8 D bl? 24 A h A h A h A h A h 3.33 A h 3.67 28 STRENGTH OF MATERIALS. 7 46. While the formulas given in Table IX will allow the safe load on a given beam to be calculated, they cannot readily be transformed to give directly the size of beam required under given conditions to carry a given load. In practice, when it is required to determine the size of beam, the length between supports is known. After the shape of cross-section of the beam has been chosen, dimen- sions for the beam may be assumed and its safe load for the given dimensions calculated. If the load thus found falls below the load the beam is to carry, larger dimensions must be chosen. In case of rolled sections, such as angle irons, T irons, channels, or I beams, catalogues of manufacturers should be consulted and standard sizes chosen. These cata- logues usually contain the area of the section for different weights and dimensions of rolled sections. 47. The values of R given in Table XI apply only when the dimension marked JL is vertical. If the beam is placed in any other position, R will have a different value, which can only be calculated by the aid of higher mathematics. Beams rigidly fixed at both ends are rarely met with in practice, and it is safer to assume that the beam merely rests on two supports. EXAMPLES FOR PRACTICE. 1. What weight will a yellow-pine rectangular beam carry safely under a steady load when used as a cantilever and uniformly loaded ? The beam is 9 feet 6 inches long, 16 inches deep, and 4 inches wide. Ans. 4,371 Ib. 2. What weight would the beam in example 1 carry safely if it rested on two supports ? Ans. 17,484 Ib. 3. What weight can safely be carried at the end of a wrought-iron round cantilever 3 inches in diameter when the load is suddenly applied 4 feet 2 inches from the support of the cantilever ? Ans. 212 Ib. 4. What steady load can be carried safely by the same beam given in the preceding example ? Ans. 726 Ib. STRENGTH OF MATERIALS. 29 SHEARING, OR CUTTING, STRENGTH OF MATERIALS. 48. The shearing strength of any material is the resistance offered by its fibers to being cut in two. Thus, the pressure of the cutting edges of an ordinary shear- ing machine, Fig. 0, causes a shearing stress in the plane a b. The unit shear- ing force may be found by dividing the force P by the area of the plane a b. Fig. 7 shows a piece in double shear; here the cen- tral piece c d is forced out FIG. 6. while the ends remain on their supports M and N. The shearing strength of any body is directly proportional to its area. ILL 49. In Table XII are given the greatest and the safe shearing strengths per square inch of different kinds of materials. 50. Rules for Shearing. In general, the force required to shear a piece of material in double shear will be twice that required for single shear. This does not apply to all cases, however; a notable exception are rivets in double shear. Experiments have determined that the force required to shear both iron and steel rivets in double shear averages about 1.85 times that required for single shear. 30 STRENGTH OF MATERIALS. ! 2 GCOD & co co co o r o*~ <^r o" z>~ 1 1 .s o 2 o O O O O O 10 O O O J < 0) 9 5 Parallel to Grain. oooooooooo OOOOOCO^C^COCO O ^ ^* ^ O c ^ to . w .b VH ^ ^-, , -S 'a^ O rC a; l-i 3 O-i O t. bo - * 5 *T te 5 ti ^ g 2 .s ! j j ! 3 O^c/2Oc^^!>*Wc/2O 7 STRENGTH OF MATERIALS. 31 Let a area of cross-section in square inches; 5 = shearing stress as given in table ; IV = load in pounds. Then, to find the safe load or the load that will shear the material : Rule 9. Multiply the area of the section by the shearing stress. Or, W=aS. In applying this rule, it should be remembered that when the safe load the material can bear is required, the value of S is to be taken from the column headed " Safe Loads," selecting the safe load corresponding to the nature of the load. When it is desired to' find the load at which the mate- rial will fail, the value of 5 is to be taken from the column headed "Ultimate Shearing Strength." EXAMPLE. If the beam in double shear shown in Fig. 7 is rect- angular and measures 4 in. X 2 in., what steady safe load would you allow if the beam were made of structural steel ? SOLUTION. According to the table, a safe load of 10,000 pounds per square inch of section may be allowed. Applying rule 9 and multi- plying by 2, since the beam is in double shear, we get W = 4 X 2 X 10,000 X 2 = 160,000 Ib. Ans. 51. To find the area required for a material subjected to a shearing stress: Rule 1O. Divide the load by the shearing stress. In applying rule 1O, it should be remembered that for double shear the result is to be divided by 2 to obtain the area of the beam. EXAMPLE. A white pine beam is subjected to a suddenly applied shearing stress by a load of 4,000 pounds, the beam being in double shear and the load being applied across the grain. What should be the area of the beam to bear this stress safely ? 32 STRENGTH OF MATERIALS. 7 SOLUTION. By the table, S= 250. Then, by rule 1O, we get a = ' = 16 square inches for single shear. 250 Dividing the result by 2 for double shear, we get 16 -i- 2 = 8 sq. in. area. Ans. EXAMPLES FOR PRACTICE. 1. A hemlock beam 8 in. X 2 in. is in single shear parallel to the grain. What load will cause it to fail by shearing ? Ans. 4,000 Ib. 2. A wrought-iron rivet 1 inch in diameter is in double shear. At what load will it be likely to shear ? Ans. 55,214 Ib., nearly. TORSIOK. 52. When a force is applied to a beam, bar, or rod in such a manner that it tends to twist it, the stress thus pro- duced is termed torsion. Torsion manifests itself in the case of rotating shafts, such as line shafts and engine shafts. LINE SHAFTING. 53. A line of shafting is one continuous run, or length, composed of lengths of shafts joined together by couplings. 54. The main line of shafting is that which receives the power from the engine or motor and distributes it to the other lines of shafting or to the various machines to be driven. Line shafting is supported by hangers, which are brackets provided with bearings, bolted either to the walls, posts, ceilings, or floors of the building. Short lengths of shafting, called countershafts, are provided to effect changes of speed and to enable the machinery to be stopped or started. 55. Shafting is usually made cylindrically true, either by a special rolling process, when it is known as cold-rolled shafting, or else it is turned up in a machine called a lathe; In the latter case it is called bright shafting. What is STRENGTH OF MATERIALS. 33 known as black shafting is simply bar iron rolled by the ordinary process and turned where it receives the coup- lings, pulleys, bearings, etc. 56. The diameter of bright turned shafting increases by inch up to about 3 inches in diameter; above this diameter it increases by inch. The actual diameter of a bright shaft is -fa inch less than the commercial diameter, it being designated from the diameter of the ordinary round bar iron from which it is turned. Thus, a length of what is called 3-inch bright shafting is really only 2f| inches in diameter. Cold-rolled shafting is designated by its commercial diam- eter; thus, a length of what is called 3-inch shafting is 3 inches in diameter. 57. In the following table is given the maximum dis- tance between the bearings of some continuous shafts that are used for the transmission of power: TABLE XIII. DISTANCE BETWEEN BEARINGS. Distance Between Bearings. Diameter of Shaft. Inches. Feet. Wrought-Iron Shaft. Steel Shaft. 2 11 11.50 3 13 13.75 4 15 15.75 5 17 18.25 6 19 20.00 7 21 22.25 8 23 24.00 9 25 26.00 11. X. I.2J 34 STRENGTH OF MATERIALS. 7 Pulleys from which considerable power is to be taken should always be placed as close to a bearing as possible. 58. The diameters of the different lengths of shafts composing a line of shafting may be proportional to the quantity of power delivered by each respective length. In this connection it is to be observed that the positions of the various pulleys in reference to the bearings must be taken into consideration in deciding upon the size of a shaft. Suppose, for example, that a piece of shafting delivers a certain amount of power; then, it is obvious that the shaft will deflect or bend less if the pulley transmitting that poAver be placed close to a hanger or bearing than if it be placed midway between the two hangers or bearings. It is impossible to give any rule for the proper distance of bear- ings that could be used universally, as in some cases the requirements demand that the bearings be nearer than in others. Wherever possible it is advisable to have the main line of shafting run through the center of the room, or at least far enough from either wall to allow countershafts to be placed on either side of it. When this is done, power may be taken off the main shaft from either side by alternate pulleys, and the deflection caused in the main shaft in one direction by one pulley will be counteracted by the deflection caused in the opposite direction by the next pulley. If the work done by a line of shafting is distributed quite equally along its entire length and the power can be applied near the middle, the strength of the shaft need be only half as great as would be required if the power were applied at one end. 59. To compute the horsepower that can be transmitted by a shaft of any given diameter : Let D = diameter of shaft ; R revolutions per minute; H '= horsepower transmitted; C constant given in Table XIV. 7 STRENGTH OF MATERIALS. TABLE XIV. 3S CONSTANTS FOR LINE SHAFTING. Material of Shaft. No Pulleys Between Pulleys Between Bearings. Bearings. Steel or cold-rolled iron 65 85 Wrought iron 70 95 Cast iron 90 120 In the above table the bearings are supposed to be spaced so as to relieve the shaft of excessive bending; also, in the third vertical column, an average number and weight of pulleys and power given off is assumed. 6O. Rules and Formulas. In determining the above constants, allowance has been made to insure the stiffness as well as strength of the shaft. Cold-rolled iron is consider- ably stronger than ordinary turned wrought iron; the increased strength is due to the process of rolling, which seems to compress the metal and so make it denser, not merely skin deep, but practically throughout the whole diameter. We have, then, the following Rule 11. The horsepower that a shaft will transmit equals the product of the cube of the diameter and the number of revolutions, divided by the value of C for the given material. Or, D* c EXAMPLE. What horsepower will a 3-inch wrought-iron shaft trans- mit which makes 100 revolutions per minute, there being no pulleys between bearings ? SOLUTION. Applying rule 11 and substituting, we have 3 X 3 X 3 X 100 70 = 38.57 H. P. Ans. 36 STRENGTH OF MATERIALS. 7 If there were the usual amount of power taken off, as mentioned above, we should take C = 95. Then, H = :: ^ = 28.42 H. P. Ans. 61. To compute the number of revolutions a shaft must make to transmit a given horsepower : Rule 12. The number of revolutions necessary for a given horsepoiver equals the product of the value of C for the given material and the number of horsepower, divided by the cube of the diameter. Or, * = % EXAMPLE. How many revolutions must a 3-inch wrought-iron shaft make per minute to transmit 28.42 horsepower, power being taken off at intervals between the bearings ? SOLUTION. Applying the rule just given and substituting, we have 62. To compute the diameter of a shaft that will trans- mit a given horsepower, the number of revolutions the shaft makes per minute being given: Rule 13. The diameter of a shaft equals t/ie cube root of the quotient obtained by dividing the product of the value of C for the given material and the number of horsepower by the number of revolutions. Or, D = EXAMPLE. What must be the diameter of a wrought-iron shaft to transmit 38.57 horsepower, the shaft to make 100 revolutions per minute, no power being taken off between bearings ? SOLUTION. By rule 13, we have x D = j ' = ^27 = 3 in. Ans. 63. As the speed of shafting is used as a multiplier in the calculations of the horsepower of shafts, it is readily seen that a shaft having a given diameter will transmit more 7 STRENGTH OF MATERIALS. 37 power in proportion as its speed is increased. Thus, a shaft that is capable of transmitting 10 horsepower when making 100 revolutions per minute will transmit 20 horsepower when making 200 revolutions per minute. We may, therefore, say the number of horsepower transmitted by a shaft is directly proportional to the number of revolutions. EXAMPLES FOB PRACTICE. 1. What horsepower will a 2^-inch wrought-iron shaft transmit when running at 110 revolutions per minute, it being used for trans- mission only ? Ans. 24.55 H. P. 2. A 6-inch cast-iron shaft transmits 150 horsepower. How many revolutions per minute must it make, no power being taken off between bearings ? Ans. 62 R. P. M. 3. What should be the diameter of a wrought-iron shaft to transmit 100 horsepower at 150 revolutions per minute, power being taken off between bearings ? Ans. 4 in., nearly. 4. The machines driven by a certain line of wrought-iron shafting take their power from various points between the bearings, and if all were working together at their full capacity, they would require 65 horsepower to drive them. What diameter should the shaft be if it runs at 150 revolutions per minute ? Ans. 3 in., nearly. A SERIES OF QUESTIONS AND EXAMPLES RELATING TO THE SUBJECTS TREATED OF IN THIS VOLUME. It will be noticed that the Examination Questions that follow have been divided into sections, which have been given the same numbers as the Instruction Papers to which they refer. No attempt should be made to answer any of the questions or to solve any of the examples until that portion of the text having the same section number as the section in which the questions or examples occur has been carefully studied. ARITHMETIC. (PART 1.) EXAMINATION QUESTIONS. (1) What is arithmetic ? (2) What is a number ? (3) What is the difference between a concrete number and an abstract number ? (4) Define notation and numeration. (5) Write each of the following numbers in words: 980; 605; 28,284; 9,006,042; 850,317,002; 700,004. (6) Represent in figures the following expressions: Seven thousand six hundred. Eighty-one thousand four hundred two. Five million four thousand seven. One hundred eight million ten thousand one. Eighteen million six. Thirty thousand ten. (7) What is the sum of 3,290, 504, 865,403, 2,074, 81, and 7? Ans. 871,359. (8) 709 + 8,304,725 + 391 + 100,302 + 300 + 909 = what? Ans. 8,407,336. (9) During a 12-hour test of a steam engine the counter showed the number of revolutions per hour to have been as follows: 12,600, 12,444, 12,467, 12,528, 12,468, 12,590, 12,610, 12,589, 12,576, 12,558, 12,546, 12,532. How many revolutions were made during the test ? Ans. 150,508 rev. 1 For notice of copyright, see page immediately following the title page. 2 ARITHMETIC. 1 (10) Find the difference between the following: (a) 50,962 and 3,338; (b) 10,001 and 15,339. . ( (a) 47,624 3 ' ( (b) 5,338. (11) (a) 70,968 - 32,975 = ? (b) 100,000 - 98,735 = ? ((*) 37,993. M(*) 1,265. (12) On a certain morning 7,240 gallons of water were drawn from an engine-room tank and 4,780 gallons were pumped in. In the afternoon 7,633 gallons were drawn out and 8,675 gallons pumped in. How many gallons remained in the tank at night, if it contained 3,040 gallons at the begin- ning of the day ? Ans. 1,622 gal. (13) Find the product of the following : (a) 526,387 X 7; (b) 700,298 X 17; (c) 217 X 103 X 67. f (a) 3,684,709. Ans. ] (b) 11,905,066. ( (c) 1,497,517. (14) If your watch ticks once every second, how many times will it tick in one week ? Ans. 604,800. (15) An engine and boiler in a manufactory are worth $3,246. The building is worth three times as much, plus $1,200, and the tools are worth twice as much as the build- ing, plus $1,875. (a) What is the value of the building and tools ? (b} What is the value of the whole plant ? I (a) $34,689. Sl (() $37,935. (16) Divide the following : (a) 962,842 by 84; (b) 39,728 by 63; (c) 29,714 by 108; (d) 406,089 by 135. f ( a ) 11,462.4047. J (b} 630.603. AnS> * (c) 275.1296. ( &> T2 TlHSy an d TT S FO- -5. .875. Ans. .15625. ,65. ,125. (3) Write out in words the following numbers: .08, .131, .0001, .000027, .0108, and 93.0101. (4) What is the sum of .125, .7, .089, .4005, .9, and .000027 ? Ans. 2.214527. (5) Add 17 thousandths, 2 tenths, and 47 millionths. Ans. .217047. (6) Work out the following examples: (a) 709.63 .8514; (b} 81.963 1.7; (c] 18 .18; (d] 1 - .001 ; (e) 872.1 - (.8721 + .008) ; (/) (5.028 + .0073) - (6.704 - 2.38). { (a) 708.7786. (b} 80.263. (c) 17.82. (d) .999. (e) 871.2199. , (/) .7113. 1 For notice of copyright, see page immediately following the title page. Ans. -< 8 ARITHMETIC. 1 (7) The cost of the coal consumed under a nest of steam boilers during a week's run was as follows: Monday, $15.83; Tuesday, $14.70; Wednesday, $14.28; Thursday, $13.87; Friday, $14.98; Saturday, $12.65. What was the cost of the week's supply of coal ? Ans. $86.31. (8) It is desired to increase the capacity of an electric- light plant to 1,500 horsepower by adding a new engine. If the indicated horsepower of the engines already in use is 482, 316, and 390f, what power must the new engine develop? Ans. 310.11$ H. P. (9) The inside diameter of a 6-inch steam pipe is 6.06 inches, and the outside diameter is 6.62 inches. How thick is the pipe ? Ans. .28 in. (10) Find the products of the following expressions: (a) . 013 X. 107; (b) 203 X 2.03 X .203; (c) (2.7x31.85) X (3. 16 -.316); (d) (107.8 + 6.541-31.96) X 1.742. (a) .001391. , (b] 83.65427. 1 (c) 244.56978. (d) 143.507702. (11) How many square feet of heating surface are in the tubes of a boiler having sixty 3-inch tubes, each 15 feet long, if the heating surface of each tube per foot in length is .728 square foot ? Ans. 677.04 sq. ft. (12) Find the values of the following expressions : 7 ... U , N 1.25 X 20 X 3 ( (a) 37$. () S ; (*) f ; <<) 87 + 88 Ans. (q .75 459 + 32 (c) 210f (13) The distance around a cylindrical boiler is 166.85 inches. If there are 72 rivets in one of the circular seams, find what the pitch (distance between the centers of any two rivets) of the rivets is. Ans. 2.317+ in. (14) A keg of * x 21" boiler rivets weighs 100 pounds and contains 133 rivets. What is the weight of each rivet ? Ans. .75+lb. 1 ARITHMETIC. 9 (15) How many inches in .875 of a foot ? Ans. 10 in. (16) What decimal part of a foot is T 3 -g- inch ? Ans. .015625 ft. (17) In a steam-engine test of an hour's duration, the horsepower developed was found to be as follows at 10-min- ute intervals: 48.63, 45.7, 46.32, 47.9, 48.74, 48.38, 48.59. What was the average horsepower ? Ans. 47.75+ H. P. s. I.2J, ARITHMETIC. (PART 4.) EXAMINATION QUESTIONS. (1) What is 25$ of 8,428 Ib.? Ans. 2,107 Ib. (2) What is |$ of $35,000? Ans. $175. (3) What per cent, of 50 is 2 ? Ans. 4$. (4) What per cent, of 10 is 10 ? Ans. 100$. (5) The coal consumption of a steam plant is 5,500 Ib. per day when the condenser is not running, or an increase of 15$ over the consumption when the condenser is used. How many pounds are used per day when the condenser is run- ning ? Ans. 4,782.61 Ib. (6) An engineer receives a salary of $950. He pays 24$ of it for board, 12$ of it for clothing, and 17$ of it for other expenses. How much of it does he save a year ? Ans. $441.75. (7) If 37$ of a number is 961.38, what is the number ? Ans. 2,563.68. (8) The speed of an engine running unloaded was 1$ greater than when running loaded. If it made 298 revolu- tions per minute with the load, what was its speed running unloaded ? Ans. 302.47 rev. per min. (9) Reduce 4 yd. 2 ft. 10 in. to inches. Ans. 178 in. 2 For notice of copyright, see page immediately following the title page. 2 ARITHMETIC. 2 (10) Reduce 3,722 in. to higher denominations. Ans. 103 yd. 1 ft. 2 in. (11) Reduce 764,325 cu. in. to cubic yards. Ans. 16 cu. yd. 10 cu. ft. 549 cu. in. (12) A carload of coal weighed 16 T. 8 cwt. 75 Ib. How many pounds did this amount to ? Ans. 32,875 Ib. (13) Reduce 25,396 Ib. to higher denominations. Ans. 12 T. 13 cwt. 96 Ib. (14) What is the sum of 2 yd. 2 ft. 3 in., 4 yd. 1 ft. 9 in., and 2 ft. 7 in. ? Ans. 8 yd. 7 in. (15) From a barrel of machine oil is sold at one time 10 gal. 2 qt. 1 pt. and at another time 16 gal. 3 qt. How much remained ? Ans. 4 gal. 1 pt. (16) If 1 iron rail is 17 ft. 3 in. long, how long would 51 such rails be if placed end to end ? Ans. 879 ft. 9 in. (17) Multiply 3 qt. 1 pt. by 4.7. Ans. 32.9 pt. (18) A main line shaft is composed of four lengths each 15ft. Sin. long; one length 14ft. Sin. long; and one length 8 ft. 10 in. long. There are six hangers spaced equally dis- tant apart, one being placed at each extremity of the shaft, 8 in. from the ends. What is the distance between the hangers ? Ans. 16 ft. 9 j in. (19) If the length of a boiler shell is 18 ft. 11 in., how many rivets should there be in one of the longitudinal seams if it is a single-riveted seam, supposing the rivets to be 1^ in. between centers and the two end rivets to be 1 in. from each end of the boiler ? Ans. 181 rivets. ARITHMETIC. (PART 5.) EXAMINATION QUESTIONS. (1) What is the square of 108 ? Ans. 11,664. (2) What is the cube of 181.25? Ans. 5,954,345.703125. (3) Find the values of the following: (a) .0133 3 ; (b) 301.011 s ; (0 (*); (d) (3f). ' (a) .000002352637. (b) 27,273,890.942264331. O A. ( d ) 52|i, or 52.734375. NO'TE. In the answers to the following examples requiring the extraction of a root, a minus sign after a number indicates that the last digit is not quite as large as the number printed. Thus, 12,497 indi- cates that the number really is 12.496 + , and that the 6 has been made a 7 because the next succeeding figure was 5 or greater. The answers to the examples requiring the extraction of cube root are exact to the last digit given. If the student is unable to get the same answer by means of the method described in Arts. 3O-34, he should then use the method described in Arts. 35 and 36. (4) Find the square root of the following: (a) 3,486,^ 784.401; (b) 9,000,099.4009. I (a) 1,867.29+. 3 ' 1 (b) 3,000.017-. (5) Extract the cube root of .32768. Ans. .6894+. (6) Extract the cube root of 2 to four decimal places. Ans. 1.2599+. 2 For notice of copyright, see page immediately following the title page ARITHMETIC. (7) Solve the following: (a) f/123.21; (b) Ans. (8) Solve the following : (a) 1/114.921. (a) 11.1. (b) 10.72+. '.0065; (b) |/. 000021. j () .18663-. 3 ' ( (J) .02759-. Find the value of ;r in the following : (9) 11.7 : 13 :: 20 : x. Ans. 22.22+. (10) (a) 20 + 7 : 10+ 8 :: 3 : x\ (b} 12 a : 100' :: 4 : x. MS " (ii) 4r 21' ' 24 16' ^ 10 100' 277.7+. 60 150 _- 600' Ans. 15 ; 45 (a) w (rf) M 12. 12. 20. 180. 40. (12) If a piece of 2-inch shafting 3 ft. long weighs 37.45 lb., how much would a piece 6f ft. long weigh ? Ans. 72.225 lb. (13) If a railway train runs 444 miles in 8 hr. 40 min., in what time can it run 1,060 miles at the same rate of speed ? Ans. 20 hr. 41.44 min. (14) If a pump discharging 135 gal. per min. fills a tank in 38 min., how long would it take a pump discharging 85 gal. per min. to fill it ? Ans. 60 T 6 T min. (15) The distances around the driving wheels of two loco- motives are 12.56 ft. and 15.7 ft., respectively. How many times will the larger turn while the smaller turns 520 times ? Ans. 416 times. (16) If a cistern 28 ft. long, 12 ft. wide, 10 ft. deep holds 798 bbl. of water, how many barrels of water will a cistern hold that is 20 ft. long, 17 ft. wide, and 6 ft. deep ? Ans. 484 bbl. MENSURATION AND USE OF LETTERS IN FORMULAS. EXAMINATION QUESTIONS. A = 5 h 200 = 10 x = 12 i = 3.5 Z> = 120 Work out the solutions to the following formulas, using the above values for the letters: (1) C =^~ An, C=8. Q,44*+J* An, 6 = Ans. v = 4. 05+. n o Switchboard. (11) A triangle has three equal angles; what is it called ? (12) If a triangle has two equal angles, what kind of a triangle is it ? (13) In a triangle ABC, angle ^4 = 23 and ,# = 32 32'; what does angle C equal ? Ans. C = 124 28'. 3 USE OF LETTERS IN FORMULAS. (14) In the figure, if AD= 10 inches, A B = 24 inches, and B C = 13 inches, how long is JD , D E being parallel toC? Ans. D=5.Q25 in. (15) An engine room is 52 feet long and 39 feet wide. How many feet is it from one corner to a diagonally oppo- site one, measured in a straight line ? Ans. 65 ft. (10) It is required to make a miter- box in which to cut molding to fit around an octagon post. At what FIG. n. angle with the side of the box should the saw run ? Ans. 67. (17) If the distance between two opposite corners of a hexagonal nut is 2 inches, what is the distance between two opposite sides ? Ans. 1.732+ in. (18) In the accompanying figure, if the distance B I is 6 inches and H K 18 inches, what is the diameter of the circle ? Ans. 19.5 in. (19) How many revolutions will a 72-inch locomotive driver make in going 1 mile ? FIG. in. Ans. 280.112 revolutions. (20) A pipe has an internal diameter of 6.06 inches; what is the area of a circle having this diameter ? Ans. 28.8427 sq. in. (21) How long must the arc of a circle be to contain 12, supposing the radius of the circle to be 6 inches ? Ans. 1.25664 in. (22) What is the area of the sector of a circle 15 inches in diameter, the angle between the two radii forming the sector being 12 ? Ans. 6.1359 sq. in. (23) (a) What would be the length of the side of a square metal plate having an area of 103.8691 square inches? 4 MENSURATION AND 3 (b) What would be the diameter of a round plate having this area ? (c] How much shorter is the circumference of the round plate than Uie perimeter of the square plate ? (a) 10.1916 in. Ans. ] () 11 J in. (c) 4. 638 in. (24) Find the area in square feet of the entire surface of a hexagonal column 12 feet long, each edge of the ends of the column being 4 inches long. Ans. 24.5774 sq. ft. (25) Find the cubical contents of the above column in cubic inches. Ans. 5,985.9648 cu. in. (26) Compute the weight per foot of an iron boiler tube 4 inches outside diameter and 3.73 inches inside diameter, the weight of the iron being taken at .28 pound per cubic inch. Ans. 5 Ib. (27) The dimensions of a return-tubular boiler are as follows: Diameter, 60 inches; length between heads, 16 feet; outside diameter of tubes, 3 inches; number of tubes, 64; distance of mean water-line from top of boiler, 18 inches. (a) Compute the steam space of the boiler in cubic feet. (b} Determine the number of gallons of water that will be required to fill the boiler up to the mean water level. Ans -f ^ 79 " 3 cu " ft> I (b) 1,246 gal., nearly. (28) The length of the circumference of the base of a cone is 18.8496 inches and its slant height is 10 inches. Find the area of the entire surface of the cone. Ans. 122.5224 sq. in. (29) If the altitude of the above cone were 9 inches, what would be its volume ? Ans. 84.8232 cu. in. (30) A square vat is 11 feet deep, 15 feet square at the top, and 12 feet square at the bottom. How many gallons will it hold ? Ans. 15,058.29 gal. (31) How many pails of water would be required to fill the vat, the pail having the following dimensions: Depth, 3 USE OF LETTERS IN FORMULAS. 5 11 inches; diameter at the top, 12 inches; diameter at the bottom, 9 inches ? Ans. 3,627.28. (32) Find (a) the area of the surface, and (b) the cubical contents of a ball 22 inches m diameter. Ans ( (a) 1,590.435 sq. in. ' ( (b) 5,964.1313 cu. in. (33) (a) What is the volume and area of a cylindrical ring whose outside diameter is 16 inches and inside diameter 13 inches ? (b} If made of cast iron, what is its weight ? Take the weight of 1 cubic inch of cast iron as .261 pound. Ans. Weight = 21 Ib. PRINCIPLES OF MECHANICS. EXAMINATION QUESTIONS. (1) Define mechanics. (2) Define (a) matter; (b) molecule; (c] atom. (3) In what three states does matter exist ? (4) Explain clearly the distinction between general prop- erties of matter and special properties of matter. (5) Define (a) motion ; (b) velocity ; (c) uniform velocity ; (d) variable velocity. (6) Define (a) acceleration; (b) retardation; (c) average velocity. (7) An ocean steamer made a run of 3,240 miles in 6 days and 16 hours. What was its average speed in miles per hour ? Ans. 20 mi. (8) How long would it take to make a tour around the world when traveling at an average speed of 3,000 feet per minute, assuming the distance around, the world to be 25,000 miles ? Ans. 30 da. 13 hr. 20 min. (9) How do we recognize the existence of a force ? (10) What conditions are necessary to 'compare the rela- tive effects of different forces on different bodies ? (11) State Newton's Laws of Motion. (12) What do you understand by the term "inertia"? (13) Define (a) dynamics; (b) statics. 4 2 PRINCIPLES OF MECHANICS. 4 (14) Why is not the weight of a given body the same at every point on the surface of the earth ? (15) Determine the mass of a body that weighs 346 pounds at a place where g is equal to 32. 174. Ans. 10. 75+. (16) How does the position of a body above or below the surface of the earth affect its weight ? (17) A locomotive weighing 30 tons has to overcome a constant force of 15 pounds per ton when it is in motion. What total force must the locomotive exert so that its speed may increase at the rate of 3 feet per second ? Ans. 6,047 Ib. (18) What will be the momentum of the locomotive in the preceding example when it has attained a velocity of 1 mile per hour? Ans. 2,736.3 Ib. (19) Define (a] work; (b) power; (c] energy. (20) What horsepower is required to raise a .body weigh- ing 66,000 pounds through a distance of 80 feet in \ hour ? Ans. 5 H. P. (21) A body weighing 6,432 pounds is moving with a constant velocity of 60 feet per second. What horsepower will be required to bring the body to rest in 3 minutes ? Ans. 3 r 7 T H. P. (22) What is the tangential pressure on the crank of an engine when the crank is on either dead center ? (23) What is friction ? (24) A body weighing 5,000 pounds rests on a horizontal surface. In order to slide the body along the surface, a horizontal force of 300 pounds must be exerted. What is the coefficient of friction in this particular case ? Ans. .06. (25) A crosshead weighing 1,000 pounds and having bronze shoes slides on a slightly greased, horizontal wrought- iron guide. If the contact area of the bronze shoe is 100 square inches, what will be (a) the total friction and (b) the friction per square inch of contact surface ? (a) 160 Ib. 1.6 Ib. 4 PRINCIPLES OF MECHANICS. 3 (20) What is the center of gravity of a body ? (27) Explain the method of finding the common center of gravity of several bodies whose weights and the dis- tances between whose centers of gravity are known. (28) Give a practical method of determining the center of gravity of a solid body. (29) Define (a) centrifugal force; (b) centripetal force. (30) What is the relation between the centrifugal and centripetal forces of a revolving body ? (31) A body weighing 10 pounds revolves at a speed of 60 revolutions per minute about a point 6 feet from its cen- ter of gravity. What is the centrifugal force tending to pull the body from the point about which it revolves ? Ans. 73.441b. (32) Name the three states of equilibrium and give examples of each. (33) In the case of a body at rest, what are the condi- tions of the forces acting on that body ? (34) How is the condition of equilibrium of the forces acting on a body affected by the position of its "line of direction " with respect to the base ? MACHINE ELEMENTS. EXAMINATION QUESTIONS. (1) Define (a) lever; (b) weight arm; (c) force arm; (d) fulcrum. (2) What is the condition necessary for the equilibrium of the lever ? (3) State the general rule that expresses the relation existing between the weight, the force, and the distances through which they move. (4) What must be the length of the weight arm in order that a force of 12 pounds at a distance of 20 inches from the fulcrum will raise a weight of 100 pounds at the end of the weight arm ? Arts. 2.4 in. (5) Into what two classes may pulleys be divided in ref- erence to their construction ? (6) What advantages have split pulleys over solid pul- leys ? (7) Explain how a crowned pulley tends to prevent the slipping off of the belt. (8) What is meant by "balancing pulleys," and how is it accomplished ? (9) Define the terms " driver " and " driven " as applied to pulleys. 5 //. 8. /. 25 2 MACHINE ELEMENTS. 5 (10) Find what diameter driver will be required which, when running at 600 revolutions per minute, will cause the driven, whose diameter is 6 inches, to run at 1,800 revolu- tions per minute. Ans. 18 in. (11) What must be the speed of a driver 12 inches in diameter in order that the driven, whose diameter is 5 inches, may make 1,600 revolutions per minute ? Ans. 666f rev. per min. (12) The distance between the centers of two pulleys, whose diameters are 6 feet and 2 feet, respectively, is 40 feet. What is the required length of an open belt ? Ans. 93 ft. (13) A single belt running at 1,650 feet per minute- is used to transmit 40 horsepower. If the arc of contact on the small pulley is 120, how wide should the belt be ? Ans. 28 in. (14) What horsepower will be transmitted by the belt in question 13 if the velocity is reduced to 1,200 feet per minute? Ans. 29.3 H. P. (15) If, in question 13, a double belt were substituted for the single belt, how wide should it be ? Ans. 20 in. (16) Which side of a leather belt should be in contact with the pulley face, and why ? (17) Should rosin be used on a belt to prevent slipping ? (18) Give the causes of flapping belts and their rem- edies. (19) State the different methods used in joining belts. Which of these makes the best joint ? (20) What precaution should be observed when using rubber belts ? (21) Define the terms "driver" and "follower" as applied to a train of gear-wheels. (22) In Fig. 19, if the diameter of A is 90 inches, that of F 30 inches, and if B has 12 teeth, C 30 teeth, D 20 teeth, and E 36 teeth, find the weight that a force of 50 pounds at /*can raise. Ans. 675 Ib. 5 MACHINE ELEMENTS. 3 (23) What is (a) circular pitch ? (b] diametral pitch ? (24) What are the most common forms of teeth used in ordinary practice ? (25) What advantages have involute teeth over epicy- cloidal teeth ? (26) Find the pitch diameter of a gear-wheel having 60 teeth and a circular pitch of 1.152 inches. Ans. 22 in. (27) What is the circular pitch of a gear-wheel 30 inches in diameter having 60 teeth ? Ans. 1.57 in. (28) What is the over-all diameter of a gear-wheel having 80 teeth with a diametral pitch of 8 ? Ans. 10^- in. (29) Find the number of teeth in a gear-wheel whose outside diameter is 7 inches and whose diametral pitch is 8. Ans. 58 teeth. (30) What distinguishes a fixed pulley from a movable pulley ? (31) In a certain combination of pulleys there are seven movable ones. Neglecting losses due to friction, how heavy a weight can a force of 150 pounds raise when applied to the free end of the rope ? Ans. 2,100 Ib. (32) How does friction affect the force required to raise a given weight by means of a rope or chain and pulleys ? (33) In an ordinary block and tackle having four mov- able pulleys, what is the probable actual force that must be applied to raise a weight of 1,500 pounds ? Ans. 312.5 Ib. (34) In what respect is the Weston differential pulley block better than the ordinary block and tackle ? (35) An inclined plane is 70 feet long and 12 feet high. What force acting parallel to the plane will be required to sustain a weight of 600 pounds on the plane ? Ans. 103 Ib. (36) What weight will a force of 36 pounds acting paral- lel to the base of the plane be able to sustain on an inclined plane having a base 50 feet long and a height of 15 feet ? Ans. 120 Ib. 1 MACHINE ELEMENTS. 5 (37) What is the probable actual weight that can be raised by means of a screw jack that has a screw 2 inches in diameter with 6 threads to the inch if a force of 50 pounds is applied at the end of a lever 20 inches from the shaft ? Ans. 2,356.2 Ib. (38) Define (a) velocity ratio; (b} efficiency. (39) What is the efficiency of the screw jack of ques- tion 37 ? Ans. 6^ per cent. MECHANICS OF FLUIDS. EXAMINATION QUESTIONS. (1) Define hydrostatics. (2) How can it be proved that liquids transmit pressure in all directions and with the same intensity ? (3) State Pascal's law. (4) In what direction does the pressure due to the weight of a body of water act in a vessel in which water is con- tained ? (5) What is the pressure on the bottom of a vessel if the base is a circle 3 inches in diameter, the height 8 inches, and the vessel is completely filled with water ? Ans. 2.045 Ib. (6) State the law for the upward pressure of a liquid on a horizontal surface submerged in the liquid. (7) Why is it that in several pipes that communicate with one another and that differ in shape and size, water will stand at the same height ? (8) Why is it that water issuing from a hose connected to a hydrant cannot be made to spout up to a level with the surface of the water in the reservoir that supplies the hydrant ? (9) Define specific gravity. 2 MECHANICS OF FLUIDS. 6 (10) Calculate the weight of a block of aluminum whose volume is 100 cubic feet. Ans. 15,605 Ib. (11) State the principle of Archimedes. (12) Explain how, by the application of the principle of Archimedes, the volume of an irregularly shaped body may be accurately determined. (13) (a) What are hydrometers ? (b) Into what two classes may hydrometers be divided ? (14) A single-cylinder pump feeds a boiler through a delivery pipe 1 inch in actual diameter. The piston speed is such as to give a velocity of flow of 400 feet per minute. How many gallons of water can be pumped into the boiler in 1 hour ? Ans. 979.2 gal. (15) What should be the commercial size of a delivery pipe from a duplex pump to deliver 936 gallons of water per hour ? Ans. 1 in. (16) Why is it important to have as few bends as possible in the suction pipe leading to a pump ? (17) What simple experiment proves that gases tend to expand and increase their volume ? (18) How high a column of mercury will the pressure of the atmosphere support ? (19) How can the degree of the vacuum in a vessel be determined ? (20) How high a column of a liquid whose specific gravity is 2.5 will the atmospheric pressure support ? Ans. 163.2 in. (21) How is the pressure of the atmosphere measured ? (22) Why does the pressure of the atmosphere decrease as the elevation above sea level increases ? (23) In what respect is the action of the pressure of the atmosphere similar to that of the pressure of a liquid ? 6 MECHANICS OF FLUIDS. 3 (24) How does the tension of a gas change with the change in volume under constant temperature ? (25) Define (a) gauge pressure ; (b) absolute pressure. (26) A metallic tube closed at one end is fitted with an air-tight movable piston. When the piston is at the open end of the tube, the pressure of the air within the tube is equal to 14.7 pounds per square inch. What will be the pressure of the air between the piston and the closed end of the tube after the former has moved toward the latter a dis- tance equal to ^ the length of the tube, the temperature remaining constant ? Ans. 73.5 Ib. per sq. in. (27) Suppose that the piston in the tube mentioned in question 26 had been moved toward the closed end until the pressure had reached 147 pounds per square inch. What fraction of the original volume would the compressed air have occupied ? Ans. -fa. (28) In pneumatics, what is an air pump ? (29) Can a perfect vacuum be produced with the air pump ? (30) Explain the operation of the dashpot of a Corliss engine. (31) Explain the principle of operating the siphon. (32) Suppose that two vessels, in one of which the water stands at a higher level than in the other, are connected by a siphon and water is siphoned from the vessel in which it stands at the higher level into the other vessel. What will happen when the water in each vessel reaches the same level ? (33) In practice, how far above the water level in the vessel from which water is siphoned can the highest point of the siphon be carried ? (34) (a) What is a pump ? (b) Into how many types may pumps be divided in reference to their mode of action ? (c) Name these types. 4 MECHANICS OF FLUIDS. 6 (35) Explain the principle on which the suction pump acts. (36) What advantage has a lifting pump over a suction pump ? (37) In what respect does a force pump differ from a lifting pump ? (38) What is the difference between a single-acting and a double-acting force pump ? STRENGTH OF MATERIALS. EXAMINATION QUESTIONS. (1) (a) Define stress, (b) Name the various kinds of stresses to which a body can be subjected. (2) A weight of 8,000 pounds rests on the top of a cubi- cal block of wood the area of each face of which is 20 square inches. What is the unit stress, in pounds per square inch, to which the block is subjected ? Ans. 400 Ib. per sq. in. (3) Define (a) strain; (ft) elasticity; (c) elastic limit. (4) Taking the ultimate tensile strength of wrought iron as 55,000 pounds per square inch, what tensile force will rupture a wrought-iron bar whose cross-sectional area is 4 square inches ? Ans. 220,000 Ib. (5) How does annealing improve old chains ? (6) Give a practical method of determining the true con- dition of a given rope, supposing that its outer surface appears to be in good condition. (7) (A) On what does the safe lifting load of a sling depend ? (V) Explain how the style of attachment to the hook of the tackle block affects the amount of load to be raised. (8) (a) For what is manila rope chiefly used ? (b) What is the object of lubricating the fibers of a rope ? 7 2 STRENGTH OF MATERIALS. 7 (9) In general, how should the size of the pulley compare with the size of manila rope used for transmitting power ? (10) What is the greatest load to which an iron-wire rope 1 inches in circumference should be subjected ? Ans. 1,350 Ib. (11) What should be the circumference of a steel-wire rope under a maximum working load of 16,000 pounds ? Ans. 4 in., nearly. (12) How does the strength of a column having both ends flat compare with that of columns both of whose ends are not flat but which in every other respect are similar to the first column ? (13) Give practical examples of the three different classes into which columns are divided with respect to the condi- tion of their ends. (14) What is the safe steady working load that a cast- iron column, having fixed ends, 14 inches in diameter and 16 feet high can sustain ? Ans. 1,291,480 Ib. (15) Would you consider a steel piston rod 6 inches in diameter of sufficient size for a 40-inch cylinder using steam at 110 pounds pressure ? If so, why ? (16) A solid yellow-pine beam 14 inches square rests on two supports 12 feet apart. What steady safe load will the beam support at its middle point ? Ans. 18,547 Ib. (17) An oak beam 4 inches wide and 6 inches deep is subjected to a sudden shearing stress of 10,000 pounds across the grain. Is this a safe load for the given conditions ? (18) What should be the area of a wrougJit-iron beam to safely support a sudden shearing stress of 400,000 pounds ? Ans. 91 sq. in., nearly. (19) What do you understand by double shear ? (20) What is the use of countershafts ? (21) What is the distinction between cold-rolled shafting and bright shafting ? 7 STRENGTH OF MATERIALS. 3 (22) How is bright shafting designated commercially with respect to size ? (23) Why is it good practice to place pulleys for trans- mitting or receiving power as near the bearings of the shaft as possible ? (24) What horsepower can be transmitted from a steel shaft 8 inches in diameter by means of pulleys between the bearings when running at a speed of 80 revolutions per minute ? Ans. 482 H. P. (25) What must be the speed of a 4-inch shaft of cold- rolled iron, having no pulleys between bearings, to transmit 75 horsepower ? Ans. 76 rev. (26) Find the diameter of a wrought-iron shaft running at 300 revolutions per minute to transmit 84 horsepower by means of pulleys between its bearings. Ans. 3 in. (27) How does a change in the speed of a shaft affect the amount of power transmitted ? (28) Does a high tensile strength in metals necessarily imply an ability on the part of the metal to safely resist repeated applications of sudden stresses ? (29) What is the chief advantage of a steel rope over an iron rope ? A KEY TO ALL THE QUESTIONS AND EXAMPLES CONTAINED IN THE EXAMINATION QUESTIONS INCLUDED IN THIS VOLUME. The Keys that follow have been divided into sections cor- responding to the Examination Questions to which they refer, and have been given corresponding section numbers. The answers and solutions have been numbered to corre- spond with the questions. When the answer to a question involves a repetition of statements given in the Instruction Paper, the reader has been referred to a numbered article, the reading of which will enable him to answer the question himself. To be of the greatest benefit, the Keys should be used sparingly. They should be used much in the same manner as a pupil would go to a teacher for instruction with regard to answering some example he was unable to solve. If used in this manner, the Keys will be of great help and assist- ance to the student, and will be a source of encouragement to him in studying the various papers composing the Course. ARITHMETIC. (PART 1.) (1) See Art. 1. (2) See Art. 3. (3) See Arts. 5 and 6. (4) See Arts. 1O and 11. (5) 980 = Nine hundred eighty. 605 = Six hundred five. 28,284 Twenty-eight thousand two hundred eighty-four. 9,006,042 = Nine million six thousand forty-two. 850,317,002 = Eight hundred fifty million three hundred seventeen thousand two. 700, 004 Seven hundred thousand four. (6) Seven thousand six hundred = 7,600. Eighty-one thousand four hundred two = 81,402. Five million four thousand seven = 5,004,007. One hundred and eight million ten thousand one = 108,- 010,001. Eighteen million six = 18,000,006. Thirty thousand ten = 30,010. 1 For notice of copyright, see page immediately following the title page. 2 ARITHMETIC. 1 (7) In adding whole numbers, place the numbers to be added directly under each other, so that the extreme right- hand figures will stand in the same 3290 column, regardless of the position of 504 those at the left. Add the first 865403 column of figures at the extreme 2074 right, which equals 19 units, or 1 ten and 9 units. We place 9 units 7 under the units column and reserve 871359 Ans. 1 ten for the column of tens, 8 + 7 + 9 + 1 = 25 tens, or 2 hundreds and 5 tens. Place 5 tens under the tens column and reserve 2 hundreds for the hundreds column. 4 + 5+2 + 2=: 13 hundreds, or 1 thousand and 3 hundreds. Place 3 hundreds under the hundreds column and reserve the 1 thousand for the thousands column. 2 + 5+3 + 1 = 11 thousands, or 1 ten-thousand and 1 thousand. Place the 1 thousand in the column of thousands and reserve the 1 ten-thousand for the column of ten-thousands. 6 + 1 = 7 ten-thousands. Place this 7 ten-thousands in the ten-thousands column. There is but one figure, 8, in the hundreds of thousands place in the numbers to be added, so it is placed in the hundreds of thousands column of the sum. A simpler (though less scientific) explanation of the same problem is the following : 7 + 1 + 4+3 + 4 + = 19; Avrite the 9 and reserve the 1. 8 + 7 + + 0+9 + 1 reserved = 25 ; write the 5 and reserve the 2. + 4 + 5 + 2+2 reserved = 13 ; write the 3 and reserve the 1. 2 + 5 + 3+1 reserved = 11 ; write the 1 and reserve 1. 6 + 1 reserved = 7 ; write the 7. Bring down the 8 to its place in the sum. (8) 709 8304725- 391 100302 300 909 8407336 Ans. 1 ARITHMETIC. 3 (9) The steam engine, during the 12-hour test, showed that the number of revolutions made were 150,508, since 12,600 + 12,444+ 12,467 + 12,528 + 12,468 + 12,590 + 12,610 + 12,589 + 12,576 + 12,558 + 12,546 + 12,532 = 150,508 rev. 12600 revolutions. 12444 revolutions. 12467 revolutions. 12528 revolutions. 12468 revolutions. 12590 revolutions. 12610 revolutions. 12589 revolutions. 12576 revolutions. 12558 revolutions. 12546 revolutions. / 12532 revolutions. 150508 revolutions. Ans. (10) In subtracting whole numbers, place the subtrahend, or smaller number, under the minuend, or larger number, (a] 5 9 6 2 so that t ^ ie right-hand figures stand 3 3 3 g directly under each other. Begin at the right to subtract. We cannot 47624 Ans. su b tract g units from 2 units, so we take 1 ten from the 6 tens and add it to the 2 units. 1 ten = 10 units, so we have 10 units + 2 units 12 units. Then, 8 units from 12 units leaves 4 units. We took 1 ten from 6 tens, so only 5 tens remain. 3 tens from 5 tens leaves 2 tens. In the hundreds column we have 3 -hundreds from 9 hundreds leaves 6 hundreds. We cannot subtract 3 thou- sands from thousands, so we take 1 ten-thousand from 5 ten-thousands and add it to the thousands. 1 ten-thousand = 10 thousands, and 10 thousands + thousands = 10 thou- sands. Subtracting, we have 3 thousands from 10 thou- sands leaves 7 thousands. We took 1 ten-thousand from 5 ten-thousands and have 4 ten-thousands remaining. Since there are no ten-thousands in the subtrahend, the 4 in the H. s. I. 26 4 ARITHMETIC. 1 ten-thousands column in the minuend is brought down into the same column in the remainder, because from 4 leaves 4. (b) 15339 10001 5338 Ans. (11) (a) 7 9 6 8 (b) I 32975 98735 37993 Ans. 1265 Ans. (12) 3040 = No. of gallons in the tank at the beginning of the day. 4780= No. of gallons pumped in during the morning. 7820 = No. of gallons in the tank after 4,780 gallons were added. 7240 = No. of gallons drawn out during the morning. 580 = No. of gallons in the tank at the beginning of the afternoon. 8675= No. of gallons pumped in during the afternoon. 9255 = No. of gallons in the tank after 8,675 gallons were added. 7633 = No. of gallons drawn out during the afternoon. 1622= No. of gallons remaining in the tank at night. Ans. (13) In the multiplication of whole numbers, place the multiplier under the multiplicand and multiply each term of the multiplicand by each term of the multiplier, writing the right-hand figure of each product obtained under the term of the multiplier which produces it. (a) 526387 7 times 7 units = 49 units, or 7 4 tens and 9 units. We write 3684709 Ans the 9 units and reserve the 4 tens. 7 times 8 tens = 56 tens; 56 + 4 tens reserved = 60 tens, or 6 hundreds and tens. Write the tens and reserve the 6 hundreds. 7x3 hun- dreds = 21 hundreds; 21 + 6 hundreds reserved = 27 hun- dreds, or 2 thousands and 7 hundreds. Write the 7 hun- dreds and reserve the 2 thousands. 7x6 thousands = 1 ARITHMETIC. 5 42 thousands; 42 + 2 thousands reserved = 44 thousands, or 4 ten-thousands and 4 thousands. Write the 4 thousands and reserve the 4 ten-thousands. 7x2 ten-thousands 14 ten-thousands; 14 + 4 ten-thousands reserved = 18 ten- thousands, or 1 hundred-thousand and 8 ten-thousands. Write the 8 ten-thousands and reserve the 1 hundred- thousand. 7 X 5 hundred-thousands =35 hundred-thousands; 35 + 1 hundred-thousand reserved = 36 hundred-thousands. Since there are no more figures in the multiplicand to be multiplied, we write the 36 hundred-thousands in the prod- uct. This completes the multiplication. A simpler (though less scientific) explanation of the same problem is the following: 7 times 7 = 49 ; write the 9 and reserve the 4. 7 times 8 = 56; 56 + 4 reserved = 60; write the and reserve the 6. 7 times 3 = 21; 21 + 6 reserved = 27 ; ' write the 7 and reserve the 2. 7 X 6 = 42; 42 + 2 reserved = 44; write the 4 and reserve 4. 7 X 2 = 14; 14 + 4 reserved = 18; write the 8 and reserve 1. 7 X 5 = 35; 35 + 1 reserved = 36; write the 36. In this case the multiplier is (b) 7 2 9 8 17 units, or 1 ten and 7 units, so 17 that the product is obtained by 4902086 adding two partial products, 700298 namely, 7 X 700,298 and 10 X 11905066 Ans. 700^98 The actual operation is performed as follows: 7 times 8 = 56; write the 6 and reserve the 5. 7 times 9 = 63; 63 + 5 reserved = 68; write the 8 and reserve the 6. 7 times 2 = 14; 14+6 reserved = 20; write the and reserve the 2. 7 times = 0; + 2 reserved = 2 ; write the 2. 7 times = 0; 0+0 reserved =0; write the 0. 7 times 7 = 49 ; 49 + reserved = 49 ; write the 49. To multiply by the 1 ten we have 1 ten times 8 units = 8 tens and units. We do not write the units, but write the 8 tens under the 8 tens in the first partial product above. We next multiply 1 ten and 9 tens = 90 tens = 9 hun- dreds + tens; write the 9 hundreds under the hundreds 6 ARITHMETIC. 1 of the first partial product above. Again, 1 ten times 2 hundreds 2 thousands; write the 2 thousands under the 2 thousands above. 1 ten times thousands = ten- thousands; write the in the ten-thousands place. 1 ten times ten-thousands = hundred-thousands; write the in the hundred-thousands place. 1 ten times 7 hundred- thousands = 7 millions; write the 7 in the millions place. This completes the second partial product. Add the two partial products; their sum equals the entire product. (c) 217 Multiply any two of the numbers 103 together, and multiply their product jj~jj~J by the third number. 000 217 22351 6 7 156457 134106 1497517 Ans. (14) If your watch ticks every second, to find how many times it ticks in 1 week, it is necessary to find the number of seconds in 1 week. 6 seconds = 1 minute. 6 minutes = 1 hour. seconds = 1 hour. 2 4 hours = 1 day. 14400 7200 86400 seconds = 1 day. 7 days 1 week. 604800 seconds in 1 week, or the number of times that your watch ticks in a week. Ans. (15) (a) If an engine and boiler are worth $3,246, and the building is worth three times as much, plus $1,200, then the building is worth 1 ARITHMETIC. $10938 = value of building. If the tools are worth twice as much as the building, plus $1,875, then the tools are worth 10938 21876 plus 1875 $23751 = value of tools. Value of building = 10938 Value of tools = 23751 $34689 = value of the building ... T , t r . and tools. Ans. (b) Value of engine and boiler = 3246 Value of build- ing and tools = 34689 $37935 = value of the whole plant. Ans. (16) (a) 84)96284 2.0 000(1146 2.4 047+ 4 Ans. 400 640 588 5 2 8 ARITHMETIC. 1 84 is contained once in 96. Place 1 as the first figure in the quotient and multiply the divisor 84 by it. Subtract the product, which is 84, from 96, leaving a remainder of 12. Bring down the next figure in the dividend, which is 2, and annex it to 12, making a new dividend of 122. 84 is contained in 122 once. Place 1 as the second figure in the quotient and multiply the divisor 84 by it. Subtract the product (84) from 122, leaving a remainder of 38. Bring down the next figure in the dividend, which is 8, and annex it to 38, making a new dividend of 388. 84 is contained in 388 4 times. Place 4 as the third figure in the quotient and multiply the divisor 84 by it. The prod- uct is 336. Subtract the product from 388, leaving a remainder of 52. Bring down the next figure, which is 4, and annex it to 52, making a new dividend of 524. 84 is contained in 524 6 times. Place 6 as the fourth figure in the quotient. Multiply the divisor 84 by it and subtract the product (504) from 524, leaving a remainder of 20. Bring down the next figure, which is 2, and annex it to 20, making a new dividend of 202. 84 is contained in 202 2 times. Place 2 as the fifth figure in the quotient. Multiply the divisor 84 by it and subtract the product (168) from 202, leaving a remainder of 34. : If it is desired to carry the quotient to 4 decimal places, annex 4 ciphers to the dividend and continue in the same way. In the quotient point off as many decimal places as there are ciphers annexed, or, in this case, 4 decimal places. (b) 63)39728.000(630.603+ Ans. 3 7 1 ARITHMETIC. 9 *63 is not contained in 38, so we place a cipher in the quotient and bring down the next figure in the dividend, which is a cipher that has been annexed, making a new divi- dend of 380. 63 is contained in 380 6 times. 6 X 63 = 378. Subtracting 378 from 380 leaves 2. Bringing down the next figure in the dividend, we have 20 for a new dividend. 63 is not contained in 20, so we place a cipher in the quo- tient and bring down the next cipher in the dividend, making a new dividend of 200. 63 is contained 3 times in 200. (c) 108)29714.0000(275.1296 Ans. 216 811 (d) 135)406089.0000(3008.0666 756 405 Ans. 554 1089* 540 1080 140 900 108 810 320 900 216 810 1040 900 972 810 680 90 648 ~32 * 135 is not contained in 10, so we place a cipher as the second figure in the quotient. Bringing down the next figure, 8, and annexing it to 10, we have a new dividend of 108. 135 is not contained in 108, so we place a cipher as the third figure in the quotient and bring down the next figure in the dividend, or 9, and annex it to 108, making a new dividend of 1,089. 135 is contained in 1,089 8 times. Write 8 as the fourth figure in the quotient. Multiply 135 by 8 and subtract the product (1,080) from 1,089, which leaves a remainder of 9. Bring down the next figure in the dividend, which is a cipher that has been annexed, and annex 10 ARITHMETIC. 1 it to the remainder 9, making a new dividend of 90. As 135 is not contained in 90, we place a in the quotient and bring down another cipher from the dividend, making a new dividend of 900. 135 is contained in 900 6 times. Write 6 as the next figure in the quotient, multiply 135 by 6, and subtract the product (810) from 900, which leaves a remainder of 90. Bring down the next figure (0) in the dividend and annex it to the remainder 90, making a new dividend of 900. 135 is contained in 900 6 times. Place 6 as the next figure in the quotient and multiply the divisor by it. It is plain that each succeeding figure of the quotient will be 6. Point off four decimal places in the quotient, since four ciphers were annexed. (17) If in 1 hour 10 pounds of coal are burned for every square foot of grate area and 9 pounds of water are evapo- rated for every pound of coal burned, then in 1 hour there would be 9 X 10 or 90 pounds of water evaporated for 1 sq. ft. of grate area; and since the grate area is 30 sq. ft., the amount of water evaporated would be 30 X 90 = 2,700 Ib. Since 2,700 Ib. of water are evaporated in 1 hour, in a day of 10 hours there would be 10 X 2,700 Ib., or 27,000 Ib. of water evaporated. (18) If a mechanic earns $1,500 a year and his expenses are $968 per year, then he would save $1500 $968, or $532 per year. 1500 968 $532 If he saves $532 in 1 year, to save $3,724 it would take as many years as $532 is contained times in $3,724, or 7 years. 532)3724(7 years. Ans. 3724 , (19) (a) (72 X 48 X 28 X 5) -f- (84 X 15 X 7 X 6). Placing the numerator over the denominator, the problem becomes 1 ARITHMETIC. 11 72 X 48 X 28 X 5 _ 84 X 15 X 7 X 6 ~ The 5 in the numerator and 15 in the denominator are both divisible by 5, since 5 divided by 5 equals 1 and 15 divided by 5 equals 3. Cross off the 5 ; also cross off the 15 and write the 3 under it. Thus, 72 x 48 X 28 X ft _ 84 X # X 7 X 6 ~ 3 72 in the numerator and 84 in the denominator are divisible by 12, since 72 divided by 12 equals 6 and 84 divided by 12 equals 7. . Cross off the 72 and write the 6 over it ; also, cross off 'the 84 and write the 7 under it. Thus, 6 JT? X 48 X 28 x ft _ ftl X tf X 7 X 6 ~ 7 3 Again, 28 in the numerator and 7 in the denominator are divisible by 7, since 28 divided by 7 equals 4 and 7 divided by 7 equals 1. Cross off\h& 28 and write the 4 0wr it; also, cross off titol. Thus, 6 4 X 48 x ffi X ft _ Again, 48 in the numerator and 6 in the denominator are divisible by 6, since 48 divided by 6 equals 8 and 6 divided by 6 equals 1. Cross off the 48 and write the 8 over it ; also, cross off \b&. Thus, 684 p X lp X J X p Again, 6 in the numerator and 3 in the denominator are divisible by 3, since 6 divided by 3 equals 2 and 3 divided by 3 equals 1. Cross offihz 6 and write the 2 over it ; also, cross 3. Thus, 12 ARITHMETIC. 2 8 4 x 48 x 28 x jfy* x J0 x /T x 7 Since there are /w# remaining numbers (one in the numerator and one in the denominator) divisible by any number except 1, without a remainder, it is impossible to cancel further. Multiply all the tincanceled numbers in the numerator together and divide their product by the product of all the uncanceled numbers in the denominator. The result will be the quotient. The product of all the uncanceled numbers in the numerator equals 2 X 8 X 4 = 64; the product of all the uncanceled numbers in the denominator equals 7. 2 __ LC6 ' # x tf X 7 X ~ ~T~ 7 ~ 7 ? (0) (80 X 60 X 50 X 16 X 14) * (70 X 50 X 24 X 20). Placing the numerator over the denominator, the problem becomes 80 X 60 X 50 X 16 X 14 ? 70 X 50 X 24 X 20 The 50 in the numerator and 70 in the denominator are both divisible by 10, since 50 divided by 10 equals 5 and 70 divided by 10 equals 7. Cross off the 50 and write the 5 over it ; also, cross off the 70 and write the 7 under it. Thus, 5 80 X 60 X 00 X 16 X 14 70 X 50 x 24 X 20 7 Also, 80 in the numerator and 20 in the denominator are divisible by 20, since 80 divided by 20 equals 4 and 20 divided by 20 equals 1. Cross off the 80 and write the 4 over it; also, cross off the 20. Thus, 1 ARITHMETIC. 13 4 5 X 60 x W x 16 x 14 W X 50 x 24 x Again, 16 in the numerator and 24 in the denominator are divisible by 8, since 16 divided by 8 equals 2 and 24 divided by 8 equals 3. Cross off the 16 and write the 2 over it ; also, cross off the 24 and write the 3 under it. Thus, 4 52 $0 X 60 X ?0 X ;0 X 14 _ W X 50 X ?4 X 20 7 3 Again, 60 in the numerator and 50 in the denominator are divisible by 10, since 60 divided by 10 equals 6 and 50 divided by 10 equals 5. Cross off the 60 and write the 6 over it ; also, cross off the 50 and write the 5 under it. Thus, 4652 $0 x 00 x w x ;0 x 14 ;0 7 x 0p x ?4 x 2 53 The 14 in the numerator and 7 in the denominator are divisible by 7, since 14 divided by 7 equals 2 and 7 divided by 7 equals 1. Cross off the 14 and write the 2 over it ; also, cross off the 7. Thus, 46522 ;0 x x w x 20 The 5 in the numerator and 5 in the denominator are divisible by 5, since 5 divided by 5 equals 1. Cross off the 5 of the dividend; also, cross off the 5 of the divisor. Thus, 46022 The 6 in the numerator and 3 in the denominator are divisible by 3, since 6 divided by 3 equals 2, and 3 divided by 3 equals 1. Cross off the 6 and place 2 over it; also, cross off the 3. Thus, 14 ARITHMETIC. 1 2 2 ;$ x # _ x?0 4 2 2 Hence, ^ * ^ X ^ * ? * fl ^ = 4 X 2 X 2 X 2 = 32. jf ^ Ans. (20) If the freight train ran 365 miles in one week, and 3 times as far, lacking 246 miles the next week, then it ran (3 X 365 miles) 246 miles, or 849 miles the second week. Thus, 365 3 1095 - 246 849 miles. Ans. (21) The distance from Philadelphia to Pittsburg is 354 miles. Since there are 5,280 feet in 1 mile, in 354 miles there are 354 X 5,280 feet, or 1,869,120 feet. If the driving wheel of the locomotive is 16 feet in circumference, in going from Philadelphia to Pittsburg, a distance of 1,869,120 feet, it will make 1,869,120 -=- 16, or 116,820 revolutions. 16)1869120(116820 rev. Ans. 1 6 ARITHMETIC. (PART 2.) (1) See Art. 1. (2) See Art. 6. (3) See Art. 3. (4) See Art. 3. (5) y is an improper fraction, since its numerator 13 is greater than its denominator 8. (6) 4; 14 T 3 7 ; 85 T V (7) To reduce a fraction to its lowest terms means to change its form without changing its value. In order to do this, we must divide both numerator and denominator by the same number until we can no longer find any number (except 1) which will divide both of these terms without a remainder. To reduce the fraction f- to its lowest terms, we divide both numerator and denominator by 4 and obtain as a 4 - 1 - 4 4-^4 result the fraction . Thus, - = \\ similarly, o ~T~ 4 J.O~r~ 4 j3_-^4_2-^-2 ^-^ 8 _4H-4_ 1 *' 33-=-4~~8-=-2~ '' 64-f-8~8-h4~~ (8) When the denominator of any number is not expressed, it is understood to be 1, so that f is the same as 6 -4- 1, or6. 1 For notice of copyright, see page immediately following the title page. 16 ARITHMETIC. 1 To reduce -f- to an improper fraction whose denominator is 4, we must multiply both numerator and denominator by some number which will make the denominator of 6 equal to 4. Since the denominator is 1, by multiplying both 6x4 terms of -f- by 4, we will have - = - 2 4 -, which has the same 1X4 value as 6, but has a different form. (9) In order to reduce a mixed number to an improper fraction, we must multiply the whole number by the denomi- nator of the fraction and add the numerator of the fraction to the product. This result is the numerator of the improper fraction, of which the denominator is the denominator of the fractional part of the mixed number. 7 means the same as 7 -f- . In 1 there are -f; hence, in 7 there are 7 X f = -*-/-. Add the | of the mixed number and we obtain -% 6 - + 1 T 3 -? which is the required improper fraction. OiXM+ . (10X+8 . .. (10) Each stroke of the engine is 18 inches in length. Since the piston makes 2 strokes for each revolution, it would pass over a distance of 2 X 18 inches = 36 inches, or 3 feet, in 1 minute, and in making 480 ft. leo revolutions it would pass over 160 X 3, X 60 or 480 ft. Since 480 ft. are passed over 2 8 8 ft. m 1 minute, in 1 hour, or 60 minutes, the distance passed over equals 60 X 480 = 28,800 ft. Since the steam engine runs " days a week and 8 hours per day, the 28800 total number of hours it runs per week 144000 = 6 X 8, or 51 hours. If the piston 1468800 passes over a distance of 28,800 ft. in 1 hour, in 51 hours it would pass over 51 X 28,800 ft., or 1,468,800 ft. Ans. (11) i+| + | = = |=l. Ans. 1 ARITHMETIC. 17 When the denominators of the fractions to be added are alike, we know that the units are divided into the same number of parts (in this case eighths) ; we therefore add the numerators of the fractions to find the number of parts (eighths) taken or considered, thereby obtaining -f, or 1, as the sum. (12) When the denominators are not alike, we know that the units are divided into unequal parts, so before adding them we must find a common denominator for the denomi- nators of all the fractions. Reduce the fractions to fractions having this common denominator, add the numerators, and write the sum over the common denominator. In this case, the least common denominator, or the least number that will contain all the denominators, is 16; hence, we must reduce all these fractions to 16ths and then add their numerators. -(- | + jV ' To reduce the fraction % to a fraction having 16 for a denominator, we must multiply both terms of the fraction by some number which will make the denomi- 1x4 nator 16. This number evidently is 4, hence, - = T 4 7 . 4: /\ 4 Similarly, both terms of the fraction f must be multiplied Q \/ O by 2 to make the denominator 16, and we have = -^. O X & The fractions now have a common denominator 16; hence, we find their sum by adding the numerators and placing their sum over the common denominator, thus : T 4 -f- T 6 T -f- T 8 ^ An, (13) When mixed numbers and whole numbers are to be added, add the fractional parts of the mixed numbers sep- arately, and if the resulting fraction is an improper fraction, reduce it to a whole or mixed number. Next, add all the whole numbers, including the one obtained from the addi- tion of the fractional parts, and annex to their sum the fraction of the mixed number obtained from reducing the improper fraction. 18 ARITHMETIC. 1 42 + 31f + 9 T \ = ? Reducing to a fraction having a 5x2 denominator of 16, we have = -fg-. Adding the two o X /* fractional parts of the mixed numbers, we have ^-g- + W _io^M_ 11 _ 11 16 The problem now becomes 42 + 31 + 9 + 1 T V ? Adding all the whole numbers and the number obtained from adding the fractional parts of the mixed numbers, we obtain 83^ as their sum. 42 3 1 8 3 T V Ans. (14) 291+501 + 41 + 69^=? | _ 10 18 , 10 , 3 _ ~~ TT- tf "T Ttr "T IT = The problem now becomes 29 + 50 + 41 + 69 + 1 T V = ? 2 9 square inches. 5 square inches. 4 1 square inches. 6 9 square inches. 1 T 9 T square inches. 1 9 0^ square inches. Ans. (15) |- T 7 -g- = ? When the denominators of fractions are not alike, it is evident that the units are divided into unequal parts ; therefore, before subtracting, reduce the fractions to fractions having a common denominator. Then, subtract the numerators and place the remainder over the common denominator. -=*. An, 1 ARITHMETIC. 19 13 7 T V = ? This problem may be solved in two ways: First : 13 = 12|-|, since -ff- = 1, and 12|f = 12 -f -J-f = 12 + 1 = 13. We can now subtract the whole numbers sepa- rately and the fractions separately, and obtain 12 7 -j n _ iy Ans. Second ' : By reducing both numbers to improper fractions having a denominator of 16. 1g _l3xl6_ ao , _(7xl6) + 7 112 + 7 119 = T - T X 16 - ^ ' 7 - ~W~ ~T6~ = W- ft/\Q _ 1 1 Q Subtracting, we have - 8 T / - W = 16 = f|, and H = 16)89(5 T 9 7 the same result that was obtained by the 80 first method. ~^ 312^ 229^ = ? We first reduce the fractions of the two mixed numbers to frac- "I r tions having a common denominator. Doing 9x2 this, we have ^ = = |f . We can now subtract the ID X & , whole numbers and fractions separately, and have 312 229 = 83 and - = "- = 83 + = 83. Ans. (16) In division of fractions, invert the divisor (or, in other words, turn it upside down) and proceed as in multi- plication. -X-V = 3 [YT- 1 F-H^ Ans. ) A-3 = iV-*-l=AXi = ^~ = A = 1 V. Ans. Y--9 = -y--f = V-x-! = ^^ = H. Ans, H. 8. I. 27 20 ARITHMETIC. 1 = 4. Ans. 448)1808(4^ 1792 1 6 _ J_ 448~28 (e) 15f -T- 4 = ? Before proceeding with the division, reduce both mixed numbers to improper fractions. Thus, lof = - 3 ff 5 -. The problem is now - 6 T 3 - -i- - 3 -/- = ? As before, invert 9 2 the divisor, multiply, and cancel ; -^ -T- *-- = -^ X -/-$ = j- H? 5 = J/ = 3|. Ans. (17) |. rvalue of the fraction, and 28 the numerator. We find that 4 multiplied by 7 = 28, so multiplying 8, the denominator of the fraction, by 4, we have 32 for the required denominator, and f f = . Hence, 32 is the required denom- inator. (18) Since these four bolts measure 2, 6|, 3 T V, and 4 inches, respectively, together they will measure IG^-g- inches, since 2| + 6 + 3 T V + 4 = 16 T V Reducing the fractions of the mixed numbers to a common denominator, we have If T V of an inch is allowed for cutting and finishing each bolt, then the allow- ance for the 4 bolts would equal 4 X T \ = || = l|f inches, -j^g T which added to 16^ inches equals 18^- inches, -p 2 the length of the piece of iron required. Ans. since - ARITHMETIC. (PART 3.) (1) A fraction is one or more of the equal parts of a unit and is expressed by a numerator and a denominator, while a decimal fraction is a number of tenths, hundredths, thousandths, etc. of a unit and is expressed by placing a period (.), called a decimal point, to the left of the figures of the numerator and omitting the denominator. (2) To reduce the fraction to a decimal, we annex one cipher to the numerator, which makes it 1.0. Dividing 1.0, the numerator, by 2, the denominator, gives a quotient of .5, the decimal point being placed before the one figure of the quotient, or .5, since only one cipher was annexed to the numerator. 5.0 0000(. 15625 3 2 180 160 200 192 80 64 160 160 To express y 6 ^ as a decimal, write the numerator and, beginning with the right-hand figure, point off towards the left as many decimal places as there are ciphers in the X For notice of copyright, see page immediately following the title page. 22 ARITHMETIC. 1 denominator, prefixing ciphers to the numerator if neces- sary, and then insert the decimal point. Proceeding thus, jfo = 65 and w& = 135. Similarly, T7r f w = . 00024. Eight hundredths. 131 = One hundred thirty-one thousandths. ill a 3 II I III! 0001 = One ten-thousandths. 5^5 s5 a 000027= Twenty-seven millionths. 111 .0108 = One hundred eight ten- thousandths. nl If 11 g o g 9 3.0 1 1 = Ninety-three and one hundred one ten- thousandths. In reading decimals, read the number just as you would if there were no ciphers before it. Then count from the 1 ARITHMETIC. 23 decimal point towards the right, beginning with tenths, to as many places as there are figures, and the name of the last figure must be annexed to the previous reading of the figures to give the decimal reading. Thus, in the first example given, the simple reading of the figure is eight and the name of its position in the decimal scale is hundredths, so that the decimal reading is eight hundredths. Simi- larly, the figures in the fourth example are ordinarily read twenty-seven; the name of the position of the figure 7 in the decimal scale is millionths, giving, therefore, the decimal reading as twenty-seven millionths. If there should be a whole number before the decimal point, read it as you would read any whole number and read the decimal as you would if the whole number were not there; or, read the whole number and then say "and" so many hundredths, thousandths, or whatever it may be, as " ninety-three and one hundred one ten-thousandths." (4) In addition of decimals, the decimal points must be placed directly .125 under each other, so that tenths will ? come under tenths, hundredths under .089 hundredths, thousandths under thou- .4005 sandths, etc. The addition is then -9 performed as in whole numbers, the .000027 decimal point of the sum being placed 2.2 1 4 5 2 7 Ans. directly under the decimal points above. (5) A JiliJ Illlll .017 .2 .000047 .217047= T^vo hundred ^and seventeen thousand and forty-seven millionths. Ans. 24 ARITHMETIC. 1 (6) (a) In subtraction of decimals, place the decimal 7096300 P intS directl y under each other and proceed as in the subtraction of ! whole numbers, placing the decimal 7 8.7 7 8 6 Ans. point in the remainder directly under the de'cimal points above. In the above example, we proceed as follows: We cannot subtract 4 ten-thousandths from ten-thousandths, and as there are no thousandths, we take 1 hundredth from the 3 hundredths. 1 hundredth = 10 thousandths = 100 ten- thousandths. 4 ten-thousandths from 100 ten-thousandths leaves 96 ten-thousandths. 96 ten-thousandths = 9 thou- sandths -|- 6 ten-thousandths. Write the 6 ten-thousandths in the ten-thousandths place in the remainder. The next figure in the subtrahend is 1 thousandth. This must be subtracted from the 9 thousandths which is a part of the 1 hundredth taken previously from the 3 hundredths. Sub- tracting, we have 1 thousandth from 9 thousandths leaves 8 thousandths, the 8 being written in its place in the remain- der. Next we have to subtract 5 hundredths from 2 hun- dredths (1 hundredth having been taken from the 3 hun- dredths leaves but 2 hundredths now). Since we cannot do this, we take 1 tenth from 6 tenths. 1 tenth = 10 hun- dredths, and 10 hundredths + 2 hundredths 12 hundredths. 5 hundredths from 12 hundredths leaves 7 hundredths. Write the 7 in the hundredths place in the remainder. Next we have to subtract 8 tenths from 5 tenths (5 tenths now, because 1 tenth was taken from the 6 tenths). Since this cannot be done, we take 1 unit from the 9 units. 1 unit = 10 tenths. 10 tenths + 5 tenths = 15 tenths, and 8 tenths from 15 tenths leaves 7 tenths. Write the 7 in the tenths place in the remainder. In the minuend we now have 708 units (1 unit having been taken away) and units in the subtrahend. units from 708 units leaves 708 units; hence, we write 708 in the remainder. (b) 8 1.9 6 3 (c) 1 8.0 (d) 1.000 1.7 .18 .001 8 0.2 6 3 Ans. 1 7.8 2 Ans. .999 Ans. 1 ARITHMETIC. 25 (e) 872.1 (.8721 + .008) = ? In this problem we are to subtract (.8721 + .008) from 872.1. First perform the operation as indicated by the Q 8 sign between the decimals enclosed by the parenthesis. - 8 8 l sum - Subtracting the sum (obtained by adding the decimals 8721000 enclosed within the parenthesis) from 8801 the number 872.1 (as required by the minus sign before the parenthesis), 8 7 1.2 1 9 9 Ans. we obtain the requ ired remainder. (/) (5.028 + .0073) (6.704 2.38) = ? First perform the operations as indicated by the signs between the numbers enclosed by the parentheses. The first parenthesis shows _1 that 5. 028 and . 0073 are to be added. This 5.0353 sum. Q ij, Q ^ gives 5.0353 as their sum. 2 3 g Q The second parenthesis shows that 2. 38 is to be subtracted from 6. 704. 4.3 2 4 diff. The difference is found to be 4 334. The sign between the parentheses indicates that the quantities obtained by performing 50353 the above operations are to be sub- 43240 tracted namely, that 4.324 is to be subtracted from 5. 0353. Performing .7113 Ans. this operation, we obtain .7113 as the final result. (7) If the cost of the coal consumed by a nest of steam . boilers amounts to $15.83 on Monday, to $14.70 on Tuesday, to $14.28 on Wednes- 4 ^ g day, to $13.87 on Thursday, to $14.98 on Friday, and to $12.65 on Saturday, then we find the total cost of the week's supply by adding the different amounts together; hence, $15.83 + $14.70 + $14.28 + $13.87 $86.31 Ans. + $14.98 + $12.65 = $86.31. (8) 482| + 316 + 390f = what ? 26 ARITHMETIC. 1 When mixed numbers are to be added, add the fractional parts of the mixed numbers separately, and if the resulting fraction is an improper fraction, reduce it to a whole or mixed number. Next, add all the whole numbers, inclu- ding the one obtained from the addition of the fractional parts, and annex to their sum the fraction of the mixed number obtained from reducing the improper fraction. First, we will reduce the fractional parts -f, ^, and % to equivalent fractions having the least common denominator. In this case the least common denominator equals the product of the denominators 5, 3, and 4, since we cannot divide any two of them by any number (except 1) without having a remainder, as can be done in the examples in Art. 26, Part 2. Hence, the least common denominator = 5 X 3 X 4 = 60. Reducing f , , and f to fractions having this least common denominator, we have 60 divided by the first denominator, 5, equals 12. Then, \ x \\ = $ . GO divided by the second denominator, 3, equals 20. Then, $ X f = f-jr- 60 divided by the third denominator, 4, equals 15. Then, f X TT TO- The sum f these fractions equals The problem now becomes 482 + 316 + 390 -f Iff, the sum of which equals 482 316 390 Iff 1 189-ff 1500 represents the actual horsepower required. 1 1 8 9 -f f represents the indicated horsepower of the engines 310 T V> or 310. llf = the H. P. to be developed by the new engine. Ans. 1 ARITHMETIC. 27 7 reduced to its equivalent decimal = 6 100 6 (9) Since the inside diameter of the steam pipe is C.06 inches and the outside diameter is 6.62 inches, there is a difference of 6.62 6.06, or .56 of an inch, in both diam- eters. But .56 of an inch is just twice the thickness of the pipe; hence, the pipe is of .56, or .28 of an inch thick. (10) (a) There are 3 decimal places in the multiplicand .107 and 3 in the multiplier; hence, there .013 are 3 + 3, or 6, decimal places in the 321 product. Since the product con- 107 tains but four figures, we prefix two ciphers in order to obtain the neces- sary 6 decimal places. (&} 203 There are 2 decimal places in the ^03 multiplier and none in the multipli- 509 cand; hence, there are 2 + 0, or 2, 000 decimal places in the first product. 406 Since in the second multiplication there are 2 decimal places in the mul- tiplicand and 3 decimal places in the multiplier, there are 3 + 2, or 5, deci- 123627 mal places in the second product. 00000 When there are one or more ciphers 82418 in the multiplier, multiply just the 83.65427 Ans. same as with the other figures. (c) First perform the operations indicated by the signs between the numbers enclosed by the parentheses, and then whatever may be required by the sign between the parentheses. 28 ARITHMETIC. 1 3 1.8 5 The first parenthesis shows that 22295 the numbers 2.7 and 31.85 are to be 6370 multiplied together. 8 5.9 9 5 The second parenthesis shows that .316 is to be taken from 3.16. 2.8 4 4 The product obtained by performing the operation indi- 8 5.9 9 5 cated by the signs within the first 2.8 4 4 parenthesis is now multiplied by the 343980 remainder obtained by performing the 343980 operation indicated by the signs within 687960 t ^ ie secon d parenthesis. 71990 2 4 4.5 69780 Ans. (d) (107.8 + 6.541-31.96) X 1.742= ? I 7.8 8 2.3 8 1 + 6.541 X 1.742 II 4.3 41 164762 - 31.96 329524 8 2.3 8 1 576667 82381 1 4 3.5 7 7 2 Ans. (11) If one 3-inch tube measures 15 ft. in length, 60 of these tubes would measure 60 X 15 ft., or 15^ 930 ft. in length. If 1 foot of tubing heats a 6 surface of .728 sq. ft., then it is evident that 900 930 ft. of tubing would hea't a surface of 930 30 X .728 ft., or 677.04 sq. ft. 930 ft. .728 930 21840 6552 6 7 7.0 4 sq. ft. 1 ARITHMETIC. 29 (12) (a) l = 7-^tV=7xY = ^-^ = H jL = 374. Ans. The heavy line indicates that 7 is to be divided by T V <" 1=1 + 1 = 1*! = = ^ An, "8" 4 , 1.25 X 20 X 3 _ , In this problem 1.25 X 20 X 3 87 -j- 88 " constitutes the numerator of the 459 -f- 32 complex fraction. 1.2 5 X 20 - - Multiplying the factors of the numerator together, we find their product to be 75. X o &*y I CG The fraction -rz- -r constitutes the denominator of -r O/i the complex fraction. The value of the numerator of this fraction equals 87 + 88 = 175. The value of the denominator of this fraction is equal to 459 -f- 32 = 491. The problem then becomes 75 3 75___1__75_^175_75 491 ?? x 491 1,473 3 175 175 1 ' 491 1 175 ;jj5 7 f 491 491 7 Ans. (13) The pitch of the rivets is the distance between the centers of the rivets. Hence, since the distance around the cylindrical boiler is 166.85 in., and there are 72 rivets in one of the seams, the pitch of the rivets equals 166.85-7-72 = 2. 317+ in. Ans. 7 2) 1 6 6.8 5 (2.3 1 7+ 144 30 ARITHMETIC. 1 (14) If a keg containing 133 boiler rivets weighs 100 pounds, then each rivet must weigh as much as 133 is contained times in 100, or . 75 of a pound. 1 3 3) 1 0.00 (.7 5+ Ans. 931 690 665 2 5 Since there are 2 decimal places in the dividend and decimal places in the divisor, we must point off 2 = 2 decimal places in the quotient, or answer. 1 foot = 12 inches. 3 I- of 1 foot = \ X ~ = ~ = 10i inches. Ans. p 1 A 2 (16) 12 inches = 1 foot. Q -J T \ of an inch = T 3 -4- 12 = ^ X ^ = -fa of a foot. 4 J_ 6~4 ) 1.0 (.0 1 5 6 2 5 Ans. 6 4 360 320 . Point off 6 decimal places in the quotient, since we annexed six ciphers to the dividend, the 1^0 divisor containing no decimal places; hence, 6 6 places 320 to be pointed off. 320 1 ARITHMETIC. 31 (17) The total horsepower developed equals 48.63 + 45. 7 + 46. 32 + 47. 9 + 48. 74 + 48. 38 + 48. 59 = 334. 26. Since the horsepower developed equals 334.26, then the average horsepower developed must equal 334.26 -f- 7, or 47.75+H. P. 48.6 3 45.7 4 6.3 2 4 7.9 48.7 4 48.3 8 4 8.5 9 7 ) 3 3 4.2 6 47.75+ Ans. ARITHMETIC. (PART 4.) (1) A certain per cent, of a number means so many htmdredths of that number. 25$ of 8,428 Ib. means 25 hundredths of 8,428 Ib. -^ = .25. Hence, 8,428 Ib. X .25 = 2,107 Ib. Ans. (2) \% means one-half of 1 per cent. Since \% is .01, \ is .005, for 2 ) 'HI- And $35,000 X .005 = 1175. Ans. $ 3 5 .005 $ 1 7 5.0 (3) If 2 is a certain per cent, of 50, then 50 multiplied by a certain rate gives a product of 2, and that rate is equal to 2 divided by 50. Dividing 2 by 50, the quotient is .04, which means that 5 ) 2.0 ( .0 4 Ans. 2 is 4$ of 50, or, since percentage = base X rate, rate = percentage -r- base = 2-^ 50 = .04, or 4#. Ans. (4) Since percentage = base X rate, rate = percentage -T- base. As percentage = 10 and base = 10, we have rate = 10 -f- 10 = 1. 2 For notice of copyright, see page immediately following the title page. ARITHMETIC. 2 But l i^-o and f = 100$; hence, the rate (1) means that 10 is 100$ of 10. (5) Since 5,500 Ib. represent an increase of 15$ over the consumption when the condenser is used, 5,500 Ib. must be the amount, .15 the rate, and the number of pounds con- sumed when the condenser is running (to be found) the base. Base = amount -^ (1 + rate) = 5,500 -h (1 + .15) = 5,500 -h 1.15 - 4,782.61 Ib., nearly. Ans. 1.1 5 ) 5 5 0.0 ( 4 7 8 2.6 1 460 5 950 920 300 230 700 690 100 115 Or, this problem could also have been solved as follows: 100$ = the number of pounds consumed when the con- denser is running. If there is a gain of 15$, then 100$ + 15$, or 115$ = 5,500 Ib., the amount used when the condenser is not running. If 115$ = 5,500 Ib., 1$ = T fg- of 5,500 = 47.8261 Ib., and 100$ = 100 X 47.8261 = 4,782.61 Ib. Ans. (6) 24 $ of $950 = 950 X .24 = $228 12$of $950 = 950 X .125= 118.75 17 $ of $950= 950 X .17 = 161.50 53$ of $950 = $508.25 The total amount of his yearly expenses, then, is $508.25; hence, his savings are $950 $508.25 = $441.75. Ans. 2 ARITHMETIC. 3 Or, as above, 24$ + 12< + 17$ = 53|#, the total percent- age of expenditures; hence, $950 X .535= $508.25, and $950 $508.25 = $441.75 = his yearly savings. Ans. (7) The percentage is 961.38 and the rate is .37. Base = percentage -f- rate = 9G1.38 -5- .375 = 2,563.68, the number. Ans. .3 7 5) 9 6 1.3 8 000 (2 5 6 3.6 8 750 2113 Another method of solv- 1875 ing is the following: o o o If 37|$ of a number is 2250 961.38, then .37^ times the number =961. 38 and the number = 961.38 -=- .37$, l l 2 5 which, as above = 2, 563. 68. 2550 Ans. 2250 3000 3000 (8) 298 revolutions per minute with the load = base, .01 = rate, and the amount (to be found) will equal the speed of the engine when running unloaded. Amount = base X (1 + rate) 298 X (1 + .015) = 302.47 rev. per min. Ans o & 9 8 X 1.0 1 5 1490 298 000 298 3 2.4 7 //. 8. 1.38 4 ARITHMETIC. 2 (9) 4 yd. 2 ft. 10 in. to inches. X 3 1 2 Since there are 3 feet in + 2 1 yard, in 4 yards there are 4x3 ^ eet feet, or 12 feet. 12 feet plus x 1 2 2 feet = 14 feet - There are 12 inches in 1 foot ; therefore, in 14 feet there are 12 X 14, or 168 inches. 168 inches 168 plus 10 inches = 178 inches. + _1_0 178 inches. Ans. (10) 1 2 ) 3 7 2 2 inches. 3)31 + 2 inches. 103 + 1 foot. Ans. = 103 yd. 1 ft. 2 in. EXPLANATION. There are 12 inches in 1 foot; hence, in 3,722 inches there are as many feet as 12 is contained times in 3,722, or 310 feet and 2 inches remaining. Write 2 inches as a remainder. There are 3 feet in 1 yard; hence, in 310 yards there are as many feet as 3 is contained times in 310, or 103 yards and 1 foot remaining. Hence, in 3,722 inches there are 103 yd. 1 ft. 2 in. (11) 1728 ) 7 6 4 3 2 5 cu. in. 27)442 + 549 cu. in. 1 6 cu. yd. + 10 cu. ft. Ans. = 16 cu. yd. 10 cu. ft. 549 cu. in. EXPLANATION. There are 1,728 cubic inches in 1 cubic foot; hence, in 764,325 cu. in. there are as many cubic feet as 1,728 is contained times in 764,325, or 442 cubic feet ^and 549 cubic inches remaining. Write the 549 cubic inches as a remainder. There are 27 cubic feet in 1 cubic yard; hence, in 442 cubic feet there are as many cubic yards as 27 2 ARITHMETIC. 5 is contained times in 442 cubic feet, or 16 cubic yards and 10 cubic feet remaining. Then, in 704,325 cubic inches there are 16 cu. yd. 10 cu. ft. 549 cu. in. (12) T. cwt. Ib. Since in 1 ton there are 16 8 7 5 20 cwt., in 16 tons there are X 2 16 X 20 = 320 cwt. 320 cwt. 320 -f 8 cwt. = 328 cwt. There are + 8 100 Ib. in 1 cwt. ; hence, in 328 cwt. 328 cwt. there are 328 X 100 X 100 = 32,800 Ib. 32,800 lb. + 75 lb. '32800 = 32,875 Ib. Ans. + 75 3 2 8 7 5 Ib. Ans. (13) 100 ) 2 5 3 9 6 Ib. 2 ) 2 5 3 cwt. + 96 Ib. 1 2 T. -f 13 cwt. There are 100 Ib. in 1 cwt. ; hence, in 25,396 Ib. there are as many cwt. as 100 is contained times in 25,396, or 253 cwt. and 96 Ib. remaining. There are 20 cwt. in 1 ton, and in 253 cwt. there are as many tons as 20 is contained times in 253, or 12 tons and 13 cwt. remaining. Hence, 25,396 -lb. = 12 T. 13 cwt. 96 lb. Ans. (14) Arrange the different terms in columns, taking care to have like denominations in the yd. ft. in. same column. We begin to add at the right-hand column. 7 + 9 419 +3 = 19 in. ; since 12 in. = 1 ft., % 7 19 in. = 1 ft. and 7 in. Place the 807 Ans. 7 in. in the inches column and reserve the 1 ft. to add to the sum of the feet. 2 + 1 + 2 + 1 (reserved) = 6 ft. Since 3 ft. = 1 yd., 6 ft. = 2 yd. and ft. remaining. Place the in the feet column and reserve the 2 yd. to add to the sum of the yards. 4 + 2 -f- 2 (reserved) = 8 yd., which we place in yards column. Ans. = 8 yd. 7 in. G ARITHMETIC. 2 (15) Since 10 gal. 2 qt. 1 pt. of machine oil is sold at one time and 16 gal. 3 qt. at another gal. qt. pt. time, together there was sold 1021 27 gal. 1 qt. 1 pt. + 1 = 1 pt. We 16 3 cannot reduce 1 pt. to any higher 27 1 1 denomination, so place it under pints column. 3 qt. +2 qt. = 5 qt. Since 4 qt. = 1 gal., 5 qt. = 1 gal. and 1 qt. remaining. Place 1 qt. under quarts column and reserve the 1 gal. to add to the gallons. 16 gal. -f 10 gal. + 1 gal. (reserved) = 27 gal. Since the barrel contained 31, or g al - qt- Pt- 31. 5 gal., and 27 gal. 1 qt. 1 pt. 31 2 were sold, there remained the dif- 27 1 1 ference, or 4 gal. 1 pt. 31.5 gal. 4 1 Ans. = 31 gal. 2 qt., since .5 = |, and | of 1 gal. = % of 4 qt. = 2 qt. 1 pt. cannot be taken from pt., so we take 1 qt. from the 2 qt. The 1 qt. taken = 2 pt. 1 pt. from 2 pt. = 1 pt. Place 1 pt. under pints column. Since we took 1 qt. from the quarts column, there remains 2 1, or 1 qt. 1 qt. from 1 qt. leaves qt. Place qt. under the quarts column. 27 gal. from 31 gal. leaves 4 gal. Place 4 gal. under the gallons column. We therefore find that 4 gal. 1 pt. of machine oil remained in the barrel. (16) In multiplication of denominate numbers, we place the multiplier under the lowest denomination of the multi- plicand, as 1 7 ft. 3 in. 5 1 8 7 9 ft. 9 in. and begin at the right to multiply. 51x3 = 153 in. Since there are 12 in. in 1 ft., in 153 in. there are as many feet as 12 is contained times in 153, or 12 ft. and 9 in. remaining. Place the 9 in. under the inches and reserve the 12 ft. 51 X 17 ft. = 867 ft. ; 867 ft. + 12 ft. (reserved) = 879 ft. 879 ft. 2 ARITHMETIC. 7 can be reduced to higher denominations by dividing by 3 ft. to find the number of yards, and by 5 to find the number of rods. 3 ) 8 7 9 ft. 9 in. 5.5 ) 2 9 3 yd. 5 3 rd. 1 yd. Then, 879 ft. 9 in. = 53 rd. Ifc yd. ft. 9 in., or 53 rd. 1 yd. 2 ft. 3 in. (17) Since 2 pt. = 1 qt., 3 qt. = 3 X 2, or 6 pt. 6 pt. + lpt. = 7 pt. 4.7 X 7= 32.9 pt. Ans. qt. pt. 7 pt. 3 1 4.7 X_2 9 6 28 +_Jl 3~2~9 pt. 7 P t. (18) If there are four lengths, each 15 ft. 5 in., 15 ft. 5 in. X 4 = 60 ft. 20 in., or the length of the four pieces. 14 8 8 10 6 ft. 2 in. 82 38, or 85 ft. 2 in. = the length of the shaft. From the length of the shaft we must subtract 8 in. X 2 = 16 in. to get the distance between the end hangers. ft. in. 82 38 1 6 82 22, or 83 ft. 10 in. Since there are six hangers, there are five spaces. The length of one space is 83 ft. 10 in -4- 5 = 16 ft. 9-J- in. Ans. 8 ARITHMETIC. 2 (19) Reducing 18 ft. 11 in. to inches, we have 227 in., or 227.25 in. ft. in. 18 Hi 1 2 3 6 1 8 216 11* 2 2 7 i in. 1 X 2, or 2^ in., for the two end rivets. is deducted from the length, leaving 225 in., which is divided into equal spaces by the rivets. 2 2 7.2 5 in. - 2.2 5 in. 225 in. The pitch of the rivets (or the distance between their centers) is 1^ in., or 1.25 in. ; hence, 1.2 5 ) 2 2 5.0 ( 1 8 225 -h 1.25 = 180 spaces between the 125 rivets. But since there will be one .. Q Q Q more rivet than the number of 1000 spaces, the number of rivets re- quired for this boiler shell will be 180 + 1 181. Ans. ARITHMETIC. (PART 5.) (1) To find the second power of a number, we must mul- tiply the number by itself once; that is, use the number twice as a factor. Thus, the second power of 108 is 108 X 108 = 11,664. 108 108 1 81.25 1 81.25 9 06 25 36250 18125 145 000 181 25 328 51.56 25 1 81.25 1642578125 6570 31250 32851 5625 2628125 000 328515625 595434 5.7 031 2 5 Ans. (2) The third power of 181. 25 equals the number obtained by using 181.25 as a factor three times. Thus, the third power of 181.25 is 181.25 X 181.25 X 181.25 = 5,954,345.703125. Since there are 2 decimal places in the multiplier and 2 in the multiplicand, there are 2 + 2 = 4 decimal places in the first product. Since there are 4 decimal places in the multiplicand and 2 in the multiplier, there are 4 + 2 = 6 decimal places in th0 7 4 6Q 1290 ARITHMETIC. 19 .1 8 6 5 ).0 3 4 8 5 2 5 4 (.1 8 6 8 7 6 + 1865 16202 14920 12825 11190 16354 14920 14340 13055 Tiio 18G 5 1 1 1 QO .186876 1660 3 ).5 5 9 8 7 6 .18662 5+, or .18663. Ans. (b) Since the number is entirely decimal, the root is entirely decimal. Pointing off into periods, we obtain . 000'021. The figures of the root will be the same as for the cube root of 21. Hence, consider the given number as 21 and annex a cipher period so that two figures of the root may be obtained from the table; in other words, we extract the cube root of 21,000. Referring to the table, 21,000 lies between 19,683, the cube of 27, and 21,952, the cube of 28; the first two figures of the root are, therefore, 27. The first difference is 21,952 19,683 2,269; the second difference is 21,000 19,683 = 1,317 ; and 1,317 ^ 2,269 =.58+. Hence, the first four figures of the root are 2758. Locating the decimal point according to the principle given in Art. 32, the root is .02758+. The result does not agree with the printed answer. Hence, proceeding as directed in the note following example 3 (see Examination Questions), we apply the method described in Arts. 35 and 36. .000021 -f- .02758 = .000761421; .000761421 -r- .02758 = .027607; .02758 + .02758 + .027607 = .082767; .082767 -f- 3 = .027589, or .02759 , correct to four significant figures. H. 8. I. 29 20 ARITHMETIC. 2 21952 21000 2 2 6 9 ) 1 3 1 7.0 (.5 8 + 19683 19683 11345 2269 1317 18250 18152 . 02758). 00002100000000(. 00076142 1-f- 19306 16940 16548 3920 2758 11620 11032 5880 5516 3640 2758 ,0 2 7 5 8 ) .0 7 6 1 4 2 1 ( .0 2 7 6 7+ 5516 20982 19306 16761 .02758 16548 .02758 .027607 21300 19306 3). 082767 1994 .0 2 7 5 8 9, or .02759- Ans 2 ARITHMETIC. 21 (9) 11.7 : 13 :: 20 : x The product of the means 11.7 x = 13 X 20 equals the product of the 11.7 x = 260 extremes. _ 260 ~~ 1 1.7 ) 2 6 0.0 ( 2 2.2 2+ Ans. 234 260 234 260 234 260 234 ~2~6 (10) (a) 20 + 7 : 10 + 8 :: 3 : x. 27 : 18 :: 3 : x 27 x = 18 X 3 27 x = 54 54 x = = 2. Ans. (b) 12 2 : 100 2 :: 4 : x 12 100 144 : 10,000 :: 4 : x 12 100 _ 40,000 ~ 144) 4 0.0 ( 2 7 7.7+ Ans. 288 1120 1008 1120 1008 22 ARITHMETIC. 2 (11) (*)== Jj is equivalent to 4:*::7:21. The product of the means equals the product of the extremes Hence, 7 x = 4 X 21 7*= 84 A x = , or 12. Ans In like manner, (b) ^7 = r is equivalent to x : 24 :: 8 : 16. 16 x = 24 X 8 16;r= 192 192 = T6~ = 2 x (c) = is equivalent to 2 : 10 :: x = 100. 10^= 2 X 100 10 .ir = 200 200 x = =20. Ans. 15 60 (d) = is equivalent to 15 : 45 :: 60 : x = 45 X 60 15 x 2, 700 2,700 * A = 180. Ans. 15 (e) = ~ is equivalent to 10 : 150 :: x : 600. 150 600 150 x 10 X 600 150 x= 6,000 6,000 ^=--=40. Ans. 2 ARITHMETIC. 23 (12) 3 ft. = 3.5 ft., since \ - .5. 6| ft. = 6. 75 ft. , since f = . 75. Consider the question "What do we wish to find?" In this case it is " pounds." We know that a piece of shafting 3.5 ft. long weighs 37.45 lb., and we wish to know the weight of a piece of shafting 6| ft. long. It is evident that the weight of a piece of shafting 6.75 ft. long bears the same relation to the weight of a piece 3.5 ft. long that 6.75 ft. bears to 3.5 ft. Letting x occupy any place in the propor- tion, we have the following, the value of x being the same in each. Thus, (a) 3.5 ft. : 6.75 ft. :: 37.45 lb. : x lb., 6.75 X 37.45 252.7875 or x = = - = 72.225 lb. Ans. o.o o.5 (b) 6.75 ft. : 3.5 ft. :: x lb. : 37.45 lb., 6.75X37.45 252.7875 orx = --- -- = = 72.225 lb. Ans. o.o o.5 (0 x lb. : 37.45 lb. :: 6.75 ft. : 3.5 ft., 6.75 X 37.45 252.7875 orx = - -- = - = 72.225 lb. Ans. O.O O.O (d) 37.45 lb. : x lb. :: 3.5 ft. : 6.75 ft., 6.75 X 37.45 252.7875 3.5 3.5 = 72.225 lb. Ans. (13) We will first reduce 8 hr. 40 min. to minutes. 8 hr. -f- 40 min. = (8 X 60 min.) + 40 min. = 520 min. In this problem we are required to find "time." We know that a railway train runs 444 miles in 520 minutes, and we want to know how long it will take it to run 1,060 miles at the same rate of speed. It is evident that the time it requires to run 1,060 miles bears the same relation to the time it takes to run 444 miles that 1,060 miles bears to 444 miles. Letting x occupy any place in the proportion, we have the following, the value of x being the same in each. Thus, 24 ARITHMETIC. 2 (a) 1,060 miles : 444 miles :: x min. : 520 min., 1,060 X 520 551,200 -^44- -444- 1060 4 4 4 ) 5 5 1 2 0.0 ( 1 2 4 1.4 4+ min. 520 , 444 21200 1072 5300 888 551200 1840 1776 640 444 1960 1776 1840 1776 6 4 Reducing 1,241.44 min. to hours by dividing by 60, we have 6 ) 1 2 4 1.4 4 ( 20 hr. 41.44 min. Ans. 120 4 1 (b] 444 miles : 1,060 miles :: 520 min. : x min. x ~ ~ 444 = 1 341 - 44: min -, or 20 hr. 41.44 min. Ans. (c] x min. : 520 min. :: 1,060 miles : 444 miles. x = 1>06 ?* 52 1,241.44 min., or 20 hr. 41.4'4 min. Ans. (d] 520 min. : x min. :: 444 miles : 1,060 miles. 1,060 X 520 444 = 1,241.44 min., or 20 hr. 41.44 min. Ans. (14) A pump discharging 135 gal. per min. fills the tank in 38 min. Therefore, a pump discharging 1 gal. per 2 ARITHMETIC. 25 min. fills it in 135 X 38 min. Hence, a pump discharging -| O K \s O Q 85 gal. per min. fills it in = 60 T 6 T min. Ans. oo 1 35 X 38 1080 405 5 ) 5130 ( 6 510 -TV 8 5~ TT (15) If a wheel measuring 12.56 ft. around it turns 520 times, a wheel measuring 1 ft. around it turns 520 X 12.56 times. Hence, a wheel measuring 15.7 ft. around it turns 520 X 12.56 -^-^ = 416 times. Ans. lo. I 520 1 2.5 6 3120 2600 1040 520 1 5.7 ) 6 5 3 1.2 (416 times. 628 251 157 942 942 (16) If a cistern 28 ft. by 12 ft. by 10 ft. holds 798 bbl. of water, 798 a cistern 1 ft. by 12 ft. by 10 ft. holds-- bbl. ; /io 798 a cistern 1 ft. by 1 ft. by 10 ft. holds bbl. ; AO X 1* 798 a cistern 1 ft. by 1 ft. by 1 ft. holds -bbl. (CO X 1< X 10 26 ARITHMETIC. 2 Therefore, by a similar course of reasoning, a cistern 20 ft. by 17 ft. by 6 ft. holds 57 w ? 798 x 20 x 17 X 6 _ ftp x ?0 X 17 X _ 969 _ 28 x 12 x 10 $$ x ;2 x ;p r 2~ = ^ ? Ans. 5 7 17 399 5 7 2)969 484 MENSURATION AND USE OF LETTERS IN FORMULAS. (1) Substituting for D, x, B, and i their values, D - x 120 - 12 108 A line between two numbers signifies that the one above the line is to be divided by the one below the line. (2) Substituting for A, //, D, and x their values, A/i + D _ (5 X 200) + 120 1,000 + 120 _ 1,120 _ 2 Jt : + 6 (2 X 12) + 6 24+6 30 37^ + D = 37 + 120 = 157i. Ans. When there is no sign between the letters, multiplication is understood. (3) Substituting for A, D, /, and B their values, _ / A D I 5 X 120 /600 3 V i&+ 1-5 ~ V (3.5 X 10) + 1.5 ~ V 36.5 = 4/16.4383 = 4. 05+. Ans. The square root sign extends over both numerator and denominator, thus indicating that the square root of the entire fraction is to be extracted. (4) Substituting for A, B, D, and h their values. _ (B AY \/h + 2 B + A _ (10 5) 2 j/200 + 2x10 + 5 A 3 (! + >) 5 3 (1 + 120) 5' - i/225" 25 - 15 125 121 4 3 Ans - For notice of copyright, see page immediately following the title page. 2 MENSURATION AND 3 (5) When one straight line meets another straight line, two angles are formed which together equal 180. Hence, if one of- the angles = 152 3', the other angle = 180 152 3', or 180 = 179 60' subtracting, 152 3' 27 57' Ans. (6) See Arts. 26-28. (7) See Art. 41. A rectangle with the same area would have the same base and altitude. (8) Since the area is to be found in square inches,' the 2 feet must be reduced to inches. 2 ft. =30 in. Area = 30 X 1H = 345 sq. in. Ans. (9) It will take 1 boards to reach lengthways of the room. Since the room is 15 feet wide and each board is 5 incites wide, it will take 15 -f- = 36 boards, laid side by side, to extend across the width of the room. Hence, num- ber of boards required = 36 X 1 = 54. Ans. (10) The total area of the floor of the station = 55 X 58 ft. = 3,190 sq. ft. 25 X 26 ft. = 650 sq. ft., the area repre- sented by the lower right-hand corner of the figure. Hence, total area of floor = 3,190 650 = 2,540 sq. ft. From this we have to deduct the following areas: 2 boilers = 2 X 8 X 19 = 304 sq. ft. Feed-pump = 2| X 5 = 12.5 sq. ft. 2 engines = 2 X 4| X 10 = 90 sq. ft. 2 dynamos = 2 X 5 X 6 = 71.5 sq. ft. Switchboard^ 1Q * 3 ' 5 = 2. 92 sq.ft. 480.92sq. ft. The unoccupied floor space, therefore, equals 2,540 - 480.92 = 2,059.08 sq. ft. Ans. (11) A triangle with three equal angles has three equal sides, and is therefore an equilateral triangle. 3 USE OF LETTERS IN FORMULAS. 3 (12) A triangle with two equal angles has two equal sides, and is therefore an isosceles triangle. (13) The sum of the three angles in any triangle = 2 right angles, or 180. In the given triangle, the sum of two angles = 23 + 32 32' = 55 32', and the third angle = 180 55 32', or 180 = 179 60' subtracting, 55 32' 124 28' Ans. (14) In Fig. I we have the propor- tion A D : D E :: A B : B C, in which AD=10 in., A = '24: in., and BC = 13$ in., to find D E. Substituting the given values, 10 : DE :: 24 : 13$, or n - 10 X 13.5 . DE = = 5.625 in. Ans. FIG. I. (15) A line drawn diagonally from one corner to the op- posite one would form the hypotenuse of a right triangle, whose two sides are 39 and 52_feet. . By rule 6, Art. 58, the length of the diagonal = /52 s + 39' = 65 ft. Ans. (16) See example, Art. 64. The process is simply to find one of the angles of the polygon, and then to divide it by 2. By rule 1O, Art. 64, one of the interior angles _ 180 X (8 - 2) _ 1350 ^ 8 (17) Since this is a regular hexagon, it may be inscribed in a circle (Fig. II), and the radius of the inscribing circle will be equal to one side of the hexagon. Since the diameter E F = 2 inches, the radii A B and A C, and the side B C each = 1 inch, and the triangle ABC is equilateral. Draw the line A D per- pendicular to the side B C\ it will bisect B C. Then, in the right-angled triangle ADB,AB=Y and B D = $', MENSURATION AND to find A D. According to rule 7, Art. 59, A D = 4/1" .5" = I/.75 = .866". Hence, the distance between two opposite sides of the hexagon A D X 2 = .866 X 2 = 1.732". Ans. (18) In Fig. Ill, we have the pro- portion B I : H I\\H I\ I A, in which B 1= 6 and HI- |of HK= = 9. tf 2 Substituting, 6 : 9 :: 9 : I A, or I A 81 = - = 13. 5 in. Hence, the diameter A B o FIG. m. = I A -\-BI- 13.5 + 6 = 19.5 in. Ans. (19) One mile = 5,280 feet. The circumference of the wheel in feet = SLSJLlii! = 18.8496. (See rule 12, Art. 1/c 77.) Number of revolutions in going 1 mile = 5,280 -f- 18.8496 = 280.112. Ans. (20) Using rule 15, Art. 8O, area = diameter squared X .7854. 6.06" = 36.7236; 36.7236 X .7854 = 28.8427 sq. in. Ans. (21) Since the radius of the circle = 6 in., its diameter = 12 in., and its circumference = 12 X 3.1416 = 37.6992 in. There are 360 in the circumference, and the length of an arc of 12 = 37.6992 X ~ = 1.25664 in. Ans. ooO (22) The area of a circle 15 in. in diameter = 15" X .7854 = 176.715 sq. in. Hence, the area of a sector of this 101 2 208 937 circle whose angle is 12^ = 176.715 X TJ~| = ouO ouO = 6.1359 sq. in. Ans. (See rule 17, Art. 82.) (23) (a) The side of a square whose area = 103.8691 sq. in. = 1/103.8691 = 10.1916 in. Ans. (b) By rule 16, Art. 81, the diameter of a circle having /103.8691 the same area =A/ f)QRA H$ m - Ans. 3 USE OF LETTERS IN FORMULAS. 5 (f) Perimeter of the square = 10.1916 X 4 = 40.7664 in. ; circumference of the circle = 11.5 X 3. 1416 = 36.1284 in.; difference = 40. 7664 36. 1284 = 4. 638 in. Ans. (24) The perimeter of the base = 4 X 6 24 in. = 2 ft. Convex area 2 X 12 = 24 sq. ft. The area of the bases is found as follows: In Fig. IV, A B = 4 in. and A C = 2 in. ; since this is a regular hexagon, A O = A B = 4 in. By rule 7, Art. 59, O C = 4/4* - 2' = /12 3.4641 in.; area of triangle AOB base = 6.9282 X 6 = 41.5692; and the area FIG. iv. of both bases = 41.5692 X 2 83.1384 sq. in. This reduced 83 1384 to square feet = ^ .5774. Hence, the area of the entire surface of the column is 24 + .5774 = 24.5774 sq. ft. Ans. (25) The cubical contents in cubic inches = area of base in square inches X altitude in inches. The area of the base in the last example was found to be 41.5692 sq. in. ; altitude = 12 X 12 = 144 in. Hence, the cubical contents 41.5692 X 144 = 5,985.9648 cu. in. Ans. (26) This example is solved by combinirig the rules for the circular ring (see example, Art. 81) and for the cylinder. To obtain the area of one end of the tube, we have 4" X .7854 = 12.5664 = area of a circle 4 inches in diameter; 3.73* X .7854 = 10.9272 = area of a circle 3.73 inches in diameter; difference = 12.5664 10.9272 1.6392 area of one end of the tube. The cubical contents = 1. 6392 X 12 = 19. 6704cu. in. ; the weight = 19.6704 X .28 = 5.5, or 5| Ib. Ans. (27) This example is done exactly like the one in Art. 92, and the solution is given here without explanation. (a) In the formula of rule 18, Art. 83, np r - -608, Ji in this case = is, and D = 60. MENSURATION AND 3 Substituting, area = 4X324 - . Q 4/3.333-. 608 = 432 X 4/2.725 lo o = 432 X 1.65 = 712.8 sq. in. This reduced to square feet = 712.8 -f- 144 = 4.95. Hence, the steam space = 4.95 X 16 = 79.2 cu. ft. Ans. (b) Total area of one end of boiler in square inches = GO 2 X .7854 2,827.44. From this is to be subtracted the area of the tube ends and of the segment found above. Area of ends of tubes 3.5 a X .7854 X 64 = G15.75 sq. in. Area of segment = 712.8 sq. in. 1,328.55 sq. in. Area of water space = 2, 827. 44 1, 328. 55 = 1, 498. 89 sq. in . Contents of water space = 1,498.89 X 16 X 12 = 287,- 786.88 cu. in., and 287,786.88 -f- 231 = 1,245.83, number of gallons, or say 1,246 gal. Ans. (28) The area of the convex surface = circumference of base X \ slant height = 18.8496 X -r- = 94.248 sq. in. (See "A rule SI, Art. 97.) The area of the entire surface 94.248 sq. in. -f- the area of the base. The diameter of the base 1 R R4-Qfi _ ICLM tfo _ 6 hence the area of the base = 6' X .7854 3.1416 = 28.2744 (rules 13 and 15, Arts. 78 and 8O) ; there- fore, the area of the entire surface = 94.248 + 28.2744 = 122.5224 sq. in. Ans. (29) Using rule 22, Art. 98, volume = area of base X -i altitude = 28.2744 X | 84.8232 cu. in. Ans. o (30) The vat has the form of an inverted frustum of a pyramid. Area of larger base = 15 2 = 225 sq. ft. ; area of smaller base = 12 2 = 144 sq. ft. Hence, by rule 24, Art. 1O2, the contents of the vat in cubic feet (225 + 144 + 4/225 X 144) = (369 + 180) X ~ = 549 X o o o = 2,013 cu. ft. This should be reduced to cubic inches by 3 USE OF LETTERS IN FORMULAS. 7 multiplying by 1,728, the number of cubic inches in a cubic foot. 2,013 X 1,728 = 3,478,464 cu. in. Since there are 231 cubic inches in a gallon, the number of gallons that the q 470 AC ' nearl y- Ans - (19) Double shear means that the body subjected to the shearing stress resists this stress at two planes, at both of which shear must occur to produce failure. See Art. 48. (20) Countershafts serve to effect changes in speed and to stop and start the machinery. See Art. 54. (21) Cold-rolled shafting is made cylindrically true by a special rolling process. Bright shafting is turned up in a lathe. See Art. 55. (22) The diameter by which bright shafting is designated is that of the bar from which it is turned. See Art. 56. (23) Pulleys should be placed as near the bearings as pos- sible so that the deflection of the shaft may be as small as possible. See Art. 58. (24) Applying rule 11, Art. 6O, after finding from Table XIV that C for this case is 85, we have H = 88 * 8 = 481.8, nearly, say 482 H. P. Ans. oo (25) From Table XIV we find C for this case to be 65. Applying rule IS, Art. 61, H. S. I.31 4 STRENGTH OF MATERIALS. 7 (26) From Table XIV, C = 95. Applying rule 13, Art. 62, = -P0,sa3in. Ans. (27) The horsepower of a shaft will vary directly with the speed. See Art. 63. (28) See Art. 11. (29) A steel rope wears better than an iron rope. See Art. 29, INDEX. NOTE. All items in this index refer first to the section (see the Preface) and then to the page of the section. Thus, "Acceleration 47" means that acceleration will be found on page 7 of section 4. A. Sec. Absolute and gauge pressure. . 6 Page. 37 1 Arc of contact of belts, To find the Sec. 5 Page. 14 15 Area H 18 " force, Constant. .. 4 Acceleration 4 16 7 Arithmetic, Definition of Atmosphere, Pressure of 1 6 4 1 28 1 gravity 4 16 Average velocity 4 7 Work of 4 24 31 Avoirdupois weight Axle 2 5 11 1 " circle 5 31 5 5 " line 5 31 Addition 1 " of decimals 1 4 39 B. Backlash Sec. 5 Page. 31 5 10 bers 2 " of fractions 1 16 28 14 Standing Balancing pulleys 5 5 g 9 9 32 2 fi 33 Aggregation, Signs of. 1 52 40 " Mercurial Base Definition of 6 12 5 3 o 10 5 5 ec. g Page. 28 See. 2 Page. 37 3 28 4 45 Shafting, Black " Bright " Cold-rolled 32 32 Standard of weight " pipe dimensions Standing balance 4 6 5 3 24 9 Statics 4 15 ^ 29 7 ' Rules for 29 " strength of materials Short column " methods of abstracting 7 o 30 15 34 Strength of chains " of columns . " of manila ho i s ting 1 7 7 16 12 7 9 7 7 7 29 5 31 52 Stress 7 1 Simple denominate numbers.. Siphon, Theory of the Slant height of cone or pyra- 2 6 T 9 44 36 " Unitof Stresses, Classification of Stretch of belts Subtraction 7 5 1 1 1 24 9 Slings 7 9 1 41 Use of 10 " of denominate Solid bodies " Center of gravity of a " Edges of a.. " Faces of a 4 4 3 3 5 2 42 32 32 7 numbers " of fractions Suction pump T. 2 1 6 Set 18 29 48 Page Special properties of matter... Specific gravity 4 6 3 15 Tangential pressure Taper of keys 4 5 30 50 5 33 cubic foot of gases . . " gravity and weight per cubic foot of liquids. " gravity and weightper 6 6 17 16 " of gear-wheels Tenacity Tensile strength of materials.. " strength, Rules and 5 4 7 29 5 2 5 ff 15 $ 36 1 24 fi 44 substances 6 18 Thread >i 53 " gravity and weightper cubic foot of woods. 6 16 " of a screw 5 4 53 19 Speed and number of teeth of 2 12 gears " of belts ... 5 5 38 16 To find the arc of contact of belts .. . 5 14 Speeds of pulleys, Rules for. . . 5 1 10 39 " find the length of a belt 5 5 13 14 g 5 32 Split pulley 5 5 32 Spur gear 5 27 7 32 Square, Definition of foot 3 fl 17 18 Transverse strength of mate- 7 23 inch " measure 3 2 18 10 28 Trapezoid, Definition of Triangle, Definition of 3 3 3 20 20 " root. Short method of " Isosceles 3 20 working 2 34 " offerees 4 26 Sec. Triangle, Right-angle 3 " Scalene 3 Troyweight 2 U. Sec. INDEX. Page. 20 Vinrnlnm See. 3 3 3 3 Sec. 6 4 4 5 2 4 2 4 2 5 5 5 5 5 5 5 7 5 5 7 6 4 4 4 4 5 6 6 Stt. 7 XIX Page. 2 33 41 33 Page. 11 3 15 11 17 11 3 11 46 1 5 25 25 25 14 13 13 56 59 20 16 20 24 20 20 29 29 18 25 Page 7 20 11 Page 6 6 1 48 33 20 18 13 - 1 46 5 6 5 19 10 Page. 29 31 29 24 2 6 34 18 6 6 Volume " of cylindrical ring.... " Unit of W. Watson-Stillman hydraulic jack Uniformly varying velocity. .. Unit, Definition of " method Weight " of work " arm " Avoirdupois.. " Lawsof " Measures of " Standard of " square United States money " pressure, General law of 6 Weston differential pulley block " pressure of a fluid 6 Use and care of belts 5 " of slings 7 V. Sec. Vacuum 6 " gauge 6 " and axle " Driven " work " work, Rules for Width of belts, To find the .... Wire rope " rope, Rules for strength of " Measurement of 6 Value of a fraction 1 Vapor 4 Variations of pressures at dif- Wooden pillars, Constants for. Woods, Specific gravity and weight per cubic foot of Work Various substances, Specific gravity and weight per cubic foot of 6 Varying velocity, Uniformly. . 4 Velocity ' of acceleration and re- tardation " power, and energy " Unit of " Average " Mean " problems, Rules for.. " ratio 7 59 6 6 6 14 36 36 13 52 Worm " wheel Wrought-iron and structural- steel pillars, Con- " Uniformly varying.. . " Variable " iron welded pipes, Standard dimen- sions of Y. Yarn... Vertex of angle 3 " of cone 3 " of pyramid 3 Vertical Hue 3 Vinculum , 1 University of California SOUTHERN REGIONAL LIBRARY FACILITY 405 Hilgard Avenue, Los Angeles, CA 90024-1388 Return this material to the library from which it was borrowed. UC SOUTHERN REGIONAL LIBRARY FACILITY A 000 572 298 8