I ? W-^m I2>2> S> 5 3) D S> ^ >> D> 3>:> ^> igS i>"ia oxro ^ ^5SP cc cc-vcc c c cccc;ccc AN ELEMENTARY TREATISE ON ELECTRIC POWER AND LIGHTING PREPARED FOR STUDENTS OF THE INTERNATIONAL CORRESPONDENCE SCHOOLS SCRANTON, PA. Volume ARITHMETIC MENSURATION MECHANICS WITH PRACTICAL "QUESTIONS AND EXAMPLES First Edition SCRANTON THE COLLIERY ENGINEER CO. 1897 235 Entered according to the Act of Congress, in the year 1897, by THE COLLIERY ENGINEER COMPANY, in the office of the Librarian of Congress. at Washington. RURR PRINTING HOUSE, FRANKFORT AND JACOB STREETS, NEW YORK. SMI Utt PREFACE. The Instruction and Question Papers which are furnished to the students of The International Correspondence Schools become so badly worn and soiled that, when a student has completed his Course, he has worn out his Instruction Papers, and they are no longer suitable for reference or re- view. Since the Instruction Papers are very valuable, es- pecially to those who have studied them, there has grown up a demand for Sets of the Instruction and Question Papers, indexed for convenient reference, and durably bound for preservation. Again, many of our students can spare but little time for study, and are, therefore, a long while passing through their Courses. These students also desire the Papers in bound volumes to use for reference. Other students begin Courses, but for- various reasons are unable to complete them, and feel that, having paid for their Scholar- ships, they ought to have the text-books, even though they can not finish their Courses. For these reasons, we have decided to publish all of the Instruction and Question Papers of our different Technical Courses in volumes bound in Half Leather, to make a small advance in our prices, and to furnish a set to each student as soon as his Scholarship is paid for, whether he has completed his studies or not. The volumes for the present Course, the Electric Power and Lighting, are five in number: Volume I contains the Instruction and Question Papers on Arithmetic, Mensuration, and Mechanics. Volume II contains the Instruction and Question Papers on Dynamos and Motors, Electric Lighting, and Electric Railways. iv PREFACE. Volume III contains the Plates and the instructions for drawing them. It forms a very complete Course in Me- chanical Drawing. Volume IV contains the Tables, Rules, and Formulas in common use. The student who has finished his Course will find these Tables of great service. All the principal for- mulas, with the definitions of the letters used in them, and the principal rules, are conveniently arranged for reference, so that the student can save the labor and time of hunting them out in the Instruction Papers. Volume V contains the answers to the questions in the Question Papers. These Instruction Papers are written from a practical standpoint, and contain only such information as the student requires in order to obtain a good working knowledge of the subjects which form the groundwork of a course in Electric Power and Lighting. There is no padding, and the student gets in a clear and concise form the exact information which he desires. To keep the student always interested in his work, we do not give him half a dozen ways of doing the same thing; neither do we enter into speculative discussions of different subjects, all of which tend to confuse the student and leave him in doubt as to which one (or any one) is correct ; we give him the best or most suitable rule, formula, or method that we know of, without mentioning any of the others. In short, the Papers are written for practical men, and all the difficulties encountered by a student studying by himself, particularly those which are due to a definition or explanation being too technical, abstract, or not clear enough to be readily understood, have been carefully considered and overcome. THE INTERNATIONAL CORRESPONDENCE SCHOOLS. CONTENTS. ARITHMETIC. PAGE Definitions, 1 Notation and Numeration, ----- i Addition, 4 Subtraction, - 9 Multiplication, 12 Division, ... 17 Cancelation, --.- ---21 Fractions, - - - . - - - - 23 Decimals, 38 Percentage, - 55 Denominate Numbers, - 61 Involution, -------- 7^3 Evolution, -" 79 Ratio, 96 Proportion, 100 MENSURATION AND USE OF LETTERS IN ALGEBRAIC . * FORMULAS. Formulas, 115 Lines and Angles, 119 Quadrilaterals, - - - - - - 123 The Triangle, 120 Polygons, 130 The Circle, 133 The Prism and Cylinder, 138 The Pyramid and Cone, - - - - 142 The Frustum of a Pyramid or Cone, ... 143 The Sphere and Cylindrical Ring, - - - 145 MECHANICS. Matter and its Properties, 149 Motion and Velocity, ------ 153 Force, - 156 v i CONTENTS. MECHANICS continued. PAGE Center of Gravity, 160 Simple Machines, ------- 165 Pulleys, - - . - " 170 Gear- Wheels, 175 Fixed and Movable Pulleys, 183 The Inclined Plane, 185 The Screw, - - - - - - - - 187 Friction, - - - . - - - - 189 Centrifugal Force, 194 Specific Gravity, 196 Work and Energy, - ' - - - - - 197 Belts, " 201 Horsepower of Gears, ------ 204 Hydrostatics, 207 Buoyant Effects of Water, 219 Pneumatics, 222 Pneumatic Machines, - - - - - 231 Pumps, - ------ 241 Strength of Materials, 247 Tensile Strength of Materials, - - - - 248 Chains, - - 251 Hemp Ropes, - - - - - 252. Wire Ropes, * - - 253 Crushing Strength of Materials, - - - - 255 Transverse Strength of Materials, " - - 261 Shearing Strength of Materials, - . -. - 266 Line Shafting, - - - - .... 268 QUESTIONS AND EXAMPLES. Arithmetic, - - - - - - - - 275 Arithmetic, continued, - - - - - - 283 Mensuration and Use of 'Letters in Algebraic Formulas, - - - ... - - 291 Mechanics, - - - 299 Mechanics, continued, - - - - - . - 309 ARITHMETIC 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 ap- plied to any object or quantity, as three, five, ten. 7. Like numbers are numbers which express units of the same kind, as 6 days and 10 days, 2 feet and 5 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 fig- ures or letters. 11. Numeration is the art of reading the numbers which have been expressed by figures or letters. 2 ARITHMETIC. 1 2. The Arabic notation is the method of expressing numbers by figures. This method employs ten different figures to represent numbers, viz. : Figures 123456789 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 one 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 express a different value at differ- ent times. 14. The value of a figure depends upon its position in relation to others. 15. Figures have simple values and local or relative values. 16. The simple value of a figure is the value it ex- presses when standing alone. 17. The local or relative value 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 one 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 The entire line of seven figures is read six millions, six hundred sixty-six tJwusands, six hundred sixty-six. ARITHMETIC. 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 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 hundreds, the 4 should have two figures on its right, and a cipher is therefore inserted in the place usually given to fens, to show that the number is composed of hundreds and units only, and that there are no tens. Four hundred five is there- fore expressed 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 hun- dreds and tens ; thus, 450. Four thousand and fifty would be expressed 4050, the first cipher indicating that there are no hundreds and the second that there are no units. NOTE. When speaking of the figures of a number by referring to them as first figure, second figure, etc., always begin to count at the left. Thus, in the number 41,625, 4 is the first figure, 6 the third figure, 5 the fifth or last figure, etc. 20. In reading figures, it is usual to point off the num- ber into groups of three figures each, beginning with the right-hand or units column, a comma (,) being used to point off these groups. Billions. Millions. Thousands. Units. A a en oS g 1 J I 3 en * " CO* g CO O r* G "3 3 d J^ 1 8 3 t> i2 3 "o s "o Ji en "o '3 V) 3 CO s f-( ? j^ a Tens of Billions. Hundred Tens of Millions. Hundred Tens of Thousan Hundred Tens of '3 432,198,765,432 4 ARITHMETIC. In pointing off these figures, begin at the right-hand figure and count units, tens, hundreds ; the next group of three figures is thousands, therefore, we insert a comma (,) before beginning with them. Beginning at the figure 5, we say thousands, tens of thousands, hundreds of thousands, and in- sert another comma ; we next read millions, tens of millions, hundreds of millions, and insert another comma; we then read billions, tens of billions, hundreds of billions. The entire line of figures would be read: Four hundred thirty-two billions, one hundred ninety-eight millions, seven hundred sixty-five thousands, four hundred thirty-two. When we thus read a 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 figures, 72,584,623, would be the notation, and the numeration would be seventy-two millions, five 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 opera- tions in Arithmetic 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 -f- 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 + 6 = 11 may be read 5 plus 6 equals 11. 25. Like numbers can be added, but unlike numbers can- not. Thus, 6 dollars can be added to 7 dollars, and the sum will be 13 dollars, but 6 dollars cannot be added to ^ feet. ARITHMETIC. 26. The following table gives the sum of any two num- bers from 1 to 12: TABLE 1. 1 and 1 are 2 2 and 1 are 3 3 and 1 are 4 4 and 1 are 5 1 and 2 are 3 2 and 2 are 4 3 and 2 are 5 4 and 2 are 6 1 and 3 are 4 2 and 3 are 5 3 and 3 are 6 4 and 3 are 7 1 and 4 are 5 2 and 4 are 6 3 and 4 are 7 4 and 4 are 8 1 and 5 are 6 2 and 5 are 7 3 and 5 are 8 4 and 5 are 9 1 and 6 are 7 2 and 6 are 8 3 and 6 are 9 4 and 6 are 10 1 and 7 are 8 2 and 7 are 9 3 and 7 are 10 4 and 7 are 11 1 and 8 are 9 2 and 8 are 10 3 and 8 are 11 4 and 8 are 12 1 and 9 are 10 2 and 9 are 11 3 and 9 are 12 4 and 9 are 13 1 and 10 are 11 2 and 10 are 12 3 and 10 are 13 4 and 10 are 14 1 and 11 are 12 2 and 11 are 13 3 and 11 are 14 4 and 11 are 15 1 and 12 are 13 2 and 12 are 14 3 and 12 are 15 4 and 12 are 16 5 and 1 are 6 6 and 1 are 7 7 and 1 are 8 8 and 1 are 9 5 and 2 are 7 6 and 2 are 8 7 and 2 are 9 8 and 2 are 10 5 and 3 are 8 6 and 3 are 9 7 and 3 are 10 8 and 3 are 11 5 and 4 are 9 6 and 4 are 10 7 and 4 are 11 Sand 4 are 12 5 and 5 are 10 6 and 5 are 11 7 and 5 are 12 8 and 5 are 13 - r > and 6 are 11 6 and 6 are 12 7 and 6 are 13 8 and 6 are 14 5 and 7 are 12 6 and 7 are 13 7 and 7 are 14 8 and 7 are 15 5 and 8 are 13 6 and 8 are 14 7 and 8 are 15 8 and 8 are 16 5 and 9 are 14 6 and 9 are 15 7 and 9 are 16 8 and 9 are 17 5 and 10 are 15 6 and 10 are 16 7 and 10 are 17 8 and 10 are 18 5 and 11 are 16 6 and 11 are 17 7 and 11 are 18 8 and 11 are 19 5 and 12 are 17 6 and 12 are 18 7 and 12 are 19 8 and 12 are 20 9 and 1 are 10 10 and 1 are 11 11 and 1 are 12 12 and 1 are 13 9 and 2 are 11 10 and 2 are 12 11 and 2 are 13 12 and 2 are 14 9 and 3 are 12 10 and 3 are 13 11 and 3 are 14 12 and 3 are 15 9 and 4 are 13 10 and 4 are 14 11 and 4 are 15 12 and 4 are 16 9 and 5 are 14 10 and 5 are 15 11 and 5 are 16 12 and 5 are 17 9 and 6 are 15 10 and 6 are 16 11 and 6 are 17 12 and 6 are 18 9 and 7 are 16 10 and 7 are 17 11 and 7 are 18 12 and 7 are 19 9 and 8 are 17 10 and 8 are 18 11 and 8 are 19 12 and 8 are 20 9 and 9 are 18 10 and 9 are 19 11 and 9 are 20 12 and 9 are 21 9 and 10 are 19 10 and 10 are 20 11 and 10 are 21 12 and 10 are 22 9 and 11 are 20 10 and 11 are 21 11 and 11 are 22 12 and 11 are 23 9 and 12 are 21 10 and 12 are 22 11 and 12 are 23 12 and 12 are 24 This table should be carefully committed to memory. Since has no value, the sum of any number and is the number itself; thus, 17 and are 17. 27. For addition, place the numbers to be added directly under each other, taking care to place units under units, tens under tens, liundreds 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 6 ARITHMETIC. of the number above it, thus bringing the unit figures of all the numbers to be added in the same vertical line. Proceed as in the following examples: 28. EXAMPLE. What is the sum of 181, 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 column, and add, mentally repeating the different sums. Thus, three and two are five and one are six and two are eight and one are nine, 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 column 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 36 9215 4 907 27 60 1500 9000 sum 10587 Ans. EXPLANATION. The sum of the numbers in the first or units column is seven and four are eleven and five are six- teen and six are twenty-two and five are twenty-seven, or 37 units; i. e., two tens and seven units. Write 27 as shown. ARITHMETIC. 7 The sum of the numbers in the second or tens column is six tens, or GO. 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 preceding results as shown. There is only one number in the fourth or thousands column, nine, which represents 9,000. 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. NOTE. It frequently happens, when adding a long column of fig- ures, that the sum of two numbers, one of which does not occur in the addition table, is required. Thus, in the first column above, the sum of 16 and 6 was required. We know from the table that 6 + 6 = 12; hence, the first figure of the sum is 2. Now, the sum of any number less than 20 and of any number less than 10 must be less than thirty, since 20 + 10 = 30; therefore, the sum is 22. Consequently, in cases of this kind, add the first figure of the larger number to the smaller number and, if the result is greater than 9, increase the second figure of the larger number by 1. Thus, 44 + 7 = ? 4 + 7 = 11; hence, 44 + 7 = 51. 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 column = 27 units, or 2 tens and 7 units. Write the 7 units as the first or right-hand figure in the sum. Reserve the two 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 = 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 -|- 9 = 10, which is written down at the left of the other figures. The second method saves space and figures, but the first js to be preferred when adding a long column. ARITHMETIC. 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. EXPLANATION. The sum of the digits in the first column equals 19 units, or 1 ten and 9 units. Write down the 9 and carry 1 to the next column. The sum of the digits in the second column-)- 1 = 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 reserved = 38. The entire sum is 3,899. 32. Rule 1. (a) Begin at the right, add each column separately, and write the sum, if it be only one figure, under the column added. (6) 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 t lie 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. \ (a) 1,134. (6) 1,875 + 3,143 + 5,826 + 10,832. . I (d) 21,676. (0 4,865-+ 2,145 + 8,173 + 40,084. " 1 (c) 55,267. (d) 14,204 + 8,173 + 1,065 + 10,043, [ (d) 33,484. ARITHMETIC 9 (*) 10,832 + 4,145 + 3,133 + 5,872. f (e) 23,982. (/) 214 + 1,231 + 141 + 5,000. I (/) 6,586. (g) 123 + 104 + 425 + 126 + 327. Ans> 1(^)1,105. (K) 6,354 + 2,145 + 2,042 + 1,111 + 3,333. [ (k) 14,985. 1. A week's record of coal burned in an engine room is as follows : Monday, 1,800 pounds ; Tuesday, 1,655 pounds ; Wednesday, 1,725 pounds; Thursday, 1,690 pounds; Friday, 1,648 pounds; Saturday, 1,020 pounds. How much coal was burned during the week ? Ans. 9,538 pounds. 2. A steam pump pumps out of a cistern in one hour 4,200 gallons ; in the next hour, 5,420 gallons, and in 45 minutes more, an additional 3,600 gallons, when the cistern becomes empty. How many gallons were in the cistern at first ? Ans. 13,220 gallons. 3. What is the total cost of a steam plant, the several items of expense being as follows : Steam engine, $900 ; boiler, $775 ; fittings and connections, $225 ; erecting the plant, $125 ; engine house, $650 ? Ans. $2,675. SUBTRACTION. 35. In Arithmetic, subtraction is the process of find- ing how much greater one number is than another. The greater of the two numbers is called the minuend. The smaller of the two numbers is called the subtrahend. The number left after subtracting the subtrahend from the minuend is called the difference or remainder. 36. The sign of subtraction is . It is read minus, and means less. Thus, 12 7 is read 12 minus 7, and means that 7 is to be taken from 12. 37. EXAMPLE. From 7,568 take 3,425. SOLUTION. minuend 75 68 subtrahend 3425 remainder 4143 Ans. EXPLANATION. Begin at the right-hand or units column and subtract in succession each figure in the subtrahend from the one directly above it in the minuend, and write the remainders below the line. The result is the entire emainder. 10 ARITHMETIC. 38. When there are more figures in the minuend than in the subtrahend, and when some figures in the minuend are less than the figures directly under them in the sttbtrahend, proceed as in the following example: EXAMPLE. From 8,453 take 844. SOLUTION. minuend 8453 subtrahend 844 remainder 7609 Ans. EXPLANATION. Begin at the right-hand or units column to subtract. We can not take 4 from 3, and must, therefore, borrow 1 from 5 in tens column and annex it to the 3 in units column. The 1 ten = 10 units, which added to the 3 in the units column = 13 units. 4 from 13 = 9, the first or units figure in the remainder. Since we borrowed 1 from the 5, only 4 remains ; 4 from 4 0, the second or tens figure. We can not take 8 from 4, so borrow 1 thousand or 10 hundreds from 8; 10 hundreds + 4 hundreds = 14 hundreds. 8 from 14 = 6, the third or hundreds figure in the re- mainder. Since we borrowed 1 from 8 only 7 remains, from which there is nothing to subtract ; therefore, 7 is the next figure in the remainder or answer. The operation of borrowing is placing 1 before the figure following the one from which it is borrowed. In the above example the one borrowed from 5 is placed before 3, making it 13, from which we subtract 4. The 1 borrowed from 8 is placed before 4, making 14, from which 8 is taken. 39. EXAMPLE. Find the difference between 10,000 and 8,763. SOLUTION. minuend 10000 subtrahend 8763 remainder 1237 Ans. EXPLANATION. In the above example we borrow 1 from the second column and place it before 0, making 10; 3 from 10 = 7\ In the same way we borrow 1 and place it before ARITHMETIC. 11 the next cipher, making 10; but as we have borrowed 1 from this column and taken it to the units column, only 9 re- mains, from which to subtract 0, 6 ' from 9 = 3. For the same reason we subtract 7 from 9 and 8 from 9 for the next two figures, and obtain a total remainder of 1,237. 40. Rule 2. Place the subtralicnd or smaller number under the minuend or larger number, in the same manner as for addition, arid proceed as in Arts. 37, 38, and 39. 41. Proof. To prove an example in subtraction, add the subtrahend and remainder. The sum sJwuld equal the minuend. If it does not, a mistake has been made, and the work should be done over. Proof of the above example: subtrahend 8703 remainder 1237 minuend 10000 EXAMPLES FOR PRACTICE. 42. From (a) 94,278 take 62,574. (b) 53,714 take 25,824. (c) 71,832 take 58,109. (d) 20,804 take 10,408. An (e) 310,465 take 102,141. (/)(81,043 + 1,041) take 14,831. (^) (20,482 + 18,216) take 21,214. (//) (2,040 +1,213 + 542) take 3,791. (a) 31,704. (b) 27,890. (f) 13,723. (d) 10,396. (e) 208,324. (/) 67,253. () 17,484. 1. A cistern is fed by two pipes which supply 1,200 and 2,250 gallons per hour, respectively, and is being emptied by a pump which delivers 5.800 gallons per hour. Starting with 8,000 gallons in the cistern, how much water does it contain at the end of an hour ? Ans. 5,650 gallons. 2. A train in running from New Vork to Buffalo travels 38 miles the first hour, 42 the second, 39 the third, 56 the fourth. 52 the fifth, and 48 the sixth hour. How many miles remain to be traveled at the end of the sixth hour, the distance between the two places being 410 miles? Ans. 135 miles. 12 ARITHMETIC. 3. On Monday morning a bank had on hand $2,862. During the day $1,831 were deposited and 2,172 drawn out; on Tuesday, $3,126 were deposited, and 1,954 drawn out. How many dollars were on hand Wednesday morning ? Ans. 3,693. MULTIPLICATION. 43. To multiply a number is to add it to itself a cer- tain number of times. 44. Multiplication is the process of multiplying one number by another. The number thus added to itself, or the number to be multiplied, is called the multiplicand. The number which shows how many times the multiplicand is to be taken, or the number by which we multiply, is called the multiplier. The result obtained by multiplying is called the product. 45. The sign of multiplication is X . It is read times or multiplied by. Thus, 9 X 6 is read 9 times 6, or 9 multi- plied by 6. 46. It matters not in what order the numbers to be multiplied together are placed. Thus, 6 X 9 is the same as 9X6. 47. In the following table, the product of any two num- bers (neither of which exceeds twelve) may be found: ARITHMETIC. 13 TABLE 2. 1 times 1 is 1 1 times 2 are 2 1 times 3 are 3 1 times 4 are 4 1 times 5 are 5 1 times 6 are 6 1 times 7 are 7 1 times 8 are 8 1 times 9 are 9 1 times 10 are 10 1 times 11 are 11 1 times 12 are 12 2 times 1 are 2 2 times 2 are 4 2 times 3 are 6 2 times 4 are 8 2 times 5 are 10 2 times 6 are 12 2 times 7 are 14 2 times 8 are 16 2 times 9 are 18 2 times 10 are 20 2 times 11 are 22 2 times 12 are 24 3 times 1 are 3 3 times 2 are 6 3 times 3 are 9 3 times 4 are 12 3 times 5 are 15 3 times 6 are 18 3 times 7 are 21 3 times 8 are 24 3 times 9 are 27 3 times 10 are 30 3 times 11 are 33 3 times 12 are 36 4 times 1 are 4 4 times 2 are 8 4 times 3 are 12 4 times 4 are 16 4 times 5 are 20 4 times 6 are 24 4 times 7 are 28 4 times 8 are 32 4 times 9 are 36 4 times 10 are 40 4 times 11 are 44 4 times 12 are 48 5 times 1 are 5 5 times 2 are 10 5 times 3 are 15 5 times 4 are 20 5 times 5 are 25 5 times 6 are 30 5 times 7 are 35 5 times 8 are 40 5 times 9 are 45 5 times 10 are 50 5 times 11 are 55 5 times 12 are 60 6 times 1 are 6 6 times 2 are 12 6 times 3 are 18 6 times 4 are 24 6 times 5 are 30 6 times 6 are 36 6 times 7 are 42 6 times 8 are 48 6 times 9 are 54 6 times 10 are 60 6 times 11 are 66 6 times 12 are 72 7 times 1 are 7 7 times 2 are 14 7 times 3 are 21 7 times 4 are 28 7 times 5 are 35 7 times 6 are 42 7 times 7 are 49 7 times 8 are 56 7 times 9 are 63 7 times 10 are 70 7 times 11 are 77 7 times 12 are 84 8 times 1 are 8 8 times 2 are 16 8 times 3 are 24 8 times 4 are 32 8 times 5 are 40 8 times 6 are 48 8 times 7 are 56 8 times 8 are 64 8 times 9 are 72 8 times 10 are 80 8 times 11 are 88 8 times 12 are 96 9 times 1 are 9 9 times 2 are 18 9 times 3 are 27 9 times 4 are 36 9 times 5 are 45 9 times 6 are 54 9 times 7 are 63 9 times 8 are 72 9 times 9 are 81 9 times 10 are 90 9 times 11 are 99 9 times 12 are 108 10 times 1 are 10 10 times 2 are 20 10 times 3 are 30 10 times 4 are 40 10 times 5 are 50 10 times 6 are 60 10 times 7 are 70 10 times 8 are 80 10 times 9 are 90 10 times 10 are 100 10 times 11 are 110 10 times 12 are 120 11 times 1 are 11 11 times 2 are 22 11 times 3 are 33 11 times 4 are 44 11 times 5 are 55 11 times 6 are 66 11 times 7 are 77 11 times 8 are 88 11 times 9 are 99 11 times 10 are 110 11 times 11 are 121 11 times 12 are 132 12 times 1 are 12 12 times 2 are 24 12 times 3 are 36 12 times 4 are 48 12 times 5 are 60 12 times 6 are 72 12 times 7 are 84 12 times 8 are 96 12 times 9 are 108 12 times 10 are 120 12 times 11 are 132 12 times 12 are 144 This table should be carefully committed to memory. Since has no value, the product of and any number is 0. 14 ARITHMETIC. 48. To multiply a number by one figure only : EXAMPLE. Multiply 425 by 5. SOLUTION. multiplicand 425 multiplier 5 product 2125 Ans. EXPLANATION. For Convenience, the multiplier is gener- ally written under the rigkt-Jiand figure of the multiplicand. On looking in the multiplication table, we see that 5x5 are 25. Multiplying the first figure at the right of the multiplicand, or 5, by the multiplier 5, it is seen that 5 times 5 units are 25 units, or 2 tens and 5 units. Write the 5 units in units place in the product, and reserve the 2 tens to add to the product of tens. Looking in the multiplication table again, we see that 5x2 are 10. Multiplying the second figure of the multipli- cand by the multiplier 5, we see that 5 times 2 tens are 10 tens, plus the 2 tens reserved, are 12 tens, or 1 hundred plus 2 tens. Write the 2 tens in tens place, and reserve the 1 hun- dred to add to the product of hundreds. Again, we see by the multiplication table that 5x4 are 20. Multiplying the third or last figure of the multiplicand by the multiplier 5, we see that 5 times 4 hundreds are 20 hundreds,//^ the 1 hun- dred reserved, are 21 hundreds, or 2 thousands //.$ 1 hundred, which we write in thousands and hundreds places, respectively. Hence, \)i\& product is 2,125. This result is the same as adding 425 five times. Thus, 425 425 425 425 425 sum 2125 Ans. EXAMPLES FOR PRACTICE. 49. Find the product of (a) 61,483X6. ( (a) 368,898. (6) 12,375x5. A = J W 61,875. (f) 10,426 X 7. ~' \ ( f ) 72,982. (<0 10,835x3. [ (d) 32,505. ARITHMETIC. 16 (e) 98,376x4 f (e) 393,504. (/) 10,873X8. (/) 86,984. (g) 71,543X9. 1 (^) 643,887. (A) 218,734X2. [ (A) 437,468. 1. A stationary engine makes 5,520 revolutions per hour. Running 9 hours a day, 5 days in the week, and 5 hours on Saturday, how many revolutions would it make in 4 weeks ? Ans. 1,104,000 revolutions. 2. An engineer earns $650 a year, and his average expenses are $548. How much could he save in 8 years at that rate ? Ans. $816. 3. The connection between an engine and boiler is made up of 5 lengths of pipe, three of which are 12 feet long, one 2 feet 6 inches long, and one 8 feet 6 inches long. If the pipe weighs 9 pounds per foot, what is the total weight of the pipe used ? Ans. 423 pounds. 5O. To multiply a number by two or more fig- ures : EXAMPLE. Multiply 475 by 234. SOLUTION. multiplicand 475 multiplier 234 1900 1425 950 product 111150 Ans. EXPLANATION. For convenience, the multiplier is gener- ally written under the multiplicand, placing units under units, tens under tens, etc. We can not multiply by 234 at one operation ; we must, therefore, multiply by the parts and then add the partial products. The parts by which we are to multiply are 4 units, 3 tens, and 2 hundreds. 4 times 475 = 1,900, the first partial prod- uct ; 3 times 475 = 1,425, the second partial product, the right-hand figure of which is written directly under the fig- ure multiplied by, or 3 ; 2 times 475 = 950, the third partial product, the right-hand figure of which is written directly under the figure multiplied by, or 2. The sum of these three partial products is 111,150, which is the entire product. 16 ARITHMETIC. 5 1 . Rule 3. (a) Write the multiplier under the multipli- cand, so that units are -under units, tens under tens, etc. (6) Begin at the right and multiply each figure of the multi- plicand by each successive figure of the multiplier, placing the right-hand figure of each partial product directly under the figure used as a multiplier, (c) The sum of the partial products will equal the required product. 52. Proof. Review the work carefully, or multiply the multiplier by the multiplicand ; if the results agree, the work is correct. 53. When there is a cipher in the multiplier, multiply the entire multiplicand by it ; since the result will be zero, place a cipher under the cipher in the multiplier. Thus, (a) (6) (c) (d) 2 15 708 XO XO X X Ans. Ans. ~0~ Ans. Ans. to CO (JO 3114 4008 31264 203 305 1002 9342 20040 62528 120240 3126400 632142 Ans, 1222440 Ans. 31326528 Ans. In examples (e], (/), and (g), we multiply by as directed above; then multiply by the next figure of the multiplier and place the first figure of the product alongside the 0, as shown. EXAMPLES FOR PRACTICE. 54. Find the product of (a) 3,842 X 26. (b) 3,716 X 45. (c) 1,817 X 124 (d) 675X38. Ans. (b) 167,220. (c) 225,308. (d) 25,650. ARITHMETIC. (*) 1,875x33. r (e) 61,875. (/) 4,836X47. I (/) 227,292. (.-) 5,682x543. (>&) 3,257 X 246. (/) 2,875 X 302. (/) 17,819 X 1,004. Ans. (k) 38,674 X 205. (/) 18,304x100. (m) 7,834 X 10. () 87,543x1,000. (o) 48,763 X 100. () 3,085,326. (h) 801,222. (/) 868,250. (f) 17,890,276. (k) 7,928,170. (/) 1,830,400. (m) 78,340. () 87,543,000. (o) 4,876,300. 1. If the area of a steam-engine piston is 113 square inches, what is the total pressure upon it when the steam pressure is 85 pounds per square inch ? Ans. 9,605 pounds. 2. A steam engine, which indicated 164 horsepower, was found to consume 4 pounds of coal per horsepower per hour. Being replaced by a new engine, which was of the same horsepower as the other, another test was made, which showed a consumption of 3 pounds per horsepower per hour. What was the saving of coal for a year of 309 days, if the engine averaged to run 14 hours a day ? Ans. 709,464 pounds. 3. Two steamers are 7,846 miles apart, and are sailing towards each other, one at the rate of 18 miles an hour, and the other at the rate of 15 miles an hour. How far apart will they be at the end of 205 hours ? Ans. 1,081 miles. DIVISION. 55. Division is the process of finding how many times one number is contained in another of the same kind. The number to be divided is called the dividend. The number by which we divide is called the divisor. The number which shows how many times the divisor is contained in the dividend is called the quotient. 56. The sign of division is -=-. It is read divided by. 54 -T- 9 is read 54 divided by 9. Another way to write 54 divided by 9 is ^. Thus, 54 ^ 9 = 6, or ~ = 6. y y In both of these cases 54 is the dividend and 9 is the divisor. Division is the reverse of multiplication. 18 ARITHMETIC. 57. To divide when the divisor consists of but one figure, proceed as in the following example : EXAMPLE. What is the quotient of 875 -H 7 ? divisor dividend quotient SOLUTION. 7)875(125 Ans. 7_ 17 14 35 35 remainder EXPLANATION.^ 7 is contained in 8 hundreds 1 hundred times. Place the one as the first, or left-hand, figure of the quotient. Multiply the divisor 7 by the 1 hundred of the quotient, and place the product 7 hundreds under the 8 hundreds in the dividend, and subtract. Beside the remainder 1, bring down the 7 tens, making 17 tens; 17 divided by 7 = 2 times. Write the two as the second figure of the quotient. Multiply the divisor 7 by the 2, and subtract the product from 17. Beside the remain- der 3, bring down the 5 units of the dividend, making 35 units. 7 is contained in 35, 5 times, which is placed in the quotient. Multiplying the divisor by the last figure of the quotient, 5 times 7 = 35, which subtracted from 35, under which it is placed, leaves 0. Therefore, the quotient is 125. This method is called long division. 58. In short division, only the divisor, dividend, and quotient are written, the operations being performed men- tally. dividend divisor 7 ) 8 ] 7 3 5 quotient 125 Ans. The mental operation is as follows: 7 is contained in 8, once and one remainder; 1 placed before 7 makes 17; 7 is contained in 17, 2 times and 3 over; the 3 placed before 5 makes 35 ; 7 is contained in 35, 5 times. These partial quo- tients placed in order as they are found, make the entire quotient 125. ARITHMETIC. 19 The small figures are placed in the example given to better illustrate the explanation; they are never written when actually performing division in this way. 59. If the divisor consists of 2 or more figures, proceed as in the following example : EXAMPLE. Divide 2,702,826 by 63. divisor dividend quotient SOLUTION. 63 )2702826(42902 Ans. 252 182 126 568 567 126 126 EXPLANATION. As 63 is not contained in the first two fig- ures, 27, we must use the first three figures, 270. Now, by trial, we must find how many times 63 is contained in 270; 6 is contained in the first two figures of 270, 4 times. Place the 4 as the first or left-hand figure in the quotient. Multi- ply the divisor 63 by 4, and subtract the product 252 from 270. The remainder is 18, beside which we write the next figure of the dividend, 2 S making 182. Now, 6 is contained in the first two figures of 182, 3 times, but on multiplying 63 by 3, we see that the product 189 is too great, so we try 2 as the second figure of the quotient. Multiplying the divisor 63 by 2, and subtracting the product 126 from 182, the remainder is 56, beside which we bring down the next figure of the dividend, making 568; 6 is contained in 56 about 9 times. Multiply the divisor 63 by 9 and subtract the product 567 from 568. The remainder is 1, and bringing down the next figure of the dividend, 2, gives 12. As 12 is smaller than 63, we write in the quotient and bring down the next figure, 6, making 126. 63 is contained in 126, 2 times, without a remainder. Therefore. 42,902 is the quo- tient. 20 ARITHMETIC. 6O. Rule 4. (a) Write the divisor at the left of the dividend, with a line between them. (bi\ Find Jiow many times the divisor is contained in the lowest number of the left-hand figures of the dividend that will contain it, and write the result at the right of the dividend, with a line between, for the first figure of the quotient. (c) Multiply the divisor by this quotient ; write the product tinder the partial dividend used, and subtract, annexing to the remainder the next figure of the dividend. Divide as before, and thus continue until all the figures of the dividend have been used. (d) If any partial dividend will not contain the divisor, write a cipher in the quotient, annex the next figure of the dividend and proceed as before. (e) If there be a remainder at last, write it after the quo- tient, with the divisor underneath. 61 . Proof. Multiply the quotient by the divisor, and add the remainder, if there be any, to the product. The result will be the dividend. divisor dividend quotient Thus, 63)4235(67| Ans. 378 ~455 441 remainder 1 4 Proof, quotient 6 7 divisor 6 3 402 * 4221 remainder 1 4 dividend 4235 ARITHMETIC. 21 EXAMPLES FOR PRACTICE. 62* Divide the following: (a) 126,498 by 58. (a) 2,181. (d) 3,207,594 by 767. (i,) 4,182. (c) 11,408,202 by 234. (c) 48,753. (d) 2,100,315 by 581. (d) 3,615. ( T S TT must e q ual i Hence, dividing both terms of a fraction by the same number does not alter its value. 84. A fraction is reduced to its lowest terms when its numerator and denominator can not be divided by the same number, as f, f , -f^-. 85. To reduce a whole number or mixed num- ber to an improper fraction : EXAMPLE. How many fourths in 5 ? SOLUTION. Since there are 4 fourths in 1 (f = IV in 5 there will be 5x4 fourths, or 20 fourths, 5 X | = -^ Ans. EXAMPLE. Reduce 8f to an improper fraction. SOLUTION. 8 x f = ^. \ 8 - + f = ^. Ans. 86. Rule 6. Multiply the whole number by the denom- inator of the fraction, add the numerator to the product, and place the denominator under the result. EXAMPLES FOR PRACTICE. 87. Reduce to improper fractions: (a) W 10 T V (<0 87*. (*5 50f ARITHMETIC. 27 88. To reduce an improper fraction to a whole or mixed number : EXAMPLE. Reduce ^ to a mixed number. SOLUTION. 4 is contained in 21, 5 times and 1 remaining; as this is also divided by 4, its value is . Therefore, 5 + , or 5, is the num- ber. Ans. 89. Rule 7. Divide the numerator by the denominator; the quotient will be the wJiolc number; the remainder, if there be any, will be the numerator of the fractional part of which the denominator is the same as the denominator of the improper fraction. EXAMPLES FOR PRACTICE. 90. Reduce to whole or mixed numbers : (a) H A - ' SI (') *f (/) W- 91. A common denominator of two or more frac- tions is a number which will contain all of the denominators of the fraction s without a remainder. The least common denominator is the least number that will contain all of the denominators of the fractions without a remainder. 92. To find the least common denominator: EXAMPLE. Find the least common denominator of ^, \, \, and -j> ff . SOLUTION. We first place the denominators in a row, separated by commas. 2 ) 4, 3, 9, 16 2 ) 2, 3, 9. 8 3 ) 1, 3, 9, 4 8 ) 1, 1, 8, 4 4)1. 1, 1, 4 1, 1, 1, 1 2x2x3x3x4 = 144, the least common denominator. Ans. EXPLANATION. Divide each of them by some prime num- ber which will divide at least two of them without a remain- der (if possible), bringing down those denominators to the row 28 ARITHMETIC. below which will not contain the divisor without a remain- der. Dividing each of the numbers by 2, the second row becomes 2, 3, 9, 8, since 2 will not divide 3 and 9 without a re- mainder. Dividing again by 2, the result is 1, 3, 9, 4. Dividing the third row by 3, the result is 1, 1, 3, 4. So continue until the last row contains only 1's. The product of all the divisors, or 2x2x3x3x4= 144, is the least common denominator. 93. EXAMPLE. Find the least common denominator of f, ^, T 7 F . SOLUTION. 3 ) 9. 12. 18 2)1. 2, 1 1, 1, 1 3X3X2X2 = 36. Ans. 94. To reduce two or more fractions to frac- tions having a common denominator : EXAMPLE. Reduce , f, and - to fractions having a common denomi- nator. SOLUTION. The common denominator is a number which will con- tain 3, 4, and 2. The least common denominator is 12, because it is the smallest number which can be divided by 3, 4, and 2 without a remain- der. Reducing f, 3 is contained in 12, 4 times. By multiplying both numerator and denominator of by 4, we find O -vy A Q Q ^ A = To- In the same wa y we find * = TS and 4 = A- o X 4 l 95. Rule 8. Divide the common denominator by the denominator of the given fraction, and multiply both terms of the fraction by the quotient. EXAMPLES FOR PRACTICE. 96. Reduce to fractions having a common denominator: () *, I . 4- f (a) |, I, i- (*) A, *> A- (6) A. II- A- 4. A, If Ans I tt- A, (') A, A- A- (') if, A, W- GO T V, W, li- I (/) tt, H. It- ARITHMETIC. 29 ADDITION OF FRACTIONS. 97. Fractions can not be added unless they have a common denominator. We can not add f to as they now stand, since the denominators represent parts of different sizes. Fourths can not be added to eighths. Suppose we divide an apple into 4 equal parts, and then divide 2 of these parts into two equal parts. It is evident that we shall have 2 one-fourths and 4 one-eighths. Now if we add these parts the result is 2 + 4 = something. But what is this something? It is not fourths, for six fourths are 1, and we had only 1 apple to begin with; neither is it eighths, for six eighths are f , which is less than 1 apple. By reducing the quarters to eighths, we have = 4 an( ^ adding the other 4 eighths, 4 + 4 = 8 eighths. The result is cor- rect, since f = 1. Or we can, in this case, reduce the eighths to quarters. Thus, -| = f ; whence, adding, 2 + 2 = 4 quarters, a correct result, since = 1. Before adding, fractions should be reduced to a common denominator, preferably the least common denominator. 98. EXAMPLE. Find the sum of i, f , and . SOLUTION. The least common denominator or the least number which will contain all the denominators is 8. | = |, f = f , and | = |. EXPLANATION. As the denominator tells or indicates the names of the/ar/^, the numerators only are added to obtain the total number of parts indicated by the denominator. 99. EXAMPLE. What is the sum of 12f , 14f , and 7^ ? SOLUTION. The least common denominator in this case is 16. 14* = sum 33 + f f = 33 + \\\ = 34}J. Ans. The sum of the fractions = f J or 1|J, which added to the sum of the whole numbers b4J. 30 ARITHMETIC. EXAMPLE. What is the sum of 17, 13 T s g , &, and 3 ? SOLUTION. The least common denominator is 32. 13j s T = 13^* 5 , 3J = ** l*fc A 3A ^; 33ff. Ans. 1OO. Rule 9. (a) Reduce the given fractions to frac- tions having the least common denominator, and write the sum of the numerators over the common denominator. (6) When there are mixed mimbers and whole numbers, add the fractions first, and if their sum is an improper frac- tion, reduce it to a mixed number and add the whole number with the other whole numbers. EXAMPLES FOR PRACTICE. 1 1 . Find the sum of () f A, * (*) t, A- W i, I, A- wk*l Ans ^ (/) il 3, S. (<") T*T- A H- (^ f , ii f- 1. The weights of a number of castings were 412f lb., 270^ lb., 1,020 lb., 75J lb., and 68 lb. What was their total weight ? Ans. 1,847 lb. 2. Four bolts are required, 2|, If, 2A, and l*f inches long. How long a piece of iron will be required to cut them from, allowing f of an inch altogether for cutting off and finishing the ends ? Ans. 9^ in. (/)!. SUBTRACTION OF FRACTIONS. 1 02, Fractions can not be subtracted without first re- ducing them to a common denominator. This can be shown in the same manner as in the case of addition of fractions. EXAMPLE. Subtract f from \\. SOLUTION. The common denominator is 16. ARITHMETIC. 31 103. EXAMPLE. From 7 take f. SOLUTION. 1 = f- ; therefore, since 7 = 6 + 1 we see that 7 6 4- f , so that 6| f = 6f . Ans. 104. EXAMPLE. What is the difference between 17 T \ and 9^f ? SOLUTION. The common denominator of the fractions is 32. 17 T * 8 = "H- minuend 17|| subtrahend 9|| difference 8^ Ans. 1 05. EXAMPLE. From 9 take 4 T 7 5 . SOLUTION. The common denominator of the fractions is 16. 9 = 9iV minuend 9^ 8? subtrahend 4 T 7 ^ or 4^ difference 4}f 4f Ans. EXPLANATION. As the fraction in the subtrahend is greater than the fraction in the minuend, it ## #/ be sub- tracted; therefore,' borrow 1, or -ff, from the 9 in the minu- end and add it to the -ft ; T \ + || = ff -ft from f $ = ||. Since 1 was borrowed from 9, 8 remains; 4 from 8 = 4; 4 + - B- 1 06. EXAMPLE. From 9 take 8 T \. SOLUTION. minuend 9 8^| subtrahend 8y\ or 8 T \ difference \\ \\ Ans. EXPLANATION. As there is no fraction in the minuend from which to take the fraction in the subtrahend, borrow 1, or ||, from 9. ^ from || = ||. Since 1 was borrowed f rom 9, only 8 is left. 8 from 8 = 0. 107. Rule 1O. (a) Reduce the fractions to fractions having a common denominator. Subtract one numerator from the other and place the remainder over the common de- nominator. (b) When there are mixed numbers, subtract the fractions and whole numbers separately, and place the remainders side by side. 33 ARITHMETIC. (c) When the fraction in the subtraliend is greater than the fraction in tlic minuend, borrow 1 from the whole number in the minuend and add it to the fraction in the minuend, from which subtract the fraction in the subtrahend. ( 36 ARITHMETIC. 1 23. EXAMPLE. How many times is 3| contained in 7 T 7 ff ? SOLUTION. 3f = ^ ; 7 T 7 ff = ^\ 9 . ^ inverted becomes T 4 g . 119 4 _ 119 X * _H9_ T6- X T5--^ X 15 ~ W- 1 24. Rule 1 2. Invert tlie divisor and proceed as in m ultiplication. 1 25. 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, -^ shows that 18 is to be divided by 3. This is also true if a fraction or a fractional expression be placed above or below a line. means that 9 is to be divided by f ; means that 16 Q I 1 \ 3 X 7 is to be divided by the value of 16 is the same as -=- 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 whether 9 is to be divided by f, or -- by 8. Whenever a heavy line is used, as in the above case, it indicates that all above the line is to be divided by all below it. EXAMPLES FOR PRACTICE. 126. Divide (a) 15by6f (b) SObyf. to 172 by f . *' Ans. 4 (a) 2i. (b) 40. (c) 215. (e) ios by 14 | - (^ ji 6. (/) 3* by 17f (J .004. 161. To reduce inches to decimal parts of a foot : EXAMPLE. What decimal part of a foot is 9 inches ? SOLUTION. Since there are 12 inches- in one foot, 1 inch is ^ of a foot, and 9 inches is 9 x r? or r ' ? of a foot. This reduced to a decimal by the above rule, shows what decimal part of a foot 9 inches is. 12)9.00(.75ofafoot. Ans. 84 60 b~ ARITHMETIC. 51 162. ' Rule 18. (a) To reduce inches to decimal parts of a foot, divide the number of inches by 12. (b) Should the resulting decimal be an unending one and it is desired to terminate the division at some point, say the fourth decimal place, carry the division one place fartJier^ and if the fifth figure is 5 or greater, increase the fourth figure by one and omit the sign -}-. EXAMPLES FOR PRACTICE. 163. Reduce to the decimal part of a foot: (a) 3 in. (6) 4|. in. (c) 5 in. Ans. (d) 6f in. (e) 11 in. 1. The lengths of belting required to connect three countershafts with the main line shaft were found with a tape measure to be 27 ft. 4 in., 23 ft. 8 in., and 38 ft. 6 in. 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 measured from the center of shaft to center of crank-pin in feet ? Ans. .5833-h foot. 3. A steam pipe fitted with an expansion joint was found to expand 1.668 inches when steam was admitted to it. How much was its expansion in decimal parts of a foot? Ans. .139 foot. TO REDUCE A DECIMAL TO A FRACTION. 164. EXAMPLE. Reduce .125 to a fraction. SOLUTION. . 125 = -j 1 ^ = T 6 5 = |. Ans. EXAMPLE. Reduce .875 to a fraction. SOLUTION. .875 = T V 5 5 5 = f $ = f . Ans. 1 65. Rule 1 9. Under the figures of the decimal, place the digit 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 denomi- nator by the same number. 52 ARITHMETIC. 166. EXAMPLES FOR PRACTICE. Reduce the following to common fractions: (a) .125. (b) .625. (c) .3125. (d) .04. (e) .06. Ans. - (/) -75. (g) .15625. (k) .875. () i- (*) f- (0 A- (<*) A- (0 A- * () A- (4) i- TO EXPRESS A DECIMAL APPROXIMATELY AS A FRACTION HAVING A GIVEN DENOMINATOR. 167. EXAMPLE. Express .5827 in 64ths. SOLUTION. -.5827 x ft = 87 '^ 28 . say |f Hence, .5827 = |J, nearly. Ans. EXAMPLE. Express .3917 in 12ths. A 7004. SOLUTION. .3917 x \\ = ^j^. say A- Hence, .3917 = T B 2 , nearly. Ans. 1 68. Rule 2O. Reduce 1 to a fraction having the given denominator. Multiply the given decimal by tiic fraction so obtained, and the result will be the fraction required. EXAMPLES FOR PRACTICE. 1 69. Express (a) .625 in 8ths. (b) .3125 in 16ths. (c) .15625 in 32ds. (d) .77 in 64ths. (e) .81 in 48ths. (/) .923 in 96ths. (/) gf. Ans. A- A- ARITHMETIC. 53 1 7O. The sign for dollars is $. It is read dollars. $25 is read 25 dollars. Since there are 100 cents in a dollar, one cent is 1-one- hundredth of a dollar; the .first two figures of a decimal part of a dollar represent cents. Since a mill is ^ of a cent, or -ruVfi- of a dollar, the third figure represents mills. Thus, $25.16 is read twenty-five dollars and sixteen cents; $25.168 is read twenty-five dollars, sixteen cents and eight mills. 171. The vine 11 1 u in , parenthesis ( ), bracket [ ], and brace { } are called symbols of aggregation, and are used to include numbers which are to be considered together; thus, 13 X 8-3, or 13 X (8 3), shows that 3 is to be taken from 8 before multiplying by 13. 13 X (8 3) = 13 X 5 = 65. Ans. 13 X 8-3 = 13 X 5 = 65. Ans. When the vinculum or parenthesis is not used, we have 13 X 8 - 3 = 104 3 = 101. Ans. 1 72. In any series of numbers connected by the signs -|-, , X, and -+, the operations indicated by the signs must be performed in order from left to right, except that no addition or subtraction may be performed if a sign of multiplication or division follow? the number on the right of a sign of addition or subtraction, until the indicated multiplication or division has been performed. In all cases the sign of multi- plication takes the precedence, the reason being that when two or more numbers or expressions are connected by the sign of multiplication, the numbers thus connected are re- garded as factors of the product indicated, and not as sepa- rate numbers. EXAMPLE. What is the value of 4 x 24 8 4- 17 ? SOLUTION. Performing the operations in order from left to right, 4x24 = 96; 96-8 = 88; 88 + 17 = 105. Ans. 173. EXAMPLE. What is the value of the following expression : 1,296 -i- 12 + 160 - 22 X 3 = ? 1 SOLUTION. 1,296-H 12 = 108; 108 + 160 = 268; here we cannot sub- tract 22 from 268 because the sign of multiplication follows 22; hence, multiplying 22 by 3$, we get 77, and 268 - 77 = 191. Ans. 54 ARITHMETIC. Had the above expression been written 1,296 ~ 12 + 160 22 X 3|- -4-7 + 25, it would have been necessary to have divi- ded 22 X 3|- by 7 before subtracting, and the final result would have been 22 X 3$ = 77; 77 -J-.7 = 11; 268 - 11 = 257; 257 -j- 25 = 282. Ans. In other words, it is necessary to per- form all of the multiplication or division included between the signs -{- and , or and -{-, before adding or subtracting. Also, had the expression been written 1,290 -4- 12 + 160 24 -4- 7 X 3 + 25, it would have been necessary to have multiplied 3^- by 7 before dividing 24, since the sign of multiplication takes the precedence, and the final result would have been 3 X 7 = 24; 24 -4- 24 == 1; 268 - 1 = 267 ; 267 + 25 292. Ans. It likewise follows that if a succession of multiplication and division signs occurs, the indicated operations must not be performed in order, from left to right the multiplication must be performed first. Thus, 24x3-=-4x2-h9x5 = f Ans. In order to obtain the same result that would be obtained by performing the indicated operations in order, from left to right, symbols of aggregation must be used. Thus, by using two vinculums, the last expression becomes 24 x 3-4-4 X 2 -=- 9X 5 = 20, the same result that would be obtained by performing the indicated operations in order, from left to right. EXAMPLES FOR PRACTICE. 1 74. Find the values of the following expressions : (a) (8 + 5 - 1) -5- 4. (b) 5 X 34 - 32. (c) 5x24 -=-15. (d) 144-5x24. (*) (1,691-540 + 559)-!- 3X57. (/ ) 2,080 + 120 - 80 x 4 - 1,670. (g) (90 + 60 -- 25) X 5 - 29. (k) 90 + 60 -- 25 X 5. (a) 3. (*) 88. (c) 8. (d) 24. (e) 10. (/) 210. 1.2. ARITHMETIC. (CONTINUED.) PERCENTAGE. 175. Percentage is the process of calculating by hundredth*. 1 76. The term per cent, is an abbreviation of the Latin words per centum, which mean by the Jiundred. A certain per cent, of a number is the number of hundreds of that number which is indicated by the number of units in the per cent. Thus, 6 per cent, of 125 is 125 X yfj- = 7.5 ; 25 per cent, of 80 is 80 X T 8 A 20 ; 43 per cent, of 432 pounds is 432 X T 4 o 3 o = 185.76 pounds. 17^7. The sign of per cent, is yd, and is read per cent. Thus, 6$ is read six per cent.; 12* is read twelve and one-half per cent. , etc. When expressing the per cent, of a number to use in cal- culations it is customary to express it decimally instead of fractionally. Thus, instead of expressing 6$, 25$, and 43$ as yl^, y 2 ^, and y 4 ^, it is usual to express them as .06, .25, and .43. The following table will show how any per cent, can be expressed either as a decimal or as a fraction : TABLE 3. Per Cent. Decimal. Fraction. Per Cent. Decimal. Fraction. 1*.... 2*.... 10*.... 25*.... 50*... . 75*.... .100*.... 125* .... .01 .02 .05 .10 .25 .50 .75 1.00 1.25 T 7 Aor| {U or 1 150 * 500 * i*. 12i*. 624*. 1.50 5.00 .0025 .005 .015 .08^ .125 .625 T\>*0 56 ARITHMETIC. 178. The names of the different elements used in per- centage are : the base, the rate per cent., the percentage, the amount, and the difference. 1 79. The base is the number on which the per cent, is computed. 180. The rate is the number of hundredths of the base to be taken. 181. The percentage is the part, or number of hun- dredths, of the base indicated by the rate ; or the percent- age is the result obtained by multiplying the base by the rate. Thus, when it is stated that 1% of $25 is $1.75, $25 is the base, 7# is the rate, and $1.75 is the percentage. 1 82. The amount is the sum of the base and percentage. 1 83. The difference is the remainder obtained by sub- tracting the percentage from the base. Thus, if a man has $180, and he earns 6# more, he will have altogether $180 + $180 X. 06, or $180 + $10. 80 = $190.80. Here $180 is the base, 6# the rate, $10.80 the percentage, and $190.80 the amount. Again, if an engine of 125 horsepower uses 16$ of it in overcoming friction and other resistances, the amount left for performing useful work is 125 125 x .16 = 125 20 = 105 horsepower. Here 125 is the base, 16# the rate, 20 the percentage, and 105 the difference. 184. From the foregoing it is evident that, to find the percentage, the base must be multiplied by the rate. Hence the following Rule 21. To find the percentage, multiply the base by the rate, expressed decimally. EXAMPLE. Out of a lot of 300 boiler tubes 76# were used in a boiler. How many tubes were used ? SOLUTION. 76^, the rate, expressed decimally, 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, we have percentage = base X rate. ARITHMETIC. 57 185. 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 y^, of some number; then, \%, or yffr, of the number is 12 -^ 6 or 2. Consequently, if 2 =1#, or yffr, 100#, or $$$ = 2 X 100 = 200. But, since the same result may be arrived at by dividing 12 by .06, for 12 -f- .06 = 200 it follows that Rule 22. When the percentage and rate are given, to find the base, divide the percentage by the rate, expressed decimally. Formula, base = percentage -4- rate. EXAMPLE. 76 of a lot of boiler tubes are used in the construction of a boiler. If the number of tubes used were 228, how many tubes were in the lot ? SOLUTION. Here 228 is the percentage, and 76, or .76, is the rate; hence, applying the rule, 228 -r- . 76 = 300 tubes. Ans. 186. When the base and percentage are given to find the rate, the rate may be found, expressed decimally, 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 the number of times that 2 is contained in 12, or 12 -=- 2 = 6$. But the same result may be obtained by dividing 12, the percentage, by 200, the base, since 12 -^ 200=. 06 = 6#. Hence, Rule 23. When the percentage and base are given, to find the rate, divide the percentage by the base, and the result ivill be the rate, expressed decimally. Formula, rate = percentage -f- base. EXAMPLE. Out of a lot of 300 boiler tubes, 228 were used. What per cent, of the total number was used ? SOLUTION. Here 300 is the base and 228 the percentage ; hence, applying rule, rate = 228 -r- 300 = . 76 = 76. Ans. EXAMPLE. What per cent, of 875 is 25 ? SOLUTION. Here 875 is the base, and 25 is the percentage ; hence, applying rule, 25 -5- 875 =.02? = 2?^. Ans. PROOF. 875 X 02f = 25. 58 ARITHMETIC. EXAMPLES FOR PRACTICE. 1 87. What per cent, of (a) 360 is 90? (&) 900 is 360? (c) 125 is 25? (d) 150 is 750? (e) 280 is 112 ? (/) 400 is 200 ? (g) 47 is 94 ? Ans. (a) 25*. (6) 40*. (c) 20*. (d) 500*. (*) 40*. (/) 50*. (g) 200*. (k) 50*. (A) 500 is 250? 1 88. The amount may be found when the base and rate are given, by multiplying the base by 1 plus the rate, ex- pressed decimally. For suppose that it is desired to find the amount when 200 is the base and 6$ is the rate. The percentage is 200 X .06 = 12, and, according to definition, Art. 182, the amount is 200+12 212. But the same result may be obtained by multiplying 200 by 1 + .06, or 1.06, since 200 X 1.06 = 212. Hence, Rule 24. When the base and rate are given, to find the amount, multiply the base by 1 plus the rate, expressed deci- mally. Formula, amount base X (1 + rate}. EXAMPLE. If a man earned $725 in a year, and the next year 10* more, how much did he earn the second year ? SOLUTION. Here 725 is the base and 10* is the rate, and the amount is required. Hence, applying the rule, 725 X 1 1 = $797. 50. Ans. 189. When the base and rate are given, the difference may be found by multiplying the base by 1 minus the rate, expressed decimally. For suppose that it is desired to find the difference when the base is 200 and the rate is 6#. The percentage is 200 X .06 = 12 ; and, according to definition, Art. 183, the difference = 200 12 = 188. But the same result may be obtained by multiplying 200 by 1 .06, or .94, since 200 X .94 = 188. Hence, Rule 25. When the base and rate are given, to find the difference, multiply the base by 1 minus the rate, expressed decimally. Formula, difference = base X (1 rate}. ARITHMETIC. 59 EXAMPLE. Out of a lot of 300 boiler tubes all but 24# were used in one boiler ; how many tubes were used ? SOLUTION. Here 300 is the base, 24 is the rate, and it is desired to find the difference. Hence, applying the rule. 300 X (1-. 24) = 228 tubes. Ans. 190. When the amount and rate are given, the base may be found by dividing the amount by 1 plus the rate. For suppose that it is known that 212 equals some number increased by 6$ of itself. Then, it is evident that 212 equals 106$ of the number (base) that it is desired to find. Con- sequently, if 212 = 106$, 1$ = fif = 2, and 100$ = 2 X 100 = 200 = the base. But the same result may be obtained by dividing 212 by 1 + .06 or 1.06, since 212-^1.06 = 200. Hence, Rule 26. When the amount and rate arc given, to find the base, divide the amount by 1 plus the rate, expressed decimally. * Formula, base = amount -t- (1 -{- rate). EXAMPLE. The theoretical discharge of a certain pump when run- ning at a piston speed of 100 feet per minute is 278,910 gallons per day of 10 hours. Owing to leakage and other defects this value is 25$ greater than the actual discharge. What is the actual discharge ? SOLUTION. Here 278,910 equals the actual discharge (base) increased by 25# of itself. Consequently, 278,910 is the amount, 25 is the rate, and applying rule, actual discharge = 278,910 -H 1.25 = 223,128 gallons. Ans. 191. When the difference and rate are given, the base may be found by dividing the difference by 1 minus the rate. For suppose that 188 equals some number less 6$ of itself. Then, 188 evidently equals 100 6 = 94$ of some number. Consequently, if 188 = 94$, 1$ = 188 -=- 94 = 2, and 100$ = 2 X 100 = 200. But the same result may be obtained by dividing 188 by 1 .06, or .94, since 188 -4- .94 = 200. Hence, Rule 27. When the difference and rate are given, to find the base, divide the difference by 1 minus the rate, expressed decimally. Formula, base = difference -f- (1 rate). 60 ARITHMETIC. EXAMPLE. From a lot of boiler tubes 76 were used in the construc- tion of a boiler. If there were 72 tubes unused, how many tubes were in the lot ? SOLUTION. Here 72 is the difference and 10% is the rate. Applying rule, 73 .5- (1 _ . 76) = 300 tubes. Ans. EXAMPLE. The theoretical number of foot-pounds of work per minute required to operate a boiler feed-pump is 127,344. If 30^ of the total number actually required be allowed for friction, leakage, etc., how many foot-pounds are actually required to work the pump ? SOLUTION. Here the number actually required is the base ; hence, 127,344 is the difference, and 30 is the rate. Applying the rule, 127, 344 -4- (1 - . 30) = 181 , 920 foot-pounds. Ans. 192. EXAMPLE. A certain chimney gave a draft of 2.76 inches of water. By increasing the height 20 feet the draft was increased to 3 inches of water. What was the gain per cent. ? SOLUTION. Here it is evident that 3 inches is the amount and that 2.76 inches is the base. Consequently, 3 -2. 76 = .24 inch is the percentage, and it is required to find the rate. Hence, applying rule 23, gain per cent. = .24 -=- 2.76 = .087 = 8.1%. Ans. 193. EXAMPLE. A certain chimney gave a draft of 3 inches of water. After an economizer had been put in the draft was reduced to 1.2 inches of water. What was the loss per cent.? SOLUTION. Here it is evident 1.2 inches is the difference (since it equals 3 inches diminished by a certain per cent, loss of itself) and 3 inches is the base. Consequently, 3 1.2 = 1.8 inches is the percent- age. Hence, applying rule 23, loss per cent. = 1.8 -5- 3 = .60 = 60. Ans. 194. To find the gain or loss per cent.: Rule 28. Find the difference between the initial and final values ; divide this difference by the initial value. EXAMPLE. If a man buys a steam engine for $1,860 and some time afterwards purchases a condenser for 25g of the cost of the engine, does he gain or lose, and how much per cent., if he sells both engine and condenser for 2,100 ? SOLUTION. The cost of the condenser was 1,860 X .25 = $465; con- sequently, the total initial value, or cost, was $1,860 + $465 = $2,325. Since he sold them for $2,100, he lost $2,325 - 2,100 = 225. Hence, applying rule, 225 -r- 2, 325 = . 0968 = 9. 68 loss. Ans. ARITHMETIC. 61 EXAMPLES FOR PRACTICE. 195. Solve the following: (a) What is 124$ of 900 ? (6) " " |$ " 627? (c) " " 33$ " 54? (d) 101 is 68f$ of what number? . (e) 784 " 83i$ " (a) $112.50. (t) 5.016. (O 18- (X) (?) 940.8. (//) 40$. (/) What $ of 960 is 160? (-) " " " $3, 606 is $450|? (X) " " " 280 is 112? 1. A steam plant consumed an average of 3,640 pounds of coal per day. The engineer made certain alterations which resulted in a sav- ing of 250 pounds per day. What was the per cent, of coal saved ? Ans. 7$, nearly. 2. If the speed of an engine running at 126 revolutions per minute should be increased 64$, how many revolutions per minute would it then make ? Ans. 134.19 revolutions. 3. The list price of an engine was $1,400 ; of a boiler, $1,150, and of the necessary fittings for the two, $340. If 25$ discount was allowed on the engine, 22$ on the boiler, and 124$ on the fittings, what was the actual cost of the plant ? Ans. $2,244.50. 4. If I lend a man $1,100, and this is 184$ of the amount that I have on interest, how much money have I on interest ? Ans. $5,945.95. 5. A test showed that an engine developed 190.4 horsepower, 15$ of which was consumed in friction. How much power was available for use ? Ans. 161.84 H. P. 6. By adding a condenser to a steam engine, the power was in- creased 14$,and the consumption of coal per horsepower per hour was decreased 20$. If the engine could originally develop 50 horsepower and required 34 pounds of coal per horsepower per hour, what would be the total weight of coal used in an hour, with the condenser, assuming the engine to run full power ? Ans. 159.6 pounds. DENOMINATE NUMBERS. 196. A denominate number is a concrete number, and may be either simple or compound, as 8 quarts, 5 feet, ten inches, etc. 197. A simple denominate number consists of units of but one denomination, as 16 cents, 10 hours, 5 dollars, etc. 62 ARITHMETIC. 198. A compound denominate number consists of units of two or more denominations of a similar kind, as 3 yards, 2 feet, 1 inch ; 3 pounds, 5 ounces. 199. In whole numbers and in decimals, the law of increase and decrease is on the scale of 10, but in com- pound or denominate numbers the scale varies. MEASURES. 200. A measure is a standard unit established by law or custom, by which quantity of any kind is measured. The standard unit of dry measure is the Winchester bushel ; of weight, the pound ; of liquid measure, the gallon, etc. 201. Measures are of six kinds: 1. Extension. 4. Time. 2. Weight. 5. Angles. 3. Capacity. 6. Money or value. MEASURES OF EXTENSION. 2O2. Measures of extension are used in measuring lengths, distances, surfaces, and solids. LINEAR MEASURE. TABLE 4. 12 inches (in.) = 1 foot . 3 feet = 1 yard . 5 yards = 1 rod . 40 rods = 1 furlong 8 furlongs = 1 mile . Abbreviation, ft. yd. rd. fur. mi. yd. rd. fur. in. ft. 36= 3 198 = 16i= 5| 7,920 = 660 = 220 = 40 =5,280 =1,760 =320= 8 SQUARE MEASURE. TABLE 5. 144 square inches (sq. in.) . 9 square feet 30 square yards 160 square rods 640 acres. sq. mi. A. 5 (sq. in.) . . = 1 square foot 1 square yard 1 square rod sq. rd. sq. yd. 1 acre . . . 1 square mile sq. ft. sq. ft. sq. yd. sq. rd. . A. sq. mi. sq. in. 1 = 640 = 102,400 = 3,097,600 = 27,878,400 = 4,014,489,600 ARITHMETIC. 63 CUBIC MEASURE. TABLE 6. 1728 cubic inches (cu. in.) . . . = 1 cubic foot cu. ft. 27 cubic feet =1 cubic yard cu. yd. 128 cubic feet = 1 cord cd. 24f cubic feet =1 perch P. cu. yd. cu. ft. cu. in. 1 = 27 = 46,656 MEASURES OF WEIGHT. AVOIRDUPOIS WEIGHT. TABLE 7. 16 ounces (oz.) = 1 pound Ib. 100 pounds = 1 hundred-weight. . . . cwt. 20 cwt., or 2,000 Ib = 1 ton T. T. cwt. Ib. oz. 1 = 20 = 2,000 = 32,000 203. The ounce is divided into halves, quarters, etc. Avoirdupois weight is used for weighing coarse and heavy articles. LONG TON TABLE. TABLE 8. 16 ounces =1 pound Ib. 112 pounds = 1 hundred-weight. . . . cwt. 20 cwt, or 2,240 Ib =1 ton T. In all the calculations throughout this and the succeeding volumes, 2,000 pounds will be considered one ton, unless the long ton (2,240 pounds) is especially mentioned. TROY WEIGHT. TABLE 9. 24 grains (gr.) = 1 pennyweight .... pwt. 20 pennyweight =1 ounce oz. 12 ounces =1 pound Ib. Ib. oz. pwt. gr. 1 = 12 = 240 = 5,760 204. Troy weight is used in weighing gold and silver ware, jewels, etc. It is used by jewelers. 64 ARITHMETIC. MEASURES OF CAPACITY. LIQUID MEASURE. TABLE 10. Pt gins (gi.) 2 pints 4 quarts 31| gallons 63 gallons hhd. bbl. gal. 1 = 2 = 63 1 quart . . qt. . . gal. . . bbl. . hhd. . = 1 gallon . = 1 barrel qt. pt. gi. = 252 = 504 = 2,016 2 pints (pt.) 8 quarts . 4 pecks. . DRY MEASURE. TABLE 11. = 1 quart . = 1 peck . =1 bushel bu. pk. qt. pt. 1 = 4 = 32 = 64 qt. pk. bu. 60 seconds (sec.) 60 minutes MEASURE OF TIME. TABLE 12. =r 1 minute 24 hours . . . 1 day 7 days . . . 365 days j 12 months j 366 days . . . 100 years . . . . = 1 week ...... = 1 common year . . . . " = 1 leap year. = 1 century. mm. hr. wk. NOTE. It is customary to consider one month as 30 days. MEASURE OF ANGLES OR ARCS. TABLE 13. 60 seconds (") = 1 minute ' 60 minutes = 1 degree * . 90 degrees = 1 right angle or quadrant [_ 360 degrees = 1 circle cir. cir. 1 = 360 = 21,600' = 1,296,000" ARITHMETIC. 65 MEASURE OF MONEY. UNITED STATES MONEY. TABLE 14. . = 1 cent ct. 1 dime d. 1 dollar $. 1 eagle E. ct. m. 1,000 :: 10,000 MISCELLANEOUS TABLE. TABLE 15. 10 mills (m.) . . 10 cents . . . . . . = 10 dimes . . . 10 dollars . . . . = E. 1 $ = 10 d. = 100 - 1 12 things are 1 dozen. 12 dozen are 1 gross. 12 gross are 1 great gross. 2 things are 1 pair. 20 things are 1 score. 1 league is 3 miles. 1 fathom is 6 feet. 1 meter is nearly 39.87 inches. 1 hand is 4 inches. 1 palm is 3 inches. 1 span is 9 inches. 24 sheets are 1 quire. 20 quires, or 480 sheets, are 1 ream. 1 bushel contains 2,150.4 cubic in. 1 U. S. standard gallon (also called a wine gallon) contains 231 cubic in 1 U. S. standard gallon of water weighs 8.355 pounds, nearly. 1 cubic foot of water contains 7.481 U. S. standard gallons, nearly. 1 British imperial gallon weighs 10 pounds. It will be of great advantage to the student to carefully memorize all of the above tables. REDUCTION OF DENOMINATE NUMBERS. 205. Reduction of denominate numbers is the pro- cess of changing their denomination without changing their value. They may be changed from a higher to a lower denomination or from a lower to a higher either is reduc- tion. As, 2 hours = 120 minutes. 32 ounces = 2 pounds. 206. Principle. Denominate numbers are changed to lower denominations by multiplying, and to Jiiglier de- nominations by dividing. To reduce denominate numbers to louver denomi- nations : 66 ARITHMETIC. 2O7. EXAMPLE. Reduce 5 yd. 2 ft. 7 in. to inches. SOLUTION. yd. ft. in. 15ft. 2ft. 17ft. 12 "34 17 2 4 in. '7 in. 211 inches. Ans. EXPLANATION. Since there are 3 feet in 1 yard, in 5 yards there are 5 X 3, or 15 feet, and 15 feet plus 2 feet - 17 feet. There are 12 inches in a foot ; therefore, 12 X 17 = 204 inches, and 204 inches plus 7 inches = 211 inches = number of inches in 5 yards 2 feet and 7 inches. Ans. 2O8. EXAMPLE. Reduce 6 hours to seconds. SOLUTION. 6 hours. 60 360 minutes. 60 21600 seconds. Ans. EXPLANATION. As there are 60 minutes in one hour, in six hours there are 6 X 60, or 360 minutes ; as there are no minutes to add, we multiply 360 minutes by 60, to get the number of seconds. 209. In order to avoid mistakes, if any denomination be omitted, represent it by a cipher. Thus, before reducing 3 rods 6 inches to inches, insert a cipher for yards and a cipher for feet ; as, rd. yd. ft. in. 3006 21 0. Rule. Multiply the number representing the high- est denomination by the number of units in the next lower ARITHMETIC. 67 required to make one of the higher denomination, and to the product add the number of given units of that lower denomi- nation. Proceed in this manner until the number is reduced to the required denomination. EXAMPLES FOR PRACTICE. 211. Reduce (a) 4 rd. 2 yd. 2 ft. to ft. () 4 bu. 3 pk. 2 qt. to qt. (<:) 13 rd. 5 yd. 2 ft. to ft. (d) 5 mi. 100 rd. 10 ft. to ft. (e) 8 Ib. 4 oz. 6 pwt. to gr. (/) 52 hhd. 24 gal. 1 pt, to pt. (g ) 5 cir. 16 20' to minutes. (K) 14 bu. to qt. Ans. (a) 74ft (b) 154 qt. (c) 231.5 ft. (d) .28,060ft. (e) 48,144gr. (/) 26,401 pt. (g) 108,980'. (K) 448 qt. To reduce lower to higher denominations: 212. EXAMPLE. Reduce 211 in. to higher denominations. SOLUTION. 1 2 ) 2 1 1 in. 3 ) 1 7 ft. + 7 in. 5 yd. -(- 2 ft. Ans. EXPLANATION. There are 12 inches in 1 foot ; there- fore, 211 divided by 12 = 17 feet and 7 inches over. There are 3 ft. in 1 yd. ; therefore, 17 ft. divided by 3 = 5 yd. and 2 ft. over. The last quotient and the two remainders con- stitute the answer, 5 yd. 2 ft. 7 in. 213. EXAMPLE. Reduce 14,135 gi. to higher denominations. SOLUTION. 4 )14135 2 ) 3533 pt. 3 gi. 4 ) 1766 qt. 1 pt. 441 gal. 2qt. 31.5)441.0(14bbl. 315 Teo 1260 EXPLANATION. There are 4 gi. in 1 pt., and in 14,135 gi. there are as many pints as 4 is contained in 14,135, or 3,533 68 ARITHMETIC. pt. and 3 gi. remaining. There are 2 pt. in 1 qt., and in 3,533 pt. there are 1,766 qt. and 1 pt. remaining. There are 4 qt. in 1 gal., and in 1,766 qt. there are 441 gal. and 2 qt. remaining. There are 31 J gal. in 1 bbl., and in 441 gal. there are 14 bbl. The last quotient and the three remainders constitute the answer, 14 bbl. 2 qt. 1 pt. 3 gi. 214. Rule 3O. Divide the number representing tJie de- nomination given, by Hie ?tumber of units of tJiis denomi- nation required to make one unit of the next higher denomi- nation. The remainder will be of the same denomination, but the quotient will be of the next higlier. Divide this quotient by the number of units of its denomination required to make one unit of the next higher. Continue until the highest denomination is reached, or until there is not enough of a denomination left to make one of the next higher. The last quotient and the remainders constitute the required result. EXAMPLES FOR PRACTICE. 21 5. Reduce to units of higher denominations: (a) 7,460 sq. in. ; (b) 7,580 sq. yd. ; () 9 mi. 13 rd. 4 yd. 2 ft. ; 16 rd. 5 yd. 1 ft. 5 in. ; 16 mi. 2 rd. 3 in. , 14 rd. 1 yd. 9 in. ? (0 3 cwt. 46 Ib. 12 oz. ; 12 cwt. 9i Ib. ; 2 cwt. 21$ Ib ? (d) 10 yr. 8 mo. 5 wk. 3 da. ; 42 yr. 6 mo. 7 da. j 7 yr. 5 mo. 18 wk. 4 da.; 17 yr. 17 da.? (e) 17 tons 11 cwt 49 Ib. 14 oz. ; 16 tons 47 Ib. 13 oz. ; 20 tons 13 cwt. 14 Ib. 6 oz. ; 11 tons 4 cwt. 16 Ib. 12 oz. ? (/) 14 sq. yd. 8 sq. ft. 19 sq. in. ; 105 sq. yd. 16 sq. ft. 240 sq. in.', 42 sq. yd. 28 sq. ft. 165 sq. in. ? (a) 86 Ib. 3 oz 16 pwt 7 gr. () 25 mi. 47 rd. 1 ft. 5 in. (0 18 cwt 2 Ib. 14 oz, Ans ' (d) 78 yr. 1 mo. 3 wk. 3 da. (e) 65 tons 9 cwt. 28 Ib. 13 oz. (/) 167 sq. yd. 136 sq. in. SUBTRACTION OF DENOMINATE NUMBERS. 221. EXAMPLE. From 21 rd. 2 yd. 2 ft. 64 in., take 9 rd 4 yd. iin. SOLUTION. rd. yd. ft. in. 21 2 2 6i 9 4 I0j 11 34 I 8i Ans. ARITHMETIC. 71 EXPLANATION. Since 10 inches can not be taken from 6 inches, we must borrow 1 foot, or 12 inches, from the 2 feet in the next column and add it to the 6f G -f 12 = 18. 18 inches 10 inches = 8J inches. Then, foot from the 1 remaining foot = 1 foot. 4 yards can not be taken from 2 yards; therefore, we borrow 1 rod, or 5^ yards, from 21 rods and add it to 2. 2 + 5 7| ; 7 - 4 = 3^ yards. 9 rods from 20 rods = 11 rods. Hence, the remainder is 11 rods 3| yards 1 foot 8 inches. Ans. To avoid fractions as much as possible, we reduce the yard to inches, obtaining 18 inches; this added to 8 inches, gives 26| inches, which equals 2 feet 2{ inches. Then, 2 feet + 1 foot = 3 feet = 1 yard, and 3 yards -f- 1 yard = 4 yards. Hence, the above answer becomes 11 rods 4 yards feet 2{ inches. 222. EXAMPLE. What is the difference between 3 rd. 2 yd. 2 ft. 10 in. and 47 ft. ? SOLUTION. 47 ft. = 2 rd. 4 yd. 2 ft. rd. yd. ft. in. 3 2 2 10 2420 3 10 or 324 Ans. To find (approximately) the interval of time be- tween two dates: 223. EXAMPLE. How many years, months, days, and hours between 4 o'clock P.M. of June 15, 1868, and 10 o'clock A.M., September 28, 1891? SOLUTION. yr. mo. da. hr. 1891 8 28 10 1868 5 15 16 23 3 12 18 Ans. EXPLANATION. Counting 24 hours in 1 day, 4 o'clock P.M. is the 16th hour from the beginning of the day, or midnight. On September 28, 8 months and 28 days have elapsed, and on June 15, 5 months and 15 days. After plac- ing the earlier date under the later date, subtract as in the previous problems. Count 30 days as 1 month. 72 ARITHMETIC. 224. Rule 32. Place the smaller quantity under the larger quantity, with like denominations under eacli other. Beginning at the right, subtract successively the number in the subtrahend in each denomination from the one above, and place the differences underneath. If the number in the minuend of any denomination is less than the number under it in the sub- trahend, one must be borrowed from the minuend of the next higher denomination, reduced } and added to it. EXAMPLES FOR PRACTICE. 225. From (a) 125 Ib. 8 oz. 14 pwt. 18 gr. take 96 Ib. 9 oz. 10 pwt. 4 gr. () 126 hhd. 27 gal. take 104 hhd. 14 gal. 1 qt. 1 pt. (c) 65 T. 14 cwt. 64 Ib. 10 oz. take 16 T. 11 cwt. 14 oz. (d) 148 sq. yd. 16 sq. ft. 142 sq. in. take 132 sq. yd. 136 sq. In. ( 3 > and s below : 3' = 3 X 3 = 9. 3 = 3 X 3 X 3 = 27. 2* = 2 X 2 X 2 X 2 X 2 = 32. 236. The root of a number is that number which, used the required number of times as a factor, produces the num- ber. In the above cases, 3 is a root of 9, since 3 X 3 are 9. It is also a root of 27, since 3x3x3 are 27. Also, 2 is a root of 32, since 2 X 2 x 2 X 2 X 2 are 32. 237. The second power of a number is called its square. Thus, 5 3 is called the square of 5, or 5 squared, and its value is 5 X 5 = 25. 238. The third power of a number is called its cube. Thus, 5 3 is called the cube of 5, or 5 cubed^ and its value is 5X5X5 = 125. To find any power of a number : 239. EXAMPLE. What is the third power, or cube, of 36 ? SOLUTION. 35x35x35, or 35 35 175 105 1225 35 6125 3675 = 42875 Ans. 78 ARITHMETIC. 24O. EXAMPLE. -What is the fourth power of 15? SOLUTION. 15x15x15x15, or 15 15 75 15 15 1125 225 3375 15 16875 3375 fourth power = 50625 Ans. 241 . EXAMPLE. 1.2 s = what ? SOLUTION. 1.2x1.2x1.2, or 1.2 1.2 1~44 1.2 "288 144 1.728 Ans. 242. EXAMPLE. What is the third power, or cube, of f ? SOLUTION.- (f )* = | X f X f =8*g*8 = &. Ans. 243. Rule 35. (a) To raise a whole number or a deci- mal to any power, use it as a factor as many times as there are units in the exponent. (b) To raise a fraction to any power, raise both the numer- ator and denominator to the power indicated by the exponent. EXAMPLES FOR PRACTICE. 244. Raise the following to the powers indicated: (a) 85 2 . f (a) 7,225. (*) (H) 4 - . I w m- (c) 6.5 s . Ans ' 1 (c) 42.25. (d) 14*. I (d) 38,416. ARITHMETIC. 79 (') (f) 3 - { (') If (/) (I) 3 - An , (/) HI- () (i) 3 - 1 (g-) 4 s - (>*) 1.4 5 . [ (A) 5.37824. EVOLUTION. 245. Evolution is the reverse of involution. It is the process of finding the root of a number which is considered as a power. 246. The square root of a number is that number which, when used twice as a factor, produces the number. Thus, 2 is the square root of 4, since 2 X 2, or 2", = 4. 247. The cube root of a number is that number which, when used three times as a factor, produces the number. Thus, 3 is the cube root of 27, since 3 X 3 X 3, or 3 3 , =27. 248. The radical sign ^/, when placed before a num- ber, indicates that some root of that number is to be found. 249. The index of the root is a small figure placed over and to the left of the radical sign, to show what root is to be found. Thus, 1/100 denotes the square root of 100. 1/125 denotes the cube root of 125. 4/256 denotes the fourth root of 256, and so on. 250. When the square root is to be extracted, the index is generally omitted. Thus, 4/100 indicates the square root of 100. Also, 4/225 indicates the square root of 225. SQUARE ROOT. 251. The largest number that can be written with one figure is 9, and 9* = 81; the largest number that can be written with two figures is 99, and 99* = 9,801; with three figures 999, and 999* = 998,001; with four figures 9,999, and 9,999 s = 99,980,001, etc. In each of the above it will be noticed that the square of the number contains just twice as many figures as the number. In order to find the square root of a number, the first step is to find how many figures there will be in the root. This 80 ARITHMETIC. is done by pointing off the number \n\.o periods of two figures each, beginning at tJie rigJit. The number of periods will indicate the number of figures in the root. Thus, the square root of 83,740,801 must contain 4 figures, since, pointing off the periods, we get 83'74'08'01, or 4 periods ; consequently, there must be 4 figures in the root. In like manner, the square root of 50,625 must contain 3 figures, since there are (5'06'25) 3 periods. 252. EXAMPLE. Find the square root of 31,505,769. root. SOLUTION. () 5 31'50'57'69 ( 561 3 Ans. (d) 5 31'50'57'69( 5613 5 (6) ^5 100 (c) 650 6 636 106 (e) 1457 6 1121 1120 33669 1 33669 1121 1 11220 3 11223 EXPLANATION. Pointing off into periods of two figures each, it is seen that there are four figures in the root. Now, find the largest single number whose square is less than or equal to 31, the first period. This is evidently 5, since 6 s = 36, which is greater than 31. Write it to the right, as in long division, and also to the left, as shown at (a). This is the first figure of the root. Now, multiply the 5 at (a) by the 5 in the root, and write the result under the first period, as sh~wn at (6). Subtract, and obtain 6 as a remainder. Bring down the next period 50, and annex it to the re- mainder 6, as shown at (c), which we call the dividend. Add the root already found to the 5 at (a), getting 10, and annex a cipher to this 10, thus making it 100, which we call the trial divisor. Divide the dividend (c) by the trial divisor () 5. (c) 12i. (d) .77J. W J. (f) 2. (*> H- (K) If- CO 10. (J) 5. (i) & (^) rV- 298. Instead of expressing the value of a ratio by a single number, as above, it is customary to express it by ARITHMETIC. 99 means of another ratio in which the consequent is 1. Thus, suppose that it is desired to find the ratio of the weights of two pieces of iron, one weighing 45 pounds and the other weighing 30 pounds. The ratio of the heavier to the lighter is then 45 : 30, an inconvenient expression. Using the frac- tional form, we have . Dividing both terms by 30, the ou consequent, we obtain -p- or 1 : 1. This is the same result as obtained above, for 1 -=- 1 = l, and 45 -=- 30 = 1|. 299. A ratio may be squared, cubed, or raised to any power, or any root of it may be taken. Thus, if the ratio of two numbers is 105 : G3, and it is desired to cube this ratio, the cube may be expressed as 105 3 : 63 3 . That this is correct is readily seen ; for, expressing the ratio in the fractional 105 . /105\ 3 105 3 form, it becomes - , and the cube is ( -- I = ^ = lOo : 63 . 63 ' \ 63 / 63 3 Also, if it is desired to extract the cube root of the ratio 105 3 : 63 3 , it may be done by simply dividing the exponents by 3, obtaining 105 : 63. This may be proved in the same way as in the case of cubing the ratio. Thus, 105' : 63 s = 3OO. Since -= it follows that 105 s : 63 s = = (|Y, it 5 3 : 3 3 (this expression is read : the ratio of 105 cubed to G3 cubed equals the ratio of 5 cubed to 3 cubed), it follows that the antecedent and consequent may always be multi- plied or divided by the same number, irrespective of any indicated powers or roots, without altering the value of the ratio. Thus, 24" : 18 J = 4 2 : 3". For, performing the opera- tions indicated by the exponents, 24 1 = 576 and 18' = 324. Hence, 576 : 324 = 1J or 1$ : 1. Also, 4' = 16 and 3* = 9; hence, 16 : 9 = IJor 1J .: 1, the same result as before. Also, , 24' /24V _ /4 V _ 4' _ , M ' 18 = 18 5 = W - \3 ) - 3 1 - 4 ' 3 ' 100 ARITHMETIC. The statement may be proved for roots in the same man- ner. Thus, f24~ 3 : tflS 3 = %~ : 4*/3l For the ^sT' 1 = 24 and ^IS 3 = 18; and, 24 : 18 = \\ or 1 : 1- Also, fiT* = 4 and $W = 3; 4 : 3 = 1J or 1 : 1. NOTE. If the numbers composing the antecedent and consequent have different exponents, or if different roots of those numbers are indicated, the operations described in Art. SOOcannotbe performed. This is evident ; for, consider the ratio 4 2 : 8 3 . When expressed in the fractional form, it becomes , which can not be expressed either as ( g- j (4 \ 3 -Q j , and, hence, can not be reduced as described above. PROPORTION. 301. Proportion is an equality of ratios, the equality being indicated by the double colon ( : : ) or by the sign of equality ( = ). Thus, to write in the form of "a proportion the two equal ratios, 8 : 4 and 6 : 3, which both have the same value 2, we may employ one of the three following forms: 8 : 4 :: G : 3 (1) 8 : 4 = G : 3 (2) H ^ 302. The first form is the one most extensively used, by reason of its having been exclusively employed in all the older works on mathematics. The second and third forms are being adopted by all modern writers on mathematical subjects, and, in time, will probably entirely supersede the first form. In this paper we shall adopt the second form, unless some statement can be made clearer by using the third form.' 303. A proportion may be read in two ways. The old way to read the above proportion was: 8 is to If. as 6 is to 3 ; the new way is : the ratio of 8 to 4 equals the ratio of 6 to 3. The student may read it either way, but we recommend the latter. 304. Each ratio of a proportion is termed a couplet. In the above proportion, 8 : 4 is a couplet, and so is 6 : 3. ARITHMETIC. 101 305. The numbers forming the proportion are called terms ; and they are numbered consecutively from left to right, thus : first second third fourth 8:4=6:3 Hence, in any proportion, the ratio of the first term to the second term equals the ratio of the third term to the fourth term. 306. The first and fourth terms of a proportion are called the extremes, and the second and third terms the means. Thus, in the foregoing proportion, 8 and 3 are the extremes and 4 and 6 are the means. 307. A direct proportion is one in which both coup- lets are direct ratios. 308. An inverse proportion is one which requires one of the couplets to be expressed as an inverse ratio. Thus, 8 is to 4 inversely as 3 is to 6 must be written 8:4 = 6 : 3 ; i. e. , the second ratio (couplet) must be inverted. 309. Proportion forms one of the most useful sections of arithmetic. In our grandfathers' arithmetics, it was called ' l The rule of three." 310. Rule 4O. In any proportion, the product of the extremes equals the product of t/ie means. Thus, in the proportion, 17 : 51 = 14 : 42. 17 X 42 = 51 X 14, since both products equal 714. 311. Rule 41. The product of the extremes divided by either mean gives the other mean. EXAMPLE. What is the third term of the proportion 17 : 51 = : 42 ? SOLUTION. Apply ing the rule, 17 X 42 = 714, and 714-*- 51 = 14. Ans. 312. Rule 42. The product of the means divided by either extreme gives the other extreme. EXAMPLE. What is the first term of the proportion : 51 = 14 : 42 ? SOLUTION. Applying the rule, 51 X 14 = 714, and 714 -H 42 = 17. Ans. 102 ARITHMETIC. 313. When stating a proportion in which one of the terms is unknown, represent the missing term by a letter, as x. Thus, the last example would be written, x : 51 = 14 : 42 and for the value of x we have x = = 17. 314. If the same (addition and subtraction excepted) operations be performed upon all of the terms of a propor- tion, the proportion is not thereby destroyed. In other words, if all of the terms of a proportion be (1) multiplied or (2) divided by the same number; (3) if all the terms be raised to the same power; if (4) the same root of all the terms be taken, or (5) if both couplets be inverted, the pro- portion still holds. We will prove these statements by a numerical example, and the student can satisfy himself by other similar ones. The fractional form will be used, as it is better suited to the purpose. Consider the proportion 8:4=6:3. Expressing it in the third form, it becomes = -, What we are to prove is that, if any of the five 4 o operations enumerated above be performed upon all of the terms of this proportion, the first fraction will still equal the second fraction. O y IV 1. Multiplying all the terms by any number, say 7, j = 6 X 7 56 42 56 , 42 = 33TT r 2~8 = 21- N W 28 6Vldently 6qUals 21' SmCC the value of either ratio is 2, and the same is true of the original proportion. q . iy 2. Dividing all the terms by any number, say 7, = 1^]; or | = i. But |-> | = 2, and f + f = S also, the same as in the original proportion. 3. Raising all the terms to the same power, say the cube, |] = ||. This is evidently true, since |[ = (|Y= 2 3 = 8, ARITHMETIC. 103 4. Extracting the same root of all the terms, say the cube root, -^= = 57=. It is evident that this is likewise true, ^8 8/8 = V 4 = 4 = > and jfl = "3 = alsa 4 3 5. Inverting both couplets, 5-=, which is true, since o b both equal \. 315. If both terms of either couplet be multiplied or both divided by the same number, the proportion is not de- stroyed. This should be evident from the preceding article, and also from Art. 294. Hence, in any proportion, equal factors may be canceled from the terms of a couplet, before applying rule 41 or 42. Thus, the proportion 45:9 = x\ 7. 1, we may divide both terms of the first couplet by 9 (that is, cancel 9 from both terms), obtaining 5: 1 = x\ 7.1, whence x = 1. 1 x 5 -T- I = 35.5. (See note in Art. 3OO.) 316. The principle of all calculations in proportion is this : Three of the terms are always given, and the remain- ing one is to be found. 317. EXAMPLE. If 4 men can earn $25 in one week, how much can 12 men earn in the same time ? SOLUTION. The required term must bear the same relation to the given term of the same kind as one of the remaining terms bears to the other remaining term. We can then form a proportion by which the required term may be found. The first question the student must ask himself in every calculation by proportion is : "What is it I want to find ?" In this case it is dollars. We have two sets of men, one set earning $25, and we want to know how many dollars the other set earns. It is evident that the amount 12 men earn bears the same relation to the amottnt that 4 men earn as 12 men bears to 4 men. Hence, we have the proportion, the amount 12 men earn is to $25 as 12 men is to 4 men; or, since either extreme equals the product of the means divided by the other extreme, we have The amount 12 men earn : $25 = 12 men : 4 men, CJOPJ v 12 or the amount 12 men earn = = $75. Ans. 104 ARITHMETIC. Since it matters not which place x or the required term occupies, the problem could be stated as any of the following forms, the value of x being the same in each : (a) 25 : the amount 12 men earn = 4 men : 12 men ; or the amount 12 men earn = * , or $75, since either mean equals the product of the extremes divided by the other mean. (b) 4 men : 12 men = $25 : the amount 12 men earn ; or the amount that 12 men 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 QQPj vx -JO amount that 12 men earn = , or 75. since either mean equals the product of the extremes divided by the other mean. 318. If the proportion is an inverse one, first form it as though it were a direct proportion, and then invert one of the couplets. EXAMPLES FOR PRACTICE. 319. Find the value of x in each of the following: (a) 16 : 64 :: x : $4. (l>) x : 85 :: 10 : 17. (c) 24 : x :: 15 : 40. (d) 18 : 94 :: 2 : x. Ans. (e) $75 : 100 = x : 100. (/) 15pwt. :jr=21:10. (g) x : 75 yd. = 15 : 5. (a) jr = $l. (6) x = 50. (c) * = 64. (d) x= lOf. (e) x = 75. (/) *=7}pwt. (g) .r = 225yd. 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 ? 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 inches. INVERSE PROPORTION. 32O. In Art. 3O8, 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 ARITHMETIC. 10o statement of the example ; in such cases the proportion can be written directly, merely inverting one of the coup- lets. But it frequently happens that oniy 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 proportion 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 third 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 re- quired term belongs. Having determined this, the student then refers to the example, and ascertains from its condi- tions whether the required term is to be larger or smaller than the other term of the same kind. If the two determi- nations agree, the proportion is direct; otherwise, it is inverse, and one of the couplets must be inverted. 321 . EXAMPLE. If A's rate of doing work is to B's as 5 : 7, and 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, referring 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 = .r : 42, from which x = r l*^ = 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 7 x 42 than 42, and the proportion would be direct, the value of x being = 58.8 days. 106 ARITHMETIC. EXAMPLES FOR PRACTICE. 322. 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 in. ; how many would be required if the pitch were 3| in. ? Ans. 40. 3. The spring hangers on a certain locomotive are 2^ in. wide and | in. thick ; those on another engine are of same sectional area, but are 3 in. wide ; how thick are they ? Ans. 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. POWERS AND ROOTS IN PROPORTION. 323. It was stated in Art. 299 that a ratio could be raised to any power or any root of it might be taken. A proportion is frequently stated in such a manner that one of the couplets must be raised to some power or some root of it must be taken. In all such cases, both terms of the coup- let so affected must be raised to the same power or the same root of both terms must be taken. 324. EXAMPLE. Knowing that the weight of a sphere varies as the cube of its diameter, what is the weight of a sphere 6 inches in diameter if a sphere 8 inches in diameter of the same material weighs 180 pounds ? SOLUTION. This is evidently a direct proportion. Hence, we write 6 3 : 8 3 = x : 180. Dividing both terms of the first couplet by 2 3 (see Art. 3OO), 3 3 : 4 s = x : 180, or 27 : 64 = x : 180 ; 27 y 1 80 whence, x = ^ = 75^ pounds. Ans. EXAMPLE. A sphere 8 inches in diameter weighs 180 pounds ; what is the diameter of another sphere of the same material which weighs 75 H pounds ? SOLUTION. Since the weights of any two spheres are to each other as the cubes of their diameters, we have the proportion 180 : 75}f = 8 3 : x*. ARITHMETIC. 107 x, the required term, must be cubed, because the other term of the couplet is cubed (see Art. 323). But, 8 3 = 512; hence, ' 7^1 5 v PCI O 180 : 75} f = 512 : x\ or x* = = 216 ; whence, x = y'aiG = 6 inches. Ans. 325. Since taking the same root of all of the terms of a proportion does not change its value (Art. 314), the above example might have been solved by extracting the cube root of all of the numbers, thus obtaining $ / ~ISQ ; ^7515 _ 8 x f = 6 inches. The process, however, is longer and is not so direct, and the first method is to be preferred. 326. If two cylinders have equal volumes, but different diameters, the diameters are to each other inversely as the square roots of their lengths. Hence, if it is desired to find the diameter of a cylinder that is to be 15 inches long, and which shall have the same volume as one that is 9 inches in diameter and 12 inches long, we write the proportion 9 : *=yi5 : 4/12. Since neither 12 nor 15 are perfect squares, we square all of the terms (Arts. 325 and 314) and obtain 81 : x* = 15 : 12; whence x* = 81 X 12 = 64.8, 1 U and x = 4/64.8 = 8.05 inches = diameter of 15-inch cylinder. EXAMPLES FOR PRACTICE. 327. Solve the following examples: 1. The intensity of light varies inversely as the square of the dis- tance from the source of light. If a gas jet illuminates an object 80 feet away with a certain distinctness, how much brighter will the object be at a distance of 20 feet ? Ans. 2 times as bright. 2. In the last example, suppose that the object had been 40 feet from the gas jet; how bright would it have been compared with its brightness at 30 feet from the gas jet ? Ans. -f g as bright. 3. When comparing one light with another, the intensities of their illuminating powers vary as the squares of their distances from the 108 ARITHMETIC. source. If a man can just distinguish the time indicated by his watch, 50 feet from a certain light, at what distance could he distinguish the time from a light 3 times as powerful ? Ans. 86.64- feet. 4. The quantity of air flowing through a mine varies directly as the square root of the pressure. If 60,000 cubic feet of air flow per minute when the pressure is 2.8 pounds per square foot, how much will flow when the pressure is 3.6 pounds per square foot ? Ans. 68,034 cu. ft. per min., nearly. 5. In the last example, suppose that 70,000 cubic feet per minute had been required; what would be the pressure necessary for this quantity ? Ans. 3.81+ Ib. per sq. ft. CAUSES AND EFFECTS. 328. Many examples in proportion may be more easily solved by using the principle of cause and effect. That which may be regarded as producing a change or alteration in something, or as accomplishing something, may be called a cause, and the change, or alteration, or thing accom- plished, is the effect. 329. Like causes produce like effects. Hence, when two causes of the same kind produce two effects of the same kind, the ratio of the causes equals the ratio of the effects; in other words, the first cause is to the second cause as the first effect is to the second effect. Thus, in the question, if 3 men can lift 1,400 pounds, how many pounds can 7 men lift ? we call 3 men and 7 men the causes (since they ac- complish something, viz., the lifting of the weight), the number of pounds lifted, viz., 1,400 pounds and x pounds, are the effects. If we call 3 men the first cause, 1,400 pounds is the first effect ; 7 men is the second cause, and x pounds is the second effect. Hence, we may write 1st cause 2d cause 1st effect 2 d effect 3:7 1,400 : x, 7 V 1 4-00 whence x = 3, 266f pounds. , or P t (read p sub. one and P major sub. one}. Third, by the use of sub. letters ; thus, p t or P t (read p sub. i and P major sub. i). In the same manner/" (read / second), p v or p 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, however, it is not convenient to use them, in which case MENSURATION. lift: primes and subs, are used instead. The prime notation may be continued as follows: /'", /'", /", etc. ; it is inad- visable to use superior figures, for example, /', /", /*, p a t etc., as they are liable to be mistaken for exponents. The main thing to be remembered by the student is that when a formula is given in which the same letters occur sev- eral times, all like letters having the same primes or subs. represent the same quantities, '<.vhile those which differ in any respect represent different quantities. Thus, in the formula w lt w and % represent the weights of three different bodies; s t , s tt and s a , their specific heats; and t^ / 2 , and / 8 , 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 lt and /, all refer to one of the three bodies ; zv^, 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 numeri- cal values of the letters, and then perform the indicated operations. Thus, suppose that the values of w,, j,, and t l are, respectively, 2 pounds, .0951, and 80; of w a , s v and / a , 7.8 pounds, 1, and 80; and of w 3 , j jt and / 3 , 3 pounds, .1138, and 780; then, the final temperature /is, substituting these values for their respective letters in the formula, 2X .0951 X 80+ 7.8 X 1 X 80 + 3J X .1138 X 780 _ 2 X .0951 + 7.8 X 1 + 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, accord- ing to the rule previously given. 118// MENSURATION. 344. The student should now be able to apply any formula 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 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 //>/-v - />/% - , or whether - - was to be divided by 25. If this latter 25 ooO 160 ~{*f*(\ result were desired, the expression would be written - . In ywO 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 7 + ^H r 25 line is not necessary, since it is impossible to mistake the operation that is required to be performed. But, since , 660 175 + 660 ., 175 + 660, , 660 7 + ^5-= 25 ' lf WC substitute ^ for 7+ , the heavy line becomes necessary in order to make the resulting expression clear. Thus, 160 160 _ 160 660 ~ 175 + 660 ~ 83F + 25 25 25 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, 4^37, 4/37, etc., these roots are indicated by 37*, 37^, 37*, 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 MENSURATION. U8e denominator is always the index of the root. Hence, instead of writing 1/37* 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, 1/37* = 37*. Likewise, {/(I + a* b)* may also be written (1 -f 2 )*, a much simpler expression. 345. 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 in which A = area of segment; * =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; and h = 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, *r*E c 3. 1416X7. 5* X 83. 46 10 2 360 40. 968 27. 95 = 13. 018 square inches, 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 y 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 ; />, the side that is 50 feet long ; and c, the side that is 46 feet long. Then substituting in the formula, 118/ MENSURATION. = 25^/441 8.25 2 = 25 |/441 68.0625 = 25 4/372.9375 = 25 X 19.312 = 482.8 square feet, 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 , 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 cau^c a long column to break is p __SA_ * ~ ~~ rr~ * 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; and G = least radius of gyration of cross-section of column. The values of S, q, 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 wrought- iron 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 = -HTTT^ for flat-ended steel columns ; A, the area of the cross-section, = .7854 (d^ -:= 30. Ans. EXAMPLES FOR PRACTICE. 1. How many seconds 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 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. MENSURATION. 123 QUADRILATERALS. 363. A plane figure is any part of a plane or flat surface bounded by straight or curved lines. 364. A quadrilateral is a plane figure bounded by four straight lines. 365. A parallelogram is a quadrilateral whose op- posite sides are parallel. There are four kinds of parallelograms: the square, the rectangle, the rhombus, and the rhomboid. 366. A rectangle (Fig. 16) is a parallelo- gram whose angles are all right angles. 367. A square (Fig. 17) is a rectangle whose sides are all of the same length. FIG. IT. 368. A rhomboid (Fig. 18) is a parallelogram whose opposite sides are equal and parallel, and whose angles are not right angles. A 369. A rhombus (Fig. 19) is a parallelogram having equal sides, and whose angles are not right angles. FIG 370. A trapezoid (Fig. 20) is a quadrilateral which has only two ^ of its sides parallel. 371. The altitude of a parallelogram is the perpen- dicular distance between the parallel sides, as shown by the dotted lines in Figs. 18, 19, and 20. 372. The base of any plane figure is the side on which it is supposed to stand. 124 MENSURATION. 373. The area of a surface is expressed by the number of unit squares it will contain. 374. 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 one inch on a side is called a square inch, and the one that measures one foot on a side is called a square foot. Square inch and square foot are abbreviated to sq. in. and sq. ft. 375. To find the area of any parallelogram: Rule 44. 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. 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. 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. Redu- cing 200 -i- 16i = 200 -r- ^ = 12.12 rods. Hence, area = 8 X 12.12 = 96.96 sq. rd. Ans. 376. To find the area of a trapezoid : Rule 45. Multiply one -half 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 ? MENSURATION. 125 SOLUTION. One-half the sum of the parallel sides = ^ = 13 inches = l feet. The length of the board corresponds to the altitude of a trapezoid. Hence, 14 x H = 21 sq. ft. Ans. 377. Having given the area of a parallelogram and one dimension, to find the other dimension : Rule 46. 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 26i feet ? SOLUTION. 212 -=- 26* = 212 H- ^ = 8 feet. Ans. The following examples illustrate a few special cases: EXAMPLE. An engine room is 22 feet by 32 feet. The engine-bed occupies a space of 3 feet by 12 feet ; the fly-wheel 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 fly-wheel 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 square feet of flooring required. Ans. EXAMPLE. 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 p- = 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 95 J sq. yd. Ans. EXAMPLE. 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 one acre, the number of acres = 416,000 -4- 160 = 2,600 acres. Ans. 12G MENSURATION. 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 thereof? . Ans. l,8401b. 3. One side of a room measures 16 ft. 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 X 15 feet and 9 feet high; the room con- tains one door 3 X 7 feet, three windows 3 X 6 feet, and a baseboard 8 inches high? Ans. 53.5 sq. yd. THE TRIANGLE. 378. A triangle is a plane figure having three sides. 379. An isosceles triangle is one having two of its sides equal; see Fig. 21. 380. An equilateral triangle (Fig. 22) is one having all of its FIG. 21. sides of the same length. 381. A scalene triangle (Fig. 23) is one having no two of its sides equal. 382. A right-angled triangle (Fig. 24) is any triangle having one right angle. The side opposite the right angle is called the hypotenuse. FlG . 34. 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 C equals two right angles, or 180. Hence, if any two angles of a tri- O angle are given and it is required to find the third angle: MENSURATION. 12? Rule 47. Add together the two given angles, and subtract their sum from 180; the result will be the third angle. EXAMPLE. If two angles of a triangle = 48 16' and 47 50' respec- tively, 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 60' = 2,820' + 50' = 2,370'. Adding the two angles together, and subtracting 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 5 034 number to degrees and minutes, ' = 83f = 83 54'. Hence, the third angle in the triangle = 83 54'. Ans. 383. In any right-angled triangle there can be but one right angle, and since the sum of all the 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-angled triangle one acute angle is known, to A c find the other acute angle : FlG - ~- Rule 48. Subtract the known acute angle from 00; the result will be the other acute angle. EXAMPLE. If one acute angle, as A, of the right-angled triangle ABC, Fig. 26, equals 30, what does the angle B equal ? SOLUTION. 90 30 = 60. Ans. 384. 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 pro- portional to the corresponding sides of the first triangle. Thus, in the triangle A B C, Fig. 27, if the line D E be drawn parallel to the side B C, the triangle FIG. 27. c A D E will be formed and we shall have (1) Side 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. 128 MENSURATION. 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), A D : 8 = 24 : 18; whence A D = *** 8 = 10J. Ans. lo 385. In any right-angled triangle, the square described on the hypotenuse is equal to the sum of the squares described upon the other two sides. If A B C, Fig. 28, is a right-angled triangle, right- angled at B, then the square described upon the hypote- nuse^ C is equal to the sum of the squares described upon the sides A B and B C. FlG - <3g " Hence, having given the two sides forming the right angle in a right-angled triangle, to find the hypotenuse: Rule 49. 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 B C= 4 inches, what is the length of the hypotenuse A C1 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 = 4/9 + 16 = |/25 = 5 inches. Ans. If the hypotenuse and one side are given, the other side can be found as follows: Rule 5O. Subtract the square of the given side from the square of the hypotenuse, and extract the square root of the remainder. EXAMPLE. The side given is 3 inches, the hypotenuse is 5 inches ; what is the length of the other side ? SOLUTION. 3 2 = 9; 5 8 = 25. 2,5 - 9 = 16, and the -j/16 = 4 inches. Ans. MENSURATION. 129 150 EXAMPLE. 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 steeple, 150 feet high; C a stake 85 feet from the foot of the steeple, and A C the rope. Here we have a right-angled triangle, right-angled at B, and A C is the hypotenuse. The square of A B = 150 2 = 22,500; of C B, 85* = 7,225. 22,500 + 7,225 = 29,725 ; '29,725 = 172.4 feet, nearly. Ans. 386. The altitude of any triangle is aline, 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. 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 FIG. 32. area of each triangle will equal one-half the area of the parallelogram, or one-half the product of the base and the altitude. Hence, to find the area of any triangle : Rule 5 1 . 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 feet 9 inches = 7f feet = -^ feet. 18 X -^ = 139$, and one-half of 139$ = 69f square feet. Ans. ' To find the altitude or base of a triangle, having given the area and the base or altitude: Rule 52. Multiply the area by 2, and divide by the given dimension. 130 MENSURATION. 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 inches. 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-angled 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 paral- lel 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-angled triangle are 92 feet and 69 feet long. How long is the hypotenuse ? Ans. 115 ft. POLYGONS. 387. 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. 388. A regular polygon is one in which all of the sides and all of the angles are equal. 389. A polygon of five sides is called a pentagon; one with six sides a bexagon; one of seven sides a hep- Pentagon. Hexagon. Heptagon. Octagon. Decagon. Dodecagon. FIG. 13. tagon, etc. Regular polygons having from five to twelve sides are shown in Fig. 33. In any polygon, the sum of all MENSURATION. 131 , Fig. 34, equals the interior angles, asA+fi+C+D + 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 53. Multiply 180 by the num- ber of sides less two, and divide the result by the number of sides; the quotient FlG will be the number of degrees in each interior angle. EXAMPLE. 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, 3 2 = 3 and 180 X 3 = 540 ; 540 H- 5 = 108 in each angle. Ans. EXAMPLE. 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 />', />' 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 the above rule. Number of sides = 6; 6 2 = 4; hence, 180 X 4 = 720, and 720 -s- 6 = 120 in each angle. Now, let M N represent the miter-box, and O S the direc- tion in which the saw should run; then, A 7> 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 A 7> C, which we have found to be 120. Hence, the saw should run at an angle of -- = 60 with the side of the miter-box. 39O. 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 54. 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. 132 MENSURATION. 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 H- 2 = 5.2; 12 X 5.2 X 6 = 374.4 sq. in. Ans. 391. 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 C D E F, Fig. 37. SOLUTION. Draw the diagonals B F and CF and the line FG perpendicular to D , 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, Are&ABF = 16X7 = 56 sq. in. Area FC D G = 14 X 10 = 140 sq. in. AreaFGE = 9 ^ =45 sq. in. Total area = 56 + 35 -f- 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 BCDEF (see Fig. 37), sup- posing each of the given dimensions to be increased to \\ 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. MENSURATION. 133 THE CIRCLE. 392. A circle (Fig. 38) 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. s* \ FIG. 38. \ 393. The diameter of a circle is a J jg straight line passing through the center \ I and terminated at both ends by the cir- \ / cumference; thus, A B (Fig. 39) is a diam- eter of the circle. 394. 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 diame- ter. The plural of radius is radii, and all radii of a circle are equal. FIG. 40. 395. An arc of a circle (see a e b, Fig. 41) is any part of its circumference. 396. 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 b, Fig. 42, is a chord of the circle whose corresponding arc is a e b. FIG. 42. 397. 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- c scribed angle, and it is measured by one- half the arc ADC. 134 MENSURATION. 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 = i of the circumference. The whole circumference = 360 ; hence, 360Xi = 72; therefore, angle A B C is an angle of 72. 398. 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 one-half a semi-circumference, or 180 -=-2 = 90. Thus, the angles A D C and ABC, Fig. 44, are right angles, since they are inscribed in a semicircle. 399. 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 per- pendicular to any chord, it bisects (cuts in halves) the chord. Thus, if the radius FlG 45 - O C, Fig. 46, is perpendicular to the chord A B, A D = D B. A EXAMPLE. If a regular pentagon be 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 penta- gon 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 a 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 -=- 5 = 5f 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 51 -f- 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. MENSURATION. 135 400. If, from any point on the circumference of a circle, a perpendicular be let fall upon a given diameter, it will di vide the diameter into two parts, one of which will be in the same ratio to the per- pendicular as the perpendicular is to the other. That is, the perpendicular will be a- 4 ' mean proportional between the two parts. If A B, Fig. 47, is the given diameter, and C any point on the circumference, then AD: C D:\ C D: D B, CD being a mean proportional between A D and D B. EXAMPLE. If H K '= 30 feet, and I fi = & 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 : IH :: IH : I A, or 8 : 15 :: 15 : I A. Therefore, I A = -^ = ?P = 28 feet, and I A + IB = 28i -t- 8 = o O 36| feet = A B the diameter of the circle. Ans. 401. 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 I is 8 feet; what is the length of the line H K ? SOLUTION. As the diameter of the circle is 36 feet, and as B I is 8 feet, 1 A is equal to 36 \ 8 = 28 feet. The two parts, therefore, are 09^ 8 and 28 feet, and their product = 8 X 28 = 8 X -5- = 225 ; the square O root of their product = ^225 = 15 feet, and as H K IH+ IK, or 2 IH, HK= 15 X 2 = 30 feet. Ans. 402. To find the circumference of a circle, the diameter being given: Rule 55. Multiply the diameter by 8.1416. EXAMPLE. What is the circumference of a circle whose diameter is 15 inches ? SOLUTION. 15 X 3. 1416 = 47. 124 inches. Ans. 403. To find the diameter of a circle, the circumference being given : 136 MENSURATION. Rule 56. Divide the circumference by S. 1416. EXAMPLE. What is ths diameter of a circle whose circumference is 65.973 inches? SOLUTION. 65.973 -*- 3.1416 = 21 inches. Ans. 404. To find the length of an arc of a circle: Rule 57. Multiply the length of the circumference of the circle of which the arc is a part by the number of degrees in the arc, and divide by 860. EXAMPLE. What is the length of an arc of 24, the radius of the arc being 18 in. ? 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 57^ = 7.54 in., or the length of the arc. Ans. ouu 405. To find the area of a circle : Rule 58. Square the diameter, and multiply by .785 '4- EXAMPLE. What is the area of a circle whose diameter is 15 inches ? SOLUTION. 15 2 = 225; and 225 X .7854 = 176.72 sq. in. Ans. 406. Given the area of a circle, to find its diameter: Rule 59. Divide the area by . 7854, and extract the square root of the quotient. EXAMPLE. The area of a circle = 17,671.5 square inches. What is its diameter in feet ? SOLUTION. |/ 17 ' 671 '^ = 150 inches. r . 7854 .7854 eet, or the diameter. Ans. EXAMPLE. What is the area of a flat circular ring. Fig. 48, v/hose outside diameter is 10 inches. and whose inside diameter is 4 inches ? SOLUTION. The area of the large circle = 10 2 X .7854 = 78.54 square inches; the area of the small circle = 4 2 X -7854 = 12.57 square inches. The area of the ring is the difference between these areas, or 78.54 12.57 = 65.97 square inches. Ans. 4O7. To find the area of a sector (a sector of a cir- cle is the area included between two radii and the circum- ference, as, for example, the area C O , Fig. 13): MENSURATION. 137 Rule 6O. Divide the number of degrees in the arc of the sector by 360. Multiply the result by the area of the circle of which 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. ?-== = ^-j-; and 12' 2 X .7854 = 113.1 square inches. ODU j&4 113.1 X HJ = 23.56 square inches, the area. Ans. 4O8. To find the area of a segment of a circle (a segment of a circle is the area included between a chord and its arc; for example, the area ABC, Fig". 49): Rule 6 1 . Divide the diameter by the height of the segment; 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 of A B CA = A? V=r ~ 608 - O II 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, 4X20* 54~~ 4 X 20* 4 X 400 = 4/2.092 = 1.447; ^~- X 1.447 = 771.7 sq. in. Ans. 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 that in Fig. 47, the diameter A B = 15 feet, and the distance B /= 3 feet. What is the length of the line HK ? Ans. 12ft 138 MENSURATION. 3. The diameter of a fly-wheel is 18 feet. What is the distance around it to the nearest 16th of an inch ? Ans. 56 ft, G T 9 5 in. 4. A carriage wheel was observed to make 71|- turns while going 300 yards. What was its diameter ? Ans. 4 ft. , nearly. 5. What is the length of an arc of 64, 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. A.ns. 84.82 sq. in. THE PRISM AND CYLINDER. 409. A solid, or body, has three dimensions: length, breadth, and thickness. The sides which enclose it are called the faces, and their intersections are called edges. 410. 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 hexag- onal prism is one having a hexagon for its base, etc. 41 1. A cylinder is a body of uniform diameter whose ends are equal parallel circles. 41 2. A parallelopipedon (Fig. 50) is a prism whose bases (ends) are parallelograms. 413. A cube (Fig. 51) is a prism .-, whose faces and ends are squares. FIG 51 All the faces of a cube are equal. In the case of plane figures, we have had to do with perimeters and areas. In the case of solids, we have to do with the areas of their outside surfaces, and with their contents or volumes. MENSURATION. 139 414. 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. 415. 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 contents is sometimes used. 416. To find the area of the convex surface of a prism or cylinder: Rule 62. Multiply the perimeter of the base by the altitude. EXAMPLE. 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; 36x24 = 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. How many square feet of sheet iron will be required for a pipe 1 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 U feet in diameter and 10 feet long. The perimeter, or circumference, of the base = 1 X 3.1416 = 1.5 X 3.1416 = 4.712 feet. The convex sur- face = 4.712 X 10 = 47.12 sq. ft. of metal. Ans. 417. To find the volume of a prism or a cylinder: Rule 63. Multiply the area of the base by the altitude. EXAMPLE. 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. 140 MENSURATION. SOLUTION. The shaft is a cylinder 16 feet 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 feet, reduced to inches = 16x12 = 192 in.; 3.1416x192 = 603.19 cu. in. = the volume. The weight = 603.19 X- 28 = 168.89 Ib. Ans. EXAMPLE. Find the cubical contents of a hexagonal prism, Fig. 52, 12 inches long, each edge of the base being one 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-angled triangle C D A, side A D = i A B, or one-half inch, and since the polygon is a hexagon, side C A = distance A B, or one inch. Hence, C A being the hypotenuse, the length of FIG. 52. side C D = /I* - (i)* = 4/1* - .5* = 1/7*75. or .866 inch. Area of triangle A CS = * x ^ 866 = .433 sq. i n . . 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. 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'\s at a dis- tance 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 'N O the altitude. First obtain the area of the segment M N S, whose height h is 16 inches, in square feet; then multiply the result by 16, the length of the boiler. By the formula previously given, the area of the segment = MENSURATION. 141 - .608 = 4/27767 = 1.663. Hence, the area = 341.33 X 1.663 = 567.63 sq. in. This, reduced to square feet, = 567.63 -H 144 = 3.942sq. 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. 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 to/at area. of one end of the boiler in square inches, from it subtracting the area of the segment M N S, and the areas of the ends of the tubes in square inches, and then by multi- plying the result by the length of the boiler in inches. Total area of one end = 54 2 X .7854 = 2,290.23 sq. in. Area of segment M NS, as found in last example, = 567.63 sq. in. Area of the end of one tube = 3.25 2 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-*- 231 = 1,018.06 gallons of water. Ans. EXAMPLES FOR PRACTICE. 1. Find the area. in square inches of the convex surface of a bar of iron 4 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. (A) Find the number of gallons of water required to fill the boiler up to the mean water line. . j (a) 56.4 cu. ft. ADS ' ( (6) 800 gallons. 143 MENSURATION. THE PYRAMID AND CONE. 418. 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. 419. A cone (Fig. 55) is a solid whose base is a circle and whose convex surface tapers uniformly to a point called the vertex. 420. The altitude of a pyramid or cone is the perpendicular distance from the vertex to the base. 421. 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. 422. To find the convex area of a pyramid or cone: Rule 64. Multiply the perimeter of the base by one-half the slant height. EXAMPLE. What is the convex area of a pentagonal pyramid, if one side of the base measures 6 inches, and the slant height = 14 inches ? SOLUTION. The base of a pentagonal pyramid is a pentagon, and, consequently, has five sides. 6 x 5 = 30 inches, or the perimeter of the base. 30 X -g- = 21 S Q- in -> or the convex area. Ans. EXAMPLE. What is the entire area of a right cone whose slant height is 1? inches, and whose base is 8 inches in diameter ? SOLUTION. The perimeter of the base = 8x3. 1416 = 25.1328 in. Convex area = 25.1328 X ^ = 213.63 sq. in. Area of base = 8 2 X .7854 = 50.27 sq. in. Entire area = 263.90 sq. in. Ans. 423. To find the volume of a pyramid or cone: Rule 65. Multiply the area of the base by one-third of the altitude. EXAMPLE. What is the volume of a triangular pyramid, each edge of whose base measures 6 inches, and whose altitude is 8 inches? MENSURATION. 143 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-angled triangle B D A, the hypotenuse B A = 6 inches, and side A D = 3 inches, to find the other side, B D 4/6* 3* 5.2 inches, nearly. Area of the base, or B A C, = = 15.6 sq. in. Hence, the volume D FIG. 50. = 15.6 x|- = 41.6 cu. in. o Ans. EXAMPLE. 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* X .7854 = 153.94 sq. in. Hence, the volume = 153.94 X -5- = 923.64 cu. in. Ans. O EXAMPLES FOR 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. 5.2 cu. 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 OR COXE. 424. 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. Flo. 58. 144 MENSURATION. 425. The upper end of the frustum of a pyramid or cone is called the upper base, and the lower end the lower base. The altitude of a frustum is the perpendicu- lar distance between the bases. 426. To find the convex surface of a frustum of a pyramid or cone: Rule 66. Multiply onc-Jialf the sum of tlie perimeters of tJie two bases by the slant heigJit of tJie frustum. EXAMPLE. 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. - 2 - = 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. 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.- - >< - X 18 = 376.99 sq. in., the area of the convex surface. Area of the upper base = 8' 2 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 + 50.27 + 113.1 = 540.30 1 -[. in. Ans. 427. To find the volume of the frustum of a pyramid or cone: Rule 67. Add together the att>as of the upper and loivcr bases, and the square root of tJie product of the two areas ; multiply the sum by one-third of the altitude. EXAMPLE. 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 = 4/25 X 144 = 60. Adding these three results, and multiplying by one-third the altitude, 25 + 144 + 60 = 229; 229 X 4r = 1.22H cu. o in. = the volume. Ans. MENSURATION. 145 EXAMPLE. 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* X .7854 = 1,809.56 sq. in. ; area of lower base = 60* X .7854 = 2,827. 44 sq. in. ; 4/1,809. 56 X 2,827.44 = 2,261.95. O/ Whence, 1,809.56 + 2,827.44 + 2,261.95 = 6,898.95; 6,898.95 X ^r = o 220,766.4 cu. in. = contents. Now, since there are 231 cu. in. in one gallon, the tank will hold 220,766.4 -*- 231 = 955.7 gallons, nearly. Ans. EXAMPLES FOR PRACTICE. 1. Find the convex surface of the frustum of a square pyramid, one edge of whose lower base is 15 inches long, one edge of whose upper base is 14 inches long, and whose slant height is one 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 gallons, nearly. THE SPHERE AND CYLINDRICAL RING. 428. 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. 429. To find the area of the surface of a sphere : Rule 68. 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 s X 3.1416 = 14 X 14 X 3.1416 = 615.75 sq. in. Ans. 146 MENSURATION. 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 in. diameter is a circle of 6 in. diameter. Any number of great circles could be described on a given sphere. 43O. To find the volume of a sphere : Rule 69.- 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 = 370.96 pounds. Ans. 431. 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 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 area: Rule 7O. 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 -| (> - Q sum of the inside and outside diameters = 5 = 10, and 10 X 3.1416 = 31.416 inches, 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 inches, and 6.2832x31.416- 197.4 sq. in., the convex area. Ans. MENSURATION. U\ 432. To find the volume of a cylindrical ring: Rule 71. The volume will be the same as that of a cylin- der whose altitude equals the length of the dotted center line D, 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 cylin- drical ring, multiply the area of an imaginary cross-section on the line A B, by the length of the center line D. FIG. i. 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 I Q , Q 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* x .7854 = 3.1416 sq. in., the area of the imaginary cross- section. And 3.1416 X 31.416 = 98.7 cubic inches, 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. in. 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 74 Inches. Ans. 107.95 sq. in. 4. Find the cubical contents of the ring in the last example. Ans. 33.73cu. in. 5. The surface of a sphere contains 314.16 square inches. What is the volume of the sphere ? Ans. 523.6 cu. in. MECHANICS. 433. Mechanics is that science which treats ot the action of forces upon bodies, and the effects which they produce ; it treats of the laws which govern the movement and equilibrium of bodies, and shows how they may be utilized. MATTER AND ITS PROPERTIES. 434. Matter is anything that occupies space. It is the substance of which all bodies consist. Matter is com- posed of molecules and atoms. 435. A molecule is the smallest portion of matter that can exist without changing its nature. 436. 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 most 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 cease to be water, and will be something else; we will call this particle a molecule. If a molecule of water be divided, it will yield two atoms of hydrogen gas, and one of oxygen gas. If a molecule of sulphuric acid be divided, it will yield two atoms of hydrogen, one of sulphur, and four of oxygen. 150 MECHANICS. It has been calculated that the diameter of a molecule is larger than TSTrroVoooo f an inch, and smaller than of an inch - 437. Bodies are composed of collections of molecules. Matter exists in three conditions or forms: solid, liqtiid, and gaseous. 438. A solid body is one whose molecules change their relative positions with great difficulty; as iron, wood, stone, etc. 439. A liquid body is one whose molecules tend to change their relative positions easily. Liquids readily adapt themselves to the shape of vessels which contain them, and their upper surface always tends to become perfectly level. Water, mercury, molasses, etc., are liquids. 440. 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. 441. A permanent gas is one which remains a gas at ordinary temperatures and pressures. 442. A vapor is a body which at ordinary tempera- tures is a liquid or solid, but when heat is applied, becomes a gas, as steam. One body may be 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 some 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 cold, all solids might be converted into gases, and all gases into solids. MECHANICS. 151 443. Every portion of matter possesses certain qualities called properties. Properties of matter are divided into two classes : general and special. 444. General properties of matter are those which are common to all bodies. They are as follows: Extension, impenetrability, weight, indestructibility, inertia, mobility, divisibility, porosity, compressibility, expansibility, and elasticity. 445. Extension is the property of occupying space. Since all bodies must occupy space, it follows that extension is a general property. 446. By impenetrability we mean that no two bodies can occupy exactly the same space at the same time. 447. 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 unconfined, they tend to move away from the earth, but, nevertheless, they will finally reach a point beyond which they can not 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 owned and kept by the Government; it weighs one pound. 448. Inertia means that a body can not put itself in motion nor bring itself to rest. To do either it must be acted upon by some force. 449. Mobility means that a body can be changed in position by some force acting upon it. 450. Divisibility is that property of matter on account of which a body may be separated into parts. 451. 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 peaches in a basket. The molecules of water are larger than those of salt; so that when salt is dissolved in water, 152 MECHANICS. 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. 452. Compressibility. This property is a natural consequence of the preceding one. Since there is space between the molecules, it is evident that by means of force (pressure) they can be brought closer together, and thus the body be made to occupy a smaller space. 453. Expansibility is the term used to denote the fact that the molecules of a body will, under certain condi- tions (when heated, for example), move farther apart, and so cause the body to expand, or occupy a greater space. 454. Elasticity is that property of matter which en- ables 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. 455. Indestructibility is the term used to denote the fact that we can not 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 ap- pearance 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 indestmctible. 456. Special properties are those which are not pos- sessed by all bodies. Some of the most important are as follows: Jiardness, tenacity, brittleness, malleability, and ductility. 457. Hardness. A piece of copper will scratch a piece of wood, steel will scratch copper, and tempered steel MECHANICS. 153 will scratch steel in its ordinary state. We express all this by saying that steel is Iiardcr 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. 458. Tenacity is the term applied to the power with which some bodies resist a force tending to pull them apart. Steel is very tenacious. 459. 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 example, good glass will bear a greater compressive force than most woods, but may be easily broken when dropped on to a hard floor; this property is called brittleness. 460. Malleability is that property which permits of some bodies being hammered or rolled into sheets. Gold is the most malleable of all substances. 461. Ductility is that property which enables some bodies to be drawn into wire. Platinum is the most ductile of all substances. MOTION A1VD VELOCITY. 462. Motion is the opposite of rest, and indicates a changing of position in relation to some object which is for that purpose regarded as being fixed. 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 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. 463. 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 which 154 MECHANICS. indicates the direction of this point for every instant that it is in motion is the path of the body. 464. 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. In all other cases it is variable. If the fly-wheel 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 40 2 , = of a mile every minute, and since equal spaces are passed over in equal times, the velocity is uniform. 465. To find the uniform velocity which a body must have to pass over a certain distance or space in a given time : Rule 72. Divide the distance by the time. EXAMPLE. The piston of a steam engine travels 3,000 feet in 5 minutes; what is its velocity in feet per minute ? SOLUTION. Here 3,000 feet is the distance, and 5 minutes is the time. Applying the rule, 3,000 -r- 5 = 600 feet per minute. Ans. CAUTION. Before applying the above or any of the suc- ceeding rules, care must be taken to reduce the values given to the denominations required in the answer. Thus, in the above example, had the velocity been required in feet per second instead of feet per minute, the 5 minutes would have been reduced to seconds before dividing. The operation would then have been, 5 min. = 5 X 60 = 300 sec. Applying the rule, 3,000 -=- 300 = 10 ft. per sec. Ans. 466. Had the velocity been required in inches per sec- ond, it would have been necessary to reduce the 3,000 feet to inches and the 5 minutes to seconds, before dividing. Thus, 3,000 ft. X 12 = 36,000 in. 5 min. X 60 = 300 sec. Now, applying the rule, 36,000 -4- 300 = 120 in. per sec. Ans. EXAMPLE. A railroad train travels 50 miles in l hours; what is its average velocity in feet per second ? MECHANICS. 155 SOLUTION. Reducing the miles to feet, and the hours to seconds, 50 miles X 5.280 = 264,000 ft. H hours X 60 X 60 = 5,400 sec. Applying the rule, 264,000 -H 5,400 = 48| ft. per sec. Ans. 467. If the uniform velocity (or the average velocity) and the time are given, and it is required to find the dis- tance which a body having the given velocity would travel in the given time: Rule 73. Multiply 'the velocity by the time. EXAMPLE. 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 ft. to miles, 1,092 H- 5,280 = J^. o,*oO Applying the rule, ^-^r, X 16 = 3.31 miles, nearly. Ans. EXAMPLE. The piston speed of an engine is 11 ft. per sec., 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 miles = p-^^ mile = velocity iX>tX&t= WX Substituting, we have 1 X 20 X 15 X 24 = IV x 4 X 5 X 4, or Ans. MECHANICS. 175 Hence, also, if W were raised 1 inch, P would fall 90 inchesj or P would have to move through 90 inches to raise W through 1 inch. 502. It is now clear that another great law has made itself manifest, and that is that whenever there is a gain in power without a corresponding increase in the initial force, there is a loss in speed ; this is true of any machine. In the last example if P were to move the entire 90 inches in one second, W would move only 1 inch in one secon'd. Instead of using the diameter or radius of a gear, the number of teeth may be used when computing the weight which can be raised, or the velocity, as in the last example. EXAMPLE. The radius of the pulley A, Fig. 83, is 40", and that of fis 12". The number of teeth in B is 9; in C, 27; in /?, 12, and in E, 36. If the weight to be lifted is 1,800 lb., how great a force at P is it necessary to apply to the belt ? SOLUTION. Let /"represent the force (power); then, by the rule, 85, P X 40 X 27 x 36 = 1,800 X 12 X 9 x 12, or P x 38,880 = 2,332,800. O QQO xi h I Hence, P = 'gg g' gQ = 60 lb. = force necessary to apply to the belt. Ans. GEAR-WHEELS. 503. A wheel that is provided with teeth to mesh with similar teeth upon another wheel is called a gear-wheel, or gear. In Fig. 84 is shown a spur gear. On spur gears the teeth are always parallel to the axis of the wheel or to its shaft. 504. In Fig. 85 is shown a pair of bevel gears in mesh, of which one is smaller than the other. 505. When both are of the same diameter they are called FIG. IM. miter gears. In Fig. 80 is shown a pair of miter gears in mesh. It 176 MECHANICS. obvious that the angle which the teeth of these gears make with the axis of the shaft must be 45. 5O6. In Fig. 87 is shown a revolving screw, or worm, as it is called, in gear; it is used to transmit motion from one shaft to another at right angles to it. As the worm is noth- ing else than a screw, each revolution given to the worm will rotate the worm-wheel a dis- tance equal to its pitch ; FIG. 85. consequently, if there are 40 teeth in the worm-wheel, a single-threaded worm will FIG. 86. have to make 40 revolutions in order to turn the wheel once. MECHANICS. 177 5O7. In Fig. 88 is shown a section of a rack and pinion, both having epicycloidal teeth. The arc C C represents part of the pitch circle ; it is on the pitch circle that all the teeth are laid out. The diameter of a gear or worm- wheel is always taken as the diameter of this circle, unless otherwise specially stated as "diameter over all," or "diameter at the root," etc. The pitch of the teeth of the gear-wheel is the distance from the edge of one tooth to the corresponding edge of the following tooth measured on the pitch circle; it is marked pitch in the figure. The length of the tooth of a gear-wheel is .7 of its pitch, .4 of it, called the root, being be- low or within the pitch circle, and .3 of it, called the ad- dendum, being above or with- out the pitch circle. Thus, if the pitch of the teeth of a gear- FIG - 87 - wheel is 2 inches, the length of a tooth below the pitch circle is 2 X .4= .8 of an inch; and its length above the FIG* pitch circle is 2 X. 3 = .6 of an inch. Consequently, we 178 MECHANICS. have only to multiply the pitch by .4 to obtain the length of the teeth below the pitch circle, and by .3 to obtain the length of the teeth above the pitch circle. The thickness of the teeth of a cast gear-wheel equals .48 X /*, that is, .48 of the pitch ; therefore, the thickness of the above teeth is .48 X 2, or .96 of an inch. A rack may be considered as a gear-wheel rolled out so as to make the pitch circle a straight line, as C' C". The teeth of racks are proportioned by the same rules as those of gear-wheels. 5O8. For the purpose of calculating the pitch diameter, number of teeth, etc., of the gear-wheels, we have the following rules: Let P = pitch ; T= number of teeth; D = pitch diameter of the wheel. To find the pitch diameter of a gear-wheel in inches, when the pitch and number of teeth are given: Rule 86. The pitch diameter equals the product of the pitch and number of teeth divided by 3. 1416. Thatis > D = EXAMPLE. What is the diameter of the pitch circle of a gear-wheel which has 75 teeth, and whose pitch is 1.675 inches ? P T SOLUTION. Formula, D = Tr^rrr^- 6. 141 b Substituting, we have D = ' 3 '? 4 vL = 40 in. Ans. To find the number of teeth in a gear-wheel when the diameter and pitch are given: Rule 87. The number of teeth equals the product of 3.1416 and the diameter divided by the pitch. rm, * ~ 3.1416/? That is, / = -p . EXAMPLE. The diameter of a gear-wheel is 40 inches, and the pitch of the teeth is 1.675 inches; how many teeth are there in the wheel ? MECHANICS. 179 3. 1416 Z? l.Ot u * 4 = 75 teeth. Ans. SOLUTION. Formula, 7 Substituting, we have T = To find the pitch of a gear-wheel when the diameter and the number of teeth are given. Rule 88. The pitch of the teeth equals the product of 8.1416 and the diameter divided by the number of teeth. That is, P= ' ' . EXAMPLE. The diameter of a gear-wheel is 40 inches, and it has 75 teeth; what is the pitch of the teeth ? SOLUTION. Formula, P = 8 ' 14 1 6Z> . Substituting, we have P = 3 - 141 r 6x4 _ 1.675 in. pitch. 1 5 Ans. 509. The forms of teeth used in ordinary practice are the epicycloidal and involute. Fig. 88 shows the epicycloidal form, which is composed of two different curves, the curve from the pitch circle to the top of the tooth being an epicycloid, and that from the pitch circle to the bottom of the tooth being a hypocycloid. In gear-wheels where this form of tooth is employed, their pitch circles must run tangent to one another. 510. In Fig. 89 is shown the involute form of teeth, or ( FIG. 89. teeth having but one curve. The outlines of the teeth shown in the rack are formed of straight lines. 180 MECHANICS. Involute teeth have two great advantages over epicycloidal teeth: 1. They are stronger for the same pitch, as they are thicker at the root. 2. They may be spread apart so that their pitch circles do not run tangent to one another without practically affecting the perfect action of the teeth. 511. To calculate the number of teeth or speed of one of two gear-wheels which are to gear together: Let N = number of revolutions per minute of the driver; n = number of revolutions per minute of the driven; T = number of teeth in the driver ; / number of teeth in the driven. Rule 89. The number of teeth in the driver equals the product of the number of teeth and number of revolutions of the driven divided by the number of revolutions of the driver. That is, T= W- EXAMPLE. The driven has 27 teeth, and will make 66 revolutions per minute ; if the driver makes 99 revolutions per minute, how many teeth are there in the driver ? SOLUTION. Formula, T=-^-. Substituting, we have T= ^ = 18 teeth. Ans. The number of revolutions per minute of the driver and driven and the number of teeth in the driver being given, to find the number of teeth in the driven : Rule 9O. The number of teeth in the driven equals the product of the number of teeth and revolutions per minute of the driver divided by the number of revolutions per minute of the driven. That is, /= n EXAMPLE. The driver has 18 teeth, and makes 99 revolutions per minute, and the driven must make 66 revolutions per minute; how many teeth must there be in the driven ? SOLUTION. Formula, / = ft MECHANICS. 181 Substituting, we have / = = 27 teeth. Ans. The number of teeth in the driver and driven and the number of revolutions per minute of the driver being given, to find the number of revolutions per minute of the driven: Rule 91. The number of revolutions per minute of the driven equals the product of the number of teeth and number of revolutions of the driver divided by the number of teeth of the driven. That is, n = . EXAMPLE. There are 18 teeth in the driver, and it makes 99 revo- lutions per minute; how many revolutions per minute will the driven make if it has 27 teeth ? T N SOLUTION. Formula, n -y-. Substituting, we have n = 18 ^" = 66 R. P. M. Ans. The number of teeth in the driver and driven and the number of revolutions per minute of the driven being given, to find the number of revolutions per minute of the driver: Rule 92. The number of revolutions of the driver equals the product of the number of teeth and revolutions of the driven divided by the number of teeth of the driver. That is, N=*-. EXAMPLE. If there are 27 teeth in the driven, and if it makes 66 revolutions per minute, how many revolutions per minute will the driver make if it has 18 teeth ? SOLUTION. Formula, N= -^. Substituting, we have N= - = 99 R. P. M. Ans. EXAMPLE. In Fig. 90, the crank-shaft makes 60 revolutions per minute ; the governor pulley is 4" in diameter, the bevel gear on the governor pulley shaft has 19 teeth ; the bevel gear which meshes with 182 MECHANICS. it and drives the governor has 30 teeth. The governor is to make 95 revolutions per minute ; what should be the size of the pulley on the crank-shaft ? FIG. 90. SOLUTION. First determine the number of revolutions of the 4* pulley in order that the governor shall turn 95 times per minute. / n ^0 "v Q^J Applying rule 92, N=-^= ^ = 150 revolutions of gear on pulley shaft = revolutions of governor pulley. Now, applying rule 8 1 , the diameter of the pulley on the crank-shaft = -jr=- = ^ = 10'. Ans. EXAMPLE. In Fig. 90, the fly-wheel is 8 feet in diameter and drives a 5-foot pulley on the main shaft. A 14" pulley on the main shaft drives a 16" pulley on the countershaft. A 12" pulley on the counter- shaft drives a 12" pulley on a shaft on which is a pinion that meshes into a large gear attached to the face plate of a large lathe, and which has 108 teeth. How many teeth must the pinion have in order that the face plate may make 9 revolutions per minute ? " SOLUTION. Applying rule 83, to find the revolutions per minute of the main shaft, = - = 96 R. P. M. Applying the same rule again o to find the revolutions of the countershaft, U * 96 = 84 R. P. M. ID Applying it once more to find revolutions of the pulley which turns the small gear, ^y| = 84 R. P. M. Applying rule 89, 108 ^ 9i = 12 teeth in pinion or driver. Ans. MECHANICS. 183 EXAMPLES FOR PRACTICE. 1. The driving pulley makes 110 R. P. M. and is 21" in diameter; what should be the size of the driven in order to make 385 R. P. M. ? Ans. 6'. 2. The main shaft of a certain shop makes 120 R. P. M. It is desired to have the countershaft make 150 R. P. M. There are on hand pulleys of 16", 24", 28", 35", and 38" in diameter. Can two of these be used, or must a new pulley be ordered ? Ans. Use the 28" and 35" pulleys. 3. The pinion (driver) makes 174 R. P. M., and follower makes 24 R. P. M. ; how many teeth must the pinion have if the follower has 87 teeth ? Ans. 12 teeth. 4. If an engine fly-wheel is 66" in diameter, and makes 160 R. P. M., what must be the diameter of the pulley on the main shaft to make 128 R. P. M.? Ans. 82*'. 5. What is the pitch diameter of a gear whose pitch is l", and has 28 teeth? Ans. 11.14". 6. How many teeth are there in a gear whose pitch is .7854", and which is 23" in diameter ? Ans. 92 teeth. 7. What is the pitch of a gear whose diameter is 20.372*, and which has 128 teeth ? Ans. i". 8. In a train of gears the drivers have 16, 30, 24, and 18 teeth, respectively ; the followers have 12, 24, 36, and 40 teeth, respectively. If the first driver makes 80 R. P. M., how many R. P. M. wilt the last follower make ? Ans. 40 R. P. M. FIXED AND MOVABLE PULLEYS. 512. Pulleys are also used for hoisting or raising loads, in which case the frame which supports the axle of the pulley is called the block. 513. A fixed pulley is one whose block is not movable, as in Fig. 91. In this case if the weight W be lifted by pulling down /-*, the other end of the cord W will evidently move the same distance upwards that P moves downwards; hence, P must equal W. 514. A movable pulley is one whose block is movable, as in Fig. 92. One end of the FIO. 91. cord is fastened to the beam, and the weight is suspended 184 MECHANICS. from the pulley, the other end of the cord being drawn up by the application of a force P. A little con- sideration will show that if P moves through a certain distance, say 1 foot, W will move through half that dis- tance, or 6 inches; hence, a pull of one pound at P will lift 2 pounds at W. The same would also be true if the free end of the cord were passed over a fixed pulley, as in Fig. 93, in which case the fixed pulley merely changes the direction in which P acts, so that a weight of 1 pound hung on the free end of the cord will bal- ance 2 pounds hung from the movable pulley. FIG. 92. FIG. 93. 515. A combina- tion of pulleys, as shown in Fig. 94, is sometimes used. In this case there are three movable and three fixed pulleys, and the amount of movement of W, owing to a certain movement of P t is readily found. It will be noticed that there are six parts of the rope, not counting the free end; hence, if the movable block be lifted 1 foot, P remaining in the same position, there will be 1 foot of slack in each of the six parts of the rope, or six feet in all. Therefore, P would have to move 6 feet in order to take up this slack, or P moves 6 times as far as W. Hence, 1 pound at P will support 6 pounds at W, since the power multiplied by the distance through which it moves equals the weight multiplied by the distance through which it moves. It will also be noticed that there are three movable pulleys, and that 3x2 = 6. FIG MECHANICS. 185 516. Law of Combination of Pulleys : Rule 93. In any combination of pulleys where one con- tinuous rope is used, a load on the free end will balance a weiglit on the movable block as many times as great as itself as there are parts of the rope supporting the load, not counting the free end. The above law is good, whether the pulleys are side by side, as in the ordinary block and tackle, or whether they are arranged as in the figure. EXAMPLE. In a block and tackle having five movable pulleys, how great a force must be applied to the free end of the rope to raise 1,250 pounds ? SOLUTION. Since there are five movable pulleys, there must be 10 parts of the rope to support them. Hence, according to the above law, a force applied to the free end will support a load 10 times as great 1 9 50 as itself, or the force = -- = 125 Ib. Ans. 517. An making an angle with a horizontal line. Three cases may arise in practice with the in- clined plane: 1. Where the power acts parallel to the pkme, as in Fig. 95. 2. Where the THE INCLINED PLANE, inclined plane is a slope, or a flat surface, W Zlbs. FIG. 95. power acts parallel to the base, as in Fig. 96. 3. Where the power acts at an angle to the plane or to the base, as in Fig. 97. 518. In Fig. 95 the relation existing between the power and FIO. 96. the weight is easily lib. 186 MECHANICS. found. The weight ascends a distance equal to c b, or the height of the inclined plane, while the power descends through a distance equal to a b, or the length of the inclined plane. There- fore, Rule 94. TAe power multiplied by the length of the in- FIG. 9r. dined plane equals t/ie weight multiplied by the height of the inclined plane; or, letting length of plane = /; base of plane = b\ height of plane = //; we have p= , ... PI and W= -=-. n EXAMPLE. The length of an inclined plane is 40 feet, and its height is 5 feet; what force P will sustain a weight IV oi 100 lb.? W h 100x5 '- -~^0~~ SOLUTION./* = lb. Ans. EXAMPLE. The length of an inclined plane is 8 feet, and its height is 15 inches; what weight will 120 lb. support ? SOLUTION. lV=- = = 768 lb. Ans. In Fig. 96 the power is supposed to act parallel to the base for any position of W\ therefore, while W is moving from the level a c to #, or through the height c b of the inclined plane, P will move through a distance equal to the length of the base a c. Hence, when the power acts parallel to the base, we have Rule 95. The power multiplied by the base of the inclined plane equals the weight multiplied by the height of the plane, . Wh Pb whence, P= r and W=-j-. MECHANICS. 187 EXAMPLE. With a base 30 feet long, and a height of 6 feet, what power will sustain a weight of 75 lb.? SOLUTION. P = = 15 lb. Ans. OU EXAMPLE. A force of 12 lb. sustains a weight on a plane whose base is 6 ft. long, and height 18 inches. Find the weight. SOLUTION. W = /" = 48 lb - Ans - For Fig. 97 no rule can be given. The ratio of the power to the weight must be determined by trigonometry for every position of W; for, as W shifts its position, the angle that its cord makes with the face of the plane varies, and the magnitude of the force P depends on this angle; the smaller the angle the less is P required to be to support a given weight on a given plane. 519. The -wedge is a movable inclined plane, and is used for moving a great weight a short distance. A common method of moving a heavy body is shown in Fig. 98. Simultaneous blows of equal force are struck on the heads of the wedges, thus raising the weight W. The laws for wedges are the same as for Case 2 of the inclined plane. THE SCREW. 520. A screw is a cylinder with a helical projection winding around its circumference. This helix is called the thread of the screw. The distance that a point on the helix is drawn back or advanced in the direction of the length of the screw during one turn, is called the pitch of the screw. 188 MECHANICS. The screw in Fig. 99 is turned in a nut a, by means of a force applied at the end of the handle P. For one complete revolution of the handle, the screw will be advanced lengthwise, an amount equal to the pitch. If the nut be fixed, and a weight be placed upon the end of the screw, as shown, it will be raised vertically a distance equal to the pitch by one revolution of the screw. During this revolution, the force at P will move through a dis- tance equal to the circumference whose radius is P F. Therefore, W X pitch of thread = P X circum- ference of P. 521. Hence, to find the weight that a given force will lift, we have Rule 96. Multiply the force by its distance from the axis, or center of screw, and by 6. 2832 '; divide this product by the pitch, and the quotient will be the weight required; PX 6.2832 X r or whence, P= 6.2832 X r EXAMPLE. It is desired to raise a weight by means of a scre.w having 5 threads per inch. The force is 40 pounds, and is applied at a distance of 14 inches from the center of the screw (FP, in Fig. 99); how great a weight can be raised ? SOLUTION. Applying the formula, ... 40x6.2832x14 W= j = 17593 lb., very nearly. Ans. MECHANICS. 189 EXAMPLE. The weight to be lifted is 4,000 pounds; the pitch of the screw is ^ (that is, there are 4 threads per inch). The force is applied at a distance of 12 inches from the center of screw; how great must it be ? SOLUTION. Applying the formula, P - g 4 ^^ X *3 = rather more than 13J Ib. , nearly. Ans. 522. Single-threaded screws of less than 1 inch pitch are generally classified by the number of threads they have in 1 inch of their length. In such cases one inch divided by the number of threads equals the pitch; thus, the pitch of a screw that has 8 threads per inch is; one of 32 threads per inch is ^, etc. 523. Velocity Ratio. The ratio of the distance through which the power moves to the corresponding dis- tance through which the weight moves is called the velocity ratio. Thus, if the power move through 12 inches while the weight moves through 1 inch, the velocity ratio is 12 to 1, or 12; that is, P moves 12 times as fast as W. If the velocity ratio be known, the weight which any machine will raise can be found by multiplying the power by the velocity ratio. If the velocity ratio is 8.7 to 1, or 8.7, W= 8.7 X P, since W X 1 = Px 8.7. NOTE. In all of the preceding cases, including the last, the effect of friction has been neglected. FRICTION. 524. Friction is the resistance that a body meets with from the surface on which it moves. 525. 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. 100, rests upon a horizontal plane, and has a cord fastened to it passing over a pulley a, from which a weight P is suspended, then, if P is just p sufficient to start W^ the ratio of P to W, or -, is the 190 MECHANICS. coefficient of friction between PFand the surface upon which it slides. The weight W\s> the perpendicular pressure, and Pis the 100 Ibs. W 10 Iba. force necessary to overcome the resistance to the motion of W, due to friction. If W=. 100 pounds and P 10 pounds, the coefficient of P 10 friction for this particular case would be -jp-= = .1. W 100 526. 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 material than between those of different materials. 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 lubricating the surfaces. 527. 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, MECHANICS. 191 Law 2 states that, no matter how small the surface may be which 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. 528. For instance, if the perpendicular pressure be- tween 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 one-half the area of the surface of the journal is 80 square inches, then, the amount of friction for each square inch of bearing is 2,000 -4- 80 = 25 pounds. If half the area of the surface 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 H- 160 = 12 pounds, just one-half as much as before, and the wear and liability to heat would be one-half as great also. TABLE OF COEFFICIENTS OF FRICTION. TABLE 10. Description of Surfaces in Contact. Disposition of Fibers. State of the Surfaces. Coefficient of Friction. Oak on oak . Parallel Drv 48 LJI y Oak on oak Parallel Soaped .16 Wrought iron on oak Parallel Dry .62 Wrought iron on oak Parallel Soaped .21 Cast iron on oak Parallel Dry .49 Cast iron on oak Parallel Soaped .19 Wrought iron on cast iron . . Slightly Unctuous .18 Wrought iron on bronze Slightly Unctuous .18 Cast iron on cast iron Slightly .15 Unctuous 192 MECHANICS. 529. The power which is required to raise a weight, or overcome an equal resistance in any machine, is thus always greater than this -weight or resistance divided by the velocity ratio of the machine. Thus, if there were no friction, a machine whose velocity ratio was 5 would, by an application of 100 pounds of force, raise a weight of 500 pounds. Now, suppose that the friction in the machine is equiva- lent to 10 pounds of -force; then, it would take 110 pounds of force to raise 500 pounds. If, in the above illustration, friction were neglected, 110 pounds X 5 = 550 pounds, or the weight that 110 pounds would raise ; but, owing to. the frictional resistance, it only raised 500 pounds. Therefore, we have for the ratio RAA between the two, = = .91. That is, 500 : 550:: .91 : 1. 5oO 530. Efficiency. This ratio between the weight ac- tually raised and the power multiplied by the velocity ratio, is called the efficiency of the machine. For example, if the weight actually raised by a machine, say a screw, is 1,600 pounds, and the power multiplied by the velocity ratio is 2,400 pounds, the efficiency of this machine is EXAMPLE. In a machine having a combination of pulleys and gears, the velocity ratio of the whole is 9.75. A force of 250 pounds just lifts a weight of 1,626 pounds. What is the efficiency of the machine ? SOLUTION. Efficiency = * t62 .? -, = .6671, or 66.71*. Ans. *OU X *. /o Since the total amount of friction varies with the load, it follows that the efficiency will also vary for different loads. 531. If the efficiency of a machine is known, the force actually required to raise a given load may be found by dividing the load by the product of the velocity ratio of the machine and the efficiency. Thus, if a certain machine has MECHANICS. 193 a velocity ratio of 10. 6, and its efficiency is G0#, the force which must actually be applied to raise a load of 840 pounds is 840 -=- 10.0 X .00 = 840 ~ G.3C = 132.1 pounds, nearly. If there had been no losses through friction, etc., the force required would have been 840 -H 10. G = 70.25 pounds, nearly. If the efficiency is known, the weight which a certain force will raise may be found by multiplying together the force, velocity ratio, and the efficiency. Thus, if a certain machine has a velocity ratio of 6, and an efficiency of 78$, a force of 140 pounds will raise a weight of 140 X 6 X .78 = 709,8 pounds. 532. When finding the force necessary to overcome the friction, the perpendicular pressure on the surface considered must always be taken. In order to find (approximately) the maximum force that is required to overcome the friction between cross-head and guides, we have Rule 97. Multiply the total piston pressure by the length of crank and by the coefficient of friction and divide by the length of main rod; or, letting P total piston pressure; / = length of main rod; r = length of crank; c = coefficient of friction between cross- head and guides; F= force required to overcome this friction, zr we have r = EXAMPLE. An engine whose piston is 16* in diameter carries a steam pressure of 80 Ib. per sq. in. If the crank is 12" long, and the connecting-rod is 66" long, what is the perpendicular pressure on the guides ? The coefficient of friction for this case being 12, what force will be required to overcome the friction ? SOLUTION. Pressure on piston = 16 X .7854 X 80 = 16.085 Ib. 1( '" S ( .' ( . ' = 2,924. 55 Ib. = perpendicular pressure. Ans. 2.924.55 X .12 = 350.95 Ib. = force required u> overcome the friction. Ans. 194 MECHANICS. EXAMPLES FOR PRACTICE. 1. How great a force must be applied to the free end of the rope of a block and tackle which has four movable pulleys, to raise a weight of 746 Ib. ? Ans. 93* Ib. 2. An inclined plane is 30 ft. long and 7 ft. high ; what force is required to roll a barrel of flour weighing 196 Ib. up the plane, friction being neglected ? Ans. 45. 7 Ib. 3. The distance from the axis of a screw to the point on the handle where the force is applied is 12". The screw has 8 threads per inch. What force is necessary to raise a weight of 1,248 Ib. ? Ans. 2.07 Ib., nearly. 4 In example 3, what should be the length of the handle to raise a weight of 5,400 Ib. by the application of a force of 20 Ib.? Ans. 5.371", nearly. 5. What is the velocity ratio (a) in example 3 ? (b) In example 4 ? Ans \ ^ 603 ' nearlv ' Ans ' ) (6) 270. 6. An engine piston is 24" in diameter. If the steam pressure is 93 Ib. per sq. in. ; the length of the connecting-rod, 8 ft. 4" ; the length of crank, 20", and coefficient of friction, 14$, what is (a) the greatest perpendicular pressure on the guides ? (b) The force required to over- come the friction ? . ( (a) 8,414.46 Ib. Ans " \ (b) 1,178 Ib. CENTRIFUGAL FORCE. 533. If a body be fastened to a string and whirled, so as to give it a circular motion, there will be a pull on the string which 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 -W^~~ *" 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 FIG. 101. a circle, the force that causes it to move in a circle instead of a straight line is exactly equal to the tension of the string. If the string were cut, the pulling MECHANICS. 195 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. 101. Since, according to the third law of motion, every action has an equal and opposite reaction, we call that force which 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. 534. The other force, or tension, of the string is called the centripetal force, and it acts towards 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. 535. The value of the centrifugal force, expressed in pounds, of any revolving body is calculated by the following rule: Rule 98. T/ie 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; N= revolutions per minute of revolving body; then F= .00034 W R N\ 536. In calculating the centrifugal force of fly-wheels, it is the usual practice to consider the rim of the wheel only, and not take the arms and hub of the wheel into account. In this case R would be taken as the distance between the center of the rim and the center of the shaft. 196 MECHANICS. EXAMPLE. What would be the centrifugal force developed by a cast-iron fly-wheel, whose outside diameter was 10 feet, width of face 20 inches, and thickness of rim 6 inches, turning at the rate of 80 revolutions per minute ? Take the weight of a cubic inch of cast iron as .261 Ib. SOLUTION. First calculate the weight of the rim. The diameter of the rim = 10 X 12 = 120 inches; the diameter of the circle midway between the inside and outside diameters of the rim = 120 6 = 114 inches. The number of cubic inches in the rim = 114 X 3.1416 X 20 X 6 = 42,977 cubic inches. 42,977 X .261 = 11,217 pounds = weight. 114 Radius = =f H- 12 = 4J feet. R. P. M. = 80. Hence, by rule 98, centrifugal force = .00034 X 11,217 X 4| X 80 2 = 115,939 pounds. Ans. SPECIFIC GRAVITY. 537. The specific gravity of a body is the ratio be- tween its weight and the weight of a like volume of water. 538. Since gases are so much lighter than water, it is usual to take the specific gravity of a gas as the ratio be- tween 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 62.5 pounds ? 450 SOLUTION. -- = 7.2. Ans. 539. The specific gravities of different bodies are given in printed tables ; hence, if it is desired to know the weight of a body that can not be conveniently weighed, calculate its cubical contents. Multiply the specific gravity of the body by the weight of a like volume of water, remembering that a cubic foot of water weighs 62.5 pounds. EXAMPLE. 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 62.5 pounds, 3,214 cubic inches weigh |i|l| X 62.5 = 116.25 pounds. 116.25x7.21 = 838.16 pounds. Ans. MECHANICS. 197 EXAMPLE. What is the weight of a cubic inch of cast iron ? SOLUTION. X 7.21 = .2608 pound. Ans. NOTE. One cubic foot of pure distilled water at a temperature of 39.2 Fahrenheit weighs 62.42 pounds, but the value usually taken in making calculations is 62 pounds. EXAMPLE. What is the weight in pounds of 7 cubic feet of oxygen ? SOLUTION. One cubic foot of air weighs .08073 lb., and the specific gravity of oxygen is 1.1056 compared with air; hence, .08073 X 1.1056 X 7 = .62479 pound, nearly. EXAMPLES FOR PRACTICE. 1. The balls of a steam-engine governor each weigh 5 pounds. If they revolve in a circle whose diameter is 14" at the rate of 80 revolu- tions per minute, what is the centrifugal force of each ball ? Ans. 6.3471b., nearly. 2. The rim of a cast-iron engine fly-wheel has an outside diameter of 15 feet. If the'rim is 8* thick and 12" wide, and the fly-wheel makes 40 R. P. M., what is the centrifugal force of the rim ? Ans. 52,785 lb. 3. If a cubic foot of a certain alloy weighs 678 pounds, what is its specific gravity ? Ans. 10.848. 4. What is the weight of (a) 12.4 cubic inches of lead ? (6) of steel ? (c) of aluminum ? f (a) 5.0964 lb. Ans. -j (t>) 3.52161b. ( (c) 1.1161b. 5. The specific gravity of an alloy of lead and zinc is 8.26; what is the weight of a cubic foot ? Ans. 516.25 lb. WORK; 540. Work is the overcoming of resistance continually occurring along tJie patli of motion. Mere motion is not work, but if a body in motion con- stantly overcomes a resistance, it does work. 541. The measure of work is one pound raised ver- tically one foot, and is called one foot-pound. All work is measured by this standard. A horse going up hill does an amount of work equal to his own weight plus the weight of the wagon and contents plus the frictional resistances reduced to an equivalent 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 198 MECHANICS. resistances equal 400 pounds, then, if the vertical height of the hill is 100 feet, the work done is equal to (1,200 -f 1,200 -f 400) X 100 = 280,000 foot-pounds. Rule 99. In all cases the force (or resistance] multiplied by tJie distance through which it acts equals the work. If a weight be raised, the weight multiplied by the vertical heig/tt of the lift equals the work. 542. The total amount of work is independent of time, whether it takes one minute or one year in which to do it; but in order to compare the work 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 the same time. 543. The common standard to which all work is reduced is the horsepower. One horsepower is 33,000 foot-pounds per minute ; in 'other words, it is 33,000 pounds raised vertically one foot in one minute, or 1 pound raised vertically 33,000 feet in one minute, or any combination that will, when multiplied together, give 33,000 foot-pounds in one minute. 544. Thus, the work done in raising 110 pounds verti- cally 5 feet in one second is a horsepower ; for, since in one minute there are 60 seconds, 110 X 5 X 60= 33,000 foot- pounds in one minute. EXAMPLE. If the coefficient of friction is .3, how many horsepower will it require to draw a load of 10,000 pounds on a level surface, a dis- tance of one mile in one hour ? SOLUTION. 10,000 X -3 = 3,000 pounds = the force necessary to over- come the resistance (resistance of the air is neglected). One mile = pr OQA 5,280 feet; one hour = 60 minutes. Therefore, -^- = 88 feet per minute. Work done = force multiplied by the space = 3,000 X 88 = 264,000 foot-pounds per minute. 264 000 Horsepower = = 8 horsepower. Ans. The abbreviation for horsepower is H. P. MECHANICS. 199 545. Energy is a term used to express the ability of an agent to do work. Work can not be done without motion, and the work that a moving body is capable of doing in being brought to rest is called the kinetic energy of the body. Kinetic energy means the actual visible energy of a body in motion. The work which a moving body is capable of doing in being brought to rest is exactly the same as the kinetic energy developed by it in falling in a vacuum through a height sufficient to give it the same velocity. Rule 1OO. The kinetic energy of a moving body in foot- pounds equals its weight in pounds multiplied by tlie square of its velocity in feet per second, and divided by 64-32. If W represents the weight of the body in pounds, and v the velocity in feet per second, Wv* kinetic energy = -^-^. If a weight is raised a certain height, a certain 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. 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 foot-pounds. Ans. 546. 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 time. EXAMPLE. A body weighing 50 pounds has a velocity of 100 feet per second; what is its kinetic energy? SoLUTioN.-Kinetic energy = - = &JT = 7 ' 773 ' 63 foot ' pounds. Ans. EXAMPLE. In the last example, how many horsepower will be required to give the body this amount of kinetic energy in 3 seconds ? SOLUTION. 1 H. P. = 83,000 pounds raised 1 foot in one minute. If 7,773.63 foot-pounds of work are done in 3 seconds, in one second 200 MECHANICS. there would be done ^ = 2,591.21 foot-pounds of work. One o horsepower = 33,000 ft.-lb. per min. = 33,000 -*- 60 = 550 ft.-lb. per sec. The number of horsepower developed will be 2.591.21 550 = 4. 71 13 H. P. Ans. 547". Potential energy is latent energy ; it is the en- ergy which 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. At a point equal to one-half the height of the fall, the po- tential and kinetic energies are equal. At the end of the first quarter the potential energy was , and the kinetic en- ergy \; at the end of the third quarter the potential energy was \, and the kinetic energy f . 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 tJien, is the energy stored within a body which may be liberated and produce motion, thus generating kinetic energy, and enabling work to be done. 548. 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 MECHANICS. 201 exists in another form. The potential energy in coal was received from the sun, in the form of heat, ages ago, and has lain dormant for millions of years. BELTS. 549. A belt is a flexible connecting band which drives a pulley by its frictional resistance to slipping at the surface of the pulley. Belts are most commonly made of leather or rubber, and united in long lengths by cementing, riveting, or lacing. 550. 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. 551. 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 as follows: Rule 1O1. The length of a belt for open-running pulleys equals 3^ times one-half the sum of the diameters of the Al- leys plus 2 times the distance between the centers of the shaft. Let D = diameter of one pulley; ->,= diameter of other pulley; L = distance between the centers of the shafts; B = length of the belt. Then, = 3( D + D *\ + 2L. 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. By formula, B = 3* ( /? ^ /?> ) + 2 L - Substituting the values given, we have, since 9 ft. 7' = 115*, + 2 x 115 = 311 in. . or 25 ft. llj in. Ans. 202 MECHANICS. 552. To find the width of a single leather belt that will transmit any given horsepower when equal pulleys are used : Rule 1O2. The width of the belt in inches equals 800 times the horsepower to be transmitted divided by the speed of the belt in feet per minute. Let W = width of belt in inches; H = horsepower to be transmitted ; S = speed of belt in feet per minute; then, EXAMPLE. What width of single leather belt is required to transmit 20 horsepower when equal pulleys are used, and the speed is 1,600 feet per minute ? SOLUTION. Formula, W -<-= . Substituting, we have . 800 X 20 i eon 10 inc hes. Ans. 553. To find the number of horsepower that a single leather belt will transmit, its width and speed being given : Rule 1O3. The number of horsepower equals the product of the width in inches and the speed in feet per minute divided by 800. -=^- EXAMPLE. If a 10" single leather belt is to be run at a speed of 1,600 feet per minute, what horsepower will it transmit ? W S SOLUTION. Formula, H = -^^-. oUv Substituting, we have H = 10 ^J)' 600 = 20 horsepower. Ans. 554. When the pulleys are of different diameters, the arc of contact must be considered. To find the number of degrees in the arc of contact, multiply the length of belt in contact on the smaller pulley by 360, and divide the product by the circumference of the pulley, calculating the result to the nearest whole number. The quotient is the arc of contact. MECHANICS. 203 Having found the arc of contact, subtract it from 180 and multiply the result by 3. Add this last result to 800 ; the number thus obtained should be used instead of 800, in rules 102 and 103. EXAMPLE. What should be the width of a single leather belt to transmit 25.24 horsepower at a speed of 1,500 feet per minute, the diameter of the smaller pulley being 24", and the belt having 30" of its length in contact with it ? SOLUTION. Arc of contact = 24 ^^ 6 = 143. (180 - 143) x 3 = 111. 800 + 111 = 911. Using rule 1O2, and 911 instead of 800, . 911 X 25.24 1,500 = 15.33", say 15f. Ans. 555. To find the width of a double belt that will trans- mit the same horsepower as a given single belt, let W^ repre- sent the width of the double belt ; then, Rule 1O4. Multiply the width of a single belt that will transmit the same horsepower by f . Or, }\\ = %W. EXAMPLE. If a single leather belt is 15* in width, and transmits 22 horsepower, what must be the width of a double belt to transmit the same horsepower ? SOLUTION. Applying the rule, 15 X f = 10 in. = width of double belt. Ans. 556. Lacing Belts. Many good methods of fasten- ing the ends of belts are employed, but lacing is generally used, as it is flexible, like the belt, and runs noiselessly over the pulleys. When punching a belt for lacing, use an oval punch, the long diameter of the hole to be parallel with the side of the belt. In a 3-inch belt there should be four holes in each end, two in each row. In a 6-inch belt, seven holes, four in the row nearest the end. A 10-inch belt should have nine holes, five in the row nearest the end. The edges of the holes should not be nearer than of an inch from the sides, and $ of an inch from the ends of the belt. The second row should be at least If inches from the end. Always begin to lace from the center of the belt, and take 204 MECHANICS. care to get the ends exactly in line. The lacing should not be crossed on the side of the belt that runs next to the pulley. Always run the hair side of the belt next to the pulley. HORSEPOWER OF GEARS. 557. To find the horsepower which can be safely trans- mitted by gears whose face, or breadth of tooth, is from 2 to 3 times their pitch : Rule 1O5. The horsepower which can be safely trans- mitted equals the continued product of the square of tlic pitch, the velocity in feet per minute and .01. Let p = the pitch ; s = circumferential speed of a point on the pitch circle in feet per minute; then, H. P. =.01 sp\ EXAMPLE. What horsepower can be safely transmitted by a gear whose pitch diameter is 66.84 in., pitch If in., and which makes 60 R. P. M.? SOLUTION. The velocity which is to be used when applying rule 1O5 is the circumferential speed of a point on the pitch circle. FIG. 102. Hence, 66.84 X 3.1416 = 209.98 in. = circumference of pitch circle = 20Q QR 20Q Qft =^p ft. ~^ X 60 = 1,049.9 = velocity in ft. per min. Now, applying formula, H. P. = .01 X 1,049.9 X 1.75* = 32.1532 horse- power. Ans. MECHANICS. 205 558. When measuring bevel gears, the diameter of the largest pitch circle should be taken, as D, Fig. 102. When calculating their horsepower, use the small, or inner, diameter, as d, Fig. 102. Either diameter rr.ay be used when calculating the revolutions per minute or number of teeth, by rules 86-92, but if the inner or outer diameter of one gear be used, the corresponding diameter of the other gear which meshes with it must also be used. EXAMPLES FOR PRACTICE. 1. How many foot-pounds of work are required to overcome for 7 minutes the friction of the cross-head of an engine which has a stroke of 4 ft., and makes 160 strokes per minute, if the coefficient of friction is 8$, and the average perpendicular pressure is 12,460 Ib. ? Ans. 4,465,664 ft.-lb. 2. In the above example, what horsepower is required ? Ans. 1 9.332 H. P. 3. A cannon ball weighing 500 Ib. is fired with a velocity of 1,600ft. per sec. ; what is its kinetic energy ? Ans. 19,900,497.5 ft.-lb. 4. An open belt drives two pulleys which are, respectively, 42" and 20" in diameter, and 23 ft. apart between their centers; what should be the length of the belt ? Ans. 652J". or 54 ft. 4f '. 5. What width of single leather belting, which has 2 ft. 9" contact on the small pulley, is required to transmit 10 horsepower at a speed of 1,500 ft. per min.? Give width to nearest half inch. Diameter of small pulley 26". Ans. 6". 6. What should be the width of the main belt of a steam engine to transmit 120 horsepower ? The engine runs at 80 R. P. M., the band- wheel is 8 ft.- in diameter, the belt is double, and has a contact of 6 ft. on the smaller pulley, which is 5 ft. in diameter. Take the speed of the belt the same as that of a point on the circumference of the band- wheel. Ans. 36r. 7. A 26" double belt runs at a speed of 2,830 ft. per min., and has a contact of 5 ft. on the smaller pulley ; what horsepower is it trans- mitting ? Diameter of small pulley is 48". Ans. 121.15 H. P. 8. What horsepower can be safely transmitted by a gear whose pitch is 2*', pitch diameter 44.66", and which makes 80 K. P. M.? Ans. 42.24 H. P. MECHANICS. (CONTINUED.) HYDROSTATICS. 559. Hydrostatics treats of liquids at rest under the action of forces. 560. Liquids are very nearly incompressible. A pres- sure of 15 pounds per square inch compresses water less than 26006 of its volume. 561. Fig. 103 represents two cylindrical vessels of exactly the same size. The ves- sel a is fitted with a wooden block of the same size as the cylinder, and can move in it ; the vessel b is filled with water, whose depth is the same as the length of the wooden block in a. Both vessels are fitted with air-tight pistons P, whose areas are each 10 sq. in. Suppose, for convenience, that the weights of the pistons, block, and water be neglected, and that a force of 100 pounds be applied to both pistons. The pressure FIG. 103. 100 per square inch will be = 10 pounds. In the vessel a this pressure will be transmitted to the bottom of the vessel, and will be 10 pounds per square inch ; it is easy to see that there will be no pressure on the sides. In the vessel b an entirely different result is obtained. The pressure on the bottom will be the same as in the other case, that is, 10 pounds per square inch, but, owing to the fact that the 208 MECHANICS. molecules of the water are perfectly free to move, this pres- sure of 10 pounds per square inch is transmitted in every direction with the same intensity ; that is to say, the pres- sure at any point c, d, e, f, g, h, etc. , due to the force of 100 pounds, is exactly the same, and equals 10 pounds per square inch. 562. This may be easily proven experimentally by means of an apparatus like that shown in Fig. 104. Let the area of the piston a be 20 sq. in. ; of b, 7 sq. in. ; of c, 1 sq. in. ; of d, 6 sq. in. ; of e, 8 sq. in., and of /, 4 sq. in. If the pressure due to the weight of the water be neglected, and a force of 5 pounds be applied at c (whose area is 1 sq. in.), a pressure of 5 pounds per square inch will be trans- mitted in all directions, and in order that there shall be no movement, a force of 6 X 5 30 pounds must be applied at d, 40 pounds at e, 20 pounds at /, 100 pounds at a, and 35 pounds at b. If a force of 99 pounds were applied to a, instead of 100 pounds, the piston a would rise, and the other pistons $, c, d, e, and f, would move inwards ; but, if the force applied to a were 100 pounds, they would all be in equilib- rium. If 101 pounds were applied at a, the pressure per square inch would be = 5.05 pounds, which would be transmitted in all directions ; and, since the pressure due to the load on c is only 5 pounds per square inch, it is now evident that the piston a will move downwards, and the pistons b, c, d, c, and/" will be forced outwards. FIG. 104. MECHANICS. 563. The whole of the preceding may be summed up as follows : The pressure per unit of area exerted anywhere upon a mass of liquid is transmitted undiminished in all directions, and acts with the same force upon all surfaces, in a direction at right angles to those surfaces. This law was first discovered by Pascal, and is the most important in hydromechanics. Its meaning should be thoroughly understood. EXAMPLE. If the area of the piston e, in Fig. 104, were 8.25 sq. in., and a force of 150 pounds were applied to it, what forces' would have to be applied to the other pistons to keep the water in equilibrium, assuming that their areas were the same as given before ? 1 ^0 SOLUTION. ITSF 18-182 pounds per square inch, nearly. 8. <) 20 X 18.182 = 363. 64 Ib. = force to balance a. 7 X 18.182 = 127.274 Ib. = force to balance b. 1 X 18.182 = 18.182 Ib. = force to balance c. Ans. 6 X 18.182 = 109.092 Ib. = force to balance d. 4 X 18.182 = 72.728 Ib. = force to balance /. 564. The pressure due to the weight of a liquid may be downwards, upwards, or sideways. 565. Downward Pressure. In Fig. 105 the pres- sure 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 , 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 bot- toms of the vessels a and b are 6 inches square, and that the FIG. 105. part c d in the vessel b is 2 inches square, and that they are filled with water. Then, the weight of one cubic inch of water is p^ pound = 210 MECHANICS. .03617 pound. The number of cubic inches in a = 6 X 6 X 24=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, or 0.868 pound per square inch. The pressure in b, due to the weight contained in the part <:is6x6xlOx .03617 = 13.02 pounds. The weight of the part contained in c d is 2 X 2 X 14 X .03617 = 2.0255 pounds, and the weight per square inch , . 2.0255 of area in c d is - = .5064 pound. 566. According to Pascal's law, this weight (pressure) is transmitted equally in all directions, therefore, an extra weight of .5064 pound is imposed on every square inch of the bottom of bc\ the area of this is 6 X 6 = 36 square inches, and the pressure on it is, therefore, 36 X .5064 = 18.23 pounds, due to water contained in c d; thus, we have a total pressure on bottom of vessel b of 13.02+18.23 = 31.25 pounds, the same as in vessel a. As a result of the above law, there is also an upward pressure of .5064 pound acting on every square inch of the top of the enlarged part be. If an additional pressure of 10 pounds per square inch were applied to the upper surface of both vessels, the total pressure on each bottom would be 31.25+ (6 X 6 X 10) = 31.25 + 360 = 391.25 pounds. If this pressure were to be obtained by means of a weight placed on each piston (as shown in Figs. 103 and 104), we should have to put a weight of 6 X 6 X 10 = 360 pounds on the piston in vessel a, and one of 2 X 2 X 10 = 40 pounds on the piston in vessel b. 567. The General Law for the Downward Pressure upon the Bottom of any Vessel : Rule 1O6. The pressure upon the bottom of a vessel con- taining a fluid is independent of the shape of the vessel, and is equal to the weight of a column of the fluid, the area of whose base is equal to that of the bottom of the vessel, and whose altitude is the distance between the bottom and the upper surface of the fluid plus the pressure per unit of area MECHANICS. 211 upon the upper surface of the fluid multiplied by the area of the bottom of the vessel. 568. Suppose that the vessel b, in Fig. 105, were in- verted, as shown in Fig. 106, the pressure upon the bottom will still be 0.8G8 pound per square inch, but it will require a weight of 3,490 pounds to be placed upon a piston at the upper surface to make the pressure on the bottom 391.25 pounds, instead of a weight of 40 pounds, as in the other case. EXAMPLE. A vessel filled with salt water, having a specific gravity of 1.03, has a circular bottom 13 inches in diameter. The top of the vessel is fitted with a piston 3 inches in diameter, on which is laid a weight of 75 pounds ; what is the total pressure on the bottom, if the depth of the water is 18 inches ? SOLUTION. Applying the rule, the weight of 1 cubic inch of the 62.5 X 1-03 water is - = .037254 Ib. 1,728 13 X 13 X -7854 X 18 X .037254 = 89.01 pounds = the pressure due to the weight of the water. 5 ^- p^-r = 10.61 pounds per square inch, due to the weight on o X X ' 854 the piston. 13 x 13 x .7854 X 10.61 = 1,408.29 pounds. Total pressure = 1.408. 29 + 89.01 = 1,497.3 pounds. Ans. 569. Upward Pressure. In Fig. 107 is represented a vessel of exactly the same size as that shown in Fig. 106. There is no upward pressure on the surface c, due to the weight of the water in the large part c d, but there is an upward pressure on c, due to the weight of the water in the small part b c. The pressure per square inch, due to the weight of the water in b c, was found to be .5064 pound (see Art. 565), the area of the upper surface c of the large part c d is (6 X C) - (2 X 2) = 36 - 4 = 32 sq. in., and FIG. 107. t ne total upward pressure, due to the weight of the water, is .5064 X 32 = 16.2 pounds. 212 MECHANICS. If an additional pressure of 10 pounds per square inch were applied to a piston fitting in the top of the vessel, the total upward pressure on the surface c would be 16.2 + (32 X 10) = 336.2 pounds. 57O. General Law for Upward Pressure: Rule 1O7. The upward pressure on any submerged hori- zontal surface equals the weight of a column of the liquid whose base has an area equal to the area of the submerged surface, and whose altitude is the distance between the sub- merged surface and the upper surface of the liquid plus the pressure per unit of area on the upper surface of the fluid multiplied by the area of the submerged surface. EXAMPLE. A horizontal surface 6 inches by 4 inches is submerged in a vessel of water 26 inches below the upper surface ; if the pressure on the water is 16 pounds per square inch, what is the total upward pressure on the horizontal surface ? SOLUTION. Applying the rule, we get 6 X 4x 26 x .03617 = 22.57 pounds for the upward pressure, due to the weight of the water, and 6 X 4 X 16 = 384 pounds for the upward pressure, due to the outside pressure of 16 pounds per square inch. Therefore, the total upward pressure = 384 + 22.57 = 406.57 pounds. Ans. Lateral (Sideways) Pressure. Suppose the top of the vessel shown in Fig. 108 is 10 inches square, and that the pro- jections at a and b are 1 inch X 1 inch and 10 inches long. The pressure per square inch on the bottom of the vessel, due to the weight of the liquid, is 1 X 1 X 18 X the weight of a cubic inch of the liquid. The pressure at a depth equal to the distance of the upper surface b is 1 X 1 X 17 X the weight of a cubic inch FIG. 108. of the liquid. Since both of these pressures are transmitted in every direction, they are also transmitted laterally (sideways), MECHANICS. 213 and the pressure per unit of area on the projection b is a mean between t/tc two, and equals 1 x 1 X 174- X the weight of a cubic inch of the liquid. To find the lateral pressure on the projection a, imag- ine that the dotted line c is the bottom of the vessel, then the conditions would be the same as in the preceding case, except that the depth is not so great. The lateral pressure per sq. in. on a is thus seen to be 1 X 1 X 1H X the weight of a cubic inch of the liquid. 572. General Law for Lateral Pressure : Rule 1O8. The pressure upon any vertical surface, due to the weight of the liquid, is equal to the weight of a column of the liquid whose base has the same area as the vertical surface, and whose altitude is the depth of the center of gravity of the vertical surface below the upper surface of 'the liquid. Any additional pressure is to be added as in the previous cases. EXAMPLE. A well 3 feet in diameter, and 20 feet deep, is filled with water; what is the pressure on a strip of the wall 1 inch wide, the top of which is 1 foot from the bottom of the well ? What is the pressure on the bottom ? What is the upward pressure per square inch, 2 feet 6 inches from the bottom ? SOLUTION. Applying the rule, the area of the strip is equal to its length ( circumference of well) multiplied by its height. The length = 36 X 3.1416 = 113.1" ; height = 1" ; hence, area of strip = 113.1 X 1 = 113.1 sq. in. Depth of center of gravity of strip = (20 1) ft. -f- | inch, the half width of strip, = 228^ inches. Consequently, the total pressure on the strip = 113.1 X 228.5 X .03617 = 934.75 Ib. Ans. The pressure on each square inch of the strip would be ' = 8.265 pounds, nearly. 36 X 36 X -7854 x 20 X 12 X .03617 = 8,836 pounds = the pressure on the bottom. Ans. 20 - 2.5 = 17.5. 1 x 17.5 X 12 X .03617 = 7.596 pounds = the upward pressure per square inch, 2 feet 6 inches from the bottom. Ans. 573. The effects of lateral pressure are illustrated in Fig. 109. In the figure, e is a tall vessel having a stop-cock near its base, and arranged to float upon the water, as shown. When this vessel is filled with water, the lateral pressures at any two points of the surface of the vessel opposite to each 214 MECHANICS. other are equal. Being equal, and acting in opposite direc- tions, they destroy each Bother, and no motion can result; but if the stop-cock be opened there will be no resistance to the pressure acting on that part, and the water will flow out; at the same time, the pressure on the cor- responding part of the opposite side of the vessel remains, and this, being no longer balanced, causes the vessel to FlG - lc9 - move backwards through the water in a direction opposite to that of the issuing jet. 574. Since the pressure on the bottom of a vessel, due to the weight of the liquid, is dependent only upon the height of the liquid, and not upon the shape of the vessel, it follows FIG. no. that if a vessel has a number of radiating tubes (see Fig. 110) the water in each tube will be on the same level, no matter MECHANICS. 215 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. 575. The above principle explains why city water reservoirs 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 pipe were attached to the nozzle whose length was equal to the vertical distance between the nozzle and the level of the water in the reservoir, the 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 from the level of the street ; what is the pressure of the water per square inch on the hydrant ? SOLUTION. Apply rule 1O6, 1 X 150 x 12 X .03617 = 65.106 pounds per square inch. NOTE. In measuring the height of the water to find the pressure which 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 in. square and 1 ft. high is 62.5 H- 144 = .434 lb., very 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 employed in practical calculations. 576. In Fig. Ill, let the area of the piston a be 1 square inch, of b, 40 square inches. According to Pascal's law, 1 pound placed upon a will balance 40 pounds placed upon b. 216 MECHANICS. 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 10 inches, and -whose altitude must be = of an inch ; that is, a movement of 4 10 inches of the piston a will cause a movement of \ of an inch in the pis- ton b. FIG - m - Here is the old principle of machines: The power multiplied by the distance through which it moves equals the weight multiplied by the distance through whicli it moves. Since, if 1 pound on the piston a represents the power P, the equivalent weight W on b may be obtained from the equation, W X i = P X 10, whence W 40 P= 40 Ib. 577. Another familiar fact is also recognized, for the velocity ratio of Pto W is 10 : , or 40; and, since in any machine the weight equals the power multiplied by the velocity ratio, W Px 40, and when P=l, W = 40. This principle is made use of in the hydraulic press rep- resented in Fig. 112. As the man depresses the lever O, he forces down the piston a upon the water in the cylinder A. The water is forced through the bent tube d into the cylinder in which the large ram, or plunger, C works, and causes it to rise, thus lifting the platform K, and compress- ing the bales. If the area of a be 1 sq. in., and that of c be 100 sq. in., the velocity ratio will be 100. If the length of the lever between the hand and the fulcrum is 10 times the length between the fulcrum and the piston a, the velocity ratio of the lever will be 10, and the total velocity ratio of the hand to the piston C will be 10 X 100 = 1,000. MECHANICS. 217 Hence, a force of 100 pounds applied by the hand will raise 100 X 1,000 = 100,000 pounds. But, if the average move- 112. ment of the hand per stroke is 10 inches, it will require 1 000 ' = 100 strokes to raise the platform 1 inch, and it is again seen that what is gained in power is lost in speed. 578. Applications of this principle are seen in the hydraulic machines used for forcing locomotive drivers on their axles, etc., and for testing the strength of boiler shells. EXAMPLE. A suspended vertical cylinder is tested for the tightness of its heads by filling it with water. A pipe whose inside diameter is i of an inch, and whose length is 20 feet, is screwed into a hole in the upper head, and then filled with water. What is the pressure per square inch on each head, if the cylinder is 40 inches in diameter -nd 60 inches long ? SOLUTION. Area of heads = 40* X .7854 = 1,256.64 sq. in. Pressure per square inch on the bottom head, due to the weight of the water in the cylinder, = 1 X 60 X .03617 = 3.17 pounds- 218 MECHANICS. (i)* X .7854 - .04909 sq. in., 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 sq. in. The pressure per square inch, due to the water in the pipe, is . x .426 = 8.68 pounds per square inch upon the upper head. Ans. The total pressure per square inch on the lower head is 8.68 + 2.17 = 10.85 pounds. Ans. EXAMPLE. In the last example, if the pipe be fitted with a piston weighing of a pound, and a 5-pound weight be laid upon it, what is the pressure per sq. in. upon the upper head ? SOLUTION. In addition to the pressure of .426 pound on the area of .04909 sq. in., there is now an additional pressure upon this area of 5 + } = 5. 25 pounds, and the total pressure upon this area is .426 + 5.25 = 5.676 pounds. The pressure per square inch is X 5.676 = 115.6 pounds. Ans. EXAMPLE. In Fig. 113, the plunger A, is 10" in diameter, and is forced 100 lb. FIG. 113. outwards by means of a small pump B, which supplies the press MECHANICS. 219 cylinder with water. The plunger <7of the pump 7? is ' in diameter. If a force of 100 Ib. be applied to C by means of the lever D, how great a weight can the plunger A raise, if the plunger itself weighs 400 Ib. ? SOLUTION. First find the pressure per sq. in. which the plunger C exerts upon the water. Area of C= (1,)* X .7854 = .19635 sq. in. Since C exerts a pressure of 100 Ib. upon an area of .19635 sq. in., it will exert a pressure of 100 X ~^^~ lb - u P on an area of 1 sq. in. This pressure is transmitted by the water in the tube E to the cylinder F, where it forces the plunger A upwards as the water is forced into F. The pres- sure per sq. in. on the bottom of A is the same as that exerted by C. 100 Hence, the total pressure is 10* X .7854 X -^^f = 40 ' 000 lb - This . lyooo result, less the weight of A, equals the load lifted. Therefore, 40,000 400 = 39,600 lb. Ans. 579. In these examples on the hydraulic press, no allow- ance has been made for the power lost in overcoming the friction between the cup leathers and the plunger; this varies according to the condition of the leathers, and, of course, the smoothness of the plunger; when the leathers are in good condition the loss is about 5$ of the total pressure on the ram ; when the leathers are old, stiff, and dirty the loss may amount to 15^ or more. BUOYANT EFFECTS OF WATER. 58O. In Fig. 114 is shown a 6-inch cube, entirely sub- merged in water. The lateral pressures are equal and in opposite directions. The upward pressure acting on the lower surface of the cube = 6 X 6 X 21 X .03617; the downward pressure acting on the top of the cube = 6 X 6 X 15 X .03617, and the difference = 6 X 6 X 6 X .03617 = the volume of the cube in cubic inches x the weight of one cubic inch of water. That is, the upward pressure 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 220 MECHANICS. acts against gravity ; that is, the water presses the body up with a greater force than it presses it down; consequently, if u body is immersed in a fluid, it will lose in weight an amount equal to the weight of the fluid it displaces. This is called the principle of Archimedes, because it was first stated by him. 58 1 . This principle may be experimentally demonstrated with the beam scales, as shown in Fig. 115. From one scale pan suspend a hollow cylinder of metal /, and below that a solid cylinder a of the same size as the hollow part of the up- per cylinder. Put weights in the other scale pan until they ex- actly balance the two cylinders. If a be im- mersed in water, the scale pan containing the weights will descend, showing that a has lost some of its weight. Now fill / with water, and the volume of water that can be poured into t will equal that dis- placed by a. The scale pan that contains the weights will gradually rise until t is filled, when the scales balance again. If a body be lighter than the liquid in which it is im- mersed, 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 im- mersed 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 MECHANICS. 221 equal, the body will remain stationary, and be in equilibrium in any position or depth beneath the surface of the liquid. 582. 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 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. EXAMPLES FOR PRACTICE. 1. Suppose a cylinder to be filled with water and placed in an upright position. If the diameter of the cylinder be 19", and its total length inside be 26", what will be the total pressure on the bottom when a pipe \" in diameter and 12 ft. 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 Ib. is applied to the water in the pipe, (a) What is the total pressure on the bottom of the cylinder ? (t>) On the upper head? j (a) 16,184 Ib. IS- 1 (b) 15.906 Ib. 4. In example 3, what is the pressure per square inch in the pipe 2* 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" 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. 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 which tends to move the cube upwards ? Ans. 1,687.5 Ib. 222 MECHANICS. PNEUMATICS. 583. Pneumatics is that branch of mechanics which treats of the properties of gases. 584. The most striking feature of all gases is their ex- treme expansibility. If we inject a portion of gas, however small it may be, into a vessel, it will expand and fill that vessel. If a bladder or foot- ball be partly filled with air, and placed under a glass jar (called a receiver), from which the air has been exhausted, the bladder or football will immediately expand, as shown in Fig. 116. The force which a gas always exerts when con- fined in a limited space is called tension. The word tension in this case means pressure, and is only used in this sense in reference to gases. 585. 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 it was proven that air really had weight. A cubic inch of air, under ordinary conditions, weighs .31 grain, nearly. At a temperature of 32 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. In Art. 58O, it was 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 be filled with a gas lighter than air, so that the total weight of the vessel and MECHANICS. gas is less than the air they displace, the vessel will rise, is on this principle that balloons are made. It 586. Since air has weight, it is evident that the enor- mous quantity of air that constitutes the atmosphere must exert a considerable pressure upon the earth. This is easily proven 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. 117. 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 glass equal to about 30 inches. This height will always be the same under the same atmospheric conditions. Now, if the atmosphere has weight, it must press upon the upper sur- face of the mercury in the glass 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 upon a portion of the surface of the mercury in the glass, 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 mercury column is .03617 X 13.6 X 30 = 14.7574 Ib. The actual height of the mercury is a little less than 30 inches, and the actual weight of a cubic inch of dis- tilled water is a little less than .03617 Ib. When these considerations are taken into account, the average weight of the mercurial 224 MECHANICS. column at the level of the sea, when the temperature is 60, is 14.69 lb., or practically 14.7 Ib. 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 lb. upon every square inch of surface. 587. Vacuum. The space between the upper end of the tube and the upper surface of the mercury is called a vacuum, meaning that it is an entirely empty space, and does not contain any substance, solid, liquid, or gaseous. If there were a gas of some kind there, no matter how small the quantity might be, it would expand, filling 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 space is then called a partial vacuum. If the mercury fell 1 inch, so that the column was only 29 inches high, we should say, in ordinary language, that there were 29 inches of vacuum. If it fell 8 inches, we should say that there were 22 inches of vacuum ; if it fell 16 inches, we should say that there were 14 inches of vacuum, and so on. Hence, when the vacuum gauge of a condensing engine shows 26 inches of vacuum, there is enough air in the condenser to produce a pressure of X 14.7 = X 14.7 = 1.96 lb. O\J - o(j per sq. in. In all cases where the mercury column is used to measure a vacuum, the height of the column in inches gives the number of inches of vacuum. Were the column only 5* high, the vacuum would be 5". If the tube had been filled with water instead of mercury, the height of the column of water to balance the pressure of the atmosphere would have been 30 X 13.6 = 408 inches = 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. 588. The barometer is an instrument used for meas- uring the pressure of the atmosphere. There are two kinds MECHANICS. 225 in general use the mercurial barometer and the aneroid barometer. The mercurial barometer is shown in Fig. 118. The principle is the same as the inverted tube shown in Fig. 117. In this case the tube and the cup at the bottom are pro- Ii^pta tected by a brass or iron casing. At the top of j!BJt the tube is a graduated scale which can be read to jiffij TTnru f an mcn by means of a vernier. Attached to the casing is an accurate thermometer for J:H^ determining 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 stand- ard temperature is usually taken at 32 F., at which temperature the height of the mercurial column is 30 inches. Another correction is made for the altitude of the place above sea-level, and a third correction for the effects of capillary attraction. 589. In Fig. 119 is an illustration of an aneroid barometer. These instruments are made in various sizes, from the size of a watch up to 8 or 10 inches in diameter. They consist of a cylindrical box of metal with a top of thin, elastic corrugated metal. The air is exhausted from the box. When the atmospheric pressure increases, the top is pressed inwards, and when it is dimin- 'FIG. us. ished, the top is pressed outwards by its own elas- ticity, aided by a spring beneath. These movements of the cover are transmitted and multiplied by a combination of delicate levers, which act upon an index hand, and cause it to move either to the right or left, over a graduated scale. These barometers are self-correcting (compensated) for vari- ations in temperature. They are very portable, occupying 226 MECHANICS. but a small space, and are so delicate that they are said to show a difference in the atmospheric pressure when trans- ferred from the table to the floor. The mercurial barometer is the standard. 59O. With air, as with water, the lower we get the greater the pressure, and the higher we get the less the pressure. At 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. 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 proven conclusively by the MECHANICS. 227 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. It is known that the height of the earth's atmosphere is at least 50 miles; hence, the air just before reaching the limit must be in an exceed- ingly 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 barometer, the heights must be calculated from the reading. 591. The atmospheric 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 con- siderable 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 Ib. ; but there is an equal pressure beneath the book which 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 tend- ing to crush the book-, so it would, but for the fact that there is a layer of air between each leaf acting upwards and downwards with a pressure of 14.7 pounds per square inch. If two metal plates be made as perfectly smooth and flat as it is possible to get them, and the edge of one be laid upon the edge of the other, so that one may be slid upon the other and thus exclude the air, it will take an immense force, compared to the weights of the plates, to separate them. This is because the full pressure of 14 7 pounds per. square inch is then exerted upon each plate, with no counteracting equal pressure between them. If a piece of flat glass be laid upon a flat surface that has been previously moistened with water, it will require con- siderable force to separate them ; this is because the water helps to fill up the pores in the flat surface and glass, and 228 MECHANICS. thus creates a partial vacuum between the glass and the sur- face, thereby reducing the counter pressure beneath the glass. 592. Tension of Gases. In Fig. 117, the space above the column of mercury was said to be a vacuum, and it was also said that if any gas or air were present, it would ex- pand, its tension forcing the column of mercury downwards. If sufficient gas be admitted to cause the mercury to stand 14.7 at 15 inches, the tension of the gas is evidently ^ = 7.35 Ib. per square inch, since the pressure of the outside air, 14 7 Ib. per square inch, now balances only 15 instead of 30 inches of mercury; that is, it balances only half as much as it would if there were no gas in the tube; therefore, the pres- sure (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 pres- sure 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 up- wards 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 Ib. 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 Ib. If the piston be shoved up, so that the space occupied by the gas is only 9 inches, in- stead 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 pressure increased two-fold, since 29.4 X 2 = 58.8 Ib. The space now occupied by the gas is 9 inches long, whereas, before the piston was moved, it was 36 inches long; as the tube was assumed to be of uniform diameter throughout its q -i 'length, the volume is now = of its original volume, and ob 4 MECHANICS. 229 58 8 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. 593. The law which states these effects is called Mariotte's law, and is as follows : Mariotte's Law. The temperature remaining the same, the volume of a given quantity of gas varies inversely as the pressure. The meaning of the law is this: If the volume of a gas be diminished to , , ^, etc., of its former volume, 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 , , jV, etc., of its original intensity, the volume of 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 movable 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 24 cubic feet; it is then ^ = 8 times its original volume, o and the pressure is of its original pressure, or GO X | = 7 lb., and 24 x 7| = 180, as in the two preceding cases. It will now be noticed that if a gas be enclosed within a con- fined space, and allowed to expand without losing any heat, t lie 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. 230 MECHANICS. 594. 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 1O9. Divide the product of the first, or original, -volume and pressure by the new volu'me; the result will be the new pressure. Or, let / = original pressure ; /, = final pressure ; v = volume corresponding to the pressure />; i>^ = volume corresponding to the pressure /,. Then, A = e EXAMPLE. At the point of cut-off in a steam engine, the amount of steam in the cylinder is 862 cu, in. The pressure at this point is 120 Ib. per sq. in. What will be the pressure of the steam when the piston has reached the end of its stroke, and the volume is 1,800 cu. in. ? SOLUTION. Applying the rule, = 57.47 Ib. per sq. in. Ans. 595. If it is required to determine the volume after a change in the pressure: Rule 11O. Divide the product of the original volume and pressure by the new pressure ; the result will be the new volume. Or, using the same letters as before, *<= . . EXAMPLE. At the commencement of compression, the volume of the steam is 380 cu. in., and the pressure is 18 Ib. per sq. in. At the end of compression, the pressure is 112 Ib. per sq. in. What is the final volume ? OQA -y 1 Q SOLUTION. Applying the rule, ^ = 61.07 cu. in. Ans. EXAMPLE. A vessel contains 10 cu. ft. of air at a pressure of 15 Ib, per sq. in., and has 25 cu. ft. of air of the same pressure forced into it; what is the resulting pressure ? SOLUTION. The original volume = 10 + 25 = 35 cu. ft. The original pressure is 15 Ib. per sq. in. The final volume is 10 cu. ft. Hence, applying rule 1O9, wf = 52/5 Ib. per sq. in. Ans. MECHANICS. 231 It must be remembered that in the preceding examples the temperature is supposed to remain constant. EXAMPLES FOR PRACTICE. 1. A vessel contains 25 cubic feet of gas, at a pressure of 18 Ib. per sq. in. ; if 125 cu. ft. of gas having the same pressure are forced into the vessel, what will be the resulting pressure ? Ans. 108 Ib. per sq. in. 2. The volume of steam in the cylinder of a steam engine at cut-off is 1.35 cu. ft., and the pressure is 85 Ib. per sq. in. The pressure at the end of the stroke is 25 Ib. per sq. in. What is the new volume ? Ans. 4.59 cu. ft. 3. A receiver contains 180 cu. ft. of gas, at a pressure of 20 Ib. per sq. in. ; if a vessel holding 12 cu. ft. be filled from the larger vessel until its pressure is 20 Ib. per sq. in., what will be the pressure in the larger vessel ? Ans. 18J 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", and the pressure of the air within is 5 Ib. per sq. in., what is the total pressure tending to crush the shell ? Ans. 9,873.42 Ib. PNEUMATIC MACHINES. 596. The Air Pump. The air pump is an instrument for removing air from a given space. A section of the principal parts is shown in Fig. 120, and the complete instru- ment in Fig. 121. The closed vessel R is called the receiver, and the space which it encloses is that from which it is desired to remove the air. It is usually made of glass, 232 MECHANICS. 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 communicating with the pump cylinder C by means of the passage / /. The pump piston fits the cylinder accurately, and has a valve V opening up- wards. Where the passage / / joins the cylinder, is another valve F, also opening upwards. When the piston is raised, the valve V closes, and, since no air can get into the cylin- der 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 t t 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. 120, 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 but- ton 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 in Fig. 121, communication between receiver and pump is made by means of the tube t. MECHANICS. 233 597. Degrees and Limits of Exhaustion. Sup- pose that the volume of R and / together is four times that of C, Fig. 121, 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 cylinder. 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 /, and the total space R -\- t + C is | ; hence, 200 X -| 100 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 / would be (200 X |) X i =-200 X (-1)' = 200 X If = 128 grains. At the end of the third stroke the weight would be [200 X (-J)*] X | = 200 X (|) 3 = 200 X T Q 2T= 103 - 4 grains. At the end of strokes the weight would be 200 X (f )". It is evident that // is impossible to remove all of the air tJiat is contained in R and t by this method. It requires an exceedingly good air pump to reduce the tension of the air in R to -fa of an 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. 598. Magdeburg Hemispheres. By means of the two hemispheres shown in Fig. 122, it can be proven that the atmosphere presses upon a body equally in all directions. They were invented by Otto Von Guericke, of Magdeburg, and are called the Magdeburg hemispheres. One of the hemispheres is provided with a stop-cock, by which it can be screwed on to an air pump. The edges fit accurately, and are well greased, so as to be air-tight. As long as the hemispheres contain air, they can be separated with no difficulty; but when the air in the interior is pumped out by means of an 234 MECHANICS. air pump, they can be separated only with great difficulty. The force required to separate them will be equal to the area of the largest circle of the hemisphere in square inches multiplied by 14.7 pounds. This force will be the same in whatever position the hem- isphere may be held, thus proving that the pressure of air upon it is the same in all directions. NOTE. A theoretically perfect vacuum is sometimes called a Torricellian vacuum. 599. The Weight Lifter. The pressure of the at- mosphere is very clearly shown by means of an apparatus like that illustrated in Fig. 123. Here a cylinder fitted with a piston is held in sus- pension by a chain. At the top of the cylinder is a plug A which can be taken out. This plug is removed, the piston pushed up (the force necessary being equal to the weight of the piston and rod B) until it touches the cylinder head. The plug is then screwed in, and the piston will remain at the top until a weight has been hung on the rod equal to the number of pounds obtained by multiplying the number of square inches in the piston by 14.7 pounds, and subtracting therefrom the weight of the piston and rod in pounds. If a force sufficiently great were em- ployed to pull the piston downwards, and then any less weight were attached, as shown in Fig. 123, the piston would ascend and carry the weight up with it. 6OO. Suppose the weight to be re- moved, and the piston to be supported, say midway of the length of the cylinder. Let the plug be removed, and air admit- FIG. 123. te d above the piston, then screw the plug back into its place; if the piston be shoved upwards, MECHANICS 235 the further up it goes the greater will be the force neces- sary to push it, on account of the compression of the air. If the piston is of large diameter it will also require a great force to pull it out of the cylinder, as a little consid- eration will show. For example, let the diameter of the piston be 20 inches, the length of the cylinder 36 inches, and the weight of the piston and rod 100 pounds. If the piston is in the middle of the cylinder, there will be 18 inches of space above it, and 18 inches of space below it. The area of the piston is 20 s X .7854 = 314.16 square inches, and the at- mospheric pressure upon it is 314.16X14.7 = 4,618 lb., nearly. In order to shove the piston upwards 9 inches, the pressure upon it must be twice as great, or 9,236 pounds, and to this must be added the weight of the piston and rod, or 9,236 -f 100 = 9,336 lb. The force necessary to cause the piston to move upwards 9 inches would then be 9,336 4,618 = 4,718 lb. Now, suppose the piston to be moved down- wards until it is just on the point of being pulled out of the cylinder. The volume above it will then be twice as great as before, and the pressure one-half as great, or 4,618 -f- 2 = 2,309 lb. The total upward pressure will be the pressure of the atmosphere less the weight of the piston and rod, or 4,618 100 = 4,518 lb., and the force necessary to pull it downwards to this point will be 4,518 2,309 = 2,209 lb. '6O1. Air Compressor*. For many purposes com- pressed air is preferable to steam or other gas for use as a motive power. In such cases air compressors are used to compress the air. These are made in many forms, but the most common one is to place a cylinder, called the air cylinder, in front of the cross-head of a steam engine, so that the piston of the air cylinder can be driven by at- taching its piston rod to the cross-head, in a manner similar to a steam pump. A cross-section of the air cylinder of a compressor of this kind is shown in Fig. 124, in which A is the piston, and B is the piston rod, driven by the cross-head of a steam engine, not shown in the figure. Both ends of the lower half of the cylinder are fitted with inlet valves D 236 MECHANICS and D' which allow the air to enter the cylinder, and both ends of the upper half are fitted with discharge valves F and FIG. 124. F' which allow the air to escape from the cylinder after it has been compressed to the required pressure. 6O2. Suppose the piston A to be moving in the direction of the arrow, then all the inlet valves D in the lower half of the left-hand end of the cylinder from which the piston is moving will be forced inwards by the pressure of the atmosphere, which overcomes the resistance of the spring C tending to keep the valve on its seat, thus allowing the air to rush into the cylinder. On the other side of the piston the air is being compressed, and, consequently, it forces the inlet valves D' in the lower half of the right-hand end of the cylinder to their seats against the resistance of their springs S. In the upper half of the right-hand end of the cylinder the discharge MECHANICS. 237 valves P are opened by the compressed air against the resist- ance of the springs ', and in the upper half of the left-hand end of the cylinder the discharge valves F are pressed against their seats by the springs E, the tension of these springs being so adjusted that the valves can not be forced from their seats until the air on the other side has been compressed to the required pressure per square inch. For example, suppose it is de- sired to compress the air to 59 pounds per square inch, and we wish to find at what point of the stroke the dis- charge valves will open, they having been set for this pressure. Now, 59 pounds per square inch = 4 atmos- pheres very nearly; hence, the vol- ume must be of the volume at the beginning of the stroke, or the valves will open when the piston has trav- eled of its stroke. The air, after being discharged from the cylinder, passes out through the discharge pipe H, and from thence is conveyed to its destination. 6O3. Hero's Fountain. Hero's fountain derives its name from its inventor, Hero, who lived at Alexandria 120 B. C. ; it is shown in Fig. 125. It depends for its operation upon the elastic FIG 125 - properties of air. It consists of a brass dish A, and two glass globes B and C. The dish communicates with the lower part of the globe C by means of a long tube D, and another tube E connects the two globes. A third tube passes through the dish A to the lower part of the globe B. This last tube being taken out thegiobe B is partially filled with water, the tube is then replaced, and water is poured into the dish. The 238 MECHANICS. water flows through the tube D into the lower globe, and ex- pels the air, which is forced into the upper globe. The air thus compressed acts upon the water and makes it jet out as represented in the figure. " Were it not for the resistance of the atmosphere and friction, the water would rise to a height above the water in the dish equal to the difference of the level of the water in the two globes. 6O4. The Siphon. The action of the siphon illustrates the effect of atmospheric pressure. It is simply a bent tube of unequal branches, open at both ends, and is used to con- vey a liquid from a higher point to a lower, over an in- termediate point higher than either. In Fig. 126, 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 B, as long as the level of the water in B is below the FIG. 120. 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 upon the surfaces 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 which is im- mersed in the water upon which the pressure is exerted. The atmospheric pressure opposed to the weight of the longer column of water will, therefore, be more resisted than that opposed to the weight of the shorter column; consequently, the pressure exerted upon the shorter column MECHANICS. 239 will be greater than that upon the longer column, and this excess pressure will produce motion. 605. Let A the area of the tube ; h = D C = the vertical distance between the surface of the water in />, and the highest point of the center line of the tube ; /*, = 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 h^ and the resultant atmospheric pressure tending to force the water up the short column is 14.7 X A .03617 X A h v . The weight of the water in the long column is .03617 A //, and the resultant atmospheric pressure tending to force the water up the long column is 14. 7 A .03617^4 h. The difference between these two is (14.7 A .03617 A //,) (14.7 A .03617 A //) = .03617 A (h //,). But // //,= E D the difference between the levels of the water in the two vessels. In the above, h and //, 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. 606. The Injector. A section of an injector is shown in Fig. 127. There are many different kinds of these in- struments, but the principle is the same in all. They are used for lifting water from a point below the discharge orifice, and forcing it into the boiler of a steam engine or locomotive. They depend for their action upon the creation of a partial vacuum by the action of steam. The valve A is opened by turning the hand wheel S, and the steam enters from F into D, and flows through the tubes C, K, M, and the outlet N, into the boiler. The valve B is opened by turning the hand wheel W, The steam, flowing through C with a high velocity, drives out the air. The air in the 240 MECHANICS. chamber G G and in the suction pipe P expands and flows into C through the conical orifice //, thus creating a partial vacuum in the chamber G and pipe P. The atmospheric pressure on the water causes it to rise in pipe P t and flow into G, and from thence into C, through the conical orifice //, whence the steam drives it through K, Jlf, and N into the boiler. 6O7. The method of operating the injector is as follows: The water valve B is opened; next the small valve R is opened by turning the handle J. The steam then flows into the passage , which connects with the chamber D through the conical tube C, and creates a partial vacuum in G, as before described. The quantity of steam admitted by the valve R not being sufficient to force all of the water which flows through the orifice H into the boiler, the water will accumulate in the chamber 7", raise the valve Z, and flow down through u and out at the overflow outlet O. As soon as the water appears at O the valve A is opened as far as possible, and the valve R closed. If the water still flows out at O close the valve B slightly until it stops. An in- jector will lift water, according to temperature and steam pressure, from 6 to 26 feet. MECHANICS. 241 PUMPS. 6O8. The Suction Pump. A section of an ordinary suction pump is shown in Fig. 128. Suppose the piston 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 vacuum behind it, and the atmos- pheric pressure upon the surface of the water in the well causes it to rise in the pipe P for the same reason that the mercury rises in the barometer tube. The water rushes up the pipe and lifts the valve V, filling the empty space in the cylinder B displaced by 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 bottom of the cylinder, and also the pipe P. The instant that the pis- ton begins its down stroke, the water in the chamber B tends to fall back into the well, and its weight forces the valve V to its seat, thus preventing any downward flow of the water. As the piston descends the water must give way to it, and, since the valve V is closed, the valves , 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 downward stroke, the weight of the water E above closes the valves ;/, All the water resting on the top of the piston is then lifted with the piston on its upward stroke, and discharged through the spout A y the valve V again opening, and the water filling the space below the piston as before. 242 MECHANICS. 6O9. It is evident that the distance between the piston when at top of its stroke and the surface of the water in the well must not exceed 34 feet, the highest column of water which the pressure of the atmosphere will sustain, since, otherwise, the water in the pipe would not rise up and fill the cylinder as the piston ascended. In practice, this dis- tance 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 weight, which, of course, acts downwards. There are many varie- ties of the suction pump, differing principally in the valves and piston, but the principle is the same in all. 61 0. The Lifting Pump. A section of a lifting pump is shown in Fig. 129 These pumps are used when water is to be raised to greater heights than can be done with the ordinary suction pump. As will be perceived, it is essentially the same as the pump previously described, except that the spout is fitted with a cock and has a pipe attached to it, leading to the point of discharge. If it is desired to discharge the water at the spout, the cock may be opened; otherwise, the cock is closed, and the water is lifted by the piston up through the pipe P to the point of discharge, the valve C preventing it from falling back into the pump, FIG 129. and the valve V preventing the water in the pump from falling back into the well. 611. Force Pumps. The force pump differs from the lifting pump in several important . particulars, but chiefly in the fact that the piston is solid ; that is, it has MECHANICS. 243 no valves. A section of a suction and force pump is shown in Fig. 130. The water is drawn up the suction pipe as before, when the piston rises; but when the piston re- verses, the pressure on the water caused by the descent of the piston opens the valve V and forces the water up the de- livery pipe P' . When the piston again be- gins its upward FIG. 130. movement, the valve V is closed by the pressure of the water above it, and the valve V is opened by the pres- sure of the atmosphere on the water below it, as in the previous cases. For an arrangement of this kind, it is not necessary to have a stuffing-box. The water may be forced to almost any desired height. The force pump differs again from the lifting pump in respect to its piston rod, which should not be longer than is absolutely necessary in order to prevent it from buckling, while in the lifting pump the length of the piston rod is a matter of indifference. 612. Plunger Pumps. When force pumps are used to convey water to great heights, the pressure of the water in the cylinder becomes so great that it becomes extremely difficult to keep the water from leaking past the piston, and the constant repairing of the piston packing becomes a nuisance. To obviate this difficulty the piston is made very long, as shown in Fig. 131, and is then called the plunger. The suction valve in this case consists of two clack valves inclined to each other and resting upon a square pin A ; they are prevented from flying back too far during the up stroke of the plunger by the two uprights /, /. During the 244 MECHANICS. down stroke of the plunger, the valves at A are closed and the valve B in the delivery pipe is open A little air is always carried into the cylinder of a pump with the entering water. In force pumps this fact becomes a serious consideration, since after repeated strokes the air accumulates, and during the down stroke of the plunger it is com- pressed. After a time it would become sufficient to entirely pre- vent the water from entering through the suction valve, the pressure on the top of the valve being greater than that of the atmosphere below. In the pump shown in the figure, the plunger is a trifle smaller than the cylin- der, and the air collects around the plunger below the stuffing- box. To remove this air a narrow passage C (shown by the dotted FIG - 131 - lines), that can be closed at its upper end by the cock Z>, connects the interior of the pump with the atmosphere when the cock is open. It is evident that this cock must not be opened except dur- ing the down stroke of the plunger, for, if it were open during the up stroke, the pressure below the plunger being less than the pressure of the atmosphere above > the air would rush in instead of being expelled. AIR CHAMBERS. 613. In order to obtain a continuous flow of water in the delivery pipe, with as nearly a uniform velocity as pos- sible, an air chamber is usually placed on the delivery pipe of force pumps as near to the pump cylinder as the con- struction of the machine will allow. The air chambers are usually pear-shaped, with the small end connected to the MECHANICS. 245 pipe. They are filled with air, which the water compresses during the discharge. During the suction, the air thus compressed expands and acts as an accelerating force upon the moving column of water, a force which diminishes with the expansion of the air and helps to keep the velocity of the moving column more nearly uniform. An air chamber is sometimes placed upon the suction pipe. These air cham- bers not only tend to promote a uniform discharge, but also to equalize the stresses upon the pump and prevent shocks due to the incompressibility of water. They subserve the same purpose on pumps that a fly-wheel does on the steam engine. Unless the pump moves very slowly, it is absolutely necessary to have an air chamber on the delivery pipe. STEAM PUMPS. 614. Steam pumps are force pumps operated by steam acting upon the piston of a steam engine directly connected to the pump, and in many cases cast with the pump. A section of a double-acting steam pump, showing the steam cylinders, with other details, is illustrated in HWp CT is the steam niston. and R the niston rod. and Fig. water 132. Here lers, wim otner aetans, is mustraiea in r j is the steam piston, and R the piston rod, 246 MECHANICS. which is secured at its other end to the pump plunger P. F is a partition cast with the cylinder, which prevents the water in the left-hand half from communicating with that in the right-hand half of the cylinder. Suppose the piston to be moving in the direction of the arrow. When the pis- ton has arrived at the end of its stroke, the water space in the left-hand half of the pump cylinder will have been in- creased by an amount equal to the area of the cross-section of the plunger, multiplied by the length of the stroke, and the volume of the right-hand half of the cylinder will have been diminished by a like amount. In consequence of this, a volume of water in the right-hand half of the cylinder equal to the volume displaced by the plunger in its forward movement will be forced through the valves f 7 ,' V, through the orifice D, into the air chamber A, and then discharged through the delivery pipe H. By reason of the partial vacuum in the left-hand half of the pump cylinder, owing to this movement of the plunger, the water will be drawn from the reservoir through the suction pipe C into the chamber K K, lifting the valves 5', S' and filling the space displaced by the plunger. During the return stroke the water will be drawn through the valves S, S into the right-hand half of the pump cylinder, and at the same time water will be dis- charged from the left-hand half through the valves F, F/out through the pipe //, as before. Each one of the four suc- tion and four discharge valves is kept to its seat, when not working, by light springs, as shown, 615. There are many varieties and makes of steam pumps, the majority of which are double-acting. In many cases two steam pumps are placed side by side, having a common delivery pipe. This arrangement is called a duplex pump. It is usual to so set the steam pistons of duplex pumps that when one is completing the stroke, the other is in the middle of its stroke. A double-acting duplex pump made to run in this manner, and having an air chamber of sufficient size, will deliver water with nearly a uniform velocity. MECHANICS. 247 In mine pumps for forcing water to great heights, the plungers are made solid, and in most cases extend through the pump cylinder. In many steam pumps, pistons are used instead of plungers, but when very heavy duty is required, plungers are preferred. STRENGTH OF MATERIALS. 616. When a force is applied to a body, it changes either its form or its volume. A force, when considered with reference 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 lengthening of the rod, which increases until the internal stress or resistance is in equilibrium with the external weight. 617. Stresses may be classified as follows: Tensile, or pulling stress. Compressive, or pushing stress. Transverse, or bending stress. Shearing, or cutting stress. Torsional, or twisting stress. 618. A unit stress is the amount of stress on a unit of area, and may be expressed either in pounds per square inch, or in tons per square foot; or, it is the load per square inch or square foot on any body. Thus, if 10 tons are suspended by a wrought-iron bar which has an area of 5 square inches, the unit stress is 2 tons per square inch, because = 2 tons. o 619. 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 yf^th of its length, or 1 per cent. 62O. Elasticity is the power which a body has of returning to its original form after the external force on 248 MECHANICS. it is withdrawn, providing the stress has not exceeded the elastic limit. Consequently, we see from this that all material is lengthened or shortened when subjected to either tensile or compressive stress, and the change of the length is directly proportional to the stress, within the elastic limit. 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. 621. The measure of elasticity of any material is the change of length under stress within the elastic limit. 622. The elastic limit is that unit stress under which the permanent set becomes visible. The elasticity of wrought iron is practically the same as that of steel; that is, each material will change an equal amount of length under the same stress within the elastic limits. The elastic limit of steel is higher than that of wrought iron; consequently, the former will lengthen or shorten more than the latter before its elasticity is injured. TENSILE STRENGTH OF MATERIALS. 623. The tensile strength of any material is the resistance offered by its fibers to being pulled apart. The tensile strength of any material is proportional to the area of its cross-section. 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 its strength per square inch, as given in the following table under the heading "Working Stress." 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. MECHANICS. 249 624. In the following table is given the average break- ing and working tensile stress of different materials: TABLE 17. Material. Breaking Stress in Pounds per Square Inch. Working Stress in Pounds per Square Inch. Timber Cast Iron Wrought Iron Steel 10,000 16,000 50,000 70 000 600 to 1,200 1,500 to 3,500 5,000 to 12,000 6 000 to 13 000 The above table shows that the tensile breaking strength of cast iron is 16,000 pounds per square inch of cross-section, and that the working strength is from 1,500 to 3,500 pounds per square inch of cross-section. 625. In machinery, such as steam engines, where the parts are subjected to shocks, or are alternately compressed and extended, it is not safe to subject cast iron to a stress of more than 1,500 pounds per square inch of section, wrought iron to more than 5,000 pounds per square inch of section, or steel to more than 6,000 pounds per square inch of section. But in structures in which the strains are constantly in one direction, as is the case with steam boilers, wrought iron may be strained with from 6,000 to 8,000 pounds per square inch of section, or steel with from 8,000 to 10,000 pounds per square inch of section. Consequently, strict attention must be given to the nature of the load the given structure has to bear, and fix the working stress accordingly. NOTE. For structures on which the load is applied suddenly, use the smaller working stresses given in the table, ana for those on which the load is applied gradually, use the larger working stresses. 250 MECHANICS. RULES AND FORMULAS FOR TENSILE STRENGTH. 626. Let W safe load in pounds; A area of minimum cross -section; 5" = working stress in pounds per square inch, as given in the foregoing table. Rule 111. The working load in pounds for any bar subjected 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. That is, W=A S. EXAMPLE. A bar of good wrought iron which is 3" square is to be subjected to a steady tensile stress ; what is the maximum load that it should carry? SOLUTION. From what has been said above in regard to the materials and to the nature of the load, it will be safe in this case to use a working stress of 12.000 pounds per square inch. Applying the rule, we have = 108,000 pounds. Ans. Rule 112. The minimum sectional area of any bar subjected 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. W That is, A = -^. o EXAMPLE. What should be the area of a wrought-iron bar to carry a steady load of 108,000 pounds, if it is to resist a" tensile stress of 12,000 pounds per square inch ? SOLUTION. Applying the rule, Rule 1 1 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 That is, S = . MECHANICS. 251 EXAMPLE. A bar of wrought iron 3" square, subjected to tensile stress, carries a load of 108,000 pounds; what is the stress per square inch : SOLUTION. Applying the rule, S = 1 f' (X f ) = 12,000 Ib. per sq. in. Ans. o X o CHAINS. 627. 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 which 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 1300 F. at the least. 628. Rules and Formulas for the Strength of Chains : Let W '= safe load in pounds; D = diameter of the iron, in inches, from which the links are made. Rule 114. The safe load in pounds of a stud-link wrought-iron chain is equal to 18,000, multiplied by the square of the diameter of the iron from which the links are made. That is, W= 18,000 D\ EXAMPLE. What is the maximum load that should be carried by a stud-link wrought-iron chain, if its links are made from J-inch round iron SOLUTION. Applying the rule, W = 18,000 D*. Substituting the value of Z?, we have W= 18,000 x J X |- = 10,125 pounds. Ans. 252 MECHANICS. Rule 115. The safe load in pounds of a close-link wrougJit-iron chain is equal to 13,000, multiplied by the square of the diameter of the iron from which the links are made. That is, W= 12,000 D\ EXAMPLE. What is the maximum load that should be carried by a close-link wrought-iron chain, if its links are made from f-inch round iron? SOLUTION. Applying the rule, W '= IS.OOOZ? 2 . O O Substituting the value of Z>*, we have W= 12,000 X -j x T = pounds. Ans. HEMP ROPES. 629. The strength of hemp ropes does not depend so much upon the quality of the material and the cross-section of the rope, as upon the method of manufacture and the amount of twisting. The ropes in common use are three-strand, shroud-laid rope, and hawser or cable-laid rope. The strongest ropes are three-strand shroud-laid, made without tar. Ropes made with tar are less flexible, and are reduced in strength about 25 per cent., but have better wearing qualities. 630. Rules and Formulas for the Strength of Hemp Ropes : Let W= maximum working load in pounds; C = circumference of rope in inches. Rule 116. The maximum -working load in pounds that should be allowed on any liemp rope is equal to the square of the circumference of the rope, multiplied by 100. That is, IV =100 C\ EXAMPLE. What is the maximum load in pounds that should be carried by a hemp rope which has a circumference of 8 inches ? SOLUTION. Substituting the value of Cin the formula, W= 100 x 8* = 6,400 Ib. Ans. MECHANICS. 253 Rule 117. The circumference of any Jiemp rope is equal to the square root of the maximum working load in pounds which it is capable of carrying, multiplied by . 1. That is, C=.l /IF. EXAMPLE. A maximum working load of 1,000 pounds is to be carried by a hemp rope; what should be the circumference of the rope ? SOLUTION. Applying the rule, C = .1 yXOOO = 3.16". Ans. When measuring ropes, the circumference is sought instead of the diameter, because the ropes are not round and the circumference therefore is not 3.1416 times the diameter. For three strands the circumference is about 2.86 d\ for seven strands, 3 d. 631. The above formulas are very convenient for use, and easily remembered, but it is well to remark that the values thus given apply to ropes of a good average quality. WIRE ROPES. 632. Wire rope is made of iron and steel wire. It is stronger than hemp rope, and, to carry the same load, is of smaller diameter. In substituting steel for iron rope, the object in view should be to gain an increase of wear from the rope, rather than to reduce the size. A steel rope to be serviceable should be of the best obtain- able quality, because ropes made from low grades of steel are inferior to good iron ropes. 633. Formulas for the Strength of Wire Ropes : Let W-= maximum of working load in pounds; C = circumference of rope in inches. Rule 118. The maximum working load in pounds that should be allowed on any iron wire rope is equal to the square of the circumference of the rope in inches, multiplied by 600. That is, W= 600 C 2 . EXAMPLE. What is the maximum load in pounds that should be carried by an iron wire rope whose circumference is 4.} inches ? SOLUTION. Applying the formula. ^=600X4.5^ = 12,1501^ Ans. 254 MECHANICS. Rule 119. The circumference of any iron wire rope in inches is equal to the square root of the maximum working load in pounds, multiplied by .0408. That is, G89 ' ( (b) $37,935. (50) How many square feet of heating surface are in the tubes of a boiler having GO 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. (51) Suppose that in one hour 10 pounds of coal are burned per square foot of grate area in a certain boiler, and that 9 pounds of water are evaporated per pound of coal burned. If the grate area is 30 square feet, how many pounds of water would be evaporated in a day of 10 hours? Ans. 27,000 Ib. (52) How many feet does the piston of a steam engine pass over in a week of 6 days, running 8 hours a day, if the length of the stroke of the engine is 1^- feet, and the number of revolutions per minute 160? Ans. 1,468,800 ft. (53) A number of boilers are constantly fed by 3 pumps; the first delivers 1J gallons per stroke, and runs at 75 strokes per minute; the second delivers $ of a gallon per stroke, and runs at 115 strokes per minute; the third delivers If gallons per stroke, and runs at 96 strokes per minute. How many gallons of water are fed to the boilers per hour? Ans. 21,022.5 gal. NOTE. If in the following examples there be a remainder, carry the quotient to four decimal places. (54) Divide the following: (a) 962,842 by 84; (b) 39,728 by 63; (c) 29,714 by 108; (d) 406,089 by 135. { (a) 11,462.4048. ; (b) 630.6032. " (c) 275.1296. (d) 3,008.0667. 280 ARITHMETIC. (55) Solve the following: (a) 35- A; (b) ^-5-3; (c) -* 9; (<*) W-*- A; (') Ans. - (a) 112. (*) A- (0 H- 375 X* ( (a) 1.75. Ans. ] (*) 1.75. ( (c) .5. T 5 ir - 125' (56) Solve the following: (a) .875-|; (b) *-s-.5; ' (c) (57) Solve the following by cancelation : (a) (72 X 48 X 28 X 5) - (84 X 15 X 7 X 6). (b) (80 X 60 x 50 X 16 X 14) (70 X50 X 24 x 20). Ans (b) 32. (58) Find the values of the following expressions: la) - (M - (,) ' f ' l 1.25X20X3 87 + 88 ' (a) 37*. Ans. (b) .75. (c) 210f 459 + 32 (59) 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. (60) A keg of |- x 2| inches boiler rivets weighs 100 pounds, and contains 133 rivets. What is the weight of each rivet ? Ans. .75 + Ib. (61) The distance around a wheel equals 3.1416 times its diameter, or the distance across it. If the distance around a fly-wheel is 56.5488 feet, what is the diameter of a wheel half as large ? Ans. 9 ft. (62) If 980,000 bricks are required to build an engine house, how many days will it take for 6 teams to draw them, each team drawing 8 loads a day, and there being 1,600 bricks to a load ? Ans. 12.76 + days. ARITHMETIC. 281 (63) If a mechanic earns $1,500 a year for his labor, and his expenses are $968 per year, in what time can he save enough to buy 28 acres of land at $133 an acre? Ans. 7 years. (64) The numerator of a fraction is 28, and the value of the fraction ; what is the denominator ? Ans. 32. (65) From 1 plus .001 take .01 plus .000001. Ans. .990999. (66) A freight train ran 305 miles in one week, and 3 times as far, lacking 246 miles, the next week; how far did it run the second week ? Ans. 849 miles. (6-7) If the driving wheel of a locomotive is 16 ft. in circumference, how many revolutions will it make in going from Philadelphia to Pittsburg, the distance of which is 354 miles, there being 5,280 feet in one mile ? Ans. 116,820 rev. (68) How many inches in .875 of a foot ? Ans. 10 in. (69) What decimal part of a foot is T 3 T of an inch ? Ans. .015625 ft. (70) If water be conducted from a tank by 7 lengths of gas pipe coupled together, each length being 12 feet 6 inches long, of an inch being added at each of the joints for coupling together, how far from the tank is the water discharged ? The first length screws into the tank of an inch. Ans. 87.8125 ft. (71) If by selling a carload of coal for $82.50, at a profit of $1.65 per ton, I make enough to pay for 72.6 ft. of fencing at $.50 a foot, how many tons of coal were there in the car ? Ans. 22 tons. (72) The connection between an engine and boiler is made up of 6 lengths of pipe, three of which are 14 feet 5 inches long, two 12 feet 6 inches long, and one 8 feet 10 inches long. If the pipe weighs 10 pounds per foot, what is the total weight of the pipe used ? Ans. 809.375 Ib. (73) Four bolts are required, 2, 6, 3^, and 4 inches long. How long a piece of iron will be required from 282 ARITHMETIC. which to cut them, allowing T \ of an inch to each bolt for cutting off and finishing ? Ans. 18 T 3 g-. (74) A double belt of a certain width can transmit 64 horsepower. How many horsepower can two single belts of the same width transmit, when running under the same con- ditions, supposing that the double belt is capable of trans- mitting y as much power as one of the single belts ? Ans. 90.3 H. P. (75) The lengths of belting required to connect four countershafts with the main line shaft were 18 feet 6 inches, 16 feet 9 inches, 22 feet 2 inches, and 20 feet 8 inches. How many feet of belting were required ? Ans. 78 ft. (76) In a steam-engine test of an hour's duration, the horsepower developed was found to be as follows, at 10- minute intervals: 48.63, 45.7, 46.32, 47.9, 48.74, 48.38, 48.59. What was the average horsepower ? Ans. 47.75+ H. P. ARITHMETIC. (SEE ARTS. 175-339.) (77) What is 25$ of 8,428 Ib. ? Ans. 2,107 Ib. (78) What is 1$ of $100 ? Ans. $1. (79) What is \% of $35,000 ? Ans. $175. (80) What per cent, of 50 is 2 ? Ans. 4$. (81) What per cent, of 10 is 10 ? Ans. 100$. (82) Solve the following: (a) Base $2,522, and percentage = $176.54. What is the rate ? (b} Percentage^ 16.96, and rate = 8$. What is the base? (c) Amount =216.7025, and base = 213.5. What is the rate ? (d) Difference = 201.825, and base = 207. (83) 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 running? Ans. 4,782.61 Ib. (84) 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 does he save a year? Ans. $441.75. (85) If 37|$ of a number is 961.38, what is the number? Ans. 2,563. 6a 284 ARITHMETIC. (86) A man owns f of a manufacturing plant. 30# of his share is worth $1,125. What is the whole property worth ? Ans. $5,000. (87) What number diminished by 35$ of itself equals 4,810? Ans. 7,400. (88) The volume of the clearance in a steam-engine cylinder is found to be 18.3cu. in., and the volume of the cylinder, neglecting the clearance, 254.5 cu. in. What per- centage of the cylinder volume is the clearance ? Ans. 7.2$, nearly. (89) The distance between two stations on a certain rail- road is 16.5 miles, which is 12$ of the entire length of the road. What is the length of the road ? Ans. 132 miles. (90) 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 revolutions per minute. (91) Reduce 4 yd. 2 ft. 10 in. to inches. Ans. 178 in, (92) Reduce 3,722 in. to higher denominations. Ans. 103 yd. 1 ft. 2 in. (93) How many seconds in 5 weeks and 3.5 days ? Ans. 3,326,400 sec. (94) Reduce 764,325 cu. in. to cu. yd. Ans. 16 cu. yd. 10 cu. ft. 549 cu. in. (95) How many gallons of water can be put into a tank holding 4 bbl. 10 gal. 3 qt. ? Ans. 136| gal. (96) A carload of coal weighed 16 T. 8 cwt. 75 Ib. How many pounds did this amount to ? Ans. 32,875 Ib. (97) Reduce 25,396 Ib. to higher denominations. Ans. 12 T. 13 cwt. 96 Ib. (98) Reduce 25,396 pt. to higher denominations. Ans. 100 bbl. 24 gal. 2 qt. (99) What is the sum of 2 yd. 2 ft. 3 in. ; 4 yd. 1 ft. 9 in. ; 2 ft. 7 in. ? Ans. 8 yd. 7 in. ARITHMETIC. 285 (100) What is the sum of 3 gal. 3 qt. 1 pt. ; 6 gal. 1 pt. ; 4 gal. 8 qt. 5 pt. ? Ans. 16 gal. 2 qt. 1 pt. (101) From 52 yd. 2 ft. 9 in. take 115 ft. Ans. 14 yd. 1 ft. 9 in. (102) 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. (103) 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. (104) Multiply 3 qt. 1 pt. by 4.7. Ans. 32.9 pt. (105) How many iron rails, each 30 ft. long, are required to lay a single railroad track 23 miles long ? Ans. 8,096 rails. (106) The main line shaft of a mill is composed of 4 lengths each 15 ft. 5 in. long, of one piece 14 ft. 8 in. long, and one piece 8 ft. 10 in. long. If there are 6 hangers spaced equally distant apart, one being placed at each extremity of the shaft, what is the distance between the hangers ? Ans. 17 ft. f in. (107) The distance around a wheel is approximately 2 Tj ? times the diameter. If the diameter of a fly-wheel is 9 ft. 6 in., find the distance around it. Ans. 29 ft. llf in. (108) If the length of a boiler shell is 18 ft. ll^ in., how many rivets should there be in one of the longitudinal seams if it is a single-riveted seam, supposing the rivets to- have a pitch of 1^ in., and the two end rivets to be 1| in. from each end of the boiler ? Ans. 181 rivets. (109) (a) What is the second power of 108 ? (b) What is the third power of 181.25 ? (c) What is the fourth power of 27.61 ? (a) 11,664. Ans. ] (&) 5,954,345.703125. (c) 581,119.73780641. 286 ARITHMETIC. (110) Solve the following : (a) 106 s ; (b) (182)'; (c) .005; (/) 67.85'; (g) 967,845 s ; (k) (A)' Ans. -I (d) .0063'; (e) 10.06 s ; (0 (i)'. () 11,236. () 33,169.515625. (c) .000025. (d) .00003969. (e) 101.2036. (/) 4,603.6225. (g) 936,7^3,944,025. Ans. (111) Solve the following : (a) 753 3 ; (b) 9S7.4 3 ; (c) .005 3 ; (d) .4044 3 ; (e) .0133 s ; (/) 301.011 s ; (g) (i) s ; (A) (3|) 3 . f () 426,957,777. (b) 962,674,279.624. (c) .000000125. (d) .066135317184. (e) .000002352637. (/) 27,273,890.942264331. (g} rnr- L (//) 52.734375. (112) What is the fifth power of 2 ? Ans. 32. (113) What is the fourth power of 3 ? Ans. 81. (114) What is the ninth power of 7 ? Ans. 40,353,607. (115) Solve the following: (a) 1.2'; (b) II 5 ; (c) 1'; (d) .01'; (e) .l\ (a) 2.0736. (b) 161,051. AnsJ (c) 1. (d) .00000001. (e) .00001. (116) In what respect does evolution differ from involution ? ARITHMETIC 287 (117) Find the square root of the following: (a) 3,486,784.401; (/>) 9,000,099.4009; (r) .001225; (d) 10,795.21; (e) 73,008,05; (/) 9; (g) .9. (a) 1,8G7.29-|-. (b) 3,000.016. (c) .035. Ans. { (d} 103.9. (e) 270.2. (/) 3. (g) .948+. (118) What is the cube root of the following: (a) .32768? (b) 74,088? (c) 92,416? (d) .373248 ? (e) 1,758.416743? (/) 1,191,016? (g) Ans. - (119) ('<) &V ? (a) .689+. (b) 42. (c) 45.2115+. (d) .72. (e) 12.07. (/) 106. (e) (A) .472+. Find the cube root of 2 to four decimal places. Ans. 1.2599+. (120) Find the cube root of 3 to three decimal places. Ans. 1.442+. (121) Solve the following: (a) i/123.21; Ans. (a) 11.1. (b) 10.72+. } (c) 709. [ (d) .0203. 288 ARITHMETIC. (122) Solve the following : (a) I/.0065; (b) y7021; (c} f/8, 036, 054, 027; (d) ^.000004096; (e) #T7. |() .186 + () .27+. (r) 2,003. (IO = 100 ; ( ^46 = ^ ; / \ 10 _ x |() * ^ 12. ^' 1~50~ 600* >; (^ ) * = 20. (d) x = 180. (,) *-40. (127) x : 5 :: 27 : 12.5. Ans. 10|. (128) . 45 : 60 :: x : 24. Ans. 18. (129) x : 35 :: 4 : 7. Ans. 20. (130) 9 : x :: 6 : 24. Ans. 36. (131) fl^OOO : vT^3l::27 : x. Ans. 29.7. (132) 64 : 81 :: 21' : x\ Ans. 23.625. (133) 7 + 8 : 7 :: 30 : x. Ans. 14. (134) 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. ARITHMETIC. 289 (135) The intensity of heat from a burning body varies inversely as the square of the distance from it. If a ther- mometer held G ft. from a stove rises 24 degrees, how many degrees will it rise if held 12 ft. from the stove ? Ans. 0. (136) If sound travels at the rate of 6. 100 ft. in 5 sec., how far does it travel in 1 min. ? Ans. 07,200 ft. (137) If a railway train runs 444 mi. in 8 hr. 40 min., in what time can it run 1,060 mi. at the same rate of speed ? Ans. 20 hr. 41.44 min. (138) If a pump discharging 135 gal. per minute fills a tank in 38 minutes, how long would it take a pump discharging 85 gal. per minute to fill it ? Ans. 60 T 6 T min. (139) If a quantity of babbitt metal contains 8 Ib. of copper, 8 Ib. of antimony, and 80 Ib. of tin, how much copper will 32 Ib. of the same metal contain ? Ans. 2f Ib. (140) The distances around the drive wheels of two locomotives are 12.50 ft. and 15.7 ft., respectively. How many times will the larger turn while the smaller turns 520 times ? Ans. 410 times. (141) If a cistern 28 ft. long, 12 ft. wide, 10 ft. deep holds 510 bbl. of water, how many barrels of water will a cistern hold that is 20 ft. long, 17 ft. wide, and ft. deep ? Ans. 309 T 9 bbl. MENSURATION AND USE OF LETTERS IN ALGEBRAIC FORMULAS. (SEE ARTS. 340-432.) A = 5 // = 200 B = 10 x = 12 z=3.5 Z>=120 Work out the solutions to the following formulas, using the above values for the letters : (142) C= D ~ X .. Ans. C=S. (143 ) 6=1 7 ' + f+/>. An, C =li. 2;r + 6 3 ' 24(iS/ < Ans. r = 187.a9+. 7^+T5- ( U0 > " = ^.000184^-^ An, = .35+. 10 j^~^) 3 . Ans. /= 12,800. Ans. g = 5. Ans. k =7.071+. yrx 292 MENSURATION. (150) T= . Ans. 7- =10. V h+!L(A*-BY (151) If one of the angles formed by one straight line meeting another straight line equals 152 3', what is the other angle equal to? Ans. 27 57'. (152) How many seconds are there in 140 17' 10". Ans. 505,030". (153) Write a definition for a degree, as applied to the measurement of angles. (154) (a) How many degrees are there in 240 minutes ? (b) How many seconds ? ( (a) 4. 3 " \ (J) 14,400". (155) Draw an obtuse angle, a right angle, and an acute angle. State the name of each angle by using letters to designate them. (150) Draw a rhombus and then draw a rectangle having the same area. (157) Can a quadrilateral be formed with lines whose lengths are 20 inches, 9 inches, 4 inches, and 7 inches ? Give reasons ? (158) A sheet of zinc measures 11 inches by 2^ feet. How many square inches does it contain? Ans. 345 sq. in. (159) If the zinc in the last example weighs 5 pounds, what is its weight per square foot? Ans. 2.19 Ib. (1GO) How many boards 1C feet long and 5 inches wide would be required to lay a floor measuring 15 X 24 feet? Ans. 54 boards. (161) A lot of land is in the shape of a trapezoid. It is 16 rods long, 9 rods wide at the front, and 6 rods wide at the back. The front and back being parallel, what part of an acre does the lot contain? Ans. of an acre. MENSURATION. 293 (163) The accompanying figure shows the floor plan of Switch an electric-light station. From the dimensions given, cal- culate the number of square feet of unoccupied floor space. Ans. 2,059.08 sq. ft. (163) A sidewalk 10 feet wide extends around a city block 528 feet long and 352 feet wide. Assuming that there is no space between the walk and the buildings, how many square yards does the walk contain? Ans. 2,000 sq. yd. (164) How many square yards of plastering will be re- quired for the four side walls of a hall 90 feet long, 50 feet wide, and 20 feet high, with four doors 5 X 10 feet and fourteen windows 5 X 11 feet? There is a baseboard 9 inches high. Ans. 490.72 sq. yd. (165) A triangle has three equal angles ; what is it called ? (166) If a triangle has two equal angles, what kind of a triangle is it ? (167) Can a triangle be formed with three lines whose lengths are 12 inches, 7 inches, and 4 inches ? Give reasons for your opinion. 294 MENSURATION. (168) (a) What is the altitude of an equilateral triangle whose sides are each G feet ? (b} What is its area ? ( (a) 5.196ft. 5 " ( (b) 15.588 sq. ft. (169) In a triangle A B C, angle A = 23, and B = 32 32' ; what does angle C equal ? Ans. C = 124 28'. (170) In the figure, if A D=W inches, A B 24 inches, and B C= 13 inches, how long is D , D R being par- allel to B Cl Ans. DE= 5.625 in. (171) An engine room is 52 feet long and 39 feet wide. How many feet is it from one corner to a diagonally op- posite one, measured in a straight line? Ans. 65 ft. (172) A ladder 24 feet long rests against a house with its upper end 8 feet from the ground. How far on the ground is the lower end of the ladder from the house ? Ans. 22 ft. 7 + in. (173) (a) Show why it is that the area of a triangle equals one-half the product of the base by the altitude, (b) Does it make any difference which side is taken as the base ? (174) (a) The area of an isosceles triangle is 200 square inches. If its altitude is 20 inches, how long is its base ? (/;) What is the length of one of its equal sides ? j (a) 20 in. 5 ' I (b) 22.36 in. (175) In an equilateral heptagon one of the sides equals 3 inches; what is the length of the perimeter ? Ans. 21 in. (176) The perimeter of a regular decagon is 40 inches; what is the length of the side ? Ans. 4 in. (177) What is one angle of a regular dodecagon equal to ? Ans. 150. (178) The area of a regular pentagon is 43 square inches. If one side is 5 inches long, what is the perpendicular dis- tance from the center to one side ? Ans. 3.44 in. MENSURATION. 205 (179) It is required to make a miter-box, in which to cut molding to fit around an octagon post. At what angle with the side of the box should the saw run ? Ans. (180) Calculate the area of the irregular polygon, Fig. 3. The dimensions are to be obtained by meas- uring. Ans. 1.78 + sq. in. (181) An angle inscrib- ed in a circle intercepts one-fourth the circumfer- ence. How many degrees are there in the angle ? Ans. 45. (182) If the distance between two opposite corners of a hexagonal nut is two inches, what is the distance between two opposite sides ? Ans. 1.732 + in. (183) In the accompanying figure, if the distance B I is G inches and H K 18 inches, what is the diameter of the circle ? B Ans. 19.5 in. (184) In the same figure, if the diameter A B = 32 feet, and the dis- tance / B 8 feet, what is the length Ans. 28 ft. FIG. 4. of the chord H Kl (185) The trunk of a tree measures 7.854 feet around it; what is its diameter ? Ans. 2 ft. (180) How many revolutions will a 72-inch locomotive driver make in going one mile ? Ans. 280.112 revolutions. (187) 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. (188) The area of a circle having a diameter correspond- ing to the internal diameter of a certain pipe is 113.0973 square inches. What is the outside diameter, the pipe being f inch thick ? Ans. 13 J inches. 296 MENSURATION. (189) How long must the arc of a circle be to contain 12, supposing the radius of the circle to be G inches ? Ans. 1.25664 in. (190) Calculate the area of a flat circular ring whose out- side diameter is 22 inches, and whose inside diameter is 21 inches. Ans. 33.7722 sq. in. (191) Find the area of a segment of a circle whose diam- eter is 56 inches, the height of the segment being 5 inches. Ans. 108.5 sq. in. (192) What are the dimensions of the end of the largest square bar that can be planed from an iron bar 2 inches in diameter ? Ans. 1.4142 in. square. (193) 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. (194) (a) What would be the length of the side of a square metal plate having an area of 103.8691 square inches ? (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 the perimeter of the square plate .? !(a) 10. 1916 in. (b) 11 in. (c) 4. 638 in. (195) A plate 46 inches long is to be rolled into a shell; what will be the diameter of the shell if 5 inches are allowed for lap? Ans. 13.05 in. (196) What is the convex area in square feet of a section of a smokestack 26 inches in diameter and 10 feet long ? Ans. 71. 4714 sq. ft. (197) 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. 24.6774 sq. ft. (198) Find the cubical contents of the above column in cubic inches. Ans. 5,985.9648 cu. in. MENSURATION. 29 ? (199) 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. (200) 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 ( (a) 79.Scu.ft ( (b) 1,246 gal., nearly. (201) The cylinders of a compound engine are 19 and 31 inches in diameter, and the stroke is 24 inches; if the clear- ance at each end of the small cylinder is 14$ of the stroke, and in the large cylinder 8$ of the stroke, (a) what is the total volume in cubic feet of the steam in the small cylinder during one stroke ? (d) In the large cylinder ? (c ) What is the ratio between the two ? (a) 4.489 cu. ft. (b) 11.321 cu. ft. Ans. ' (c) Ratio = 2. 522 : 1. (202) Find the volume of a triangular pyramid 10 inches high; the base is an equilateral triangle, and one edge measures 10 inches. Ans. 144.336 cu. in. (203) The slant height of a square pyramid is 25 inches, and one edge of its base is 16 inches. Find its altitude by the principle of the right-angled triangle. Ans. 23.6854 in. (204) 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. (205) If the altitude of the above cone were 9 inches, what would be its volume ? Ans. 84.8232 cu. in. (206) 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. 298 MENSURATION. (207) How many pails of water would be required to fill the vat, the pail having the following dimensions: Depth, 11 inches; diameter at the top, 12 inches; diameter at the bottom, 9 inches? Ans. 3,627.28. (208) Required the area of the convex surface of the frustum of a hexagonal pyramid whose slant height is 32 inches, the perimeter of the lower base measuring 48 inches, and of the upper base 36 inches. Ans. 1,344 sq. in. (209) It is desired to find the number of gallons that a certain tank will hold. By measuring with a tape measure, it is found to be 190 inches in circumference at the bottom and 170^- inches in circumference at the top. The depth of the tank is 7 feet, and the thickness of the sides 1 inches. Make the calculation in round numbers. Ans. 861 gal. NOTE. To obtain the area of the upper and lower bases, first find the outside diameters and then deduct the thickness of the walls, or 2 X li inches. (210) Find (a) the area of the surface, and (b) the cubical contents of a ball 22 inches in diameter. Ans ( (a) 1,590. 435 sq. in. 3 ' ( (&] 5,964.1313 cu. in. (211) What is the volume of a ball whose surface has an area of 201.0624 square inches ? Ans. 268.0832 cu. in. (212) (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. (213) The volume of a certain cylindrical ring is 144.349 cu. in., its length before bending (see line D in Fig. 61, Art. 432) was 20.42 inches; what is the area of its surface ? Ans. 192.45 sq. in. MECHANICS. (SEE ARTS. 433-558.) (214) (a) What is a molecule ? (b) An atom ? (215) If a body has an average velocity of 40 feet per second, how far will it travel in 14 minutes ? Ans. 6 T 4 T miles. (216) Show how to represent a force by a line. (217) In Fig. 78, Art. 491, let the distance F c be 21', and F b 3"; what weight will a force of 85 Ib. applied at P raise ? Ans. 510 Ib. (218) What must be the speed of the driver pulley in order that the driven may make 80 R. P. M. and be 28* in diameter, the diameter of the driver being 21" ? Ans. 106f R. P. M. (219) The number of teeth in a spur gear is 50 and the pitch is II" ; (a) what is the pitch diameter ? (b) What is the outside diameter? . j (a) 23.87". ( (b) 24.77". (220) The driving gear has 45 teeth and the driven 180; if the driver makes 212 R. P. M., how many will the driven make ? Ans. 53 R. P. M. (221) What pressure can be exerted by a force of 24 Ib. on a half-inch screw which has 13 threads per inch, the dis- tance from the center of the screw to the point on the handle where the force is applied being 11" ? Ans. 21, 563. 94 Ib. (#22) A ball weighing 5 Ib. revolves in a circle whose radius is 32" at the rate of 350 R. P. M. ; what is the pull on the support caused by the ball ? Ans. 555 Ib. (223) A body weighing 2 Ib. has a velocity of 600 ft. per sec. ; what is its kinetic energy ? Ans. 11,194 ft.-lb 300 MECHANICS. What should be the width of a double leather belt to transmit 150 horsepower, when the belt has a velocity of 3,000 feet per minute, and has 7 feet of its length in contact with the smaller pulley, whose diameter is 63" ? Give width to nearest '. Ans. 29.5*. (225) (a) What are the three states of matter ? (/;) Name some of the general properties of matter; (<:) some of the specific properties. (226) If a man could run a mile at the average rate of 100 yards in 12 seconds, how long would it take him ? Ans. 3 min. 31.2 sec. (227) What is meant by "center of gravity " ? (228) (a) Why is crowning usually given to the face of a pulley ? (b) Why should high-speed pulleys be balanced ? (229) At what speed must the engine run when the di- ameter of the band-wheel is 13 feet and of the main pulley 91', if the speed of the main shaft is to be 108 R. P. M. ? Ans. 63 R. P. M. (230) The pitch of a gear is 24-', and the number of teeth is 192; what is the pitch diameter ? Ans. 152.79*. (231) How many revolutions per minute must the driv- ing gear make if it has 18 teeth, and the driven has 81 teeth and makes 80 R. P. M. ? Ans. 360 R. P. M. (232) The nuts on a cylinder head are tightened by means of a wrench 26" long. The threads in the nuts are 8 to the inch, and the efficiency of the screw is 40$. What pressure will the nut exert against the head when a force of 60 Ib. is applied to the end of the wrench ? Ans. 31,365.7 Ib. (233) What do you understand by specific gravity ? (234) If the center of gravity of a section of an engine fly-wheel rim is 6 ft. If in. from the center of the shaft, and the weight of the rim is 13,000 pounds, what is its kinetic energy when making 150 R. P. M.? Ans. 1,883,661.7 ft. -Ib. (235) What should be the width of a single leather belt to transmit 2^ horsepower when the belt has a velocity of MECHANICS. 301 2,000 feet per minute ? The diameter of the smaller pulley is 14", and the belt has 18" of its length in contact with it. Ans. If. (236) (a) What is meant by inertia ? (/;) By weight ? (c) How is weight measured ? (237) The speed of a certain belt is 3,000 ft. per min. ; if it drives a 48" pulley, how long will it take the pulley to make 100 revolutions ? Ans. 25.13 sec., nearly. (238) Find the point of suspension of a rectangular cast- iron lever 4 ft. 6 in. long, 2 in. deep, and in. thick, having weights 47 and 71 pounds hung from each end, in order that there may be equilibrium. Take the weight of a cubic inch of cast iron as .261 Ib. Ang j Short arm = 22.343*. 5 ' 1 Long arm = 31.657*. (239) When two pulleys are used to transmit power, which is called the driven and which the driver ? (240) The driver is 2 feet in diameter and the driven 32" ; if the driven makes 63 R. P. M., how many must the driver make ? Ans. 84 R. P. M. (241) A certain gear has a pitch of 1", and its pitch diameter is 11.48"; how many teeth are in the gear ? Ans. 32 teeth. (242) A fly-ball governor is designed to run at 88 R. P. M. The speed of the engine is 200 R. P. M. The diameter of the governor pulley is 8" ; the number of teeth in the bevel gear which it 'turns, 44, and the number of teeth in the other bevel gear, 75; what must be the diameter of the pulley on the crank-shaft which drives the governor belt ? Ans. 6*. (243) A bookbinder has a press, the screw of which has 4 threads to the inch. It is worked by a lever 15" long, to which is applied a force of 25 Ib. ; (a) what will be the pres- sure if the loss by friction is 5,000 Ib. ? . (b) What would be the theoretical pressure ? A . j (a) 4,424.8 Ib. b> 1 (b) 9,424. 8 Ib. (244) A cubic foot of a certain kind of wood weighs 51 Ib. ; what is its specific gravity ? Ans. -816, 302 MECHANICS. (245) The piston of an engine weighs 325 lb., including the piston rod; what is its kinetic energy when moving at the rate of 660 ft. per min. ? Ans. 611.4 ft.-lb., nearly. (246) What horsepower can be safely transmitted by a gear whose pitch is 1", a point on the pitch circle having a velocity of 1,200 ft. per min.? Ans. 12 H. P. (247) (a) What is motion ? (b) Velocity ? (c) Rest ? (d) Can a body be in motion with respect to one object and at rest with respect to another ? Explain fully. (248) (a) What is force ? (b} Name several kinds of forces. (249) Find by measurement the center of gravity of a triangle whose sides are 4", 5", and 6" long. Ans. Ij 3 / from 6" side. (250) The driving pulley makes 40 R. P. M. ; the driven makes 60 R. P. M., and is 36" in diameter. What is the diameter of the driver ? Ans. 54". (251) The diameter of the band-wheel of an engine is 12 ft. ; of the main pulley, 8 ft. ; of a driving pulley on the main shaft, 20"; of the driven pulley on the countershaft, 6" ; of the driver on the countershaft, 6", and of the driven on an emery-wheel spindle, 4"; if the engine makes 80 R. P. M., (a) what is the speed of the main shaft ? (b) Of the countershaft ? (c] Of the emery wheel ? ( (a) 120 R. P. M. Ans. | (b) 400 R. P. M. ( (c) 600 R. P. M. (252) The pitch diameter of a gear is 34.15"; if the pitch is If", how many teeth has the gear ? Ans. 78 teeth. (253) In order to raise a weight, a combination of a fixed and movable pulley is used; if a force of 225 lb. be applied to the free end of the rope, what load will it raise ? Ans. 450 lb. (254) If the power moves through a distance of 5 ft. 6 in. while the weight is moving 6 in., (a) what is the velocity MECHANICS. 303 ratio of the machine ? (b) What weight would a force of 5 Ib. applied to the power arm raise ? . (()!!. S ' I (b) 55 Ib. (255) In the last example, if the efficiency were 65#, what weight could the machine raise ? Ans. 35.75 Ib. (256) How many cubic inches of platinum will it take to weigh 10 Ib. ? Ans. 12.80 cu. in., nearly. (257) (a) For what are belts used ? (b} What is a single belt ? (c) A double belt ? (258) What horsepower can be safely transmitted by a gear whose pitch is 1.57", pitch diameter 30", and which makes 100 revolutions per minute? Ans. 19.30 H. P. (259) (a) What is uniform motion ? (b} What is vari- able motion ? (c) If a body moves 10 feet the first second, 12 feet the second second, 15 feet the third second, etc., is its motion uniform or variable, and why ? (200) What three conditions are required to be known in order to compare forces ? (201) A steel rod $* in diameter has on one end a cast- iron spherical ball 5" in diameter; if the length of the rod is 40" between the ball and the end, where is the center of gravity ? SUGGESTION. First calculate the weights of the ball and rod by the aid of the specific gravity tables. Ans. 0.427" from the center of the ball. (202) The driving pulley makes 240 R. P. M., and the driven pulley 180 R. P. M. ; if the diameter of the driven pulley is 30", what is the diameter of the driver ? Ans. 22*. (203) In a train of gears used to raise a weight of 6,000 Ib. in a manner similar to that shown in Fig. 83, Art. 5(K), the diameters of the drivers and belt pulley are 18", 12*, 15", and 12", and of the pinions and drum 6", 5", 8", and 3*; what force must be applied to the belt to raise the weight, if 20$ of the total force is lost through friction ? Ans. 138J Ib. (204) The pitch diameter of a gear is 24.16", and the number of teeth is 38; what is the pitch ? Ans. 1.997*. 304 MECHANICS. (265) It is required to raise a load of 1,890 Ib. by means of a block and tackle which has four fixed and four movable pulleys ; what force is required to be applied to the free end of the rope ? Ans. 236 Ib. (266) In a block and tackle, the theoretical power neces- sary to raise a weight of 1,000 Ib. is 50 Ib. ; (a) what is the velocity ratio ? (b) If the actual power necessary to raise the load is 95 Ib., what is the efficiency ? Ang j (a) 20. ' ( (*) 52.63#. (267) A piece of lead is %* in diameter and 10* long; how much does it weigh ? Ans. 12.91 oz. (268) If the distance between the centers of the crank- shaft and main shaft is 38 ft., and the diameters of the band- wheel and main pulley are 11 ft. and 7 ft., respectively, what must be the length of the main belt ? Ans. 105 ft. 3*. (269) The entire solar system is moving through space at the rate of 18 miles per second; (a) what is its velocity in miles per hour ? (b) How far will it go in one day ? Ans -f ( fl ) 64,800 mi. per hr. 5 ' I (b) 1,555,200 mi. (270) (a) What is a path of a body ? (b) What line do we measure when we wish to find the distance a body has traveled. (271) (a) How are forces measured ? (b) What kind of an effect do forces always tend to produce ? (272) It is required to raise a weight of 1,500 Ib. by means of a lever like that shown in Fig. 71, Art. 484. The length of the lever is 4 ft., and the distance from the fulcrum to the weight is 4* ; what force will it be necessary to apply ? Ans. 136 T 4 T Ib. (273) Had the lever in the above example been like that shown in Fig. 72, Art. 484, what force would have been required ? Ans. 125 Ib. (274) The band-wheel of an engine is 10 feet in diameter ; what must be the diameter of the pulley on the main shaft in order to make 110 R. P. M., if the band-wheel makes 88 R. P. M.? Ans. 8ft. MECHANICS. 305 (275) (a) What is a spur gear ? (b) A miter gear ? (c) A bevel gear ? (276) The pitch diameter of a gear is 30.56", and the number of teeth is 42; what is the pitch ? Ans. 2.735*. (277) The length of an inclined plane is 400 feet, and the height is 45 feet; what force acting parallel to the plane will be required to pull up the plane a weight of 4,000 Ib. ? Ans. 450 Ib. (278) (a) What is meant by the velocity ratio of a machine ? (b} By the efficiency ? (279) If a coil of brass wire weighs 10 Ib. , and the diam- eter of the wire is T y, how Long is the wire ? Ans. 806 ft., nearly. (280) The diameters of two pulleys are 14* and 18*, and the distance between their centers is 14 feet; what must be the length of a belt to drive these pulleys ? Ans. 32 ft. 4". (281) A velocity of 30 miles per hour corresponds to how many feet per second ? Ans. 44 ft. per sec. (282) The stroke of a steam engine is 28*, and it makes 1,500 strokes in 6 minutes; what is the velocity of the piston in feet per second ? Ans. Off ft. per sec. (283) State the three relations between force and motion. (284) A pulley on the main shaft is 40* in diameter, and makes 120 R. P. M. ; what must be the diameter of a pulley on the countershaft that is to make 160 R. P. M. ? Ans. 30'. (285) (a) What is a rack ? (b) A worm-wheel ? (c) A worm ? (286) (a) What distinguishes the epicycloidal teeth from the involute teeth ? (b} Name two advantages which the latter possess over the former. (287) An inclined plane has a length of 1,200 feet and a height of 125 feet. It is required to pull a load of 50,000 Ib. up this plane. A block and tackle having 6 fixed and 6 movable pulleys is stationed at the top of the plane, and the weight end of the rope is attached to the load. If the rope which connects the block to the load is parallel to the plane. 306 MECHANICS. what force will it be necessary to exert on the free end of the rope to pull up the load, no allowance being made for friction? Ans. 434 Ib. (288) (a) What do you understand by centrifugal force ? (b) By centripetal force ? (289) Define (a) work; (b) horsepower; (c) kinetic energy ; (d) potential energy. (290) Two pulleys have diameters of 8" and 20"; the dis- tance between their centers being 19 ft. 3", what must be the length of a belt to drive them ? Ans. 42 ft. 3*. (291) Two bodies starting from the same point, move in opposite directions, one at the rate of 11 feet per second, and the other, 15 miles per hour; (a) what will be the distance between them at the end of 8 minutes; (b) How long before they will be 825 feet apart ? Ang j (a) 3 miles. ' 1 (b) 25 seconds. (292) A railroad train runs 2 miles in 2 minutes and 10 seconds ; what is its average velocity in feet per second ? Ans. 81.23 ft. per sec. (293) Why is it difficult to jump from a rowboat into the water ? (294) A compound lever, similar to the one shown in Fig. 77, Art. 49O, is required to lift a weight of 1,250 Ib. The lengths of the power arms P F are 30*, 20", 10", and 15", of the weight arms W F 6", 5", 4", and 7"; what force will be required? Ans. llf Ib. (295) The driving pulley is 20" in diameter and driven 16"; if the driver makes 150 R. P. M., how many will the driven make ? Ans. 187 R. P. M. (296) How is the diameter of a gear measured ? (297) The driving gear makes 100 and the driven, 40 R. P. M. ; if the driven has 60 teeth, how many has the driver ? Ans. 24. (298) The base of an inclined plane is 80 feet long, and the height of the plane is 50 feet; what force exerted parallel to the base will raise a load of 750 Ib. ? Ans. 468| Ib. MECHANICS. 307 (299) What is the centrifugal force of the counterweight of a steam engine, the counterweight weighing 128 Ib. , and its center of gravity being 8f from the center of the shaft ? The crank makes 180 R. P. M. Ans. 1,028.16 Ib. (300) How much work can be done by 20 cubic feet of water falling from a height of 50 feet ? Ans. G2,500 ft.-lb. (301) What horsepower can be transmitted by a single leather belt 5" wide, which runs at the rate of 1,960 ft. per min. ? The diameter of the smaller pulley is 15" and the length of the arc of contact is 21" ? Ans. 11.4 H. P., nearly. (302) It is required to raise a weight of 18,000 Ib. by means of a screw having 3 threads per inch ; if the length of the handle is 15", and there is a loss of 10,000 Ib. due to friction, etc., what force will it be necessary to apply to the handle ? Ans. 99 Ib. , nearly. (303) The fly-wheel of an engine is 9 feet in diameter (outside); if the fly-wheel makes 100 R. P. M., how many miles will a point on the rim travel in 1^ hours ? Ans. 40.16 miles. (304) Suppose that an air gun can throw a ball with a velocity of 100 feet per second, and that a man standing on a railroad train, which is moving at the rate of 100 feet per second, were to fire the gun in a direction exactly opposite to that in which the train is moving, what would become of the ball ? Why ? (305) If the distance between the center line of the handle and the axis of the drum shown in Fig. 79, Art. 49 1 , is 1 4", and the diameter of the drum is 5", what load will a force of 30 Ib. exerted on the handle raise ? Ans. 174 Ib. (306) A pulley on the main shaft is 42' in diameter, and makes 108 R. P. M. ; what will be the speed of the counter- shaft if the driven pulley is 36" in diameter ? Ans. 126 R. P. M. (307) What is (a) the pitch circle ? (If) The pitch of a gear? 308 MECHANICS. (308) The driving gear makes 360 R. P. M., and the driven 170 R. P. M. ; if the driver has 34 teeth, how many has the driven ? Ans. 72 teeth. (309) Name some particular use in the engine room or shop to which you have seen the inclined plane put. (310) The mean diameter of the rim of an engine cast- iron fly-wheel is 9 ft. lOf, its width is 22", and its thickness 2"; what is the centrifugal force when running at 210 R. P. M.? Ans. 397,450 Ib. (311) Assuming the average pressure upon the piston of a steam engine to be 41.38 pounds per square inch, what is the horsepower ? The diameter of the piston is 10* ; the stroke 16", and the number of strokes per minute 450. Ans. 59.091 H. P., nearly. (312) What horsepower can be transmitted by a 20* double leather belt which has a velocity of 2,800 ft. permin. ? The diameter of the smaller pulley is 4 ft., and the belt has 5 ft. 9 in. of its length in contact with it. Ans. 99.4 H. P. MECHANICS. (SEE ARTS. 559-655.) (313) State Pascal's law. (314) A cylinder fitted with a piston is used as a lifting cylinder by passing a rope over a pulley and fastening one end to the piston rod. The piston is moved by means of water obtained from the city reservoir, and a gauge attached to a pipe near the cylinder shows the pressure to be 90 Ib. per sq. in. The diameter of the cylinder is 19 in. and of the pipe in. If friction be neglected, how great a weight can be raised ? Ans. 25,517.0 Ib. (315) Give Mariotte's law. (31G) What do you understand by (a) the tensile strength of a material ? () The working stress ? (31?) A close-link wrought-iron chain is made from f* iron; what is the greatest safe load that it will carry f Ans. 1,087.5 Ib. (318) What is the allowable working load for a steel-wire rope 5i* in circumference ? Ans. 27,502.5 Ib. (319) What steady force is required to shear a steel crank- pin which is 6" in diameter ? Ans. 1,090,404 Ib. (320) What must be the diameter of a wrought-iron shaft to transmit 40 horsepower when running at 110 revolutions per minute, no power being taken off between bearings. Ans. 2.89'. A shaft 2}|* in diameter would be used, that is, 3* round iron turned down. (321) In Fig. 104, Art. 562, suppose that the area of the piston c is .G sq. in. ; of d, 3 sq. in. ; of e, sq. in. ; of/, 2 sq. in. ; of a, 14 sq. in., and of b, 9 sq. in. ; if a force of 310 MECHANICS. 24 lb. be applied to c, what must be the forces applied to the other pistons to counterbalance the force at c, neglecting the weight of the water and the weight of the pistons ? f At d, 120 Ib. At e, 240 Ib. Ans. J At/, 80 Ib. At a, 5GO Ib. [ At b, 360 Ib. (322) The upper base of a cylinder submerged in water is 40 ft. below the surface. The diameter of the cylinder is 20 inches, the altitude 36 inches, and the bases are parallel. If the bases are horizontal, what is (a) the upward pressure of the water on the cylinder ? (b) The downward pressure ? Ans \ <*) 5 ' 863 ' 39 lb ' I (b) 5,454.32 lb. (323) (a) What is a barometer ? (b} What is the essential difference between a mercurial and an aneroid barometer ? (324) The smallest section of a connecting-rod is 3.5 sq. in. ; what is the unit stress when subjected to a tensile stress of 12,400 lb. ? Ans. 3,543 lb. per sq. in., nearly. (325) What load may be safely carried by a hemp rope 4* in circumference ? Ans. 1,600 lb. (326) What load can be safely sustained by a round wooden pillar, 8" in diameter and 10 ft. long, having both ends flat ? Ans. 13 tons. (327) What force is required to punch a I" hole through a wrought-iron plate T 7 T * thick ? Ans. 54,978 lb. (328) What horsepower will a 1* wrought-iron shaft transmit when running at 180 revolutions per minute, an average amount of power being taken off between the bearings ? Ans. 12.49 H. P. (329) Does the shape of the vessel have any effect in regard to the pressure exerted by a liquid upon its bottom ? (330) In Fig. Ill, Art. 576, suppose that the diameter of the piston a is 2", and of b 7$'; if a weight of 400 lb. be MECHANICS. 311 laid upon b, what weight must be applied to a to balance the weight on b ? Ans. 28.44 + Ib. (331) A vessel contains 42 cu. ft. of coal gas having a tension of 20 Ib. per sq. in. ; what will be the new tension when allowed to communicate with a perfectly empty vessel whose volume is 14 cu. ft.? Ans. 15.19 Ib. per sq. in., nearly. (332) What safe steady load can be sustained by a 1* round wrought-iron bar, the load producing a tensile stress ? Ans. 21,205.2 Ib. (333) What load can a hemp rope 0" in circumference carry with safety ? Ans. 3,600 Ib. (334) What load will a hollow cast-iron pillar support with safety if the pillar is 20 ft. long, outside diameter 14", inside diameter 11^-*, and both ends are fixed ? Ans. 219.24 tons. (335) What force is required to punch a hole \\" in diameter through a f steel plate ? Ans. 212,058 Ib. (330) A vessel is filled with water to the depth of 18"; if the area of the bottom is 46 sq. in., what is the total pressure on the bottom ? Ans. 29.95 Ib. (337) Why does water seek its level ? (338) The volume of steam in an engine cylinder is feu. ft. at cut-off, and its pressure is 94.7 Ib. per sq. in.; what will be its pressure when the volume has increased to 2 cu. ft.? Ans. 23.675 Ib. per sq. in. (339) A bar of steel having a cross-section of If* X 3' is subjected to a tensile stress; if the stress is suddenly applied, what is the greatest load that it will safely carry ? Ans. 31,500 Ib. (340) A load of 2,400 Ib. is to be raised by means of a hemp rope; what should be the circumference of the rope ? Ans. 4.9'. (341) Regarding the piston rod of a steam engine as a pillar which has one end flat and the other round, what should be the greatest diameter of the piston if the rod is 4 ft. 8 in. long, 3 in. in diameter, and the greatest steam 312 MECHANICS. pressure is not to exceed 100 Ib. per sq. in. ? The rod is made of wrought iron. Ans. 25*, nearly. (342) (a) What is cold-rolled shafting? (b) Bright shafting ? (c) Black shafting ? (343) A tank contains cylinder oil having a specific gravity of say .92. This tank is connected to a four-gallon can by means of a pipe whose internal diameter is f . If the vertical distance between the level of the oil in the tank and the end of the pipe is 8 ft. 7 in., what is the pressure per square inch at the end of the pipe ? Ans. 3.427 Ib. per sq. in. (344) In Fig. 113 (see examples following Art. 578), suppose that the diameter of the plunger A is 12", and of the plunger C of the pump B %" ; if a force of 100 Ib. be applied to C by means of the lever D, how great a weight can the plunger A raise if the weight of A is 600 Ib. ? Ans. 58, 382. 4 Ib. (345) The volume of steam in an engine cylinder at the beginning of compression is 1.11 cu. ft., and its pressure is 18 Ib. per sq. in. At the end of compression the volume is .3 cu. ft. ; what is the final pressure ? Ans. 66.6 Ib. per sq. in. (346) What should be the least area of one of the 14 wrought-iron cylinder head stud bolts if the diameter of the cylinder is 19', and the greatest steam pressure is 180 Ib. per sq. in. ? Assume that the studs are subjected to shocks. Ans. .729 sq. in. (347) What should be the circumference of a hemp rope to safely sustain a load of 4,200 Ib.? Ans. G". (348) Regarding the connecting-rod of a steam engine as a pillar with two round ends, what is the greatest force that may be exerted on the cross-head if the connecting-rod is made of wrought iron, is 10 ft. long, and has a rectangular cross-section 6* by Z* ? Ans. 35,489 Ib. (349) If you were to order some 2* bright-turned shafting, what size would you expect to get ? MECHANICS. 313 (350) A tank having the shape of a frustum of a cone is filled with water. If the diameter of the large end is 8 ft., of the small end 6 ft., and the perpendicular distance be- tween the two ends is 10", (a) what is the pressure on the bottom when the large end is down ? (b] When the small end is down ? . ( (a) 2,618 Ib. S ' ( (If) 1,473 Ib., nearly. (351) What is the total pressure on all of the six sides of a cube which has been sunk in the water until its top is 50 ft. below the surface ? One edge of the cube measures 2 ft., and its upper base is parallel with the water level. Ans. 70,502 Ib. (352) A vessel contains 25 cu. ft. of air having a pres- sure of 45 Ib. per sq. in. When allowed to communicate with a second vessel which is entirely empty, the pressure falls to 15 Ib. per sq. in. What is the volume of the second vessel ? Ans. 50 cu. ft. (353) A steam cylinder is 44" in diameter and sustains a steam pressure of 100 Ib. per sq. in. The diameter of the cylinder head studs is If and the area at the bottom of the thread is 1.057 sq. in. How many wrought-iron studs are required ? Assume that the studs are subjected to shocks. Ans. 29 studs. (354) An iron-wire rope 4* in circumference is used on a crane for hoisting loads; what is the greatest load that the rope will sustain with safety ? Ans. 9, GOO Ib. (355) A cast-iron rectangular cantilever beam having a cross-section of 1* wide by 2" deep is 4 ft. 8* long; how great a weight will the beam sustain at its end ? Ans. 201 Ib., nearly. (350) What horsepower will a 2 T V steel shaft transmit, when running at 120 revolutions per minute, pulleys being carried between the bearings to distribute the power along the line ? Ans. 20.445 H. P. , ,(357) A board 8* X 20* and If thick is so placed in water that its flat sides are horizontal; if the distance from the 314 MECHANICS. level of the water to the top of the board is 5 feet, what is the total upward pressure on the board ? Ans. 355.91 -f- Ib. (358) Will any solid body whose specific gravity is greater than that of water, sink in it to any depth ? Why ? (359) The volume of steam in an engine cylinder at cut- off is 1.6 cu. ft., and its pressure is 90 Ib. per sq. in. ; what must be its volume at release when the pressure has fallen to 21 Ib.? Ans. Gf cu. ft (360) What should be the least diameter of a wrought- iron bolt that is to resist a sudden pull of 12,000 Ib. ? Ans. 1.74"+. (361) A steel-wire rope is 41" in circumference ; what load will it safely sustain ? Ans. 22,562.5 Ib. (362) A white-pine beam supported at both ends has a rectangular cross-section 8" wide by 10" deep; if the beam is 28 ft. long, what total uniform load will it support in safety ? Ans. 6,857| Ib. (363) What horsepower can a 10" wrought-iron crank- shaft transmit, when running at 200 revolutions per minute ? Ans. 2,857|. (364) A vertical cylinder having a diameter of 20" and a length inside of 36" is filled with water ; a pipe having a diameter of |" is screwed into the upper head and fitted with a piston weighing 10 oz., on which is laid a weight of 25 Ib. ; if the end of the pipe is 10 ft. above the level of the water in the cylinder, (a) what is the pressure per square inch on the top of the cylinder ? (b] On the bottom ? (c) What equivalent weight laid on the lower cylinder head would replace the pressure it sustains? , (^) 330.45 Ib. per sq. in. Ans. | () 237.75 Ib. per sq. in. ( (c) 74,691. 54 Ib. (365) If, in the last example, a hole one inch in diameter is drilled through the cylinder wall midway of its length, and covered by a flat plate in such a manner that the water can not leak out, what would be the pressure against the plate ? Ans. 186.22 Ib. MECHANICS. 315 (3GG) The vacuum gauge of a condensing engine indi- cates 20" of vacuum ; what is the pressure in the condenser ? Ans. 4.9 Ib. per sq. in. (367) The volume of the receiver of an air compressor is 300 cu. ft., and the pressure of the air which is contained in it is 52 Ib. per sq. in. ; if 120 cu. ft. of air, having a pres- sure of 52 Ib. per sq. in., are removed from the receiver, what will be the pressure of the air which remains ? Ans. 31.2 Ib. per sq. in. (368) What is the greatest safe load that may be applied to a stud-link wrought-iron chain if the diameter of the iron from which the link is made is " ? Ans. 4,500 Ib. (369) It is desired to handle loads up to 14,000 Ib. by means of an iron-wire rope; what should be its circum- ference ? Ans. 4.83", nearly. (370) What is the greatest load that a bar of wrought iron 2" in diameter and 6 ft. long can safely sustain in the middle ? The bar is merely supported at its ends. Ans. 480 Ib. (371) What must be the diameter of a cast-iron crank- shaft to transmit 1.000 horsepower at 80 revolutions per minute ? Ans. 10.4 in. (372) (a) What must be the height of the mercury column to indicate a vacuum of 12"? (b) Of 18*? (373) (a) What is a stress? (b) A strain? (c) A unit stress? (374) The links in a stud-link wrought-iron chain are made from iron |f m diameter; what is the greatest safe load that the chain can handle? Ans. 11,883 Ib. (375) A steel-wire rope is used to haul loads up an inclined plane ; the greatest stress in the rope is 8,000 Ib. ; what should be its circumference? Ans. 2.83*. (376) What uniform load can be safely sustained by a steel beam 20 ft. long, 2" wide, and 6" deep? Ans. 4,608 Ib. (377) A 4" steel shaft is to transmit 80 horsepower; how many revolutions per minute must it make if used for trans- mission only (that is, no power being taken off at intermediate points)? Ans. 81 R. P. M. 316 MECHANICS. (378) Define tension as applied to gases. (379) (a) What is elasticity? (b) Elastic limit? (c) What is meant by set? (380) What safe load may be carried by a close-link wrought-iron chain whose links are made from f iron? Ans. 4,687.5 Ib. (381) What is the allowable working load for an iron-wire rope 6" in circumference? Ans. 21,600 Ib. (382) What force is required to shear a wrought-iron strip 4 ft. long and " thick? Ans. 960,000 Ib. (383) A 7" wrought-iron crank-shaft is to transmit 200 horsepower; how many revolutions per minute must it make? Ans. 40.8 rev., nearly. INDEX. A. PAGE PAGE Abstract number .... i Area of frustum of pyramid . M4 Acute angle 120 " " irregular polygon I3 Addition 4 " " parallelogram -134 " of decimals. 40 " " prism . . . . : " " denominate numbers . 68 " " pyramid .... 143 " " fractions 29 " " regular polygon . 3 " Proof of .... 8 " " sector of circle . n " Rule for 8 " " segment of circle 137 Sign of .... 4 " " trapezoid .... 104 table 5 " " triangle .... tag Adjacent angle . . . . . 120 Atmosphere, Pressure of . 223 Aeriform bodies .... '5 Atoms >49 Aggregation, Symbols of 53 Avoirdupois weight .... 63 Air chamber 244 " compressors .... 235 B. PAGE " pump 231 Balanced pulleys .... '-> Altitude of cone .... 142 Barometer MM " " frustum of cone or cyl- " Aneroid .... tat inder .... 144 " Mercurial waj " " parallelogram 123 Base (in percentage) .... yi " " pyramid 142 " of parallelogram 123 " " triangle .... 129 " " plane figure .... 133 Amount (in percentage) . 56 Beams, Transverse strength of rfa Aneroid barometer .... 225 Bearings for line shafting, Distance Angle 120 apart of ". " Acute 120 Belts 201 " Adjacent 120 " Double . i " Inscribed 133 " Horsepower of ... t * " Measured by arc . 121 " Lacing 1 .; " Obtuse 120 " Length of .- I " Right 120 " Rules for .... " Vertex of 120 " Single | I] Angles or arcs. Measures of . 64 " Width of Antecedent (of a ratio) 97 Bevel gears T- Arabic notation 2 Black shafting .- Arc of circle ..... '33 Block (pulley) I8 3 " " " To find length of 136 Bodies, Aeriform .... ' Archimedes, Principle of . 220 " Gaseous .... .. , Area of a surface .... 124 how composed [90 " " circle 1 3 6 Liquid ' - " " circular ring I 3 6 Solid IV " " cone 142 Brace 53, 116 " " cylinder .... '39 Bracket 53, 116 " " cylindrical ring . 1 4 6 Bright shafting u " frustum of cone . 144 Brittleness . i ; Buoyant effect of water . Bushel, Cubic inches in . Butt joint . . . Cancelation " Rule for Cantilever " Strength of Capacity, Measures of Cause and effect . Center of gravity . PAGE .219 .65 .316 PAGE 21 . 23 261 . 263 .64 . 108 . 160 " " " of irregular plane figure . . 162 " " " " parallelogram. 162 " " " " regular p 1 a n e figure . . 162 " " " " solid . . .164 " " " " system of bod- ies . . . 161 " " " " triangle . . 162 Centrifugal force .... 194 " " Rule for . . 195 Centripetal force .... 195 Chains ....... 251 " Rules for strength of . . 251 " Treatment of ... 251 Chamber, Air ..... 244 Chord of circle ..... 133 Cipher ..... Circle ...... 120, 133 " Arc of .... 121, 133 " Area of ..... 136 " Center of ... 120, 133 " Chord of . . . . . 133 " Circumference of . . 120, 133 " Diameter of . . . . 133 " Divisions of . . . .121 " Radius of . . . . . 133 " Sector of ..... 136 " Segment of . . . .137 Circular ring. Area of . . '. 136 Circumference of circle . . . 133 Coefficient of friction . . . 189 Table of . 192 Cold-rolled shafting . . . .268 Combination of pulleys . . . 184 " " " Law of . 185 Common denominator ... 27 " " Least . . 27 Composite numbers .... 21 Compound denominate numbers . 62 lever . . .167 " proportion . . . 109 Compressibility . . . .152 Compressive strength of materials. 255 Compressor, Air Concrete number Cone " Altitude of " Convex area of " Frustum of " Slant, Height of " Vertex of " Volume of Consequent Conservation of energy . PAGE 235 . 142 . 142 . 142 143 . 142 . 142 . 142 97 . 200 Constants for cast-iron pillars . 258 " " line shafting . . 270 " " transverse strength . 261 " " wooden pillars . . 259 " wrought-iron pillars . 257 Convex area of cone .... 142 " " " cylinder . . . 139 " " " cylindrical ring . 146 " " " frustum . . . 144 " " prism . . .139 " " pyramid . . . 142 " " " sphere . . . 145 " " " surface of solid . 139 Countershafts 268 Couplet (ratio and proportion) 97, 100 Cross-head friction in guides . . 193 Crowning of pulleys . . . 171 Crushing strength of materials . 255 "Table of 256 Cube (solid) 138 " of a number . . .. 77 " root 79, 86 " " Proof of .... 92 " " Rule for . ... 90 Cubic measure 63 Cubical contents .... 139 Curved line 119 Cylinder, Convex surface of . . 139 " " Volume " . . 139 Cylindrical ring, Convex area of . 146 " " Volume of . . 147 D. PAGE Decimal 38 " number, how read . . 38 " part of a foot. To reduce inches to . . . .50 " point 38 " To express as a fraction with a given denomi- nator . . . .52 " To reduce a fraction to . 50 " " " to a fraction - 51 Decimals, Addition of . . 40 " Division of . -45 PAGE Decimals, Multiplication of . .43 " Subtraction of . . .42 PAGE Expansibility 152 Exponent 77 " and limits of exhaustion . 233 " Measures of ... o* 6 Denominate numbers 61 Extremes of a proportion tat " " Addition of . 68 " " Compound . 62 F. PAGE " " Division of . 73 Factor -.-I " " Multiplication " Prime 9i of 72 Figure, Plane 123 " " Reduction of. 65 Figures a " " Simple . 61 Fixed pulley 183 " " Subtraction of 70 Follower Denominator 23 Foot-pound X 9K " Least common . 27 Force . 156 Diagonal of parallelogram 129 " Centrifugal .... '94 Diameter of circle .... 133 " Centripetal .... 195 Difference 9 " Direction of .... 156 " (in percentage) 56 " Point of application of . . 156 Digits 2 " Pump 343 Direct proportion 101 " Representation of . H',,, " ratio 97 " required to punch plates 067 Direction of force .... 156 Formula us Dividend . . '7 " How applied ixa Divisibility 'Si " Symbols used in "5 Division i? Fraction 3 " of decimals. 45 " Improper , . . . -5 " " denominate numbers . 73 " Proper .... t " " fractions. 34 " Terms of .... 5 Proof of .... 20 " To in vert a . 35 " Rule for 20 " " reduce a decimal to a . S* " Sign of .... i? " " " " whole or mixed Divisor. . . . . '7 number to a . 06 Double belts 201 " " " to a decimal SP " shear 266 " " " " lowest terms . >6 Double-acting pump 245 " Value of a ... 94 Downward pressure of water . 209 Fractions, Addition of .'i Driver (pulleys) '73 " Division of ... 34 Dry measure . . 64 " Multiplication of . )a Ductility 'S3 " Reduction of . 25 Duplex steam pump .... 246 " Roots of . 94 " Subtraction of 3 E. PAGE " To find least common de- Efficiency of machine 192 nominator of 7 Elastic limit 248 " " reduce to common de- Elasticity 152, 247 nominator iS " Measure of ... 248 Friction i8g Energy '99 " between cross-head and " Conservation of . 200 guides .... 193 Kinetic 109 " Coefficient of ... Bg " Potential .... 200 Laws of .... 90 Epicycloidal teeth .... 179 " Table of coefficients . 99 Equality, Sign of . 4 Frustum of cone or pyramid . 43 Equilateral triangle .... 126 Altitude of . -i i Evolution 79 " Convex surface of 44. Exhaustion of air .... 233 " Volume of . ft INDEX. PAGE PAGE Fulcrum .... . 165 Involute teeth .... . 179 Fundamental principle of machines 166 Involution . 76 G. PAGE Isosceles triangle . 126 Gain or loss per cent. . 60 K. PAGE Gallon, Cubic inches in . 65 Kinetic energy .... . 199 Gallon of water. Weight of . 65 Gas, Permanent . 150 L. PAGE " Tension of . 222, 228 Lacing belts .... . 203 Gaseous body . mo Lateral pressure of water . 212 Gases, Application of Mariotte's Law of Mariotte . 229 law to .... 230 " " Pascal . . 209 Cear wheel's Laws of friction. Gears, Bevel '75 !75 " " inertia .... 157 " Horsepower of 204 " " liquid pressure . . 209 Miter J 7S " " motion, Newton's '57 " T? 1 f Lever ific " Spur . ... 178, 180 175 " Compound 105 . i6 7 Gravity, Center of . . . 160 " Fulcrum of . . i6 5 " Specific . 196 " Law of . . i6 S Guides, Friction on . *93 Lifting pump .... . 242 Like number .... i H. PAGE Line, Curved .... . 119 Hardness .... . 152 " Horizontal . 119 Head (of water) . . 215 " Perpendicular to another . 119 Helix . . 187 " shafting .... . 268 Hemp ropes . . . . 252 " " Bearings for . 269 " " Strength of . . 252 " " Horsepower of . . 270 Heptagon . . . 130 " " Rules for . . 271 Hero's fountain. 237 " Straight . . . 119 Hexagon " Vertical .... Horizontal line . . 119 Linear measure. . 62 Horsepower . . 198 Lines Parallel I TO ." of belts. . 202. Liquid body .... . 150 " " gears. . 204 " measure 6 4 " " shafting, Fori nulas pressure, Laws of. . 20g for . . 270 Liquids, Incompressibility of . . 207 Hydraulic press. . 216 Local value of a figure . 2 Hydrostatics . 207 Long ton table . . 6 3 Hypotenuse . 126 Loss per cent - 60 I. PAGE M. PAGE Impenetrability. . 151 Machines, Pneumatic . 231 Improper fraction 25 " Principle of . 166 Inclined plane . . 185 Simple . 165 " " Rule for . 186 Magdeburg hemispheres . 233 Incompressibility of liquids . 207 Magnitude of a force T 57 Indestructibility, Law of . 152 Malleability . 153 Index of a root . 79 Mariotte's law . . . . . 229 Inertia , 151. 158 Materials, Crushing strength of 255 Injector . . . . . 239. " Shearing " " . 266 " How operated . 240 " Tensile " . 248 Inscribed angle . '33 " Transverse " " . 261 polygon . *34 Matter M9 Integer . . . " Properties of . IS' Inverse proportion . 101, 104 " Special properties of . 152 " ratio 97 Means (of a proportion) . . 101 PAGE O. PAGE Measure, Cubic . . 6, Obtuse angle lao Dry . . 64 " Linear . . 62 P. P AGE " Liquid . 64 Parallel lines 119 " of money . . 65 Parallelogram i3 " " time 64 Altitude of 133 " Square 62 " Area of ... 124 Measures, Miscellaneous . . 65 " Center of Gravity of 162 of angles and arcs . 64 " Diagonal of ... i*9 " " capacity 64 Parallelopipedon .... 138 " " extension 62 Parenthesis 53 , 116 " " weight . 63 Partial vacuum . Mechanics .... T 49 Pascal's law . ... Mensuration . 119 Path of a body in motion. ' 1 Minuend .... 9 Percent ' I ' 53 Minus 9 " " Gain Miscellaneous measures . 65 " " Loss 6O Miter gears. 175 " " Sign of .... 55 Mixed number . . - . 25 Percentage 55 , 5 6 Mobility .... 151 Rules of. 56 Molecule .... M9 Perimeter of polygon Money, Measure of . . - 65 Permanent gas . 150 United States . . 65 Perpendicular lines .... "9 Motion .... '53 Pillars, Constants for . . 257 " Newton's laws of . '57 " Formulas for strength of . 256 Movable pulley . . 183 Pinion . . J 73 Multiplicand Pitch circle '77 Multiplication 12 " of gear teeth .... 177 " of decimals 43 " " screw thread 187 " " denominate num- Plane figure "3 bers . 72 " " Center of gravity of . i6a " " fractions 32 " Inclined 95 " Proof of . 16 " Law of . . i Rule for . . 16 Plunger pump ... 43 Sign of 12 Pneumatic machines. 931 table . 13 Pneumatics Multiplier . . . f 12 Point of application of force . ,.'. N. PAGE Polygon 1 - Naught .... " Area of .... IOJ Newton's laws of motion. 157 Inscribed .... 134 Notation I. 4 " Perimeter of ... 130 " Arabic 2 " Regular .... 130 Number " Sum of interior angles of 131 " Abstract I Porosity ' Jl " Composite . 21 Potential energy " Concrete 1 Power arm .65 " Denominate . 61 " of a number .... ;' " Like . . . i " " " ratio .... 99 " Mixed ... 25 Powers and roots in proportion " Prime . . -. 21 Press, Hydraulic IT " Reciprocal of a . 97 Pressure of air .-; Unlike , " " atmosphere . ,.; " Unit of a I " liquid .... g Numeration . I, 4 " " water .... PAGE R. PAGE Pressure of water, Lateral 212 Radical sign n ' Upward 211 Radius of circle i a Prime factor 21 Rate (in percentage) .... 56 " number 21 Ratio . 96 Principle of Archimedes . 220 " Couplet of a a " " machines 1 66 " Direct 91 Prism 1 3 8 " How expressed 9 Convex surface of '39 " Inverse 97 " Volume of '39 " Reciprocal .... '~<7 Product ' 12 Symbol of .... <-/> Partial ..... 15 " Terms of a .... V Proof of addition .... 8 " Value of a .... 91 " " division .... 20 " Velocity 189 " " multiplication :6 Reading numbers .... 3 " " subtraction .... n Reciprocal of a number . 7 Proper fraction 25 " " ratio 91 Properties of matter .... IS' Rectangle 1*3 Proportion 100 Reduction of decimals to fractions 51 " Compound 109 " " denominate numbers .-, " Couplet of ... 100 " " fractions 25 " Direct .... 101 " " " to decimals SO " Extremes of . 101 Regular polygon .... 130 " how read. 100 Remainder 9 " Inverse . . . 101, 104 Rhomboid 123 " Means of a 101 Rhombus 1*3 Powers and roots in 106 Right angle 120 Rules for ... 101 Right-angled triangle u6 " Simple .... 109 Ring, Circular 1*6 Pulley, Fixed 183 " " Area of ... i;'' Movable .... . '83 " Cylindrical, Area of 146 Pulleys 170 " " Volume of . 14? " Balanced . . . . - . 171 Root 77 " Combination of . . " . 184 " Cube. . . . 79 86 " Crowning of. 171 " Index of 19 " Law of .85 " of gear tooth .... '77 " Rules for .... 172 " " ratio I'M Pump, Air 23' ' Square 79 " " chamber for 244 Roots of fractions . . . ^ . 94 " Force 242 11 other than square and cube . 95 " Lifting 242 Rope, Hemp j:j " Plunger . 243 " Strength of . t$9 Steam 245 " Wire .... S3 Suction 241 " Strength of . .... S3 Punching holes in plates, Force re- quired for 267 S. PA GE i 'fj Altitude of . . . . 142 Screw " Convex area of . 142 " Law of . ... t88 ** Frustum of * v Pitch of ,x- " Slant height of . 43 142 " thread . . . ' . ' . 7 Vertex of ... 142 Sector, Area of 3I " Volume of . 142 Segment, Area of 37 Semicircle ":4 Q. PAGE Semi-circumference .... i ;4 Quadrilateral 123 Shafting, Black .... M Quotient 17 " Bright .... tU INDEX. PAGE PAGE Shafting, Cold-rolled 268 Subtraction 9 " Distance between bear- of decimals . 42 ings of .... 269 " " denominate num- " Horsepower of . 270 bers 7 " Line 268 " " fractions . 3 " Rules for .... 271 Proof of ii Shear, Double 266 Rule for ... ii " Single 266 Sign of . 9 Shearing strength .... 266 Subtrahend 9 Rule for . ' . 267 Suction pump 241 Table of . 266 Surface of solid .... 139 Sign of addition .... 4 Symbol of ratio 96 " " division .... '7 Symbols of aggregation . 53 " " dollars 53 " " equality .... 4 T. i 'AGE ' " multiplication 12 Table, Addition .... 5 " " per cent 55 " Multiplication 13 " " subtraction .... 9 " of coefficient of friction 192 " Radical 79 " " constants for cast-iron pil- Simple denominate number . 61 lars . 258 " machines .... 165 Table of constants for transverse " proportion .... 109 strength . 261 " value of figure 2 " " " " wooden pil- Single belts ..... 201 lars . 259 Siphon 2 3 8 " " " " wrought-iron Slant height of cone or pyramid 142 pillars 257 Solid body I 3 8 " " crushing strength 256 " Center of gravity of l6 4 " " distances between shaft Specific gravity .... I 9 6 bearings .... 269 Sphere M5 " " shearing strength 266 " Area of surface of 145 " " tensile strength 249 " Volume . 14 6 Teeth of gears 177 Spur gears 175 " " " Addendum of . '77 " " Horsepower of 204 " " " Epicycloidal 179 Square ir>3 " " " Involute . 179 " foot . . . . . 124 " " " Pitch of ... '77 " measure 62 Tenacity '77 '53 " of a number .... 77 Tensile strength .... 248 " root 79 " " of chains 251 " " Rule for ... 84 " " " materials . 249 " " Proof of . 84 " " " ropes 252 " " Short method for 85 Rules for . 250 Steam pump 245 Table of 249 Strain 247 Tension of gases . ._ .222 , 228 Strength, Compressive 2 55 Terms, Higher 25 " of beams .... 262 " Lower 26 " " belts .... 202 " of a fraction .... 25 " " chains .... 2S' " " " proportion 101 " " materials 247 " " " ratio .... 97 " " pillars, Formulas for . 256 Thread of screw .... 187 " " ropes (hemp and wire' 252 Time, measures of .... 64 " Shearing .... 266 Torricellian vacuum 234 " Tensile .... 248 Train '73 Stress 247 Transverse strength 261 " Unit 247 " " of beams 261 Transverse strength of beams, PAGE Vertex of angle .... PAGE . 120 Rules fo r 262 " ' cone .... '44 " " cantilevers 263 Vertical line .... . nq " " " constants . 261 Vinculum S3. 1.6 Trapezoid ... "3 Volume of circular ring . . 147 " Area of . . . 124 " " cone .... . 142 Triangle , 126 " " cylinder . 139 Altitude of . 129 " " cylindrical ring 147 " Center of gravity of . 162 " " ' " pyramid 742 " Equilateral . 126 ' " prism . 139 " Hypotenuse of . . 126 ' " pyramid . . 142 Isosceles . 126 " " sphere . 146 " Right-angled . 126 Unit of . . 139 " Scalene . 126 Troy weight .... 63 W. PAGE Water, Buoyant effect of . 219 U. PAGE " Gallons in a cubic foot of 65 Uniform velocity '54 " Incompressibility of . 207 Unit I " Pressure of . . . 209 " of a number I Wedge . 187 " square .... . 124 Weight 'SI United States money . 65 " Arm .... . 165 Unlike number I " Avoirdupois . 63 P P " Measures of . 234 63 V. PAGE Troy .... . 63 Vacuum . 224 Wheel and axle .... . 169 Partial , 224 Wheelwork .... 173 " Torricellian 234 Wheelwork, Rules for '74 Value of a fraction . . . . 24 Wire rope, Strength of 253 " ratio 97 Work . 197 Vapor 5 " Measure of ... 197 Variable velocity 54 " Unit of .... . 197 Velocity 54 Worm and worm-wheel . . 176 " ratio .... . 89 " Uniform 54 Z. PAGE variable 54 Zero * 2 5?S>3 DZ> >I> D >^SDO 3^.d . ^>K)^ ' ^)^lI)2>C>jffi>> ! >>^ ^i> > > lbB> - ; 3^>^>1^ O ^?DD> ^ CCX AA 000513599 wccggx. CSCCCJK mOTOf recede "<2CCCCCC