IRLF SB tlfl 5m r'CCC :ccc c cTc< o cc ^ C c < C '< -: C c CC < 2 c cc d r c (OC cccc< CC C C4 'CC.O . c cc cc c c cccrr: TJ; c cc cccc^ 2ac**'*'* 2. A unit is one, or a single thing, asa$?/ tone bojv She** ' horse, one dozen. 3. A number is a unit or a collection of units, as one, tliree apples, five boys. 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. 5. A concrete number is a number applied to some particular kind of object or quantity, as three horses, five dollars, ten pounds. 6. An abstract number is a number that is not applied to any object or quantity, as three, five, ten. 7. like numbers are numbers which express units of the same kind, as six days and ten days, two feet and five feet. 8. Unlike numbers are numbers which express units of different kinds, as ten months and eight miles, seven dol- lars and five feet. NOTATION AND NUMERATION. 9. Numbers are expressed in three ways: (1) by words; (2) by figures ; (3) by letters. 1C. 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 12. The Arabic notation is the method of expressing numbers by figures. This method employs ten different ligures to represent numbers, viz. : Figures 0123456789 Names naught, one tivo three four five six seven eight nine cipher, or zero The first character (0) is called naught, cipher, or zero, li^ii feiiiidjng alone has no value. .The. pjtiier. nine- ^figures are called digits, and each has a Any whole number is called an integer. 1 3. As there are only ten figures used in expressing num- bers, each figure must have a different value at different times. 14. The value of a figure depends upon its position in relation to others. 15. Figures have simple values and local, or rela- tive, values. 16. The simple value of a figure is the value it ex- presses when standing alone. 17. The local, or relative, value of a figure is the increased value it expresses by having other figures placed on its right. For instance, if we see the figure 6 standing alone, thus ...................... 6 we consider it as six units, or simply six. Place another 6 to the left of it ; thus ...... 66 The original figure is still six units, but the sec- ond figure is ten times 6, or 6 tens. If a third 6 be now placed still one place further to the left, it is .increased in value ten times more, thus making it 6 hundreds ......... ... 666 A fourth 6 would be 6 thousands ........ 6666 A fifth 6 would be 6 tens of thousands, or sixty thousands ............ ...... 66666 A sixth 6 would be 6 hundreds of thousands . 666666 A seventh 6 would be 6 millions . . 6666666 ARITHMETIC. The entire line of seven figures is read six millions six hundred sixty-six thousands six hundred sixty-six. 18. The increased value of each of these figures is its local, or relative, value. Each figure is ten times greater in value than the one immediately on its right. 19. The cipher (0) has no value in itself, but it is useful in determining the place of other figures. To represent the number four hundred five, two digits only are necessary, one to represent four hundred, and the other to represent five units; but if these two digits are placed together, as 45, the 4 (being in the second place) will mean 4 tens. To mean 4 hundreds, the 4 should have two figures on its right, and a cipher is therefore inserted in the place usually given to tens, to show that the number is composed of hundreds and units only, and that there are no tens. Four hundred five is 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 hundreds and tens; thus, 450. Four thousand and fifty would be expressed 4050, the first cipher indicating that there are no units and the second that there are no hundreds. 20. In reading numbers that have been represented by figures, it is usual to point off the number 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. rA d ^ 3 gj . ^ o .2 00 *s tc 3 d 3 go o a c3 'S PQ ^ d ^H en ^1 p oj *8 3 *8 3 e O 09 *s 'S Hundreds PQ o t/3 1 Billions. Hundreds S '->-! O a (U Millions. Hundreds e "S 09 d 0) Thousand; Hundreds Tens of U 1 4 3 2,1 9 8,7 6 5,4 3 2 4 ARITHMETIC. 1 In pointing off these figures, begin at the right-hand figure and count units, tens, hundreds; the next group of three figures is thousands; therefore, we insert a comma (,) before beginning with them. Beginning at the figure 5, we say thousands^ tens of thousands, hundreds of thousands, and insert another comma. We next read millions, tens of mil- lions, hundreds of millions (insert another comma), billions, tens of billions, hundreds of billions. The entire line of figures would be read: four hundred thirty-two billions one 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 following figures, 72,584,623, would be the notation, and the numeration would be sev- enty-two millions five hundred eighty-four thousands six hun- dred twenty-three. 21. NOTE. It is customary to leave the s off the words mil- lions, thousands, etc., in cases like the above, both in speaking and writing; hence, the above would usually be expressed seventy -two million five hundred eighty-four thousand six hundred twenty-three. 22. The four fundamental processes of arithmetic are addition, subtraction, multiplication, and division. They are called fundamental processes because all operations in arithmetic are based upon them. ADDITION. 23. Addition is fo& process of finding the sum of two or more numbers. The sign of addition is -j- . It is read plus, and means more. Thus, 5 -\- 6 is read 5 plus 6, and means that 5 and 6 are to be added. 24-. The sign of equality is = . It is read equals or is equal to. Thus, 5 + 6 11 may be read 5 plus 6 equals 11. ARITHMETIC. 5 25. Like numbers can be added, but unlike numbers cannot be added. Thus, G dollars can be added to 7 dollars, and the sum will be 13 dollars; but 6 dollars cannot be added to 7 feet. 26. The following table gives the sum of any two num- bers from 1 to 12: 1 and 1 is 2 2 and 1 is 3 3 and 1 is 4 4 and 1 is 5 1 and 2 is 3 2 and 2 is 4 3 and 2 is 5 4 and 2 is 6 1 and 3 is 4 2 and 3 is 5 3 and 3 is 6 4 and 3 is 7 1 and 4 is 5 2 and 4 is 6 3 and 4 is 7 4 and 4 is 8 1 and 5 is 6 2 and 5 is 7 3 and 5 is 8 4 and 5 is 9 1 and 6 is 7 2 and 6 is 8 3 and 6 is 9 4 and 6 is 10 1 and 7 is 8 2 and 7 is 9 3 and 7 is 10 4 and 7 is 11 1 and 8 is 9 2 and 8 is 10 3 and 8 is 11 4 and 8 is 12 1 and 9 is 10 2 and 9 is 11 3 and 9 is 12 4 and 9 is 13 1 and 10 is 11 2 and 10 is 12 3 and 10 is 13 4 and 10 is 14 1 and 11 is 12 2 and 11 is 13 3 and 11 is 14 4 and 11 is 15 1 and 12 is 13 2 and 12 is 14 3 and 12 is 15 4 and 12 is 16 5 and 1 is 6 6 and 1 is 7 7 and 1 is 8 8 and 1 is 9 Sand 2 is 7 6 and 2 is 8 7 and 2 is 9 8 and 2 is 10 5 and 3 is 8 6 and 3 is 9 7 and 3 is 10 Sand 3 is" 11 5 and 4 is 9 6 and 4 is 10 7 and 4 is 11 8 and 4 is 12 5 and 5 is 10 6 and 5 is 11 7 and 5 is 12 8 and 5 is 13 5 and 6 is 11 6 and 6 is 12 7 and 6 is 13 8 and 6 is 14 5 and 7 is 12 6 and 7 is 13 7 and 7 is 14 8 and 7 is 15 5 and 8 is 13 6 and 8 is 14 7 and 8 is 15 8 and 8 is 16 5 and 9 is 14 6 and 9 is 15 7 and 9 is 16 8 and 9 is 17 5 and 10 is 15 6 and 10 is 16 7 and 10 is 17 8 and 10 is 18 5 and 11 is 16 6 and 11 is 17 7 and 11 is 18 8 and 11 is 19 5 and 12 is 17 6 and 12 is 18 7 and 12 is 19 8 and 12 is 20 9 and 1 is 10 10 and 1 is 11 11 and 1 is 12 12 and 1 is 13 9 and 2 is 11 10 and 2 is 12 11 and 2 is 13 12 and 2 is 14 9 and 3 is 12 10 and 3 is 13 11 and 3 is 14 12 and 3 is 15 9 and 4 is 13 10 and 4 is 14 11 and 4 is 15 12 and 4 is 16 9 and 5 is 14 10 and 5 is 15 11 and 5 is 16 12 and 5 is 17 9 and 6 is 15 10 and 6 is 16 11 and 6 is 17 12 and 6 is 18 9 and 7 is 16 10 and 7 is 17 11 and 7 is 18 12 and 7 is 19 9 and 8 is 17 10 and 8 is 18 11 and 8 is 19 12 and 8 is 20 9 and 9 is 18 10 and 9 is 19 11 and 9 is 20 12 and 9 is 21 9 and 10 is 19 10 and 10 is 20 11 and 10 is 21 12 and 10 is 22 9 and 11 is 20 10 and 11 is 21 11 and 11 is 22 12 and 11 is 23 9 and 12 is 21 10 and 12 is 22 11 and 12 is 23 12 and 12 is 24 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 is 17. 27. For addition, place the numbers to be added directly under each other, taking care to place units under units, tens under tens, hundreds under hundreds, and so on. 6 ARITHMETIC. 1 When the numbers are thus written, the right-hand figure of one number is placed directly under the right-hand figure of the one above it, thus bringing" units under units, tens under tens, etc. Proceed as in the following examples : 28. EXAMPLE. What is the sum of 131, 222, 21, 2, and 413 ? SOLUTION. 131 222 2 1 2 413 sum 789 Ans. EXPLANATION. After placing the numbers in proper order, begin at the bottom of the right-hand, or imits, col- umn, 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 1 ARITHMETIC. 7 units, column is seven and four are eleven and five are six- teen and six are twenty-two and five are twenty-seven, or 27 units; i. e., two tens and seven units. Write 27 as shown. The sum of the numbers in the second, or tens, column is six tens, or 60. Write 60 underneath 27, as shown. The sum of the numbers in the third, or hundreds, column is 15 hundreds, or 1,500. Write 1,500 under the two preceding results as shown. There is only one number in the fourth, or thousands, column, 9, 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 30, 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 is 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, is 8, which is written down as the second figure in the sum. There is nothing to carry to the next column, because 8 is less than 10. The sum of the numbers in the next column is 15 hundreds, or 1 thousand and 5 hundreds. Write down the 5 as the third, or hundreds, figure in the sum and carry the 1 to the next 1-2 8 ARITHMETIC. 1 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 is to be preferred when adding a long column. 31. EXAMPLE. Add the numbers in the column below: SOLUTION. 890 82 90 393 281 80 770 83 492 80 383 84 191 sum 3899 Ans. 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 4- 1 is 109 tens, or 10 hundreds and 9 tens. Write down the 9 and carry the 10 to the next column. The sum of the digits in this column plus the 10 reserved is 38. The entire sum is 3,899. 32. Rule. I. Begin at the right, add each column sep- arately, and write the sum, if it be only one figure, under the column added. II. If the sum of any column consists of two or more figures, put the right-hand figure of the sum under that column and add the remaining figure or figures to the next column. 33. Proof. To prove addition, add each column from top to bottom. If you obtain the same result as by adding from bottom to top, the work is probably correct. gl ARITHMETIC. EXAMPLES FOR PRACTICE. 34. Find the sum of: (a) 104 + 203 + 613 + 214. (b) 1 , 875 + 3, 143 + 5, 826 + 10, 832. (c) 4,865 + 2,145 + 8,173 + 40,084. (d) 14,204 + 8,173 + 1,065 + 10,042. . (e) 10,832 + 4,145 + 3,133 + 5,872. (/) 214 + 1,231 + 141 + 5,000. Or) 123 + 104 + 425 + 126 + 327. (//) 6,354 + 2,145 + 2,042 + 1,111 + 3,333. (a) 1,134. () 21,676. (0 55,267. (d) 33,484. (e) 23,982. (/) 6,586. (-) 1,105. (//) 14,985. SUBTRACTION. 35. In arithmetic, subtraction is the process of finding- 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 7568 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 remain- ders below the- line. The result is the entire remainder. 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 subtra- hend, proceed as in the following example : EXAMPLE. From 8,453 take 844. SOLUTION. minuend 8453 subtrahend 844 remainder 7609 Ans. 10 ARITHMETIC. I EXPLANATION. Begin at the right-hand, or units, column to subtract. We cannot 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 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 cannot take 8 from 4, and must, therefore, borrow 1 from 8 in thousands column. Since 1 thousand = 10 hundreds, 10 hundreds + 4 hundreds = 14 himdreds, and 8 from 14 ='C, the third, or hundreds, figure in the remainder. 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 performed by mentally placing 1 before the figure following the one from which it is borrowed. In the above example the 1 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 the next cipher, making 10 ; but as we 'have borrowed 1 from this column and have taken it to the units column, only remains from which to subtract 6 ; 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. Place the subtrahend (or smaller number] under the minuend (or larger number), in the same manner as for addition, and proceed as in Arts. 37, 38, and 39. ARITHMETIC. 11 41. Proof. To prove an example in subtraction, add the subtrahend and the remainder. The sum should 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 8763 remainder 1237 minuend 10000 42. w (/) (*) (*) EXAMPLES FOR PRACTICE. From: 94,278 take 62,574. .53,714 take 25,824. 71,832 take 58,109. 20,804 take 10,408. . 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. (d) w (/") Or) 31,704. 27,890. 13,723. 10,396. 208,324. 67,253. 17,484. 4. MULTIPLICATION. 43. To multiply a number is to add it to itself a certain number of times. 44. Multiplication is the process of multiplying one mimber 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 multi- plicand 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 multiplied by 6. 46. It matters not in what order the numbers to be multi- plied together are placed. Thus, 6 X 9 is the same as 9 X 6. ARITHMETIC. . In the following table, the product of any two num- bers (neither of which exceeds 12) may be found : 1 times 1 is 1 2 times 1 is 2 3 times 1 is 3 1 times 2 is 2 2 times 2 is 4 3 times 2 is 6 1 times 3 is 3 2 times 3 is 6 3 times 3 is 9 1 times 4 is 4 2 times 4 is 8 3 times 4 is 12 1 times 5 is 5 2 times 5 is 10 3 times 5 is 15 1 times 6 is 6 2 times 6 is 12 3 times 6 is 18 1 times 7 is 7 2 times 7 is 14 3 times 7 is 21 1 times 8 is 8 2 times 8 is 16 3 times 8 is 24 1 times 9 is 9 2 times 9 is 18 3 times 9 is 27 1 times 10 is 10 2 times 10 is 20 3 times 10 is 30 1 times 11 is 11 2 times 11 is 22 3 times 11 is 33 1 times 12 is 12 2 times 12 is 24 3 times 12 is 36 4 times 1 is 4 5 times 1 is 5 6 times 1 is 6 4 times 2 is 8 5 times 2 is 10 6 times 2 is 12 4 times 3 is 12 5 times 3 is 15 6 times 3 is 18 4 times 4 is 16 5 times 4 is 20 6 times 4 is 24 4 times 5 is 20 5 times 5 is 25 6 times 5 is 30 4 times 6 is 24 5 times 6 is 30 6 times 6 is 36 4 times 7 is 28 5 times 7 is 35 6 times 7 is 42 4 times Sis 32 5 times 8 is 40 6 times 8 is 48 4 times 9 is 36 5 times 9 is 45 6 times 9 is 54 4 times 10 is 40 5 times 10 is 50 6 times 10 is 60 4 times 11 is 44 5 times 11 is 55 6 times 11 is 66 4 times 12 is 48 5 times 12 is 60 6 times 12 is 72 7 times 1 is 7 8 times 1 is 8 9 times 1 is 9 7 times 2 is 14 8 times 2 is 16 9 times 2 is 18 7 times 3 is 21 . 8 times 3 is 24 9 times 3 is 27 7 times 4 is 28 8 times 4 is 32 9 times 4 is 36 7 times 5 is 35 8 times 5 is 40 9 times 5 is 45 7 times 6 is 42 8 times 6 is 48 9 times 6 is 54 7 times 7 is 49 8 times 7 is 56 9 times 7 is 63 7 times 8 is 56 8 times 8 is 64 9 times 8 is 72 7 times 9 is 63 8 times 9 is 72 9 times 9 is 81 7 times 10 is 70 8 times 10 is 80 9 times 10 is 90 7 times 11 is 77 8 times 11 is 88 9 times 11 is 99 7 times 12 is 84 8 times 12 is 96 9 times 12 is 108 10 times 1 is 10 11 times 1 is 11 12 times 1 is 12 10 times 2 is 20 11 times 2 is 22 12 times 2 is 24 10 times 3 is 30 11 times 3 is 33 12 times 3 is 36 10 times 4 is 40 11 times 4 is 44 12 times 4 is 48 10 times 5 is 50 11 times 5 is 55 12 times 5 is 60 10 times 6 is. 60 11 times 6 is 66 12 times 6 is 72 10 times 7 is 70 11 times 7 is 77 12 times 7 is 84 10 times 8 is 80 11 times 8 is 88 12 times 8 is 96 10 times 9 is 90 11 times 9 is 99 12 times 9 is 108 10 times 10 is 100 11 times 10 is 110 12 times 10 is 120 10 times 11 is 110 11 times 11 is 121 12 times 11 is 132 10 times 12 is 120 11 times 12 is 132 12 times 12 is 144 This table should be carefully committed to memory. Since has no value, the product of and any number is 0. 1 ARITHMETIC. 13 48. To multiply a number by one iigure 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 right-hand 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 multi- plicand, 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 multiplicand by the multiplier, 5, we see that 5 times 2 tens are 10 tens, and 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 hundred 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 hun- dreds are 20 hundreds, and 20 hundreds plus the 1 hundred re- served are 21 hundreds, or 2 thousands and 1 hundred, which we write in thousands and hundreds places, respectively. Hence, the product is 2,125. This result is the same as adding 425 five times. Thus, 425 425 425 425 425 sum 2125 Ans. EXAMPLES FOB PRACTICE. 49. Find the product of : (a) 61,483 X 6. (&) 12,375 X 5. (c) 10,426 X 7. (d) 10,835X3. Ans. (a) 368,898. () 61,875. (c) 72,982. (d) 32,505. 14 ARITHMETIC. ( remainder EXPLANATION. 7 is contained in 8 hundreds, 1 hundred times. Place the 1 as the first, or left-hand, figure of the quotient. Multiply the divisor, 7, by the 1 hundred of the 1 ARITHMETIC. 17 quotient, and place the product, 7 hundreds, under the 8 hundreds in the dividend, and subtract. Beside the re- mainder, 1, bring down the next, or tens, figure of the dividend, in this case 7, making 17 tens; 7 is contained in 17, 2 times. Write the 2 as the second figure of the quo- tient. Multiply the divisor, 7, by the 2 in the quotient, and subtract the product from 17. Beside the remainder, 3, bring down the units figure 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 1 7 3 5 quotient 125 Ans. The mental operation is as follows: 7 is contained in 8, once and 1 remainder; imagine 1 to be placed before 7, making 17; 7 is contained in 17, 2 times and 3 over; imag- ine 3 to be placed before 5, making 35 ; 7 is contained in 35, 5 times. These partial quotients, placed in order as they are found, make the entire quotient 125. 59. If the divisor consists of two or more figures, pro- ceed as in the following example : EXAMPLE. Divide 2,702,826 by 63. divisor dividend qttotient SOLUTION. 63)2702826(42902 Ans. 252 18 ARITHMETIC. 1 EXPLANATION. As 63 is not contained in the first two figures, 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. Multiply 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, 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 divi- sor, 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 prod- uct, 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 quotient. 6O. Rule. I. Write the divisor at the left of the divi- dend^ ivith a line between them. II. Find- how 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. III. Multiply the divisor by this quotient; write the prod- uct under 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. IV. 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. ARITHMETIC. 19 V. If there be at last a remainder, write it after the quotient, ivith the divisor underneath. 61. Proof. Multiply the quotient by the divisor and add the remainder, if there be any, to the product. The result ivill be the dividend. Thus, divisor dividend quotient 63 ) 4235 ( 67| Ans. 378 PROOF. remainder quotient divisor remainder dividend 455 441 14 67 63 201 402 4221 14 4235 EXAMPLES FOB PRACTICE. 63. Divide the following: (a) 126,498 by 58. (a) 2,181. (*) 3,207,594 by 767. w 4,182. to 11,408,202 by 234. to ^8,753. (d) 2,100,315 by 581. . ( i x I = TT>> and > if I =-yV> A must equal f Hence, dividing both terms of a fraction by the same number does not alter its value. 1 ARITHMETIC. 25 87. A fraction is reduced to its lowest terms when its numerator and denominator cannot both be divided by the same number without a remainder ; for example, f , f , \\, T 8 T . EXAMPLES FOB PRACTICE 88. Reduce the following: (a) T \ to 128ths. () ^ to its lowest terms. to its lowest terms. Ans. (d) f to 49ths. (e) jf to 10,000ths. (') (') If- 89. To reduce a whole number or a mixed number to an improper fraction : EXAMPLE. How many fourths in 5 ? SOLUTION. Since there are 4 fourths in 1 ( = 1), in 5 there will be 5 X 4 fourths, or 20 fourths ; i. e. , 5 X f = \- Ans. EXAMPLE. Reduce 8f to an improper fraction. SOLUTION. 8 X | = -7-- \ 2 - + l = - 3 5 . Ans. 90. Rule. Multiply the wJiole number by the denomina- tor of the fraction, add the numerator to the product and place the denominator under the result. If it is desired to reduce a ivhole number to a fraction, multiply the whole number by the denominator of the given fraction, and write the result over the denominator. EXAMPLES FOR PRACTICE. 91. Reduce to improper fractions : (a) (b) 5. w (d) 37f. Ans< * (e) 60|. (/) Reduce 7 to a fraction whose denominator is 16. I (/) 92. To reduce an improper fraction to a \vhole or a mixed number: EXAMPLE. Reduce ^ to a mixed number. SOLUTION. 4 is contained in 21, 5 times and 1 remaining (see Art. 75) ; as this is also divided by 4, its value is . Therefore, 5 + J, or 5, is the number. Ans. 26 ARITHMETIC. 1 93. Rule. Divide the numerator by the denominator, the quotient will be the whole 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. 94. Reduce to whole or mixed numbers: () JL t JL - (') It- (/) W- (a) 24. (b) 61f Ans " ^ (d) 49f. (/) 5. 95. A common denominator of two or more fractions is a number which will contain (i. e. , which may be divided by) the denominator of each of the given fractions without a remainder. The least common denominator is the least number that will contain each denominator of the given fractions without a remainder. 96. To find the least common denominator: EXAMPLE. Find the least common denominator of , ^, , and T ^. SOLUTION. We first place the denominators in a row, separated by commas. 2 ) 4, 3, 9, 16 2 ) 2, 3, 9, 8 3 7T ^ 9^ 4 3 ) 1, 1, 3, 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 remainder (if possible), bringing down those denominators to the row below which will not contain the divisor without a remainder. Dividing each of the numbers by 2, the sec- ond row becomes 2, 3, 9, 8, since 2 will not divide 3 and 9 without a remainder. Dividing again by 2, the result is 1, 3, 9, 4. Dividing the third row by 3, the result, is 1, 1, 1 ARITHMETIC. 27 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. 97. EXAMPLE. Find the least common denominator of f, ^, T 7 . SOLUTION. 3 ) 9, 12, 18 3 ) 3, 4, 6 2)1, 4. 2 2 ) 1, 2, 1 1, 1, 1 3X3X2X2 = 36. Ans. 98. To reduce two or more fractions to fractions having a common denominator: EXAMPLE. Reduce f , f , and to fractions having a common denom- inator. 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 remainder. 2 _ s 39 i _ 6 3 ~i~s> t T> Ti~- Reducing f (see Art. 84), 3 is contained in 12, 4 times. By multi- plying both numerator and denominator of f by 4, we find 2X4 8 3X4 = 12- Intl 99. Rnle. Divide the common denominator by the denominator of the given fraction, and multiply botJi terms of the fraction by the quotient. EXAMPLES FOR PRACTICE. 1OO. Reduce to fractions having a common denominator; () f . f . f Ans (') -A. A. A- (/) TVU.fi (') if. A. if ADDITION OF FRACTIONS. 1O1. Fractions cannot be added unless they have a com- mon denominator. We cannot add f to \ as they now stand, since the denominators represent parts of different sizes. Fourths cannot be added to eighths. 28 ARITHMETIC. 1 Suppose we .divide an apple into 4 equal parts, and then divide 2 of these parts into 2 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 = 6 something. But what is this something ? It is not fourths, for 6 fourths are 1|-, and we had only 1 apple fo begin with; neither is it eighths, for 6 eighths are f , which is less than 1 apple. By reducing the quarters to eighths, we have f = |-, and adding the other 4 eighths, 4 + 4 = 8 eighths. This result is correct, since -J = 1. Or we can, in this case, reduce the eighths to quarters. Thus, |- = \ ; 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. 1O2. EXAMPLE. Find the sum of , f, and f. SOLUTION. The least common denominator, or the least number which will contain all the denominators, is 8. i = t, 1 = 1. and | = f . EXPLANATION. As the denominator tells or indicates the names of the parts, the numerators only are added, to obtain the total number of parts indicated by the denominator. Thus, 4 one-eighths plus 6 one-eighths plus 5 one-eighths = 1O3. EXAMPLE. What is the sum of 12|, 14f, and 7 T 5 ? SOLUTION. The least common denominator in this case is 16. 12* = 12H 14f = 14ft 7* = _*&_ sum ' 33 + fl = 33 + 1ft = 34ft. Ans. The sum of the fractions = f or 1ft, which added to the sum of the whole numbers = 34ft. EXAMPLE. What is the sum of 17, 13 T \, &, and 8J? SOLUTION. The least common denominator is 32. 13^ = Si = 8. 17 18* _ sum 33||. Ans. 1 ARITHMETIC. 29 1O4. Rule. I. Reduce the given fractions to frac- tions having tJie least common denominator, and write the sum of the numerators over the common denominator. II. When there are mixed numbers and whole numbers, add tJie fractions first, and if their sum is an improper fraction, reduce it to a mixed number and add the whole number with the other whole numbers. EXAMPLES FOR PRACTICE. 1O5. Find the sum of: () f A,f- (*) *, T^ If- (0 *,.!, fV (<0 i ii if- A (') if.sV If. (/) ft, iiif () A. A. if <*) f.if. f (b} IA- (0 ii 3 ^- W iff- (/) if- (/o i!" SUBTRACTION OF FRACTIONS. 1O6. Fractions cannot be subtracted ivithout 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. 13-6 J(f Ans. 107. EXAMPLE. From 7 take |. SOLUTION. 1 = f ; therefore, since 7 = 6 + 1, 7 = 6 + f = 6f , or | = 6f. Ans. 108. EXAMPLE. What is the difference between 17^ and 9|| ? SOLUTION. The common denominator of the fractions is 32. 17^ minuend stibtrahend difference Ans. 30 ARITHMETIC. 1 1O9. EXAMPLE. From 9 take 4 T V SOLUTION. The common denominator of the fractions is 16. 9^ 9 iV minuend 9 T \ or 8f subtrahend 4^ 4 T 7 ^ dijference ~4f ~4f. Ans. EXPLANATION. As the fraction in the subtrahend is greater than the fraction in the minuend, it cannot be sub- tracted; therefore, borrow 1, or if, from the 9 in the minuend and add it to the T \; -j^-hyf TIP yV fr m |f = if. Since 1 was borrowed from 9, 8 remains; 4 from 8 = 4; 4 + lf = 4if. 110. EXAMPLE. From 9 take 8 T \. SOLUTION. minuend 9 or subtrahend 8 T \ difference f| f. Ans. EXPLANATION. As there is no fraction in the minuend from which to take the fraction in the subtrahend, borrow 1, or if, from 9. T 3 g- from if if. Since 1 was borrowed from 9, only 8 is left. 8 from 8 = 0. 111. Rule. I. Reduce the fractions to fractions having 1 a common denominator. Subtract one numerator from the other and place the remainder over the common denominator. II. When there are mixed numbers, subtract the fractions and whole numbers separately, and place the remainders side by side. III. When the fraction in the subtrahend is greater than the fraction in the minuend, borrow 1 from the ivhole number in the minuend and add it to the fraction in the minuend, from which subtract the fraction in the sub- trahend. IV. When the minuend is a whole number, borrow 1; reduce it to a fraction whose denominator is the same as the' denominator of the fraction in the subtralicnd, and place it over that fraction for subtraction. 81 ARITHMETIC. 31 EXAMPLJE8 FOll PRACTICE. 5. Subtract: (a) from f (&) ^ from ^f () 4 7 from T 5 ^. (rf) ^ from ff (*) if from f f (/) 13 from 30. () 12 from 27. (/*) 5i from 30. (d) (*) to (d?> to 14*. MUIjTIPilCATION OF FRACTIONS. 113. /;/ multiplication of fractions it is not necessary to reduce tJie fractions to fractions having a common denominator. 114. Multiplying the numerator or dividing the denom- inator multiplies the fraction, EXAMPLE. Multiply f by 4. SOLUTION. x4 = X4 = J = 3. Ans. Or, f X4 = ^_ 4 == f = 3. Ans. The word "of," when placed between two fractions, or between a fraction and a whole number, means the same as X , or times. Thus, | of 4 = |X4 = 3. EXAMPLE. Multiply f by 2. SOLUTION. 2X1 = 3 = I = f Ans. o o Or, 2 X I = g ^ o = Ans - 115. EXAMPLE. What is the product of T \ and | ? SOLUTION. A X I 4 V 7 = i X i = A\ = A. Ans. Or, by cancelation, 4X8 =. A- Ans - 116. EXAMPLE. What is f of f of || ? SOLUTION. t x 3 x n _ 3. 8X^X3? "~ 8X2 = A. Ans. ARITHMETIC. EXAMPLE. What is the product of 9f and 5f ? SOLUTION. 9f = ^ ; 5| = \ 5 -. X = f *f = HF = 54M- Ans ' 118. EXAMPLE. Multiply 15| by 3. SOLUTION. 15 15| _3 or 3 47f 45 + \L = 45 + 2f = 47f . Ans. 119. Rule. I. Divide the product of the numerators by the product of the denominators. All factors common to the numerators and denominators should* first be cast out by cancelation. II. To multiply one mixed number by another, reduce them both to improper fractions. III. To multiply a mixed number by a whole number, first multiply the fractional part by the multiplier, and if the prod- uct is an improper fraction, reduce it to a mixed number and add the whole-number part to the product of the mul- tiplier and the whole number. EXAMPLES FOR PRACTICE. 1 2O. Find the product of : () 7 X T V (b) 14 X T V if. 4. Ans. (/) 17HX7. (jsr) ill X 32. (/) 125. DIVISION OF FRACTIONS. In division of fractions it is not necessary to reduce the fractions to fractions having a common denominator. 122. Dividing the numerator or multiplying the denom- inator divides the fraction. EXAMPLE. Divide f by 3. SOLUTION. When dividing the numerator, we have ^8 = 3 = = Ans. 1 ARITHMETIC. 33 When multiplying the denominator, we have A iK 3 = g x3 = -& = \. Ans. EXAMPLE. Divide T 3 ^ by 2. o SOLUTION. & -4- 2 = ^ ^ g = ^V Ans. EXAMPLE. Divide ^f by 7. 14 7 SOLUTION. ~^ 7 = = = Ans - 123. To invert a fraction is to /^r^ zV upside down ; that is, make the numerator and denominator change places. Invert f and it becomes f . 124. EXAMPLE. Divide T 9 W by T V SOLUTION. 1. The fraction T 3 is contained in T 9 ^, 3 times, for the denominators are the same, and one numerator is contained in the other 3 times. 2. If we now invert the divisor, T 3 ^, and multiply, the solution is 9 16_!xtf_ 8 Ang 16 X 3 - W><$ This brings the same quotient as in the first case. 125. EXAMPLE. Divide f by . SOLUTION. We cannot divide f by , as in the first case above, for the denominators are not the same ; therefore, we must solve as in the second case. l^f=fXf =|^=|orli. Ans. 2 126. EXAMPLE. Divide 5 by $. SOLUTION. -^ inverted becomes ^. 127. EXAMPLE. How many times is 3f contained in SOLUTION. 3f = ^; 7 T 7 = Vir- 2^- inverted equals T ^. 119 4 119 X^ 119 16 X 15~^ X 15~ 60 = 4 128. Rule. Invert the divisor and proceed as in mul- tiplication. 34 ARITHMETIC. 1 129. 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. 9 3x7 - means that 9 is to be divided by ; means that 8 -T-4 16 3 X 7 is to be divided by the value of ?~ . 2 is the same as i -=- . 8 It will be noticed that there is a heavy line between the 9 and the . This is necessary, since otherwise there would be nothing to show as to whether 9 was to be divided by , or f was to be divided by 8. Whenever a heavy line is used, as shown here, it indicates that all above the line is to be divided by all below it. EXAMPLES FOR PRACTICE. 130. Divide: (a) 15by6f. (*) 30byf. (c) 172 by f . ( regarding the fraction 9X4 above the heavy line as the numerator of a fraction whose denominator is 9, \ , as before. J X 4 J X 4 1 ARITHMETIC. 35 9 9x4 2. - = 5 = 12. The proof is the same as in the first f o case. t5x4 = Q o = IT- For, regarding % as the numerator ^ o x y of a fraction whose denominator is , | = - - ; and if x y o x y 5 X4 5X4 4 This principle may be used to great advantage in cases . Reducing the mixed numbers to r .. . T, T fractions, the expression becomes 4 12 31 - . Now 40 X Y X -g- transf erring the denominators of the fractions and canceling, 3 n s 3 1x310x27x72x2x6 40X9X31X4X12 Greater exactness in results can usually be obtained by using this principle than by reducing the fractions to deci- mals. The principle, however, should not be employed if a sign of addition or subtraction occurs either above or below the dividing line. DECIMALS. 132. Decimals are tenth fractions; that is, the parts of a unit are expressed on the scale of ten, as tenths^ hun- dredths, thousandths, etc. 133. The denominator which is always ten or a multiple of ten, as 10, 100, 1,000, etc., is not expressed, as it would 36 ARITHMETIC. 1 be in common fractions, by writing- it tinder the numerator with a line between them, as y 3 ^, T f ^ -j-^, but is expressed by placing a period ( .), which is called a decimal point, to the left of the figures of the numerator , so as to indicate that the number on the right is the numerator of a fraction whose denominator is 10, 100, 1,000, etc. 134. The reading of a decimal number depends upon the number of decimal places in it, or the number of figures to the right of the decimal point. One decimal place expresses tenths. Two decimal places express hundredths. Three decimal places express thousandths. Four decimal places express ten-thousandths. Five decimal places express hundred-thousandths. Six decimal places express millionth*. Thus: .3 = A- = 3 tenths. .03 = rihr = 3 hundredths. .003 = -j-^pj- = 3 thousandths. .0003 = TTn nnr = 3 ten-thousandths. .00003 = -nnnnnr = 3 hundred-thousandths. .000003 y^nmnnr = 3 million ths. We see in the above that the number of decimal, places in a decimal equals the number of ciphers to the right of the figure 1 in the denominator of its equivalent fraction. This fact kept in mind will be of much assistance in reading and writing decimals. Whatever may be written to the left of a decimal point is a whole number. The decimal point merely sep- arates the fraction on the right from the whole number on the left. When a whole number and decimal are written together, the expression is a mixed number. Thus, 8. 12 and 17.25 are mixed numbers. g 1 ARITHMETIC. 37 The relation of decimals and whole numbers to each other is clearly shown by the following table : The figures to the left of the decimal point represent whole numbers; those to the right are decimals. In both the decimals and whole numbers, the units place is made the starting point of notation and numeration. Both whole numbers and decimals decrease on the scale of ten to the right, and both increase on the scale of ten to the left. The first figure to the left of units is tens, and the first figure to the rigJit of units is tenths. The second figure to the left of units is hundreds, and the second figure to the ' rigJit is hundredths. The third figure to the left is thousands ^ and the third to the right is thousandths^ and so on ; the iv hole numbers on the left and the decimals on the right. The figures equally distant from units place correspond in name, the decimals having the ending ths, to distinguish them from whole numbers. The following is the numeration of the number in the above table : nine hundred eighty-seven million, six hundred fifty-four thousand, three hundred twenty-one and twenty-three million, four hundred fifty-six thousand, seven hundred eighty-nine hundred-millionths. The decimals increase to the left, on the scale of ten, the same as whole numbers; for, if you begin at the 4 in thousandths place in the above table, the next figure to the left is hundredths, which is ten times as great, and the next tenths, or ten times the hundredths, and so on through both decimals and whole numbers. 38 ARITHMETIC. 1 135. Annexing^ or taking aivay, a cipher at the right of a decimal, does not affect its value. .5 is ^; .50 is ^V but T 5 o = f^r ; therefore, .5 =.50. 136. Inserting a cipher between a decimal and the decimal point, divides the decimal by 10. 5 = A; A-*- 10 = ^=.05. 137. Taking away a cipher from the left of a decimal, multiplies the decimal by 10. lO - =.5. 138. In some cases it is convenient to express a mixed decimal fraction in the form of a common (improper) frac- tion. To do so it is only necessary to write the entire num- ber, omitting 1 the decimal point, as the numerator of the fraction, and the denominator of the decimal part as the denominator of the fraction. Thus, 127.483 = ^-f^f 1 ; for, 127.483 = ADDITION OF DECIMALS. 139. Addition of decimals is similar in all respects to addition of whole numbers units are placed under units, tens under tens, etc. ; this, of course, brings the decimal points in line, directly under one another. Hence, in pla- cing the numbers to be added, it is only necessary to take care that the decimal points are in line. In adding whole numbers, the right-hand figures are always in line; but in adding decimals, the right-hand figures will not be in line unless each decimal contains the same number of figures. wltole numbers decimals mixed numbers 342 .342 342.032 4234 .4234 4234.5 26 .26 26.6782 3_ .03 3.06 sum 4605 Ans. sum 1.0 5 5 4 Ans. sum 4 6 6.2 7 2 Ans. ARITHMETIC. 39 14O. EXAMPLE. What is the sum of 242, .36, 118.725, 1.005, 6, and 100.1? SOLUTION. 242. .36 1 1 8.7 2 5 1.005 6. 100.1 sum 468.190 Ans. 141. Rule. Place the numbers to be added so that the decimal points will be directly under each other. Add as in wJiole numbers, and place the decimal point in the sum directly under the decimal points above. EXAMPLES FOR PRACTICE. Find the sum of: (a) .2143, .105, 2.3042, and 1.1417. () 783.5, 21.473, .2101, and .7816. (c) 21.781, 138.72, 41.8738, .72, and 1.413. (d) .3724, 104.15, 21.417, and 100.042. (*) 200.172, 14.105, 12.1465, .705, and 7.2. (/) 1,427.16, .244, .32, .032, and 10.0041. () 2,473.1, 41.65, .7243, 104.067, and 21.073. (/i) 4,107.2, .00375, 21.716, 410.072, and .0345. w M (<*) M 00 te) .(*) 3.7652. 805.9647. 204.5078. 225.9814. 234.3285. 1,437.7601. 2,640.6143. 4,539.02625. SUBTRACTION OF DECIMALS. 143. As in subtraction of whole numbers, units are placed under units, tens under tens, etc., bringing- the decimal points- under each other, as in addition of decimals. EXAMPLE. Subtract .132 from .3063. SOLUTION. minuend .3063 subtrahend .132 difference .1743 Ans. . 144. EXAMPLE. What is the difference between 7.895 and .725 ? SOLUTION. minuend 7.8 9 5 subtrahend .725 difference 7.1 70 or 7.1 7. Ans. 40 ARITHMETIC. 1 145. EXAMPLE. Subtract .625 from 11. SOLUTION. minuend 1 1.0 subtrahend .625 difference 1 0.3 7 5 Ans. 146. Rule. Place the subtrahend under the minuend, so that the decimal points will be directly under each other. Subtract as in whole numbers, and place the decimal point in the remainder directly under the decimal points above. When the figures in the decimal part of the subtrahend extend beyond those in the minuend, place cipliers in the min- uend above them and subtract as before. EXAMPLES FOR PRACTICE. 147. From: (a) 407.385 take 235.0004. () 22. 718 take 1.7042. (c) 1,368. 17 take 13. 6817. (d) 70.00017 take 7.000017. ( r -f? of a foot. This reduced to a decimal by the above rule shows what decimal part of a foot 9 inches is. 1 2 ) 9.0 ( .7 5 of a foot. Ans. 84 60 6_0 168. Rule. I. To reduce inches to a decimal part of a foot, divide the number of inches by 12. II. 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 further, and if the fifth -figure is 5 or greater increase the fourth figure by 1, omitting the signs -f- and . (*) .25 ft. (^) .375 ft. Ans. w .4167 ft. (rtT) .5521 ft. w .9167 ft. EXAMPLES FOR PRACTICE. 160. Reduce to the decimal part of a foot: (a) 3 in. () 4 in. (c) 5 in. (d) 6f in. (e) 11 in. TO REDUCE A DECIMAL TO A FRACTION. 1 7O. EXAMPLE. Reduce .125 to a fraction. SOLUTION. .125 = ^ffo = ^ = . Ans. EXAMPLE. Reduce .875 to a fraction. SOLUTION. .875 = /ow = ft = I- Ans. 171. Rule. Under the figures of the decimal, place 1 with as many ciphers at its right as there are decimal places in the decimal, and reduce the resulting fraction to its lozvest terms by dividing both numerator and denominator by the same number. 48 ARITHMETIC. 172. EXAMPLES FOR PRACTICE. Reduce the following to common fractions: (a) .125. .625. .3125. .04. (e) .06. (/) -75. (-) .15625. (A) .875. (d) Ans. () (*) (') f (') iV (/) f teO A" (*) f 173. To express a decimal approximatively as a fraction having a given denominator: 174. EXAMPLE. Express .5827 in 64ths. SOLUTION. .5827 X f = 87 'ff 28 , say f|. o4 Hence, .5827 = f|, nearly. Ans. EXAMPLE. Express .3917 in 12ths. 4 7004 SOLUTION. .3917 X if = ' 2 , say T 5 . Hence, .3917 = T 5 ^, nearly. Ans. 175. Rule. Reduce 1 to a fraction having the given denominator. Multiply the given decimal by the fraction so obtained, and the^result will be the fraction required. EXAMPLES FOR PRACTICE. 176. Express: (*) M .625 in 8ths. .3125 in 16ths. .15625 in 32ds. .77in64ths. .81 in 48ths. .923 in 96ths. Ans. (a) (*) (d) ft 177. The sign for dollars is $. It is read dollars. 125 is read 25 dollars. Since there are 100 cents in a dollar, 1 cent is 1 one-hun- dredth of a dollar ; the first two figures of a decimal part of g 1 ARITHMETIC. 49 a dollar represent cents. Since a mill is y 1 ^ of a cent, or TTJ ? T5 . Tr of a dollar, the third figure represents mills. Thus, $25.16 is read twenty-Jive dollars and -six teen cents; $25.168 is read tzventy-five dollars sixteen cents and eight mills. SYMBOLS OF AGGREGATION. 178. The vinculum , parenthesis ( ), brackets [ ], and brace { } are called symbols of aggregation, and are used to include numbers which are to be considered together ; thus, 13X8 3, or 13 X (8 3), shows that 3 is to be taken from 8 before multiplying by 13. 13x(8-3) = 13X5 = 65. 13X8^ = 13X5 = 65. When the vinculum or parenthesis is not used, we have 13X8-3 = 104-3 = 101. 179. In any series of numbers connected by the signs -f-, , X, and -T-, 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 follows the number on the rigJit of a sign of addition or subtraction until the indicated multiplication or division has been performed. In all cases the sign of multiplication 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 regarded as factors of the product indicated, and not as separate numbers. EXAMPLE. What is the value of 4 X 24 8 + 17? SOLUTION. Performing the operations in order from left to right, 4X24 = 96; 96-8 = 88; 88 + 17 = 105. Ans. 180. EXAMPLE. What is the value of the following expression: 1,296 -4- 12 + 160 - 22 X 3 = ? SOLUTION. 1,296-*- 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. 50 ARITHMETIC. 1 Had the above expression been written 1,296-^-12 + 160 22 X 3 -4- 7 -f- 25, it would have been necessary to have divided 22 X 3^ by 7 before subtracting, and the final result would have been 22 X3 = 77; 77-*- 7 = 11; 268-11 = 257; 257 + 25 = 282. Ans. In other words, it is necessary to perform all the indicated multiplication or division included between the signs -j- and , or and -j- , before adding or subtracting. Also, had the expression been written 1,296 -T- 12 + 160 24| -+ 7 X 3| + 25, it would have been necessary to have multiplied 3^ by 7 before dividing 24-J-, since the sign of multiplication takes the precedence, and the final result would have been 3|X7 = 24|; 24|-+24| = 1; 268 1 = 267; 267 + 25 = 292. Ans. It likewise follows that if a succession of multiplication and division signs occur, the indicated operations must not be performed in order, from left to right the multiplication must be performed first. Thus, 24x3-=-4x2-+9x5 = . 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 iised. Thus, by using two vinculums the last expression becomes 24x3-f-4x2-=-9x5 = 20, the same result that would be obtained by performing the indicated operations in order, from left to right. EXAMPLES VOR PRACTICE. 181. Find the values of the following expressions : (a) (b) 5X24-32. (c) 5 X 24 -*- 15. (tt) 144 _ 5 X 24. (e) (1,691 -540 + 559)-- 3X57. Ans> (/) 2,080 + 120-80X4-1,670. (g} (90 + 60 -r- 25) X 5 - 29. (//) 90 + 60 -s- 25X5. (a) 3. (b) 88. (0 8. (d) 24. (e) 10. (/) 210. 1.2. ARITHMETIC. (CONTINUED.) PERCENTAGE. 1. Percentage, is the process of calculating by hun- dredths. 2. The term per cent, is an abbreviation of the Latin words per centum, which mean by the hundred. A certain per cent, of a number is the number of hundredths of that number which is indicated by the number of units in the percent. Thus, 6 per cent, of 126 is 125 X^Q = 7.5; 25 per cent, of 80 is 80 X T 2 ^- = 20 ; 43 per cent, of 432 pounds fs 432XiV\r = 185.76 pounds. 3. The sign of per cent, is fa and is read per cent. Thus, 6$ is read six per cent.; V&\% 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 Qfa 25$, and 43$ as yJ 7 , T 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 : Per Cent. Decimal. Fraction. Per Cent. Decimal. Fraction. 1$ 01 T^TT 150$ 1 50 44 or 14 2% 02 ITR) 2 orTTf 500$ 5 00 TTTTT or 5 5$ 05 j> or -rV 1$ 0025 \__ or T-irj 10$ 10 10 or l 1$ . .005 . i_. or -sijc 25$ .25 25 or 1 u$ .015 Ji_ or * A 50$. . .50 T 5 nn or i 8i$.. .081 2L or JL 75$ .75 T 7 A or | 12^$ .125 I (TIT 12 -^Eor i 100$ 1 00 4S or 1 16 8 % 16| THO 16 i or i 125< 1 25 *%* or 14 624$ 625 Ttffr ^ 62 ior | iTJff Wi A w-g/v. . Tirti * 2 ARITHMETIC. 8 4. The names of the different elements used in percent- age are: the base, the rate per cent., the percentage, the amount, and the difference. 5. The base is the number on which the per cent, is comptited. 6. The rate is the number of hundredths of the base to be taken. 7. The percentage is the part, or number of him- drcdtlis, of the base indicated by the rate ; or, the percentage is the result obtained by multiplying the base by the rate. Thus, when it is stated that 7# of $25 is $1.75, $25 is the base, 7# is the rate, and $1. 75 is the percentage. 8. The amount is the sum of the base and percentage. 9. 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 obtaining useful work is 125 125 X.I 6 = 125 20 = 105 horsepower. Here 125 is the base; 16$, the rate; 20, the percentage; and 105, the difference. 10. From the foregoing it is evident that to find the percentage, the base must be multiplied by the rate. Hence, the following Rule. To find the percentage, multiply the base by the rate expressed decimally. EXAMPLE. Out of a lot of 300 bushels of apples 76# were sold. How many bushels were sold ? SOLUTION. 76$, the rate, expressed decimally, is .76; the base is 300; hence, the number of bushels sold, or the percentage, is, by the above rule, 300 X .76 = 228 bushels. Ans. Expressing the rule as a Formula, percentage = base^rate. 2 ARITHMETIC. 3 11. When the percentage and rate are given, the base may be found by dividing the percentage by the rate. For, suppose that 12 is G#, or -j-J^, of some number; then \%, or j-J^, of the number, is 12 -i- 6, or 2. Consequently, if 2 .= Itf, or y^, 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. 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. Bought a certain number of bushels of apples and sold r them. If I sold 228 bushels, how many bushels did I buy ? SOLUTION. Here 228 is the percentage, and 76, or .76, is the rate; hence, applying the rule, 228 -f- .76 = 300 bushels. Ans. 12. When the base and percentage are given, to find the rate, the rate may be found, expressed decimally, by divi- ding the percentage by the base. For, suppose that it is desired to find what per cent. 12 is of 200. 1# of 200 is ^>0 x .01 = 2. Now, if \% is 2, 12 is evidently as many per cent, as the number of times that 2 is contained in 12, or 1 2 -j- 2 = . 6^. But the same result may be obtained by dividing 12, the percentage, by 200, the base, since 12 -f- 200 = .0fj = 6#. Hence, Rule. When the percentage and base are given, to find the rate, divide the percentage by the base, and the result will be the rate expressed decimally. Formula, rate = percentage -5- base. E x AMPLE. Bought 300 bushels of apples and sold 228 bushels. What per cent, of the total number of bushels was sold ? SOLUTION. Here 300 is the base and 228 is the percentage; hence, applying rule, rate = 228-^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 -- 875 = .OSf = 2f*. Ans. PROOF. 875 X .<&f = 25. ARITHMETIC. 2 EXAMPLES FOR PRACTICE. 13. What per cent, of: (a) (*) (') 360 900 125 150 is is is is 90? 360? 25? 750? (*) 280 is 112? (/) 400 is 200? te) 47 is 94 ? 500 is 250 ? (e) 40#. (/) 50#. (A'') 200#. 14. The amount may be found, when the base and rate are given, by multiplying- the base by 1 plus the rate, expreSvSed decimally. For, suppose that it is desired to find the amount when 200 is the base and G$ is the rate. The percentage is 200 X- 06 = 12, and, according to definition, Art. 8, the amount is 200 + 12 = 212. But, the same result may be obtained by multiplying 200 by 1-f.OG, or LOG, since 200 X LOG = 212. Hence, Rule. When the base and rate are given, to find the amount, multiply the base by 1 plus the rate expressed decimally. Formula, amount = basexQ -\-ratc). 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, 725X1.10 = $797.50. Ans. 15. When the base and rate are given, the difference may be found by multiplying the base by 1 minus the rate expressed decimally. For, suppose that it is desired to find the difference when the base is 200 and the rate is 6#. The percentage is 200 X. 06 = 12; and, according to definition, Art 9, the difference = 200 12 = 188. But, the same result may be obtained by multiplying 200 by 1 .06, or .94, since 200 X. 94 = 188. Hence, Rule. When the base and rate are given, to Jin d the differ- ence, multiply the base by 1 minus the rate expressed decimally. Formula, difference = base X(L rate). ' 2 ARITHMETIC. 5 EXAMPLE. Bought 300 bushels of apples and sold all but 24$ of them. How many bushels were sold ? 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 bushels. Ans. 16. 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 G$ of itself. Then, it is evident that 212 equals 106$ of the number (base) that it is desired to find. Conse- 010 quently, if 212 = 10G$, 10 = = 2, and 100$ = 2x100 = 200 = the base. But the same result may be obtained by dividing 212 by l + .OG, or LOG, since 212 4- LOG = 200. Hence, Rule. When the amount and rate are given^ to find the base, divide the amount by 1 plus the rate expressed decimally. Formula, base = amount -f- (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, and 25$ is the rate. Applying rule, actual discharge = 278,910 -* 1.25 = 223,128 gallons. Ans. 17. 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 G$ of itself. Then, 188 evidently equals 100 6 = 94$ of some number. Consequently, if 188 = 94$, 1$ = 188-^-94 = 2, and 100$ = 2x100 = 200. But the same result may be obtained by dividing 188 by 1 . OG, or . 94, since 188 -f- . 94 = 200. Hence, Rule. When the difference and rate are given, to find the base, divide the difference by 1 minus the rate expressed decimally. Formula, base difference -4- (1 rate). 1-5 6 ARITHMETIC. 2 EXAMPLE. Bought a certain number of bushels of apples and sold 76$ of them. If there were 72 bushels left unsold, how many bushels did I buy ? SOLUTION. Here 72 is the difference and 76$ is the rate. Applying rule, 72 -*- (1 - .76) == 300 bushels. Ans. ' EXAMPLE. The theoretical number of foot-pounds of work per min- ute 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 -T- (1 - .30) = 181,920 foot-pounds. Ans. 18. EXAMPLE. A certain chimney gives 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 per- centage, and it is required to find the rate. Hence, applying the rule given in Art. 12, gain per cent. = .24-4-2.76 = .087 = 8.7$. Ans. 19. EXAMPLE. A certain chimney gave a draft of 3 inches of water. After an economizer had been put in, the draft was reduced to 1.2 inches of water. What was the loss per cent.? SOLUTION. Here it is evident that 1.2 inches is the difference (since it equals 3 inches diminished by a certain per cent, loss of itself), and 3 inches is the base. Consequently, 3 1.2 = 1.8 inches is the percent- age. Hence, applying the rule given in Art. 12, loss per cent. = 1.8 -*- 3 = .60 = 60$. Ans. 20. To flntl the gain or loss per cent. : Rule. Find the difference between the initial and the final value; divide this difference by the initial value. EXAMPLE. If a man buys a house for $1,860, and some time after- wards builds a barn for 25$ of the cost of the house, does he gain or lose, and how much per cent, if he sells both house and barn for $2,100? SOLUTION. The cost of the barn-was $1,860 X-25 = $465; conse- quently, the initial value, or total cost, was $1,860 + $465 = $2,325. Since he sold them for $2,100 he lost $2,325 $2,100 = $225. Hence, applying rule, 225-^-2,325 = .0968 = 9. 68$ loss. Ans. (a) $112.50. 5.016. 940.8. 2 ARITHMETIC. EXAMPLES FOB PRACTICE. J31 Solve the following : (a) What is 12|$ of $900 ? () What is |$ of 627 ? (;) What is 33$ of 54 ? (a?) 101 is 68f $ of what number ? (e) 784 is 83i$ of what number ? (/) What $ of 960 is 160 ? (g) What $ of $3,606 is $450f ? (>&) What $ of 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 saving of 250 pounds per day. What was the per cent, of coal saved ? Ans. 7$, nearly. 2. If the speed of an engine rtrnning at 126 revolutions per minute should be increased 6|$, how many revolutions per minute would it then make ? Ans. 134.19 rev. 3. The list price of a lot of silk goods is $1,400, of some laces $1,150, and of some calico $340. If 25$ discount was allowed on the silk, 22$ on the laces, and 12^$ on the calico, what was the actual cost of the purchase? Ans. $2,244.50. 4. If I loan a man $1,100, and this is 18|$ 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 increased 14$ and the consumption of coal per horsepower per hour was decreased 20$. If the engine could originally develop 50 horsepower, and required 3^ 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. DE3STOMIKATE LUMBERS. 22. A denominate number is a concrete number, and may be either simple or compotind; as, 8 quarts; 5 feet; 10 inches, etc. 23. A simple denominate number consists of units of but one denomination; as, 1C cents; 10 hours; 5 dollars, etc. 8 ARITHMETIC. 2 24. A compound denominate number consists of units of two or more denominations of a similar kind ; as, 3 yards, 2 feet, 1 inch ; 34 square feet, 57 square inches. 25. 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. 26. 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. 27. Measures are of six kinds : 1. Extension. 4. Time. 2. Weight. 5. Angles. 3. Capacity. 6. . Money or value. MEASURES OF EXTENSION. 28. Measures of extension are used in measuring lengths, distances, surfaces, and solids. LINEAR MEASURE. TABLE. Abbreviation. 12 inches (in.) - 1 foot . . ft. 3 feet ... =1 yard . yd. 5. 5 yards . . = 1 rod . . rd. 40 rods . . . = 1 furlong fur. 8 furlongs . = 1 mile . mi. in. ft. yd. rd. fur. mi. 36= 3 =1 198 = 16| =5.5 = 1 7,920 = 660 =220 =40 = 1 63,360 = 5,280 = 1,760 = 320 = 8 = 1 SURVEYOR'S LINEAR MEASURE. TABLE. 7.92 inches = 1 link li. 25 links = 1 rod rd. 4 rods ) 100 links [ 1 clmm -<& 80 chains = 1 mile mi. mi. ch. rd. li. in. 1 = 80 = 320 = 8,000 = 63,360 29. The linear unit, generally used by surveyors, is Gunter's chain, which is equal to 4 rods, or 66 feet. 2 ARITHMETIC. 9 3O. An engineer's chain, used by civil engineers, is 100 feet long, and consists of 100 links. In computations, the links are written as so many hundredths of a chain. SQUARE MEASURE. TABLE. 144 square inches (sq. in.) . . . 9 square feet . 30 square yards 160 640 square rods acres sq. mi. A. sq. rd. sq. yd. 1 square foot .... sq. ft. 1 square yard . . . . sq. yd. 1 square rod . . . . sq. rd. 1 acre A. 1 square mile .... sq. mi. sq. ft. sq. in. 1 = 640 = 102,400 = 3,097,600 = 27,878,400 = 4,014,489,600 SURVEYOR'S SQUARE MEASURE. TABLE. 625 square links (sq. li.) . . . . = 1 square rod . . 16 square rods = 1 square chain 10 square chains = 1 acre . . . . 640 acres = 1 square mile . . 36 square miles (6 mi. square) . . = 1 township . . . sq. mi. A. sq. ch. sq. rd. sq. li. 1 = 640 = 6,400 = 102,400 = 64,000,000 sq. rd. sq. ch. . A. sq. mi. . Tp. CUBIC MEASURE. TABLE. . = 1. cubic foot , . = 1 cubic yard 1,728 cubic inches (cu. in.) . . 27 cubic feet 128 cubic feet = 1 cord 24f cubic feet = 1 perch cu. yd. cu. ft. cu. in. 1 = 27 = 46,656 cu. ft. cu. yd. . cd. P. MEASURES OF WEIGHT. AVOIRDUPOIS WEIGHT. TABLE. = 1 pound . . 16 ounces (oz.) 100 pounds ......... = 1 hundredweight 20cwt., or 2,000 Ib ...... = 1 ton .... T. cwt. Ib. oz. 1 = 20 = 2,000 = 32,000 Ib. cwt. T. 10 . ARITHMETIC. 2 31. The ounce is divided into halves, quarters, etc. Avoirdupois weight is used for weighing coarse and heavy articles. One avoirdupois pound contains 7,000 grains. LONG TON TABLE. 16 ounces = 1 pound Ib. 112 pounds = 1 hundredweight . . . cwt. 20cwt, or 2,240 Ib. " ......= 1 ton T. 32. In all the calculations throughout this and the fol- lowing sections, 2,000 pounds will be considered 1 ton, unless the long ton (2,24=0 pounds) is especially mentioned. TROY WEIGHT. TABLE. 24 grains (gr.) = 1 pennyweight .... pwt. 20 pennyweights = 1 ounce oz. 12 ounces .' = 1 pound Ib. Ib. oz. pwt. gr. 1 =-. 12 = 240 = 5,760 33. Troy weight is" iised in weighing gold and silver- ware, jewels, etc. It is used by jewelers. MEASURES OF CAPACITY. LIQUID MEASURE. TABLE. 4 gills (gi.) = 1 pint pt. 2 pints ._= 1 quart qt. 4 quarts = 1 gallon gal. 81-J- gallons 1 barrel bbl. 2 barrels, or 63 gallons . = 1 hogshead hhd. hhd. bbl. gal. qt. pt. gi. 1 = 2 = 63 = 252 = 504 = 2,016 DRY MEASURE. TABLE. 2 pints (pt.) = 1 quart ....... qt. 8 quarts = 1 peck pk. 4 pecks = 1 bushel bu. bu. pk. qt. pt. 1 = 4 = 32 = 64 ARITHMETIC. 11 MEASURE OF TIME. TABLE. 60 seconds (sec ) . . . 1 minute .... min. GO minutes . . . . 1 hour . . hr. 24 hours . . . . 1 day . . da. . . 1 week 365 days [ . . . 1 common year . . yr. 12 months j 366 days . . . . . 1 leap year. 100 years . . 1 century. NOTE. It is customary to consider one month as 30 days. MEASURE OF ANGLES Oil ARCS. 60 seconds (") 60 minutes 90 degrees 360 degrees . TABLE. . . .....= 1 minute ' , . = 1 degree . = 1 right angle or quadrant |__. = 1 circle . cir. 1 cir. = 360 = 21,600' = 1,296,000" 10 mills (m.' 10 cents . 10 dimes 10 dollars MEASURE OF MOKEY. UNITED STATES MONEY. TABLE. . = 1 cent . E. 1 = ct. . . . . = 1 dime d. . . . . = 1 dollar $. . . . . = 1 eagle E. $ d. ct. m. 10 = 100 = 1,000 = 10,000 MISCELLANEOUS TABLE. 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 39.37 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 the above tables. 12 ARITHMETIC. 2 REDUCTION OF DENOMINATE NUMBERS. 34. Reduction of denominate numbers is the process of changing- their denomination without changing their value. They may be changed from a higher to a lower denomina- tion, or from a lower to a higher either is reduction. As 2 hours = 120 minutes. 32 ounces = 2 pounds. 35. Principle. Denominate numbers are changed to lower denominations by multiplying, and to higher denom- inations by dividing. To reduce denominate numbers to lower denom- inations : 36. EXAMPLE. Reduce 5 yd. 2 ft. 'fin. to inches. SOLUTION. yd. ft. in. 527 3 15ft. 17ft. 1 2 "34 1 7 2 4 in. 7 in. 211 inches. Ans. EXPLANATION. Since there are 3 feet in 1 yard, in 5 yards there are 5 X3 or 15 feet, and 15 feet plus 2 feet = 17 feet. There are 12 inches in a foot; therefore, 12x17 = 204 inches, and 204 inches plus 7 inches = 211 inches = num- ber of inches in 5 yards 2 feet and 7 inches. 37. EXAMPLE. Reduce 6 hours to seconds. SOLUTION. 6 hours. 60 360 minutes. 60 21600 seconds. Ans. ARITHMETIC. 13 EXPLANATION. As there are 60 minutes in 1 hour, in 6 hours there are 6 X 60, or 360, minutes ; as there are no min- utes to add, we multiply 360 minutes by 60, to get the number of seconds. 38. 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 39. Rule. Multiply the number representing the high- est denomination by the number of units in the next loiver 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 FOB PRACTICE. 4O. Reduce: (a) 4 rd. 2 yd. 2 ft. to ft. (b) 4 bu. 3 pk. 2 qt. to qt. (c) 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. (h) 14 bu. to qt. Ans. - (a) (If) w 74ft. 154 qt. 231.5ft. 28,060 ft. 48,144 gr. 26,401 pt. 108,980'. 448 qt. To reduce lower to higher denominations : 41. EXAMPLE. Reduce 211 inches to higher denominations. SOLUTION. 12 ) 2 1 1 in. 3) 17ft. +7 in. 5 yd. + 2 ft. Ans. EXPLANATION. There are 12 inches in 1 foot; therefore, 211 divided by 12 = 17 feet and 7 inches over. There are 3 feet in 1 yard; therefore, 17 feet divided by 3 = 6 yards 14 ARITHMETIC. 2 and 2 feet over. The last quotient and the two remainders constitute the answer, 5 yards 2 feet 7 inches. 42. EXAMPLE. Reduce 15,735 grains Troy weight to higher denominations. SOLUTION. 24)15735 gr ? (655 pwt. . 144 133 1 20 135 120 - 1 5 gr. 20)655 pwt. ( 3 2 oz. 60 55 40 1 5 pwt. 1 2 ) 3 2 oz. ( 2 Ib. 24 8oz. EXPLANATION. There are 24 grains in 1 pennyweight, and in 15,735 grains there are as many pennyweights as 24 is con- tained in 15,735, or 655 pennyweights and 15 grains remain- ing. There are 20 pennyweights in 1 ounce, and in 655 pennyweights there are 32 ounces and 15 pennyweights remaining. There are 12 ounces in 1 pound, and in 32 ounces there are 2 pounds and 8 ounces remaining. The last quotient and the three remainders constitute the answer, 2 pounds 8 ounces 15 pennyweights 15 grains/ The above problem is worked out by long division, because the numbers are too large to solve easily by short division. The student may use either method. 43. Rule. Divide the number representing the denomi- nation given by the number of units of this denomination required to make one unit of the next higher denomination. The remainder will be of the same denomination, but the quotient will be of the next higher. Divide this quotient by the number of units of its denomination required to make one tinit of the next higlier. Continue until the highest ARITHMETIC. 15 denomination is reached, or until there is not enough of a denomination left to make one of the next higher. The last quotient and t/ie remainders constitute tJie required result. EXAMPLES FOB PRACTICE. 44. Reduce to units of higher denominations: (a) 7,460 sq. in.; () 7,580 sq. yd.; (c) 148,760 cu. in.; () 2_5 (it) 100 (c) 650 6 636 1 6 (,) 1457 6 1121 1120 88669 1 38669 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 2 = 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 shown at (#). Subtract, and obtain 6 as a remainder. Bring down the next period, 50, and annex it to the remainder, 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 ( and write 9 as the third figure of the root. Complete the work as before. 9 7 . EXAMPLE. What is the cube root of 78,292. 892952 ? SOLUTION. root 4 16 78'292.892'952(42.78 4 32 64 8 4800 14292 4 244 10088 120 5044 4204892 _2 248 3766483 122 529200 438409952 2 8869 438409952 124 538069 2 8918 1260 54698700 7 102544 1267 . 54801244 7 1274 7 EXPLANATION. Since the above is a mixed number, begin at the decimal point and point off periods of three figures 36 ARITHMETIC. 2 each, in both directions. The first period contains but two figures, and the largest number whose cube is less than 78 is 4 ; consequently, 4 is the first figure of the root. The remainder of the work should be perfectly clear. When dividing the dividend by the trial divisor for the third figure of the root, the quotient was 8 -f- > but, on trying it, it w r as found that 8 was too large, the complete divisor being con- siderably larger than the trial divisor. Therefore, 7 was used instead of 8. 98. EXAMPLE. What is the cube root of 5 to five decimal places ? SOLUTION. root 1 1 5.0 O'O O'O O'O O'O ( 1.70997+ 1 2_ 1_ - 2 300 4000 1 259 3913 30 559 87000000 7 308 78443829 37 8670000 8556171000 7 45981 7889992299 44 8715981 666178701000 7 46062 614014317973 5TOO 876204300 52164383027 9 461511 5109 876665811 9 461592 5118 87712740300 9 3590839 51270 87716331139 9 51279 512970 7 512977 2 ARITHMETIC. 37 EXPLANATION. In the above example, we annex five periods of ciphers, of three ciphers each, to the 5 for the decimal part of the root, placing the decimal point between the 5 and the first cipher. Since it is easy to see that the next figure of the root will be 5, we increase the last figure by 1, obtaining 1.70998 for the correct root to 5 decimal places. Ans. 99. EXAMPLE. What is the cube root of .5 to four decimal places ? SOLUTION. root 7 49 , .500'000'000'000(.7937 + 7 98 343 14 14700 157000 7 1971 150039 210 16671 6961000 9 2052 5638257 219 1872300 1322743000 9 7119' 1321748953 228 1879419 994047 9 7128 2370 188654700 3 166579 188821279 EXPLANATION. In the above example, we annex two ciphers to the .5 to complete the first period, and three periods of three ciphers each. The cube root of 500 is 7 ; this we write as the first figure of the root. The remainder of the work should be perfectly plain from the explanations of the preceding examples. 1-7 38 ARITHMETIC. 1OO. EXAMPLE. What is the cube root of .05 to four decimal places ? SOLUTION.- 3 _3 6 J 90 6 96 6 103 6 1080 388800 8704 397504 8768 4 6 2 7 2.0 44176 40671376 root .0 5 O'O O'O O'O (.3 6 8 4 + 27 23000 19656 3344000 3180032 163968000 162685504 1282496 101. Proof. To prove cube root, cube the result ob- tained. If the given number is an exact poiver, the cube of the root will equal it; if not an exact power, the cube of the root will very nearly equal it. 102. Rule. I. Arrange tJie work in three columns, placing the number, w/iose cube root is to be extracted, in tJic third, or rigJit-liand, column. Begin at units place, and separate the number into periods of three figures each, pro- ceeding from the decimal point towards the right with the decimal part, if there is any. II. Find the greatest number whose cube is not greater than the number in the first period. Write this number as the first figiire of the root ; also, write it at the head of the first column. Multiply the number in the first column by the first figure in the root, and write the result in the second columti. Multiply the number in the second column by the first figure of the root ; subtract the product from the first period, and 2 ARITHMETIC. 39 annex the second period to the remainder for a new dividend ; add the first figure of the root to the number in the first column for the first correction. Multiply the first correction by the first figure of the root, and add the product to the number in the second column. Add the first figure of the root to the first correction to form the second correction. Annex one cipher to the second correction and two ciphers to the last number in the second column ; the last number in the second column is the trial divisor. III. Divide the dividend by the trial divisor to find the second figure of the root. Add the second figure of the root to the number in the first column, multiply the sum by the second figure of the root, and add the result to the trial divisor to form the complete divisor. Midtiply the complete divisor by the second figure of the root, subtract the result from the dividend in the third column, and annex the third period to the remainder for a new dividend. Add the second figure of the root to the number in the first column to form the first correction ; multiply the first correction by tJie second figure of the root, and add the product to the complete divisor. Add the second figtire of the root to the first correction to form the second correction. Annex one cipher to the second correction and tivo ciphers to the last number in the second column to form the new trial divisor. IV. If there are more periods to be brought doivn, proceed as before. If there is a remainder after tJie root of the last period has been found, annex ciplier periods, and proceed as before. The figures of the root thus obtained will be decimals. V. If the root contains an interminable decimal, and it is desired to terminate the operation at some point, say, the fourth decimal place, carry the operation one place further, and if the fifth figure is 5 or greater, increase the fourtJi figure by 1 and omit the sign -{- . 1O3. Art. 91 can be applied to cube root (or any other root) as well as to square root. Thus, in the exam- ple, Art. 98, there are to be 5 + 1 = 6 figures in the root. Extracting the root in the usual manner to G -^ 2 = 3, 40 ARITHMETIC. 2 say 4, figures, we get for the first four figures 1,709. The last remainder is 8,556,171, and the next trial divisor with the ciphers omitted is 8,762,043. Hence, the next two figures of the root are 8,556,171 -r- 8, 762, 043 =.976, say .98. Therefore, the root is 1.70998. ROOTS OF FRACTIONS. 104. If the given number is in the form of a fraction, and it is required to find some root of it, the simplest and most exact method is to reduce the fraction to a decimal and extract the required root of the decimal. If, however, the numerator and denominator of the fraction are perfect powers, extract the required root of each separately, and write the root of the numerator for a new numerator, and the root of the denominator for a new denominator. 105. EXAMPLE. What is the square root of -fa ? / n A/ q SOLUTION. r 64 = /^ = *' Ans ' 106. EXAMPLE. What is the square root of f ? SOLUTION. Since f =.625, |/f = |/.625 =.7906. Ans. 107. EXAMPLE. What is the cube root of f| ? SOLUTION.- = - = *. Ans. 108. EXAMPLE. What is the cube root of ? SOLUTION. Since =.25, ^ - ^725 =.62996 + . Ans. 109. Rule. Extract the required root of the numer- ator and denominator separately ; or, reduce the fraction to a decimal, and extract the root of the decimal. EXAMPLES FOR PRACTICE. HO. Find the cube root of : (a) f (b) 2 to five decimal places. (f) 4, 180, 769, 192. 462 to five decimal places. A ^ r* (/) 513,229.783302144 to three decimal places. (b) 1. 25992 + . (c) 1,610.96238. (d) .8862 + . (e) .7211 + . (/) 80.064. 2 ARITHMETIC. 41 TO EXTRACT OTHER ROOTS THAN THE SQUARE AND CUBE ROOTS. 111. EXAMPLE. What is the fourth root of 256 ? SOLUTION. 4/256 = 16. 4/16 = 4. Therefore, /f/256 = 4. Ans. In this example, f56, the index is 4, which eqiials 2x2. The root indicated by 2 is the square root ; therefore, the square root is extracted twice. EXAMPLE. What is the sixth root of 64 ? SOLUTION. 4/64 = 8. ^8 = 2. Therefore, -j/64 = 2. Ans. In this example, {/(M, the index is G, which equals 2 X 3. The root indicated by 3 is the cube root; therefore, the sqtiare and cube roots are extracted in succession. 113. Rule. Separate the index of the required root into its factors (2's and 3's), and extract, successively, the roots indicated by the several factors obtained. The final result zvill be the required root. 114. EXAMPLE. What is the sixth root of 92,873,580 to two decimal places ? SOLUTION. 6 = 3x2. Hence, extract the cube root, and then extract the square root of the result, f 92, 873, 580 = 452.8601, and V45278601 = 21.28-H. Ans. 115. It matters not which root is extracted first, but it is probably easier and more exact to extract the cube root first. EXAMPLES FOR PRACTICE. 116. Extract the (a) Fourth root of 100. f (a) 3. 16227 (b) Fourth root of 3,049,800,625. Ans. j (b) 235. (f) Sixth root of 9,474,296,896. (V) 46. 42 ARITHMETIC. 2 BATIO. 117. Suppose that it is desired to compare two num- bers, say 20 and 4. If we wish to know how many times larger 20 is than 4, we divide 20 by 4 and obtain 5 for the quotient; thus, 20 -f- 4 = 5. Hence, we say that 20 is 5 times as large as 4, i. e. , 20 contains 5 times as many units as 4. Again, suppose we desire to know what part of 20 is 4. We then divide 4 by 20 and obtain J; thus, 4 -=-20 |, or .2. Hence, 4 is ^ or .2 of 20. This operation of com- paring two numbers is termed finding the ratio of the two numbers. Ratio, then, is a comparison. It is evident that the two numbers to be compared must be expressed in the same unit; in other words, the two numbers must both be abstract numbers or concrete numbers of the same kind. For example, it would be absurd to compare 20 horses with 4 birds, or 20 horses with 4. Hence, ratio may be denned as a comparison between two numbers of the same kind. 118. A ratio may be expressed in three ways; thus, if it is desired to compare 20 and 4, and express this compari- 20 son as a ratio, it may be done as follows : 20 -f- 4, 20 : 4, or . All three are read the ratio of 20 to 4. The ratio of 4 to 20 4 would be expressed thus: 4^20, 4:20, or . The first yv\/ method of expressing a ratio, although correct, is seldom or never used ; the second form is the one of tenest met with, while the third is rapidly growing in favor, and is likely to supersede the second. The third form, called the fractional form, is preferred by modern mathematicians, and possesses great advantages to students of algebra and of higher mathe- matical subjects. The second form seems to be better adapted to arithmetical subjects, and is the one we shall ordinarily adopt. There is still another way of expressing a ratio, though seldom or never used in the case of a simple ratio like that given above. Instead of the colon, a straight vertical line is used ; thus, 20 I 4. 2 ARITHMETIC. 43 119, The terms of a ratio are the two numbers to be compared ; thus, in the above ratio, 20 and 4 are the terms. When both terms are considered together they are called a couplet ; when considered separately, the first term is called the antecedent, and the second term, the conse- quent. Thus, in the ratio 20 : 4, 20 and 4 form a couplet, and 20 is the antecedent, and 4, the consequent. 120. A ratio may be direct or inverse. The direct ratio of 20 to 4 is 20 : 4, while the inverse ratio of 20 to 4 is 4 : 20. The direct ratio of 4 to 20 is 4 r-20, and the inverse ratio is 20 : 4. An inverse ratio is sometimes called a reciprocal ratio. The reciprocal of a number is 1 divided by the number. Thus, the reciprocal of 17 is ; of f is 1 -f- 1 = | ; i.e., the reciprocal of a fraction is the frac- tion inverted. Hence, the inverse ratio of 20 to 4 may be expressed as 4 : 20 or as - : . Both have equal values ; for, /cO 4 121. The term vary implies a ratio. When we say that two numbers vary as some other two numbers, we mean that the ratio between the first two numbers is the same as the ratio between the other two numbers. 122. The value of a ratio is the result obtained by per- forming the division indicated. Thus, the value of the ratio 20 : 4 is 5 ; it is the quotient obtained by dividing the antecedent by the consequent. 123. By expressing the ratio in the fractional form, for 20 example, the ratio of 20 to 4 as , it is easy to see, from the laws of fractions, that if both terms be multiplied or both divided by the same number it will not alter the value of the ratio. Thus, 20 20X5 100 ,20 20 -=-4 5 . _ _ _ - - 4X5 ' "20 4 " ' 4-J-4 " 1' ARITHMETIC. It is also evident, from the laws of fractions, that multiplying 1 the antecedent or dividing the consequent mul- tiplies the ratio, and dividing the antecedent or multiplying the consequent divides the ratio. 125. When a ratio is expressed in words, as the ratio of 20 to 4, the first number named is always regarded as the antecedent and the second as the consequent, without regard to whether the ratio itself is direct or inverse. When not otherwise specified, all ratios are understood to be direct. To express an inverse ratio the simplest way of doing it is to express it as if it were a direct ratio, with the first num- ber named as the antecedent, and then transpose the ante- cedent to the place occupied by the consequent and the consequent to the place occupied by the antecedent ; or if expressed in the fractional form, invert the fraction. Thus, to express the inverse ratio of 20 to 4, first write it 20 : 4, and then, transposing the terms, as 4 : 20 ; or as , and 4 then inverting, as . Or, the reciprocals of the numbers may be taken, as explained above. transpose its terms. To invert a ratio is to EXAMPLES FOR PRACTICE. 126. What is the value of the ratio of: (a) w w (<0 w (h) (0 01 (*) 98 : 49 ? $45 : $9 ? 6J:f? 3.5:4.5? The inverse The inverse The inverse The inverse The ratio of The ratio of The ratio of The ratio of ratio of 76 to 19 ? ratio of 49 to 98 ? ratio of 18 to 24 ? ratio of 9 to 15 ? 10 to 3, multiplied by 3 ? 35 to 49, multiplied by 7 ? 18 to 64, divided by 9 ? 14 to 28, divided by 5 ? Ans. () 2. (c) (O PJ vx 1 2 or the amount 12 men earn = -j = $75. Ans. 50 ARITHMETIC. 2 Since it matters not which place .r, or the required term, occupies, the problem could be stated in 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 cj9K V 1 2 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 that 12 men earn ; or the (COR *y 1 O 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 amount that 12 men earn = -? , or $75, since either mean equals the product of the extremes divided by the other mean. 147. 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. 148. Find the value of x in each of the following: (a) $16 : $64 :: x : $4. () x: 85:: 10: 17. (c) 24 : x :: 15 : 40. (d) 18 : 94 :: 2 : x. Ans. ( Vat tya, etc. signify the square root, cube root, fourth, root, etc. of a. 16. The use of the parenthesis, bracket, brace, and vin- culum is explained in Arithmetic. These symbols are called symbols of aggregation, meaning that the quantities enclosed within them are aggregated, or collected, into one quantity. 17. The terms of an algebraic expression are those parts which are connected by the signs -{- and . Thus, x~, 2,rj/, and j/ 2 are terms of the expression x* 2xy-\-y 2 . When a term contains both figures and letters, the part con- sisting of letters is called the literal part of the term; thus, xy is the literal part of the term ^ 18. like terms are those which differ only in their numerical coefficients; all others are unlike terms. Thus, 2## 2 and zab* are like terms; 5ab and Sab* are unlike terms, because one contains b and the other P. 6 ELEMENTARY ALGEBRA AND 3 19. A monomial is an expression consisting of only one term; as, 4afc, 3;r 2 , 2rt.r 3 , etc. 2O. A binomial is an expression consisting of two terms; as, a-\-b^ 2# + 5^, etc. 21. A trinomial is an expression consisting of three terms; as, a* \-'bab + b*, (a + x)* *l(a + x)y+y>, etc , the expression (a + x) being treated as one quantity. 22. A polynomial is an expression consisting of more than one term. The name is usually applied only to an expression consisting of four or more terms. 23. The polynomial a + cfb + 2 8 3a*& a 5 is said to be arranged according to the increasing powers of a, because the exponents of a increase in each term from left to right, the exponent of the first a being 1 understood. (Art. 12.) The polynomial a*b 3 + <*&* + ^ a "b -f- 1 is arranged according to the decreasing powers of b, the exponents of b decreasing in order from left to right. 24. The arrangement of the terms of a polynomial does not affect its value. Thus, x* -\- ^xy +/ 2 has the same value as %xy + y* + tf' 2 , just as 2 + 6 + 4 has the same value as READING ALGEBRAIC EXPRESSIONS. 25. Quantities like , x, # 2 , etc. are read *'#," u .r," " b square," etc. In reading monomials in which multipli- cation is indicated, the word "times "is not used. Thus, abc is read " abc "; lad*b* is read *' lad square b cube." 26. The polynomial a + a*b + Za 3 3a*b a* is read ", pltis a square b, plus 2a cube, minus 3a fourth ^, minus a fifth. " Considerable care is required when reading expres- sions containing polynomials. Thus, if 4(# b} were read ' ' 4<7 minus ^, " the binomial ka b would be understood. It should 'be read "4 times a b" or "4 times the parenthesis 3 TRIGONOMETRIC FUNCTIONS. 7 a minus b, " in which case it will be understood that 4 multi- plies the whole quantity a b, since the word "times" is not used with monomials. Again, m(ii? -f- "Zmn -f- // a ) and ;//(///- -{- 2 mil} -f- /f2 should each be so read that there can be no doubt as to whether the ;/* is to be multiplied by ;// or not. Let the distinction to be made in reading the following be observed : / A/ and A/ m-\ ^ In the first case, the whole quantity ;// + ;/ * s divided by x y, and it would be clear to say, " the square root of the fraction m-\-n over x j /2 . " In the second case, where the ;/ only is divided by JIT /*, it may be read, "the square root of the quantity, in plus the fraction n over x j' 2 . " The word "quantity" shows that the square root of the whole expression is taken, and the word " fraction " after " plus " shows that only the n is divided by x / 2 . 27. When a polynomial is affected by an exponent, it should be indicated clearly. Thus, should be read, "3a d square, times the square of 3a d, times the square of Za d square. " 28. Sometimes expressions like A', R'\ c', d" ', T, *. Ans. 6' + 2 X 6 X 5 + 5 a = 121. 7. 2 a + Me - 5. Ans. 72 + 60-5 = 1 27. 8. 2ac 5 a\a + b). Ans. 11,892. 9. abc* + ab*c a*bc. Ans. 360. When x 8 and j/ = 6, what do the following equal : 10. (.r +y) (.r -y) - .{/ ^- ? Ans. (8 + 6) (8 - 6) - 1/ -5^ = 26. 11. \/(jc +y*) (,r 2 + y) - (.1 - ADDITION AND SUBTRACTION. PRELIMINARY IDEAS. 36. Suppose we take a point, as A on the line shown in Fig. 1, and lay off equal distances in opposite directions. T M T f f T M ] f T f f f T M FIG. 1. Now, let us call distances to the right -f-, or positive, and distances to the left , or negative. Let us also call a move- ment to the right positive, and one to the left negative. Suppose the positive direction east, the negative west, and the distances to represent miles. g 3 TRIGONOMETRIC FUNCTIONS. 11 Suppose a man starts from A and walks east 6 miles, and after a pause walks 3 miles farther east. His distance from A is + () + (+ 3) = +9 miles, the plus sign being taken because the motion is in the positive direction. Suppose, however, the man starts from A and walks west 3 miles, and after a pause walks 5 miles farther west. His distance from A is 8 miles west of A, or 8 miles in a negative direction; that is, 3 -|- ( 5) = 8. As a third case, imagine the man to walk 6 miles east, and then turn around and walk 4 miles west. Counting 4 west from 6, we see that he would still be 2 miles east of A , or 2 miles in a positive direction ; that is, +C + ( 4) = 2. If, instead of walking back 4 miles, he had walked back 10 miles, we find by counting 10 miles west from 6 that he would have been 4 miles west of A, or 4 miles in a negative direction, +G + ( 10) = 4. For reference, the above results are collected : ( + 3) = +9 _3 + (-5) = -8 -f-6-K-lO) = -4 The student should observe carefully that in each of these additions the signs immediately before the numbers denote their positive or negative character, while the -f- s ig n m front of the parenthesis denotes the operation of addition. 37. From these illustrations we have the following important principle : If all the terms to be added arc positive, the sum is positive ; if all are negative, the sum is negative. If one term is positive and the other is negative, the sum has the sign of the numerically greater. If there are several terms to be added, part of which are positive and part nega- tive, the sum is positive if the sum of the positive terms is numerically greater than the sum of the negative terms. When the terms have the same sign, the numerical sum is that which would be obtained if the signs were disregarded. Thus, in the first case above, the signs of and 3 are the same, therefore the sum is numerically + 3 = 9 ; likewise 12 ELEMENTARY ALGEBRA AND 3 in the second, the signs of the 3 and the 5 are alike and the sum is numerically 3 -|- 5 = 8. When, however, one term is positive and the other is negative, the sum is the numerical difference between the terms. Thus, in the third case above, the signs of 6 and 4 are different, and the sum, 2, is the numerical difference; likewise in the fourth case, the and the 10 have different signs, and the sum, 4, is the numerical difference. 38. To add like terms containing letters, we simply add the coefficients, having regard for the proper signs, and annex the literal part. Thus, the sum of ax*y and ^ax^y is 9a#y- t the sum of %ab and \\ab is ab\ and the sum of 9/ a // 3/;/ 2 #, 8;// 2 //, and 2m*n is 6;/r;/. ADDITION OF MONOMIALS. 39. Like Quantities. To add like quantities having the same sign : Rule I. Add t lie coefficients, give the sum the common sign, and annex the common literal part. To add like quantities having different signs : Rule II. Add the positive and the negative coefficients separately, and from the greater sum subtract the less. Give the remainder the sign of the greater sum, and annex the common literal part. EXAMPLE. Find the sum of %abxy, abxy, *&abxy, and abxy. SOLUTION. The sum of the coefficients is 12 (remember that the coefficient of abxy is 1), and the common sign is . The common literal part, abxy, annexed to these gives as the' result VZab.iy. (Rule I.) EXAMPLE. Combine ,ry 2 , 2.rj/ 2 , 8_ry 2 , and 4.ry 2 . SOLUTION. The sum of the coefficients of the positive terms is 9, and of the negative terms, 6. Their difference is 3, and the sign of the greater sum is +. The common literal part, xy*, annexed to these gives as the result 3^/ 2 . (Rule II.) 3 TRIGONOMETRIC FUNCTIONS. 13 SUBTRACTION OF MONOMIALS. 40. Referring 1 again to Fig". 1, suppose two men, C and D, start from point A and travel eastward. At the end of a certain time, C has walked 8 miles and D has walked 5 miles. Now, the distance between C and D is 3 miles; to pass from -j- 5 to + 8, we must walk + 3 miles east, or in a positive direction ; or -f- 8 ( + 5) = + 3- Observe that the minus sign here denotes subtraction, while the three plus signs denote that the three quantities which they precede are positive. The difference between 5 and 8 means how far, and in what direction we must go to pass from 5 to 8 or from 8 to 5. If we pass from -f- 5 to -f- 8, we move in a positive direction, and we say +5 from +8 is equal to + 3. Sup- pose, however, we pass from + 8 to + 5 ; we move westward, or in a negative direction, a distance of 3 miles. Hence, we say that +8 from +5 is -3, or +5 - ( + 8) = -3. We shall always consider the point we pass from, as the subtra- hend, or quantity to be subtracted, and the point we approach as the minuend, or quantity we subtract from. Suppose C has walked 7 miles east and D has walked 4 miles west, how far apart are they ? To pass from D to C we must travel 11 miles east, or in a positive direc- tion. Therefore, 7 ( 4) = +11. To pass from C to D, we travel 11 miles west, or in a negative direction; -4-( + 7) = -11. The following exercises may be studied in connection with Fig. 1: From + 3 to + 8 is + 5, or +8- ( + 3) = + 5. From + 10 to + 6 is 4, or + 6 ( + 10) = 4. From 5 to + 4 is + 9, or + 4 ( 5) = +9. From - 9 to 2 = + 7, or - 2 - ( - 9) = + 7. From 3 to 7 . = 4, or 7 ( 3) = 4. From + 2 to - 4 = - 6, or - 4- ( + 2) = - 6. 41. In every case, the difference is what must be added to the subtrahend to obtain the minuend. Thus, if I am 3 miles east of A, Fig. 1, how far must I go to be 8 miles east of A ? Evidently 5 miles, since 5 miles added to 3 miles gives 8 miles. Similarly, the difference between 5 and 14 ELEMENTARY ALGEBRA AND 3 -|-4 is -f- 9, since -f~9 must be added to 5 to obtain -|- 4; that is, we must travel 9 miles east in passing from 5 to 4:2. We have seen that if we add -f- 6 and 4 we obtain + 2 as the sum. (Art. 36.) If we subtract -+- 4 from -f 6, the difference is -f- 6 ( -f- 4) = + 2, since -f- 2 must be added to + 4 to make -j- G. If we add + 6 and 10, the sum is 4. If we subtract +10 from -{ 6, the difference is 4. Therefore, it appears that if we wish to subtract one quantity from another, we obtain the same result if we change the sign of the quantity to be subtracted and add it to the other quantity. Thus, + 7 subtracted from -|- 12 is the same as 7 added to + 12, the result being 5 in either case. 3mn subtracted from kmn gives the same result as -f- Sinn added to 4;;z;z. The following exercises are given as illustrations: = +7. = -6 + (-3) = -9. + 2a ( + a) = + 2a + ( a) = + a. + xy ( + IS.ry) = + xy + ( \xy} \xy. 43. To subtract like quantities : Rule. Change the sign of the subtrahend, and proceed as in addition. EXAMPLE. From ZaPx take lab**. SOLUTION. Changing the sign of the subtrahend, lab*x, and add- ing, we have ^ab^x + ( lab**) = IQab^x. Ans. 44. Unlike Quantities. In arithmetic, unlike num- bers, as 5 books and 3 dollars, cannot be added or sub- tracted. So, in algebra, unlike terms, as 3## 2 , xy, 2m, etc., cannot be combined or subtracted, except by indicating the operations by signs. Expressions in algebra are composed of quantities between which operations of addition, multiplication, etc. are indi- cated. The trinomial m z %mn -\- n*, for example, is the indicated sum of ;;z 2 , %mn, and a , and it is to be considered 3 TRIGONOMETRIC FUNCTIONS. 15 as one quantity, in the same way that an arithmetical sum, obtained by actually performing the addition, is considered. EXAMPLE. What does led* %cx cd* + adx + 2cd* equal ? SOLUTION. In this case, part of the terms are like and part unlike. Combining like terms, led* + 2ed* ed 3 = 8ed'\ Connecting the unlike terms with this result by their respective signs, we have as the final result Zed* Sex + Qadx. Ans. EXAMPLE. Subtract 2m 3 from 1m + %xy. SOLUTION. As in the example above, we have like and unlike terms. Subtracting like terms, Art. 43, we have 1m 2m = 5m. We must now connect the unlike terms by their respective signs. Since 3 is in the subtrahend, its sign will be changed, giving us 5m + Zxy -h 3. Ans. EXAMPLES FOR PRACTICE. 45. Find the sum of the following. 1. - 60 2 , 2a\ - 50 2 , 4# 2 , - 3 2 , and a\ Ans. - 2. 2a*t>, -a*&, lla*&, -5a*t>, 4a 2 6, and - 9 2 . Ans. 3. 2x*, %xy, x'\ 8y 2 , 5xy, and 7j/ 2 . Ans. x* Zxy +y'\ NOTE. Combine like terms and connect with respective signs. 4. a>bc, -2#V, Zabc-, -IcPbc, and 5aV. Ans. %ab *c ZcPbc +- abc*. Solve the following : 5. From \la take \\a. Ans. 2Sa, 6. From \\a take Via. Ans. 2Sa. 1. Subtract 5cd from cd. Ans. $cd. 8. Subtract - 10 2 from - 10^ 2 . Ans. 0. 9. What quantity added to \xy will produce \%xy ? Ans. 22.ry. 10. What, then, does IQxy subtracted from \2xy equal ? Ans. 2%xy. ADDITION AND SUBTRACTION OF POLYNOMIALS. 46. To add polynomials: Rule. Write the expressions underneath one another, with like terms in the same vertical column. Add each column separately, and connect the sums by their proper signs. 1C ELEMENTARY ALGEBRA AND 3 EXAMPLE. Find the sum of 5# 2 + Gac 3^ 2 2xy, 7 ac 3a* + 4l>' 2 + 3,ry, and 4.tj 5 2 + Sac a' 2 . SOLUTION. Writing like terms in the same vertical column, we have - a' 2 + Sac - sum a 2 + 21ac 4^- + 5.ij. Ans. EXAMPLE. Find the sum of a?x ax* X*, ax x^a 1 , 2a* - 2a' 2 .r - 2ax\ and 3rt 2 - 3a\r + %ax\ SOLUTION. a^x ax"* x* sum a?x + 2.r a + + ax - ax-a*x-2x\ Ans. (Arts. 23 and 34.) 47. To subtract one polynomial from another: Rule. Write tJic subtrahend underneath the minuend, with like terms in the same vertical column. Change the sign of each term of the subtrahend, and proceed as in addi- tion. EXAMPLE. From Sac 2b subtract ac b d. SOLUTION. Sac Zb ac + b -\-d, subtrahend with signs changed. difference 2ac b + d. Ans. EXAMPLE. From 2.t 3 3.t 2 / + 2,ry a subtract .r 3 aj 2 +_y 3 . SOLUTION. 2.r 3 3.r 2 ^/ + 2.i;/ 2 - .i 3 + xy* y*, subtrahend with signs changed. difference ,r 3 EXAMPLES FOR PRACTICE. 48. Find the sum of the following: 1. ax + 2bx + by %ay, 2ax + bx + 2ay by, and 4ax + %by. Ans. lax + %bx + by ay. 2. a x + 4y 3^ + w t z + 3a 2.v y w, and x +y + z. Ans. 4# 3 TRIGONOMETRIC FUNCTIONS. 17 3. 2a 30 + 4 + 3c-(-3a + 2t>-c). Ans. 80-6+4* and SOLUTION. - 9 5 + 3a*&'* - 4 2 3 - b b . Ans. 62. When both factors are polynomials: Rule. Multiply each term of one polynomial by each term of the other, and add the partial products. EXAMPLE. Multiply Qa 4 by 4 2. SOLUTION. Write the multiplier under the multiplicand, and begin to multiply at the left instead of at the right, as in arithmetic, since polynomials are always written and read from the left, and there are no numbers to carry. 6 - 4 (1) 4a - 26 Multiplying (1) by 4rt gives 24a* I6a& (2) Multiplying (1) by - 2 gives _ -12d + 8 2 (3) Adding (2) and (3) gives 240 2 - 28a& + 8 2 . Ans. It will be noticed that the like terms, \ab and I2a&, are written under each other, so that it will be easier to add them. EXAMPLE. Multiply x* x + 1 + x'* by 1 x 1 + x. SOLUTION. With a view to bringing like terms in the same columns, arrange both multiplicand and multiplier either according to the increasing or the decreasing powers of the same letter. (Art. 23.) Arranging in this case according to the increasing powers of x, we have l-x + x 2 + x* (1) Multiplying (1) by 1 gives 1 x + x* + x 3 (2) Multiplying (1) by + ogives x x* + x* + x 4 (3) Multiplying (1) by x* gives ,r 2 + x* x* x' a (4) Adding (2), (3), and (4) gives 1 -.r 2 + 3^r 3 -x\ Ans. 63. Multiplication is frequently indicated by enclosing each of the quantities to be multiplied in a parenthesis. The sign of multiplication is not placed between the paren- theses, multiplication being understood. When the quanti- ties are multiplied together, the expression is said to be expanded. For example, in the expression (in 2n) (2 ;# //), the 3 TRIGONOMETRIC FUNCTIONS. 23 binomial m 2n is to be multiplied by the binomial 2m n. Performing the multiplication, the product is 2;/z 2 5mn which is the expanded form of the expression. EXAMPLES FOB PRACTICE. 64. Multiply the following : 1. x* + 2xy +y* by x +y. Ans. .t 3 2. 3 a / 8 + 40V - 2 by aWm*. Ans. 3. ^ _ ^ by ^ + ^ 4. .r 4 + .r 2 j 2 +j/ 4 by ^r 2 j/ 2 . Ans. ^- 6 j 6 . 5. 3 2 -7rt + 4by 2 2 + 9-5. Ans. 6 4 + 13^ 3 - 70 2 + lla -20. Expand the following : 6. (2a &) (4 3). Ans. 8 12^r 6 2 + 9^. 7. (.r + 2) (JT - 2) (-r 2 + 4). Ans. .r 4 ~ 16. 8. .r^ 2 - 2 NOTE. The expressions in the brackets reduce to x* xy* 2 and ^3 + X y* + 2. The product of these is x 6 x^ xy* 4. Ans. THREE IMPORTANT EXAMPLES. 65. Let a and b be any two quantities; we wish to find the forms of the following products: (a + b}\ (a-b)\ and (a + b)(a-b). By actual multiplication we find (a + by = c? + ^ab + b\ (1.) (a - by = a 2 - %ab + b\ (2.) ( a + b)(a-b} = a*-b\ (3.) Hence : 66. The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first and the second, plus the square of the second. The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first and the second, plus the square of the second. The product of the sum and difference of tivo quantities is equal to the difference of their squares. 24 ELEMENTARY ALGEBRA AND 3 67. The foregoing statements should be committed to memory, since their use will frequently save tedious calcula- tion. The student is advised to practice until he is certain that he knows them perfectly. EXAMPLE. Square 3a' 2 + 5. % SOLUTION. The square of the first term is 3.t 2 X 3.r 2 = 9.r 4 ; twice the product of the terms is 30.*" ; and the square of the last term is 25. Hence, by formula 1, letting a = 3.t 2 and b = 5, (3.r a + 5) 2 = 9x* + 30.r 2 + 25. Ans. EXAMPLE. Square fad x. SOLUTION. The square of the first term is IG^: 2 ^ 2 ; twice the product of the first and the second is 8cd.r ; and the square of the last term is .r 2 . Hence, by formula 2, letting a = 4^/and b = x, (fadxf - IGcW Scdx + x*. Ans. EXAMPLE. Expand (.r 2 + 3) (.r 2 - 3). (See Art. 63.) SOLUTION. The square of the first term is .v*, and of the second, 9. Hence, by formula 3, letting a = x* and b = 3, (.t 2 + 3) (,r 2 3) = .r 4 9. Ans. EXAMPLES FOR PRACTICE. 68. Square the following : 1. m + n. Ans. 2. 4.1- + 2. Ans. IG.t- + 16.V + 4. 3. 3-5. Ans. 9# 2 - 30^ + 25 2 . Expand the following : 4. (m + 1) (m 1). Ans. ; a 1. 5. ( x *+y*)(jc*-y*). Ans. x*-y*. 6. (4rt + 4 2 )(4tf-4 2 ). Ans. 16 2 - 16 4 . 7. Square 2c* c + d. NOTE. First separate 2^ 2 c + d into two terms by enclosing c + d in parenthesis ; then the expression becomes 2^ 2 (c d*), and consid- ering this as a binomial we find the square to be 4c* fa" 2 (c d) + (c-d)\ -fa'\c-d} = _4^ (c-df = c*-2c Adding these results to 4/ 2 2' 9^ 3 j/^ 2 333^ 3 ^ 2 ^ 2 by 3^ 3 j2-. Ans. 3. iO(jr + yY - 5a (* +y) + 5 2 (* +y) b y 5 (-^ +^)- Ans. 3 TRIGONOMETRIC FUNCTIONS. 27 76. When the divisor is a polynomial: Rule. Arrange both dividend and divisor according to the ascending or descending powers of some letter. Divide the first term of the dividend by the first term of the divisor for the first term of the quotient. Subtract from the dividend the product of the divisor and this term of the quotient. Treat the remainder as a new dividend, and proceed as before, until there is no remainder, or until the final remain- der contains no term which is divisible by the first term of the divisor. EXAMPLE. Divide x* + x* 9-r 2 16.r 4 by x* + 4.r + 4. SOLUTION. quotient. + ^3_ 9^_i6jr_4j: a -3.r-l. Ans. _ x>- 4.r-4 00 The first term x* of the divisor is contained in x*, the first term of the dividend, x* times; hence, x* is the first term of the quotient. The whole divisor multiplied by this term gives x* -f- 4;tr 3 + 4^r 2 as a product, which subtracted from the dividend gives as a remainder, 3x 3 13x* 16^ 4. It is not necessary here to bring down the 4, since only three terms are required to contain the divisor. The first term x* of the divisor is contained in 3;r 3 , the first term of the new dividend, 3x times. Multiplying the divisor by this new term of the quotient, we have 3x 3 ISx* \%x. Subtracting this from the first remainder, we obtain x* kx 4 for a new remainder, the 4 being brought down from the original dividend. The first term of the divisor is contained in the first term of the new remain- der or dividend, 1 times. Multiplying the divisor by this, we get x* x 4, which subtracted from x* x 4, the last remainder, leaves a difference of zero. The work 28 ELEMENTARY ALGEBRA AND 3 ends here, since there are no more terms in the dividend to be brought down. ExAMPLE.-^Divide 9.r 2 j 2 + .r 4 - 4y 4 - Gx 3 y by ^ + 2y* - Zxy. SOLUTION. First arrange the dividend and divisor according to the descending powers of x. * 4j/ 4 ( .r 2 3.ry 2y 2 . Ans. EXAMPLES FOR PRACTICE. 77. Divide the following : 1. , r _ ix + 12 by x 3. Ans. .r 4. 2. .r 2 + x - 72 by .r + 9. Ans. x - 8. 3. 2.v 3 .r 2 + 3.v 9 by 2.r 3. Ans. a- 2 + ,r + 3. 4. .v 4 + II* 8 - 12.v - 5.r 3 + 6 by 3 + .r 2 - 3.r. Ans. .r a - 2.r + 2. 5. ,r 4 6.17 9,i- 2 y 1 by ,r a +_K + 3.i . Ans. .r' 2 3.r y. 6. *" 1 by x 1. Ans. * s + .r 4 + .r 3 + .i-' 2 + .v + 1. FACTORING. 78. Factoring is the process of finding the factors of a quantity, that is, the quantities or numbers which will .divide that quantity without a remainder. 79. Expanding- 60 3 (2#-frt), we have hence, Mb and (2# + tf) are the factors of l^Y? 2 -f 6 3 . The monomial factor Mb may be further resolved into .3 X 2 X tf X a X #. In solving examples in factoring-, it is not customary to write out the factors of a monomial, since they are generally apparent. That the student may be able to recognize factors without the labor of actual division, several methods of readily discovering factors are here given. 3 TRIGONOMETRIC FUNCTIONS. 29 80. Equal factors are those whose terms have the same letters, and whose letters have the same exponents and the same signs. Thus, 5a(%y x) and 5a(2y x) are equal factors of 5#(2j x) X 5 a (by - .v) = 25a*(2y - x}* ; but 5#(2j ;r) and 5a(2y x) are unequal factors, since the signs of 5a are not the same in both expressions. 81. A product of two equal factors is a perfect square. Either of the equal factors of a quantity is called its square root. 82. A product of three equal factors is a perfect cube. Any one of the equal factors of a quantity is called its cube root. 83. In factoring, it is important to be able to easily dis- ,tinguish quantities that are perfect squares and cubes, and to determine their roots. By definition, 9# 2 $ 2 is a perfect square, because 3ab X %ab = 9# 2 # 2 , and %ab is its square root. Also, 8a e> is a perfect cube because 2 3 -8^ 5 + 8. Ans. 5. 4x*jr 12*y* + S.rj/ 3 . Ans. 6. 49 2 V 4 - 63 3 V 4 + 7rt 4 V 3 . Ans. 7 CASE II. 86. To factor a trinomial which is a perfect square: Any trinomial is a perfect square when the first and the last term are perfect squares and positive, and the second term is twice the product of their square roots. 3 TRIGONOMETRIC FUNCTIONS. 31 Thus, let a and b represent any two quantities whatever, and we have the general forms of the square as follows : 2 + 2tf + 2 = (a+b)(a + b) = (a + b)\ (4.) a* - % a & 4- {,* = ( a - b) (a - b) = (a- b)\ (5.) These, it will be seen, are simply the inverse of formulas 1 and 2, Art. 65. The sign of the second term of the square always determines the sign of the second term of the root, b in this particular case. 87. Since a may represent one quantity and b any other quantity, it is evident that any trinomial having the form # a + 2ab -\- b* or a* %ab -f- 2 is a perfect square. Rule. Extract the square roots of the first and the last term of the trinomial, and connect the results by the sign of the second term. EXAMPLE. Factor _r 2 4- %xy +y*. SOLUTION. We first see if the trinomial has the form stated in Art. 86. The first and the last term we see to be perfect squares, and their roots to be x and^. The second term is also twice the product of the roots x and y, and, since it has the plus sign, the binomial root must be x +y. Hence, we have a square of the form a? + %ab -f- b*, and x" 1 + 2xy +y = (x +y) (x +y) = (x +y)\ Ans. EXAMPLE. Factor 16w 4 + 9 6 24w 2 3 . SOLUTION. The first term of the expression is a perfect square, but the last term is not. Inspecting the second term, we find it to be the square of 3 3 , and the third term to be twice the product of 3# 3 and the square root, 4w 2 , of the first term. Arranging the trinomial so that the first and the last term are perfect squares, we get Wm* 24m*n 3 + 9 6 (a square of the form fl 2 2ab -+- 2 ), and we have = (4; 2 -3 3 ) 2 . Ans. EXAMPLE. Factor 4x 2 + x^y" 1 SOLUTION. Arranging the trinomial so that the first and the last term are perfect squares, we have 4,r 2 + 2.r 2 _y + .i' 2 ^ 2 . Now, although the first and the last term are perfect squares with roots Zx and xy, respectively, the second term is only equal to the product of the roots ; hence, the trinomial is not a perfect square, and can only be factored by Case I. Each term contains .r 2 , and we have 4.i - 2 + x*y* = .r 2 4 + + 2. Ans. 32 ELEMENTARY ALGEBRA AND 3 88. Should two of tlie terms of a trinomial be perfect squares, and have like signs, and the other term be twice the product of their roots, the trinomial is a perfect square. Compare this statement with Art. 86. Thus, %ab a* b'\ if divided by 1, becomes 2at> -f- a 2 + b* = a* 2ab + b* ; hence, 2al>-a*-l>* = - (a* - Zab + b*} = -( a -b)\ EXAMPLE. Factor pq 4/ 2 q'\ SOLUTION. Dividing first by 1 we have 4/ty+4/ 3 + rS = 4/ 2 pq + ?* (Zp q} 2 . Hence, pq 4/ 2 q' 2 (4/ 2 pq + q*) = -(%p-qy. Ans. EXAMPLE. Factor 16r 2 .? 2 + 16r 4 + 4s 4 . SOLUTION. The expression contains three squares, but, by careful inspection, we see that the first term is also twice the product of the square roots of the other two. Thus, 16rV + 16r 4 + 4^ 4 = 16r 4 + 16r V + 4.y 4 = (4r 2 + 2j 2 ) 2 . Ans. EXAMPLES FOU PRACTICE. 8O. Factor the following trinomials: 1. -r* - 16.r + 64. Ans. (.r-8) 2 . 2. ;z 6 - 26 3 + 169. Ans. ( 3 - 13) 2 . 3. 25.r 2 + IQxyz + 49y 2 2- 2 . Ans. (5.r + lyz)*. 4. 16^ 2 + ^ 2 -8^. Ans. (4c-W. 5. %mx m* ,r 2 . Ans. (m .r) 2 . 6. 2 ^V 6 -2^V 3 + 1. Ans. (ab*c* - I) 8 . CASE III. 9O. To factor an expression which is the difference between two perfect squares: This case is the inverse of formula 3, Art. 65, and may be expressed by the formula -b). (6.) 91. Since a may represent one quantity and b any other quantity, it is evident from formula 6 that any expression which is the difference between two perfect squares may be factored by the following 1 g 3 TRIGONOMETRIC FUNCTIONS. 33 Ilule. Extract the square roots of the first and the last term. Write the first root phis the second for one factor, and the first root minus the second for the other. EXAMPLE. Factor 9.*^" 4. SOLUTION. The square roots of the first and the last term are 3.t 4 y* and 2. The sum of these roots is 3.i' 4 j/ 3 + 2, and the second subtracted from the first is 3.r 4 j/ 3 2. Hence, by formula 6, letting a = 8-r 4 j> 3 and b = 2, _ 4 = (3.* V + 2) (3*y 3 - 2). Ans. EXAMPLE. Factor (a + by m' 2 n'\ SOLUTION. The square roots of the first and the last term are a + b and mn. The sum of these roots is a + b + mn, and the second sub- tracted from the first is a + b mn. Hence, by formula 6, letting a = a + b and b = mn, (a + by m^ri* = (a + b + mn) (a + b mn). Ans. EXAMPLES FOR PRACTICE. 9. Factor the following expressions: 1. a' 2 16. Ans. (a + 4) (a 4). 2. a' 2 49 4 -l. . Ans. (9.r 3 / 2 + 1) (9.ry - 1). 4. (ax + by)'* 1. Ans. (ax + by + 1) (ax -\-by- 1). 5. 25.r 4 / J - (bx + I) 2 . Ans. [5.r 2 j + (bx + 1)] [5*y - (bx + 1)] = (5*y + bx + 1) (5*y -bx- 1). 6. l-169.rV. Ans. 1 93. In example 5, the expression (^r-fl) 2 should be regarded as a single term ; in fact, any number of terms may be regarded as a single term by enclosing them in parenthe- sis and operating on them as though they were a single letter. 94. When solving any examples requiring the applica- tion of the rule in Art. 91, first ascertain if the numerical coefficients of the two terms are perfect squares ; if not, there is no use of examining further. 34 ELEMENTARY ALGEBRA AND 3 FRACTIONS. DEFINITIONS. 95. A fraction, in algebra, is considered as an expression indicating division. The sign -i- is seldom used, it being more convenient to write the dividend, or quantity to be divided, above a horizontal line, with the divisor below it, in the form of a fraction. Thus the fraction . means that a -\- b is to be divided c d by c d, and is the same as (a-\-b) -f- (c d). It is read * * a + b divided by c */" or " a + b over c d." All fractions are read in this way in algebra, except simple numerical fractions, as ^-, -|f , etc. , which are read as in arithmetic. 96. The quantities above and below the line are called the numerator and the denominator, respectively, as in the case of numerical fractions. They are known as the terms of a fraction. 97. Since dividing any quantity by 1 does not change its value, we may write any quantity as a fraction by making the quantity itself the numerator, and 1 the denominator. Thus, 7*y may be written -^-, and not be altered in value. 98. A reciprocal of a quantity is 1 divided by that quantity. It is not necessarily written as a fraction, but when so written has the quantity for the denominator, and 1 for the numerator. Thus, the reciprocal of 5 is , but the reciprocal of is ^ = 5; the reciprocal of -r'+j 2 = 2 a ; and the reciprocal of g = !> Hence, the recipro- cal of a fraction may be obtained by inverting the fraction. 99. The three signs of a fraction are : the sign before the dividing line, which affects the entire fraction ; the sign of the numerator; and the sign of the denominator. When any one of these signs is omitted, it is understood to be plus. 3 TRIGONOMETRIC FUNCTIONS. 35 Any t^tvo signs of a fraction may be changed without altering its value, btit if any one, or all three, be changed, the value of the fraction will be changed from -\-to- or from to -j-. When either the numerator or the denominator has more than one term, it should be enclosed in a parenthesis when performing" operations affecting it as a whole. The paren- thesis may be removed after the operations are completed. Take the fraction - ^ ; placing numerator and denom- inator in parentheses, we have - -- -J-. The signs of the numerator and denominator are each -f and that of the fraction . Let the quotient oia b-^-c d = q\ then : ' =-<+*>=-* REDUCTION OF FRACTIONS. 1OO. To reduce a fraction is to change its form without changing its value. Thus, and have different o 1U forms, but like values, since 10^-^5 and 20^-^10 are each equal to 2;r. 36 ELEMENTARY ALGEBRA AND 3 The terms of a fraction may botJi be multiplied, or may both be divided by the same quantity without changing tJieir value. 1O1. To reduce a fraction to its simplest form: Rule. Resolve each term into its factors, and cancel fac- tors wJiidi appear in both. 1 O2. In per forming all operations on fractions, the student must learn to use a polynomial factor as a single quantity, like a monomial factor. This is illustrated in the following examples, where there are polynomial factors in both numerator and denominator that can be canceled. t 2 + 2xy + y 2 EXAMPLE. Reduce - - , to its simplest form. x y SOLUTION. Factoring both numerator and denominator, y' z _ (x +y) (x Canceling the common factor x +y from both, gives as the result, 3r 5 6 rV EXAMPLE. Reduce 77- .. , to its simplest form. 22 8 - SOLUTION. = = ../ . = ^ ; - ~, when factored. xy\x 2y) Canceling the common factors, we have as the result, __ *? ~ ' 2 1O3. Sometimes the whole numerator is contained in the denominator, or the denominator in the numerator. The numerator or denominator will then reduce to the number 1. EXAMPLE. Reduce + ^ a to its simplest form. = Ans ' TRIGONOMETRIC FUNCTIONS. 37 EXAMPLE. Reduce -^ - to its simplest form. _ ..- SOLUTION. ^g-^i = - ^,- i - = i = *+! Ans. (Art. 97.) 1O4. From the last example it will be seen that division may sometimes be performed by cancelation. Thus, x 1 means (x* 1) -j- (-r 3 1), and the divisor x* 1 canceled from the dividend x* 1 gives the quotient x* -f- 1. A factor must be common to each term of the numerator and to each term of the denominator in order to be canceled. Thus, the factor x cannot be canceled from - - because x + 4;// it is not common to both terms of the denominator. EXAMPLES FOR PRACTICE. 1O5. Reduce the following to their simplest form .* Ans. a" 2 b-' 'a If' 2. 'z ^-5. Ans, .r 2 +}>"*. Qab^c Ans. -r . Ans. - 5. -= - A -. Ans. 106. When fractions are to be added or subtracted, it is necessary to so reduce them that all the denominators will be alike. This is called reducing them to a common denominator. 107. To reduce fractions to a common denominator: Rule. Resolve each denominator into its factors. Take each factor as many times as it occurs in any one denominator, and find the product of these factors. 38. ELEMENTARY ALGEBRA AND 3 Divide this product by each of the denominators. Multiply the corresponding numerators by these quotients, for new numerators. Write each new numerator with the common denominator beneath it. la Zab 2b EXAMPLE. Reduce , >a _ a , and - - to a common denomi- nator. SOLUTION. Factoring the denominators, we have x+y not factora- ble. ,r 2 y* = (.v +y) (x y), and (x + y)'* (x +y) (x +y). Now we have two separate factors, x +y and x y, of which x +y occurs twice in (x +yY. Hence, our common denominator is (x +y) (x +y) (x y} (x +yf (x y}. Dividing this product by x +y we have (x +y) (x y) = x"* y" 2 as our quotient. Hence, our first new numerator is 1a(x* - y 2 ) and the new fraction is , ~ . Similarly, (x+yY(x-y) *' becomes - \, . J ' r, and ; - becomes ' - , . r, ; - - - - , r. (x +yY (x -y)' (x +yy (x +yY (x -y} The student should note that this denominator can be written in several different ways, and he should not become confused if his work does not always agree with the answer. Besides (x+y) (x+f) (xy) and (x+y)' 1 (xy), it may be written (x* /) (x+y), (x* + %xy + y*) (xy), or x 3 -f- x*y xy* y 3 . These five expressions have exactly the same value. The student should . prove this statement by substituting numbers for x and y. EXAMPLES FOR PRACTICE. 1O8. Reduce the following to a common denominator: 2. 8. 4. R 2x ay 42 x*y xy z . lyz* -&< is- and ir 23 4 \2xyz \Zxyz . 3.r 2 j/ 2xyz Ans - IT- -W-- An- 3 Sa ' 12xyz* . lax* and . a 3 x 3 2mn + 2 a 3 x 3 ' ax 3 ' * [ a'x m + n m n m n in + n 2 3 2x-l nnn m 2 n 1 m' 2 - n 2 d ' L 3 TRIGONOMETRIC FUNCTIONS. 39 ADDITION AND SUBTRACTION OF FRACTIONS. 1O9. To add or subtract fractions: Rule. Reduce the fractions, if necessary, to a common denominator. Add or subtract the numerators, and write the result over the common denominator. EXAMPLE. Find the sum of = and - A . 5 4 SOLUTION. = and -. , reduced to a common denominator, 5 4 4{2a -b] , become v - and -, which are equal, respectively, to ,*0 i*u /& and ^ . Adding the numerators, we have 8a 4 + 5a + b = 13# -|- b. The result written over the common denominator gives as The work is written as follows: 2a b a+b Sa b 5a + the sum, p^ . The work is written as follows: 20 54 20 20 Sa - b -+- 5a + 5b ISa + b 20 EXAMPLE. Subtract 5-7 from - . ou 4/tt SOLUTION. Reducing the fractions to a common denominator, 2Y ~ ~~~ ~~ '* removing parentheses and combining. 2 a a TRIGONOMETRIC FUNCTIONS. 41 - R __ I __ Arm 4;;/ 2 " 12m a "* 12;? ' ' y y i ' y( X - xx 3.r .r O. 5 T. .T\.llrt. MUI/TIPLICATIOX OF FRACTIONS. Multiplication, in fractions, is the process of finding a fractional part of a fraction. Thus, fx means i of f. One-half of f inch, for example, is f inch; f of f inch is ^, or T 6 2, inch. The result in each case is the same as that which would be obtained by finding 1 the product of the numerators and writing it over the product of the denominators. 113. Hence, to multiply fractions: Rule. Multiply the numerators together for the numera- tor of the product, and the denominators together for the denominator of the product. 114. Any number of fractions may be multiplied together. The operation may be very much shortened by resolving the terms of the fractions into their factors, and canceling. The product should be reduced to its simplest form. , 6rt 2 Zab , Zac EXAMPLE. Find the product of - , -^ , and -75-. 5 oc b SOLUTION. The product of the numerators is 6 2 X %ab X 2ac = 24a*frc, and of the denominators, 5 X &: X !>'* 150V. Writing Za*bc over 150V. Z4a*bc 8a* we have for the product, = -^-, when reduced to its lowest terms. The work is written as follows : 42 ELEMENTARY ALGEBRA AND 3 _ 4 r -I- 4- - EXAMPLE. Find the product of y _ ,. .* 2 - 1, and - SOLUTION. First make x* \ a fraction by writing 1 for its denom- x -i _ i inator, thus, ^ ; then, factoring both terms of each fraction, we have *r # EXAMPLE. Find the product of = -- , and - H . 3 a 1 + 2ac I 4^ SOLUTION. Performing the subtraction, = -- = ' a* a EXAMPLES FOR PRACTICE. 115. Multiply the following: 1. = r~r> by -^ -, . Ans. . abc >abc c 2. by 2\xy. Ans. 15^ 2 _^ 2 . Find the product of: 3^ 2 y Sy 2 ^ , 12jr 2 . 15.r 3 - i-rr 2 -flTr. and o-... 2 Ans - ^r^r- .r 2 j/ 2 c d , .r 3 + y 3 A ^ 2 aj +j/ 2 4. -T ^. - r, and - = . Ans. -5 - , ^ t . ) J x y c 2 + cd + d* 4y 16 1 5. -*- -- , and - -- j. Ans. x xy DIVISION OF FRACTIONS. 116. Division, in fractions, is the reverse of multipli- cation, and is the process we use when, given one of two fractions and their product, we are required to find the other. For example, we are required to divide - by -. We wish 3 TRIGONOMETRIC FUNCTIONS. 43 to find such a fraction that, multiplied by -, will give -. r ,. . a a I a A1 x x 7 . This fraction is -, for ^X^ = 7. Also -=-^--^ = ^ since A & Z 4: 5 7 7 x x - x -^ = . If, in this case, we had inverted the divisor and 57 5 x 7 7 multiplied, we should have had X - = -. 5 x 5 117. Hence, to divide by a fraction: , Rule. 'Invert the divisor, and proceed as in multiplication. EXAMPLE. Divide -^- by ~ 5.i-y J w.i-y * SOLUTION. The divisor inverted = . . _ _ Hence ' 5P?^IOP? = 5J> X "9^ r - XX^>y "8^ 3 ' EXAMPLE. Divide ^ 2 -4- 2_t + 1 by "* ' 17 SOLUTION. By Art. 97, (.r' + S.t+l) - ,T T l) _ -^-1. Ans. EXAMPLES FOR PRACTICE. 118. Divide the following: I- ^^ by *f . Ans. 5^L'. _ ab bx . ac ex b 2. - by . Ans. -. a + z a + z c a l-^ + 16^ by j_-4g^ Ans. 8(1-2J). 4. ^cPcd abcd by = ^ 77,. Ans. a? b*. 7 a* + aA+6* 44 ELEMENTARY ALGEBRA AND 3 MIXED QUANTITIES AND COMPLEX FRACTIONS. 119. An integral expression is one containing neither fractions nor negative exponents. The expression c? 1 3 is integral, but the expressions 2 -f ~ - %a~*, a , are not. o The expression 2 etc - ab abn* alrk a 5y~ l 5x% In the last, the positive exponent 1 of the y is not written. 3 TRIGONOMETRIC FUNCTIONS. 49 EXAMPLE. Express, with positive exponents, SOLUTION. Since these terms may be taken as fractions, with 1 for the denominators, we have, by transferring the letters with negative exponents to the denominators, Ans ' 134. The student must note very carefully that factors of an entire term only can be changed from numerator to denominator, or vice versa, and that when thus changed they become factors of the whole of the other term. Thus, in _ a c~* cannot be transferred to the numerator by merely changing the sign of the exponent. The exponent may, however, be made positive by multiplying both terms T . ' thus ' -* = ' In ' lf we trans - fer the c"*, it becomes a/ , , .., c* becoming a factor of the fty+a) entire denominator. EXAMPLE. Clear *y-*2r l + -jg-^- 3 ^ ^' o f negative expo- nents. SOLUTION. Treat each term of the expression separately. x*y-*z~* x iy-i 2 -\ = -- j - ; changing the factors with negative exponents to the denom- inator, and at the same time changing the signs of the exponents, we ~ l ^ S n0t E ^ actor ^ the w ^l e denominator, so y v~ l we must multiply both terms of the fraction by the reciprocal of y~ l or y ; of the entire numerator, so we write them as factors of the entire denominator, with the signs of the exponents changed ; thus, ' nce ' 50 ELEMENTARY ALGEBRA AND 3 EXAMPLE. Solve the following: 2n + c" _ Write the answers with positive exponents. SOLUTION. rt'x^" 1 = rt 3 + ( ~'> = rt :t ~ J = a\ Ans. = '+<-*> = '"* = *. Ans. 9 Ans. n 2n n _r^ ^ Ans . 135. A letter may be raised to any power by multi- plying its exponent by the index of the power. Thus, (a*)i = a, (')-* = a~\ etc. EXAMPLE. Find the values of the following: (a ')~i; (r^ 2 )l; (x*)-**-^-*)-*. SOLUTION. In the first, multiplying the exponents, we have IX I =\- Hence, (flr 1 )"* = *, or 4/. Ans. In like manner, = c*rf- 5 , Ans., since 1 X f = |, and 2 X | = 5. In the next one, (.r)~ b = ^~ 6 and (j;- f ')~ 6 = x? b . Dividing, 136. A root of a letter affected with an exponent is extracted by dividing the exponent by the index of the root. Thus, Vy" = y j/ 12 * 4 = y\ From this principle, the fol- lowing rule may be deduced. To extract any root of a monomial : Rule. Extract the required root of the numerical coeffi- cient, and divide the exponent of each letter by the index of the root. Make the sign of every even root of a positive quan- tity , and the sign of every odd root of any quantity the same as that of the quantity. 3 TRIGONOMETRIC FUNCTIONS. 51 EXAMPLE. Find the value of 2 SOLUTION. The 4th root of 256 is 4. The exponent of a in the root is 4 -f- 4 1 ; of b, 12 -;- 4 = 3 ; and of c, 8 -4- 4 = 2. As this is an even root of a positive quantity, the sign should be . Hence, . Ans. EXAMPLE. Find the value of SOLUTION. $27m 3 x' J = 3w.r 3 ; -$VV a = rt s V. The quantity is positive, and, as this is an odd root, its sign must be the samo, or positive. Hence, = . Ans. EXAMPLES FOR PRACTICE. 137. Clear the following of negative exponents : 1. jtV" 2 .^. Ans. 2, 3. : Express the following without radical signs: 4. itfTr*. Ans. (- 2 )* or < Find the values of^ the following : 6. n? X m "*. Ans. 7. 2a$ X *. Ans. 2 w 8. ^ 2 -T- |/^-. Ans. 9. 2,r~ 2 -j- (,r 2 )"^ Ans. 2, 10. (a/~J X W 4 "- Ans. ^ EQUATIONS. DEFINITIONS. 138. As defined in Art. 5, an equation is a statement of equality between two expressions, as x + = 14. Every equation has two parts, called the first and the 52 ELEMENTARY ALGEBRA AND 3 second member. The first member is the part on the left of the sign of equality, and the second member the part on the right of that sign. In x-\-6 = 14, x-\- is the first member, and 14 is the second member. 139. Equations usually consist of known and unknown quantities ; that is, of quantities whose values are given, and of quantities whose values are not given, but are to be found. Thus, in x + 6 = 14, 6 and 14 are known quantities, and x is unknown ; but since by the statement of the equa- tion, x-\- must equal 14, x must have such a value that when added to 6 the sum will be 14. Hence, the value of x is fixed for this particular case, and in a similar manner the value of a single unknown quantity in any equation is fixed by the relations that it bears to the known quantities, and this value can usually be found. 140. To solve an equation is to find the value of the unknown quantity. This is done by a series of transforma- tions by which the first member becomes the unknown quantity, and the second member becomes a known quantity, which is, therefore, the value of the unknown quantity. TRANSFORMATIONS . 141. In transforming an equation, the equality of its members must be preserved ; otherwise the existing rela- tions between the known and unknown quantities will be destroyed. Transformations are based upon the following principles : 142. In any equation : I. The same quantity may be added to both members. For example, if 2 be added to both members of x* = 16, the members of the resulting equation, x* -\-2 = 18, will be equal. II. The same quantity may be subtracted from both members. Thus, if .r 2 = 16, then x* 2 = 14. 3 TRIGONOMETRIC FUNCTIONS. 53 III. Both members may be multiplied or both divided by the same quantity. Thus, if x* = 16, then 2x* = 32 and IV. Both members may be raised to the same power. Thus, if x* = 16, then x* = 256. V. Like roots of both members may be extracted. Thus, if x* = 16, then x = 4. A little thought will show that none of these operations will destroy the quality of the members. In the equation 16 = 16, for example, by I, 16 + 2 = 16 -f 2 ; by II, 16-2 = 16 2 ; by III, 16x2 = 16X2, etc. It is to be observed, however, that after any transformation, the members do not equal their original values. 143. To transpose a term in an equation is to change it from one member of an equation to the other. A term may be transposed to the other member of an equation, if its sign be changed. Thus, in the equation 2^ + 5 = 13, let it be required to transpose + 5 to the second member; chang- ing its sign, we have %x = 13 5, or 2;r = 8. For, subtract 5 from both members and we have 2;r + 5 5 = 13 5, or %x = 8. Suppose we had %x 5 13; changing the sign of 5, and placing it in the second member, we have %x = 18; for 2^ 5 + 5 = 13 + 5, or %x = 18. 144. When the same term appears in both members of an equation, with the same sign, it may be dropped from each. This is called cancelation. Thus, in x-{-a = 16 -\-a we may cancel a, and have x = 16, for subtracting a from both members x-\-a a = 16 -\- a a; hence, x = 16. In x a = 16 a, x a-\-a = \ a-\-a, and x = 16. But, in x a = 16 + 0, the #'s will not cancel, since they have different signs. 145. Changing Signs. It is sometimes desirable to change the sign of a term in one of the members of an equa- tion. This may be effected by multiplying both members 54 ELEMENTARY ALGEBRA AND 3 by 1. This gives the same result as changing the signs of all the terms of both members; thus, jr-j-^ + S _ a __ x _ 7 may be changed to x a 3 a-\-x + 1. According to Art. 142, III, this transformation does not destroy the equality of the members. 14G. Clearing of fractions is usually necessary before performing any operations. Rule. To clear an equation of fractions, multiply each term of the equation by the common denominator of all the fractions. EXAMPLE. Clear of fractions x -f ^ + - f = 10. 246 SOLUTION. The common denominator of all the fractions is 12; multiplying each term by 12, we have 12.r + 6* + 9* - 4r = 120. 2* 1 3* + 2 EXAMPLE. Clear of fractions = o~~~2 T- SOLUTION. The common denominator is 2(.r 2 4); multiplying through by this we get 4*(* 2) = (* 2 4) 2(3* + 2) ; removing the parentheses this becomes 4* 2 8* = * 2 4 6* 4. Where a fraction is preceded by a minus sign, care must be taken to change the sign of every term of the numerator when clearing of fractions. (See Art. HO.) EXAMPLES FOR PRACTICE. Clear the following equations of fractions: 1. , r + + = 16 _ . Ans. 28* 2 + 21^ 2 -f 20* = 448* - 56. 2. - A -"^ = Ans. 3*- 6* + 18 = 2a. 4 A O o x a-b O. - j - * - r - 1. a b a+b Ans. ax + bx a?x + Px a? 2ab + b'* a 1 + fr*. 1 . * a + b o n LI 4. ; = - -. -- . Ans. * = * a a? + P. a b a b x 3 TRIGONOMETRIC FUNCTIONS. 55 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY. 148. Simple equations, when reduced to their simplest form, contain the first power only of the unknown quantity. When there is but one unknown quantity, it is usually rep- resented by x. Numbers and the first letters of the alphabet are used for known quantities. 149. To solve simple equations with one unknown quantity: Hule.^-C/ear of fractions. Remove all signs of aggrega- tion. Transpose all terms containing tJie unknown quantity to tJie first member; all others to tJie second member. Com- bine the first member into one term, and simplify tJie second member. Divide both members by tJie coefficient of tlie unknown quantity. In some cases, this order of operations may be changed to advantage. 150. To verify the result, substitute the value of the unknown quantity in the original equation, which should then reduce so that both members will be alike. When this occurs the equation is said to be satisfied. EXAMPLE. Solve 2.r + 5 = 25 Sx. SOLUTION. Transposing the unknown quantities to the first member, wehave 2.r + 3.r + 6 = 25.. Transposing the known quantities to the second member, we have 2.r + 3.t = 25-5. Combining like terms, 5.r = 20. Dividing by 5, x = 4. Ans. Now, if the substituting of 4 for x will satisfy the original equation, we know our answer is correct. Thus, substituting 4 for x> we have 2X4 + 5 = 25-3X4. 8 + 5 = 25-12. 13 = 13. Hence the equation is satisfied, and our result is correct. 56 ELEMENTARY ALGEBRA AND 3 EXAMPLE. Solve 16 x \ 7,r \x (x 8* 6,r)] } = 0. SOLUTION. Removing the symbols of aggregation (Art. 52), 16 x Ix + Sx $x + Sx x = 0. Transposing 16 to the second member, x lx + $x %x + Sx Qx = 16. Combining like terms, 12x = 16. Dividing both members by 12, x = -if = li. Ans. 1 8-Gx 2(6.^ + 7) EXAMPLE. Solve f + -, = - -- h M . A 4 o SOLUTION. Simplifying the first and the last term of the equation, it becomes x + l + - = _ -- 1 -- Z . 454 Clearing of fractions, we have 20.r + 20 + 5 = 32 -24.r + 3(hr + 35. Transposing, 2(Xr + 24.* 30.* = 32 + 3520 5. Combining, 14;r = 42. Dividing by 14, x = 3. 001 EXAMPLE. Solve - -- - -- h- - =. = 0. 1 x l + x 1 x* SOLUTION. Clearing of fractions by multiplying by 1 ;r 2 , = 0. = 0. Uniting terms, x = 2. x .4. Ans. NOTE. multiplied or divided by any number rr 0. 151. literal equations are those in which the known quantities are represented partially or wholly by letters. In solving such equations, we cannot always combine the unknown quantities into one term. EXAMPLE. Solve %ax %b = x + c Sax. SOLUTION. Transposing the terms containing the unknown quanti- ties to the first member and the remaining terms to the second mem- ber, and combining like terms, Factoring Sax x to bring x alone in the first member, (5a - l)x = Zb + c. The coefficient of x is now 5^ 1, this being considered as one quantity. T\- ' *' 3^ + c . Dividing by 5^ 1, x = _ . Ans. 3 TRIGONOMETRIC FUNCTIONS. 57 PROOF. Since the original equation is equivalent to 5a.i x = %b 4- c, it will be sufficient to satisfy this equation. Hence, substituting the value of x, Canceling the 5a 1, 3 -f c = 3x + l Sbx EXAMPLE. Solve - ^- = -=- - SOLUTION. Clearing of fractions, (3* + 1) [b(x + !)-] = (x + 1) (Zbx - 2a -K c), or bx(x + 1) - 3ax +b( x + \)-a = 3bx(x + 1) - (2a - c} (x + 1). Canceling 3bx(x + 1) from both members, Sax -\-bx + b a = 2ax + cx Transposing and uniting terms, ax + bx ex = a b + c. Changing signs and factoring, (a b + c}x = a + b c. a + b c EXAMPLES FOR PRACTICE. 5. Solve the following: 1. 16 3;t = 13 Gx. Ans. x = 1. 2. 3(4. r _ 5) + 6 = 1 + 2x. Ans. .r = 1. 3. 6(5 -2x) = Q-2(x-2). Ans. x = 2. Ans. x = 60. Ans. * = 41. Ans. * = 7. Ans. * = 4. > Ans. ;r = jr+1 ^r+4 . 5. = = lo 6x x 5x ' ^~~^ 7" O OX i 5 _ 3-2*. = 0. 8. 2x 4a = a* 3a + 2 Ans. .r = 5a* + Wa SUGGESTION. Transposing the second term to the second member, ax + 2x _ a' 2 + 4a + 4: _ (a + 2) 2 5a 58 ELEMENTARY ALGEBRA AND 3 Multiplying both sides by 5#, ,+ = **+). 40 Solving for x, - - afc* + ,r 2 ) , ax . . c 10. - - = ab -\ -- . Ans. x T. ex c b PROBLEMS LEADING TO SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY. 153. There are two steps in the solution of problems by algebra : I. The relations which exist between the known and the unknown quantities, that is, between those whose values are given in the problem and those whose values are required, must be stated by one or more equations. This is called the statement of the problem. II. The resulting equation or equations must be solved, giving the values of the required quantities. It will thus be seen that by the algebraic method, the answer to a problem is used in the solution and operated upon as though it were a known quantity, which is one great advantage over the arithmetical method. 154. The ability to state a problem by means of an equation depends upon the ingenuity of the operator and his ability to reason, rather than upon his knowledge of algebra. No definite rule can be given for making the statement, but in general, where there is only one unknown quantity in a problem : Decide what quantity it is whose value is to be found. This will be the unknown quantity, or the answer. Then represent the unknown quantity by x and form an equation that will indicate the relations between the known and the unknown quantities as stated in the problem. The equation will also indicate the operations that would be performed in proving the statement made in the problem g 3 TRIGONOMETRIC FUNCTIONS. 59 were the answer known. Hence, the equation may often be formed by noticing what operations would be performed upon the answer in proving. 155. The following examples are illustrations of the statement and the solution of algebraic problems ; they should be studied carefully. EXAMPLE. Find such a number that, when 14 is added to its double, the sum shall be 30. SOLUTION. The quantity whose value is required is the number itself. As this is the unknown quantity, let x = the number, whence 2x must be double the number. Now the problem states that when 14 is added to double the number the sum will be 30. In other words, when 14 is added to 2.r, the sum will be 30. Hence, the statement of the problem in the form of an equation is 2x + 14 = 30 ; whence, solving, x = 8. Ans. EXAMPLE. Find a number which, when multiplied by 4, will exceed 40 as much as it is now below 40. SOLUTION. Let x the required number, which, when multiplied by 4, becomes 4.r. According to the conditions of the problem, the amount by which 4 times the required number, or 4.r, exceeds 40 is equal to the amount that the number itself, or x, is below 40. But 4x 40 is the amount by which 4x exceeds 40, and 40 x is the amount by which x is below 40. Hence, by the conditions, we have the statement, 4.r-40 - 40 -.r. Transposing and uniting, 5,r = 80, or x = 16. Ans. EXAMPLE. Two loads of brick together weigh 4,000 Ib. ; but if 500 Ib. be transferred from the smaller to the larger load, the latter will weigh 7 times as much as the former. How much does each load weigh ? SOLUTION. If the weights of the two loads were known and it was desired to prove the correctness of the example, we should add 500 Ib. to the weight of the larger load and subtract 500 Ib. from the weight of the smaller load, as stated in the example. The larger load should then weigh 7 times as much as the smaller. To obtain the equation the same thing is done by letting x = the weight of one load, whence 4,000 .v is equal to the weight of the other load. Let x the weight of the smaller load. 60 ELEMENTARY ALGEBRA AND 3 Then, 4,000 - x = the weight of the larger load. Also, x 500 = the weight of the smaller load after transferring 500 Ib. And 4,000 ^ + 500 = the weight of the larger load after transfer- ring 500 Ib. By the conditions, the larger load now weighs 7 times as much as the smaller. Hence, l(x - 500) = 4,000-.* + 500. Solving, 7.* -3, 500 = 4,500-jr, or 8x = 8,000; whence, x 1,000 Ib. = weight of smaller load, j and 4,000 - x = 3,000 Ib. = weight of larger load, i ' PROOF. 1,000 - 500 = 500 = weight of the smaller load, and 3,000 + 500 = 3,500 = weight of the larger load after the 500 pounds have been transferred ; 3,500 -r- 500 = 7. Until the student has obtained considerable proficiency in solving problems of this kind, it is a good plan to prove all problems. EXAMPLE. The circumference of the fore wheel of a carriage is 10 feet, and of the rear wheel, 12 feet. What distance has the carriage traveled, when the fore wheel has made 8 more turns than the rear wheel ? SOLUTION. In this example the distance traveled is not known, but is required to be found. Suppose that the distance is known, and that it is equal to x feet, and that we wish to see whether the statement is true that the fore wheel makes 8 more revolutions than the rear wheel in passing over x feet. The number of revolutions of the fore wheel Jf 3C is evidently y^, and of the rear wheel, -^. The example states that the difference between them is equal to 8. Solving for x, 12* - 10* = 960, or 2x = 960, and x = 480 feet. Ans. 480 PROOF. ^- = 48 = revolutions of fore wheel. 480 - = 40 = revolutions of rear wheel. 48 40 = 8. Compare this proof with (1). EXAMPLE. A water cistern connected with three pipes can be filled by one of them in 80 minutes, by another in 200 minutes, and by the third in 300 minutes. In what time will the cistern be filled when all three pipes are open at once ? 3 TRIGONOMETRIC FUNCTIONS. 61 SOLUTION. Here the unknown quantity is the number of minutes required to fill the cistern by all three pipes together. Supposing this to be x minutes, the example may be proved by noticing that the sum of the quantities of water flowing through each pipe separately in a given length of time, as 1 minute, must be equal to the quantity flow- ing through all three together in the same length of time. According to the problem, the quantity discharged by the first pipe in 1 minute would be ^, by the second ^\^, and by the third ^^ of the contents of the cistern. In like manner the quantity discharged by all three at once in 1 minute would be -. Then, if the example is stated correctly, we must have -+ + ~- 80 "*" 200^300 x' Clearing of fractions, ;r(30 + 12 + 8) = 2,400, or 50.* = 2,400; whence, x = 48 minutes. Ans. EXAMPLE. A man rows a boat a certain distance with the tide, at the rate of 6| miles an hour, and returns at the rate of 3i miles an hour, against a tide half as strong. If the man is pulling at a uniform rate, what is the velocity of the stronger tide ? SOLUTION. If the following statement is not clear, the student should reason it out for himself in a manner similar to that used in the last three examples. Let x = number of miles per hour that the stronger tide is running ; then = number of miles per hour that the weaker tide is running. & V Hence, 6| x and 3i + ^- are expressions for the rate at which the man is pulling. But, as he is pulling at a constant rate all the time, these expressions must be equal. Hence, 20 10 x or -s- - x -^ + -. o o a Clearing of fractions, 40 6.r = 20 + 3.v, or 9.r = 20; whence, x = 2| miles per hour. Ans. EXAMPLES FOR PRACTICE. 156. Solve the following examples: 1. The greater of two numbers is four times the lesser number, and their sum is 400 ; what are the numbers ? Ans. 80 and 320. 62 ELEMENTARY ALGEBRA AND 3 2. A farmer has 108 animals, consisting of horses, sheep, and cows. He has four times as many cows as horses, lacking 8, and five times as many sheep as horses, lacking 4 ; how many has he of each kind ? f 12 horses. Ans. j 40 cows. [ 56 sheep. 3. A can do a piece of work in 8 days, and B can do it in 10 days ; in what time can they do it working together ? Ans. 4f days. 4. Find five consecutive numbers whose sum is 150. Ans. 28, 29, 30, 31, and 32. 5. A boat whose rate of sailing is 6 miles per hour in still water moves down a stream which flows at the rate of 3 miles per hour, and returns, making the round trip in 8 hours ; how far did it go down the stream ? Ans. 18 miles. THE TRIGONOMETRIC FUNCTIONS. DEFINITIONS. 157. Plane trigonometry treats of the solution of plane triangles. Every triangle has six parts three angles and three sides. When three of these parts are given, if one part at least is a side, the other three can be found. This process of finding the unknown parts of a triangle is the solution of the triangle. 158. The complement of an angle is the difference between 90 and the angle. Thus, the complement of an angle of 35 is an angle of 55, because 90 - 35 = 55. In a right-angled triangle, the right angle is 90 ; since the sum of the three angles of the triangle is 180, the sum of the two acute angles is 180 90 = 90. Therefore, each acute angle of a right-angled triangle is the complement of the other acute angle. 159. The supplement of an angle is the difference between 180 and the angle. Thus, the supplement of an angle of 35 is an angle of 145, because 180 35 = 145. g 3 TRIGONOMETRIC FUNCTIONvS. (>3 160. The solution of a triangle is accomplished by means of the trigonometric functions. These functions are the ratios of the sides of a right-angled triangle ; the niost important of these functions are the sine, cosine, tangent, and cotangent. These are abbreviated to sin, cos, tan, and cot. 161. In the right-angled triangle ABC, Fig. 2, the sides a, b, and c are opposite, respect- ively, to the angles A, B, and C. The hypotenuse is c, and C is the right angle. The short side b is adjacent to angle A and opposite angle B ; the short side a is opposite to angle A and adja- cent to angle B. Then the trigonometric functions are defined as follows: 162. The sine of an angle is the quotient, of the opposite side divided by the hypotenuse. Thus, in Fig. 2, sin A = - c . 163. The cosine of an angle is the quotient of the adja- cent side divided by the hypotenuse. Thus, cos A = -. 164. The tangent of an angle is the quotient of the opposite side divided by the adjacent side. Thus, tan A = * b 165. The cotangent of an angle is the quotient of the adjacent side divided by the opposite side. Thus, cot A = b -. a 166. From the definitions of the functions, we have : sin B = - ; cos B = - ; tan B - ; and cot B = 7 . Com- c c a b paring the functions of angle B with those of angle A, 64 ELEMENTARY ALGEBRA AND 3 sin B = cos A (since each is equal to -), cos B = sin A, tan B cot A, and cot B = tan A. It has been shown (Art. 158) that angle A is the complement of angle B. There- fore, the sine of an angle is equal to the cosine of its comple- ment, and the tangent of an angle is equal to the cotangent of its complement. For example, sin 36 = cos (90 36) = cos 54; tan 28 = cot 62; etc. TRIGONOMETRIC TABLES. 167. Every angular function has a different value for each of the angles between and 90. The numerical values of these functions are called natural sines, cosines, etc., and are given in the tables of Natural Sines, Cosines, Tangents, and Cotangents. In many tables, both natural and logarithmic functions are given. The student should not attempt to use the latter until he thoroughly under- stands logarithms. The table of natural functions, and its use, will now be explained. 168. Given an Angle, to Find Its Functions. EXAMPLE. Let it be required to find the sine, cosine, and tangent of an angle of 37 24'. SOLUTION. Look in the table of Natural Sines along the tops of the pages and find 37. The left-hand column is marked ('), meaning that the minutes are to be sought in that column, and begin with 0, 1, 2, 3, etc., up to 60. Glancing down this column until 24 is found, find opposite this 24 in the column marked sine, and headed 37, the number .60738; then .60738 = sin 37 24'. In exactly the same manner, find in the column marked cosine and headed 37, the number .79441, which corresponds to cos 37 24' ; or cos 37 24' = .79441. So, also, find in the column marked tangent and headed 37, and opposite 24', the number .76456; hence, tan 37 24' = .76456. 169. In most of the tables published, the angles run only from to 45 at the top of the page; to find an angle greater than 45, look at the bottom of the page and glance upivards, using the extreme right-hand column to 3 TRIGONOMETRIC FUNCTIONS. 65 flml minutes, which begin with at the bottom and run upwards, 1, 2, 3, etc. to 60. EXAMPLE. Find the sine, cosine, and tangent of 77 43'. SOLUTION. Since this angle is greater than 45, look in the tables along the bottom of the page, until the column marked 77 is found. Glancing up the column of minutes on the right, until 43' is found, find opposite 43' in the column marked sine (and 77) at the bottom, the number .97711 ; this is the sine of 77 43', or sin 77 43' = .97711. Sim- ilarly, in the column marked cosine, find opposite 43' in the right-hand column, the number .21275; this is the cosine of 77 43', or cos 77 43' = .21275. So, also, find that 4.59283 is the tangent of 77 43', or tan 77 43' = 4.59283. 170. Let it be required to find the sine of 14 22' 26". EXPLANATION. The sine of 14 22' 26" lies between sin 14 22' and sin 14 23'. Sin 14 22' = .24813; sin 14 23' = .24841 ; difference = .00028. Neglect for the moment the fact that the functions are decimal fractions; then the difference between sin 14 22' and sin 14 23', that is, the dif- ference between 24841 and 24813, is 28, or, corresponding to a difference of V in the angle, there is a difference of 28 in the sine. Now, since I' = 60 ", the difference between sin 14 22' and sin 14 22' 26", that is, the difference corresponding to a difference of 26" in the angle, must be f f X 28 = 12.1. Since .1 is less than .5, omit it, and we have 12 as the differ- ence to be added to 24813. 24813 + 12 = 24825; taking account now of the fact that the function is a decimal frac- tion, sin 14 22' 26" = .24825. In all work with the tables it will always be found most convenient to neglect for the moment the decimal point, consider the difference a whole number, and afterwards locate the decimal point in its proper position. 171. Reference to the table of functions shows that, as the angles increase in magnitude, the sines and tangents increase, while the cosines and cotangents decrease. In the above example, therefore, had it been required to find the cosine of 14 22' 26", the correction for the 26" would have been subtracted from the cosine of 14 22 ', instead of being added to it. 60 ELEMENTARY ALGEBRA AND 3 EXAMPLE. Find the sine, cosine, and cotangent of 56 43' 17". SOLUTION. Sin 56 43' = .83597. Sin 56 44' = .83613. Since 56 43' 17" is greater than 56 43' and less than 56 44', the value of the sine of the angle lies between .83597 and .83613 ; neglecting the decimal character of the functions, the difference = 83613 83597 = 16; multi- plying this by the fraction ^, 16 X H = 4 - 53 - Since .53 exceeds .5, we take 5 as the difference to be added to 83597. Adding, 83597 + 5 = 83602; hence, sin 56 43' 17" = .83602. Ans. Cos 56 43' = .54878; cos 56 44' = .54854; expressed as a whole number, the difference = 54878 - 54854 = 24, and 24 X \l = 7, nearly. Now, since the cosine is desired, we must subtract this correction from 54878; subtracting, 54878 -.7 = 54871. Hence, cos 56 43' 17" = .54871. Ans. Cot 56 43' = .65646; cot 56 44' = .65604; difference = 65646-65604 = 42, and 42 X \\ 12, nearly. Now, since the cotangent decreases as the angle increases, we must subtract this correction from 65646 ; thus, 65646-12 = 65634. Hence, cot 56 43' 17" = .65634. Ans. 172. Given the Function, to Find the Correspond- ing Angle. This is the reverse of the process .for finding the function of the angle. If the angle corresponding to the given function is an exact number of degrees and minutes, the function will be found in the table. In siich a case, we have simply to find the given function, and take the degrees from the end of the column, and the minutes at the end of the row. If the name of the function is at the top of the column, the number of degrees will be found there also, and the minutes at the left. If the name is at the bottom, the degrees will be at the bottom, and the minutes at the right. CASE I. 173. The function is found exactly in the table. EXAMPLE. Find the angle whose sine is .24982. SOLUTION. In the table of Sines, we find .24982 in the column under 14. Since the name of the function is at the tqp of the column, we take the number of degrees from the top, and the minutes from the left; thus, 14 at the top, and 28' at the left. Hence, the angle whose sine is .24982 is 14 28'. Ans. 3 TRIGONOMETRIC FUNCTIONS. 67 EXAMPLE. Find the angle whose cotangent is .68557. SOLUTION. In the table of Tangents, we find .68557 under 34. The name of the function is, however, at the bottom of the page, and we must take the degrees 55, from that end of the column. The minutes, 34', are. found in the right-hand column. Hence, the angle whose cotangent is .68557 is 55 34'. Ans. CASE II. 174. The function is not found exactly in tJie table. Let it be required to find the angle whose sine is .42531. EXPLANATION. Referring to the table of Sines, this num- ber is found to lie between .42525, the sine of 25 10', and .42552, the sine of 25 11'. Neglecting decimals, the differ- ence between these two sines = 42552 42525 = 27 ; the difference between 42525, corresponding to the sine of 25 10', and 42531, corresponding to the sine of the given angle = 42531-42525 = 6. Since 27 is the difference for 1', a difference of 6 corresponds to -fa of 1'; hence, the angle whose sine = .42531 = 25 10^'. Since V = GO", -/ r of a minute = ^ 8 T X60 = 13.3". There- fore, the angle whose sine is .42531 = 25 10' 13.3". The given function should always be compared with the function of the angle next lo^ver, and the correction in sec- onds should be added to that angle. In the case of the sine and tangent, we take the difference between the given func- tion and the next smaller function appearing in the tables; but with the cosine and cotangent, we take the difference between the given function and the next larger function which appears in the tables. EXAMPLE. Find the angle whose cosine is .27052. SOLUTION. Looking in the table *of Cosines, the given function is found to belong to an angle greater than 45, and, hence, must be sought for in the columns marked cosine at the bottom of the page. It is found between the numbers .27060 = cos 74 18' and .27032 = cos 74 19'. The difference between the two is .27060 .27032 = .00028, or 28, neg- lecting decimals. The cosine of the smaller angle, or 74 18', is .27060, and the difference between this and the given cosine is .27060 .27052 .00008, or 8, neglecting decimals. Hence, / X60 = 17.14", nearly, and the angle whose cosine is .27052 = 74 18' 17.14", or cos 74 18' 17.14" = .27052. Ans. 68 ELEMENTARY ALGEBRA AND 3 EXAMPLE. Find the angle whose tangent is 2.15841. SOLUTION. 2.15841 falls between 2.15760 = tan 65 8', and 2.15925 = tan 65 9'. The difference = 2.15925-2.15760 = .00165, or 165, con- sidered as a whole number. 2.15841 2.15760 = 81, neglecting the decimal. T y r X 60 = 29.5", nearly, and the angle whose tangent is 2.15841 = 65 8' 29.5", or tan 65 8' 29.5" = 2.15841. EXAMPLES FOR PRACTICE. 175. Solve the following examples : 1. Find the (a) sine, (b) cosine, and (c) tangent of 48 17'. C (a) .74644. Ans. { (b) .66545. ((c) 1.12172. 2. Find the (a) sine, (b} cosine, and (c) cotangent of 13 11' 6". ((a) .22810. Ans. J (b) .97364. [ (c) 4.26855. 3. Find the (a) sine, (b) cosine, and (c) tangent of 72 0' 1.8". f (a) .95106. Ans. -j (b) .30901. [ (c) 3.07777. 4. (a) Of what angle is .26489 the sine, (b) of what is it the cosine, and (c) of what is it the cotangent ? ( (a) 15 21' 37.2". Ans. \ (b) 74 38' 22.8". [ (c) 75 9' 49". 5. (a) Of what angle is .68800 the sine, (b} of what the cosine, and (c) of what the tangent ? f () 43 28' 20". Ans. \ (b} 46 31' 40". \(c) 34 31' 40. 5". SOLUTION OF TRIANGLES. RIGHT-ANGLED TRIANGLES. 176. When any side and one of the acute angles of a right-angled triangle (also called right triangle) are given, the remaining sides and angles may be found ; also, when any two sides are given, the remaining parts may be found. 3 TRIGONOMETRIC FUNCTIONS. 69 177. The following relations between the sides and angles of a right triangle are derived directly from the definitions of the trigonometric functions: I. Side opposite an angle = hypotenuse X sine of angle. II. Side adjacent = hypotenuse X cosine. III. Side opposite = side adjacent x tangent. IV. Side adjacent = side opposite X cotangent. TT TT side opposite V. Hypotenuse = - . sine rj. side adjacent VI. Hypotenuse = 4 . cosine These relations are sufficient to find the sides. The angles may be found from the sides by the relations given in Arts. 162 to 165. To show the application of these relations to the solution of triangles, a number of examples are given. 178. Case I. The hypotenuse and an acute angle being given, to find the remaining parts : EXAMPLE. In Fig. 3, the hypotenuse A B of the right-angled tri- angle A C B is 24 feet and the angle A is 29 31' ; to find the sides A C and B C and the angle B. NOTE. When working examples of this kind, construct the figure, and mark the known parts. This is a great help in solving the example. Hence, in the figure draw the angle A to represent an angle of 29 31', and complete the right-angled triangle A C B, right-angled at C, as shown. Mark the angle FlG - 3 - A and the hypotenuse, as is done in the figure. SOLUTION. Angle B = 90 - 29 31' = 60 29'. A Cis the side adja- cent to angle A. Hence, from relation II, Art. 177, A C, or side adjacent = hypotenuse X cosine =' 24 X cos 29 31'. In the table of Natural Cosines we find the cosine of 29 31' to be .87021. Therefore, A C = 24 X .87021 = 20.89 feet, nearly. To find the side B C, we use relation I, Art. 177. B C, or side opposite = hypotenuse X sine = 24 X sin 29 31'. The sine of 29 31' is .49268; hence, B C = 24 X .49268 = 11.82 feet, nearly. f Angle B - 60 29'. Ans. J Side A C = 20.89 ft. [Side BC = 11.82ft. 70 ELEMENTARY ALGEBRA AND 3 179. Case II. An acute angle and one of the short sides being given, to determine the remaining parts : EXAMPLE. In Fig. 3, suppose the side A C to be 75 feet and the angle A to be 32 24'. The sides B C, A B, and the angle B are required. SOLUTION. Angle B = 90 - 32 24' = 57 36'. To find B C, we have relation III, Art. 177, B C, side opposite = side adjacent X tangent = 75 X tan 32 24'. Referring to the table of Natural Tangents, the tangent of 32 24' = .63462. B C = 75 X .63462 = 47.6 ft, nearly. To find the hypotenuse A B, we use relation VI, Art. 177, hypotenuse _ side adjacent . _ A C 75 75 _ cosine " cos A ~ cos 32 24' ~ .84433 ~ f Angle B = 57 36'. Ans. J Side B C = 47.6 ft. [ Side A B = 88.81 ft. 180. Case III. Two sides being given, to find the third side and the acute angles : EXAMPLE. In the right-angled triangle A B C, Fig. 4, right-angled B at C, A C = 18 and B C - 15; to find A B and the angles A and B. SOLUTION. According to the definition of the tangent, Art. 164, . side opposite 15 tangent A = -. - = ~ = .83333. side adjacent 18 Looking in the table of Tangents, the angle whose tangent is nearest to .83333 is 39 48'. IS "" Hence, angle A 39 48', nearly. Angle B FIG. 4. = 90 _ 39 48' = 50 12'. To find the hypotenuse, we use relation V, Art. 177, side opposite B C A B. hypotenuse = f = -= -r. s*me sm A The sine of 39 48' is .64011. Hence, A B = j^r = 23.43. f Angle A = 39 48'. Ans. J Angle B - 50 12'. [ Side A B = 23.43. EXAMPLE. In the right-angled triangle ABC, Fig. 5, right-angled at C, A C = .024967 mile and A B .04792 mile ; to find the other parts. TRIGONOMETRIC FUNCTIONS. 71 SOLUTION. According to the definition of the cosine, Art. 163, . side adjacent '.024967 COS A = hypotenuse = 1>4W = - 53101 - Referring to the table, the angle whose cosine is .52101 is 58 36'. Therefore, angle A = 58 36'. Angle B = 90 - 58 36' = 31 24'. To find side B C, relation III, Art. 177, is used. Side opposite A side adjacent X tan A, or B C = A C X tan 58 36' = .024967 X 1.63826 = .0409 mile. j' Angle A = 58 36'. Ans. -j Angle B = 31 24'. \ B C = .0409 mile. .024967 FIG. 5. EXAMPLE. In the right-angled triangle ABC, Fig. 6, right-angled at C, A B =. 308 feet and B C = 234 feet ; to find the other parts. SoLUT,ON.-Sin A = hypotenuse = !* = .75974. 308 (Art. 163.) The angle whose sine is .75974 = 49 26|', nearly, = angle A. Referring to the table of Sines, the sine of 49 26' is .75965, and the sine of 49 27' is .75984. Since the value obtained, .75974, lies nearly half way between these val- ues, the angle lies midway between the above angles, and is 49 261'. Angle B = 90 - 49 261' = 40 33|'. To find A C, use relation IV, Art. 177. Side adjacent A = side opposite X cot A, or A C = 234 X .85586 = 200.27 feet. f Angle A 49 261'. Ans. J Angle B = 40 331'. [AC = 200.27 ft. FIG. 6. EXAMPLES FOR PRACTICE. 1 8 1 . Solve the following examples : 1. In a right triangle ABC, right-angled at C, the hypotenuse A B = 40 inches and angle A = 28 14' 14". Solve the triangle. [Angle B = 61 45' 46". Ans. \ A C = 35.24 in. C= 18.92 in. 72 ELEMENTARY ALGEBRA AND 3 2. In a right triangle ABC, right-angled at C, the side B C = 10 feet 4 inches. If angle A = 26 59' 6", what do the other parts equal ? (Angle B = 63 0' 54". A B = 22 ft. 9 in. , nearly. A C = 20 ft. 3| in., nearly. 3. In a right triangle A B C, right-angled at C, the hypotenuse A B = 60 feet and the side A C = 22 feet. Solve the triangle. [ Angle A = 68 29' 22. 2". Ans. \ Angle B = 21 30' 37.8". [ C = 55.82ft. 4. In a right triangle ABC, right-angled at C, side A C = .364 foot and side B C = .216 foot. Solve the triangle. f Angled = 30 41' 7.5". Ans. \ Angle j9 = 59 18' 52.5". i A B = .423 ft. OBLIQUE-ANGLED PLANE TRIANGLES. 182. We will give here the method of solving any oblique-angled plane triangle, when (1) two sides and an angle opposite one of them are given, and (2) when two angles and a side opposite one of them are given. Let the student bear in mind, however, that he may use this method of solu- tion only when he knows the general form of the triangle of which he desires the values of some of the parts. This is necessary because in some cases two solutions are possible, resulting in the determinations of triangles of quite different forms. We do not think it necessary to go so deeply into the explanation of this point that the student can detect the cases where two solutions are possible. 183. The solution of the triangle depends upon the fol- lowing principle : In any triangle, the sides are proportional to the sines of the opposite angles. Thus, referring to Fig. 7, the following proportions are true : a : b sin A : sin B. a \ c sin A : sin C. b : c = sin B : sin C. 3 TRIGONOMETRIC FUNCTIONS. 73 The method of solving 1 the triangle is shown in the follow- ing examples: 184. Case I. Two sides and an angle opposite one of them are given : EXAMPLE. In the triangle ABC, Fig. 7, having given the side a = 1,686 feet, the side b = 960 feet, and the angle A = 33 35' ; to find the angle B. SOLUTION. We have a : b = sin A : sin B. Substituting the known values, we have 1,686 : 960 = sin 33 35' : sin B. From the table of Natural Sines we get .55315 as the sine of 33 35'. Substituting this value in the proportion and solving, we find that .31496 is the sine of the angle B, which, by consulting the table of Natural Sines, we find to correspond nearly with the angle 18 22'. Ans. 185. Case II. Two angles and a side opposite one of them are given : EXAMPLE. In the triangle ABC, given the angle A = 33 35', the angle B = 18 22', and the side a 1,686 feet; to find the side b. SOLUTION. We have sin A : sin B = a : b. Substituting known values, we get .55315 : .31509 = 1,686 : b. Solving the proportion, we find b = 960 feet. Ans. TABLES OF NATURAL SINES, COSINES, | TANGENTS, AND COTANGENTS GIVING THE VALUES OF THE FUNCTIONS FOR ALL DEGREES AND MINUTES FROM . O TO QO NATURAL SINES AND COSINES. 77 j C o I 2 3 o 4 Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine .00000 01745 .99985 .03490 99939 .05234 99863 .06976 99756 60 1 .00029 .01774 .99984 03519 .99938 .05263 .9986! .07005 99754 59 2 .00058 .01803 99984 03548 99937 .05292 .99?-6o .07034 99752 58 3 .00087 .01832 .99983 3577 .99936 .05321 .99858 .07063 9975 57 4 .00116 .01862 .99983 .03606 99935 0535 .99857 .07092 .99748 56 5 .00145 .01891 .99982 03 6 35 99934 05379 99855 .07121 .99746 55 6 .00175 .01920 .99982 .03664 99933 .05408 .99854 .07150 99744 54 7 . 00204 .01949 .99981 03693 99932 05437 99852 .07179 99742 53 8 .00233 .01978 .99980 03723 9993 1 .05466 .99851 .07208 .99740 52 9 .00262 .02007 .99980 03752 .99930 05495 .99849 .07237 99738 5 1 10 .00291 .02036 99979 03781 .99929 05524 .99847 .07266 99736 50 ii .00320 99999 .02065 99979 .03810 99927 05553 .99846 .07295 99734 49 12 .00349 .99999 .02094 .99978 03839 .99926 05582 .99844 07324 9973 1 48 13 .00378 99999 .02123 99977 .03868 .99925 .05611 .99842 07353 .99729 47 14 .00407 99999 .02152. 99977 .03897 99924 .05640 .99841 .07382 99727 46 15 .00436 99999 .02181 .99976 .03926 99923 .05669 99839 .07411 99725 45 16 .00465 99999 .02211 .99976 03955 .99922 .05698 .99838 .07440 99723 44 17 .00495 99999 .02240 99975 .03984 . 9992 i .05727 .99836 .07469 .99721 43 18 .00524 99999 .02269 99974 .04013 .99919 05756 .99834 .07498 .99719 42 *9 oo553 .99998 .02298 99974 .04042 .99918 05785 99833 .07527 .99716 4 1 20 .00582 99998 .02327 99973 .04071 .99917 .05814 .99831 07556 .99714 40 21 .00611 .99998 .02356 99972 .04100 .99916 .05844 .99829 07585 .99712 39 22 .00640 .99998 .02385 99972 .04129 99915 05873 .99827 .07614 .99710 38 2 3 .00660 .99998 .02414 .99971 .04159 999!3 .05902 .99826 .07643 .99708 37 24 .00698 9999 s .02443 .99970 .04188 .99912 05931 .99824 .07672 99705 36 25 .00727 99997 .02472 .99969 .04217 .99911 .05960 .99822 .07701 99703 35 26 .00756 99997 .02501 .99969 .04246 .99910 .05989 .9982? .0773 .99701 34 27 .00785 99997 .02530 .99968 .04275 .99909 .06018 .99819 07759 .99699 33 28 .00814 .99997" .02560 .99967 .04304 .99907 .06047 .99817 .07788 .99696 32 29 .00844 .99996 .02589 .99966 4333 .99906 .06076 .99815 .07817 99694 3i 3 fc .00873 99996 .02618 .99966 .04362 99905 .06105 .99813 .07846 .99692 3 3 1 . 00902 .99996 .02647 .99965 .04391 .99904 06134 .99812 07875 .99689 29 32 .00931 .99996 .02676 .99964 .04420 .99902 .06163 .99810 .07904 .99687 28 33 .00960 99995 .02705 .99963 .04449 .99901 .06192 .99808 07933 .99685 27 34 .00989 99995 02734 .99963 .04478 .06221 . 99806 .07962 .99683 26 9 .01018 .01047 99995 99995 .02763 .02792 .99962 .99961 .04507 0453 6 99897 .06250 .06279 .99804 .99803 .07991 .08020 .99680 .99678 2 5 24 37 .01076 99994 .02821 .99960 04565 .99896 .06308 .99801 .08049 .99676 23 38 .01105 99994 .02850 99959 .04594 .99894 06337 99799 .08078 99673 22 39 .01134 99994 .02879 99959 .04623 .99893 .06366 99797 .08107 .9967: 21 40 .01164 99993 .02908 .99958 04653 .99892 .06395 99795 .08136 .99668 20 4i .01193 99993 .02938 99957 .04682 .99890 .06424 99793 .08165 .99666 19 42 .01222 99993 .02967 .99956 .04711 .99889 06453 .99792 .08194 .99664 18 43 .01251 99992 .02996 99955 .04740 .99888 .06482 .99790 .08223 .99661 17 44 .OI28o .99992 .03025 99954 .04769 .99886 .06511 .99788 .08252 99659 !6 45 .01309 .99991 .03054 99953 .04798 .99885 .06540 .99786 .08281 99657 15 46 .01338 .99991 '.03083 .99952 .04827 99883 .06569 .99784 .08310 99654 14 47 .01367 .99991 .03112 99952 .04856 .99882 .06598 .99782 08339 99652 13 48 .01396 .99990 .03141 99951 .04885 .99881 .06627 .99780 .08368 99649 12 49 .01425 .99990 .03170 .99950 .04914 .99879 .06656 9977 8 .08397 99647 II 50 .01454 99989 .03199 99949 .04943 .99878 .06685 .99776 .08426 '99644 10 Si .01483 .99989 .03228 .99948 .04972 .99876 .06714 99774 08455 .99642 9 52 OI 5 I 3 .99989 03257 99947 .05001 99875 .06743 .99772 .08484 99639 8 53 .01542 .99988 .03286 .99946 .05030 99873 .06773 .99770 08513 99637 7 54 .01571 .99988 .03316 99945 05059 .99872 .06802 .99768 .08542 99635 6 55 .01600 .99987 3345 99944 .05088 .99870 .06831 .99766 .08571 .99632 5 56 .01629 .99987 3374 99943 .05117 .99869 .06860 99764 .08600 .99630 4 57 .01658 .99986 03403 .99942 .05146 .99867 .06889 .99762 .08629 .99627 3 58 .01687 .90986 03432 .99941 5i75 .99866 .06918 .99760 .08658 99625 2 59 .01716 .99985 .03461 .99940 .05205 .99864 .06947 99758 .08687 .99622 I 60 01745 99985 .03490 99939 .05234 .99863 .06976 99756 .08716 .99619 O Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine / r 8 ? 8 s 8 7 8 5 . .0 78 NATURAL SINES AND COSINES. 5 6 7 8 9 f Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine .08716 .99619 10453 .99452 .12187 99255 .13917 .99027 15643 .98769 ~6o I .08745 .99617 . 10482 99449 . 12216 99251 13946 .99023 .15672 .98764 59 2 08774 .99614 .10511 99446 .12245 .99248 13975 .99019 .15701 .98760 58 3 .08803 .99612 10540 99443 .12274 .99244 .14004 .99015 1573 98755 57 4 .08831 .99609 . 10569 .99440 . 12302 .99240 14033 .99011 15758 98751 56 5 .08860 .99607 10597 99437 .12331 99237 . 14061 .99006 15787 .98746 55 6 .08889 .99604 . 10626 99434 . 12360 99233 .14090 .99002 .15816 .98741 54 7 .08918 .99602 . 10655 99431 12389 99230 .14119 .98998 15845 98737 53 8 .08947 .99599 . 10684 .99428 .12418 .99226 .14148 .98994 > 15873 .98732 52 9 .08976 99596 .10713 .99424 12447 .99222 .14177 .98990 .15902 .98728 5 1 10 .09005 99594 . 10742 .99421 .12476 .99219 . 14205 .98986 15931 .98723 So ii .09034 9959 1 .10771 .99418 . 12504 .99215 14234 .98982 15959 .98718 49 12 .09063 .99588 .10800 99415 12533 .99211 .14263 .98978 .15988 .98714 48 13 .09092 .99586 . 10829 .99412 .12562 .99208 .14292 98973 .16017 .98709 47 14 .09121 99583 .10858 99409 12591 .99204 .14320 .98969 .16046 .98704 46 15 .09150 9958o .10887 .99406 . 12620 .99200 14349 .98965 .16074 .98700 45 16 .09179 99578 .10916 .99402 .12649 99197 14378 .98961 .16103 98695 44 17 .09208 99575 10945 99399 .12678 99193 . 14407 98957 .16132 .98690 43 18 09237 99572 10973 99396 . 12706 .99189 .14436 98953 . 16160 .98686 42 19 .09266 99570 .11002 99393 12735 .99186 .14464 .98948 .16189 .98681 4 1 20 .09295 99567 .IIO31 .99390 . 12764 .99182 14493 98944 .16218 .98676 40 21 .09324 .99564 . Ilo6o .99386 12793 .99178 .14522 .98940 .16246 .98671 39 22 09353 .99562 .11089 99383 .12822 99175 14551 .98936 .16275 .98667 38 2 3 .09382 99559 ,llll8 .90380 .12851 99171 .14580 .98931 .16304 .98662 37 24 .09411 99556 .11147 99377 .12880 99167 . 14608 .98927 16333 98657 36 25 .09440 99553 .11176 99374 .12908 99163 14637 .98923 .16361 .98652 35 26 .09469 99551 . II2O5 9937 12937 .99160 .14666 .98919 .16390 .98648 34 27 .09498 .99548 .11234 99367 99156 .14695 .98914 .16419 .98643 33 28 09527 99545 .11263 .99364 12995 99152 14723 .98910 '. 16447 .98638 3 2 29 09556 99542 .11291 .99360 .13024 .99148 14752 .98906 .16476 98633 30 09585 .99540 .11320 99357 13053 99144 .14781 .98902 . 16505 .98629 3 31 .09614 99537 "349 99354 .13081 .99141 .14810 .98897 16533 .98624 29 32 .09642 99534 .11378 99351 .13110 99137 . 14838 .98893 .16562 .98619 28 33 .09671 99531 .11407 99347 13139 99133 . 14867 .98889 .16591 .98614 27 34 .09700 .99528 .11436 99344 .13168 .99129 .14896 .98884 .16620 .98609 26 35 .09729 .99526 .11465 99341 13197 99125 .14925 .98880 .16648 .98604 25 36 .09758 99523 .11494 99337 . 13226 .99122 14954 .98876 .16677 .98600 24 37 09787 .99520 "523 99334 13254 .99118 .14982 .98871 .16706 98595 23 38 .09816 99517 "552 99331 13283 .99114 .15011 .98867 16734 .98590 22 39 09845 99514 .11580 .99327 13312 .99110 .15040 .98863 .16763 .98585 21 40 .09874 995" .11609 .99324 13341 .99106 .15069 .98858 .16792 .98580 20 4i .09903 .99508 .11638 .99320 .1337 .99102 15097 .98854 .16820 98575 *9 4 2 .09932 .99506 .11667 99317 13399 .99098 .15126 .98849 . 16849 .98570 18 43 .09961 99503 .11696 99314 13427 99094 15155 .98845 .16878 98565 17 44 .09990 99500 .11725 .99310 13456 .99091 .15184 .98841 .16906 98561 l6 45 . 10019 99497 "754 99307 13485 .99087 . 15212 .98836 16935 98556 15 46 . 10048 99494 . 11783 9933 13514 .99083 .15241 .98832 .16964 98551 M 47 .10077 .99491 .11812 .99300 13543 9979 .15270 .98827 .16992 98546 48 .10106 .99488 .11840 99297 13572 99075 15299 .98823 .17021 .98541 12 49 .5 . 10135 . 10164 .99485 .99482 . i i 869 .11898 99293 .99290 . 13600 . 13629 .99071 .99067 15327 15356 .98818 .98814 .17050 .17078 98536 98531 11 IO 5I .10192 99479 .11927 .99286 .13658 .99063 .15385 .98809 .17107 98526 9 S 2 . IO22I .99476 .11956 .99283 13687 .99059 15414 .98805 17136 .98521 8 53 . 10250 99473 .11985 99279 .13716 99055 .15442 . 98800 .17164 98516 7 54 . 10279 .99470 .12014 .99276 .13744 .99051 15471 .98796 17193 .985" 6 .10308 99467 .12043 .99272 13773 .99047 15500 .98791 .17222 .98506 5 56 10337 .99464 .12071 .99269 .13802 .99043 15529 .98787 .17250 .98501 4 57 . 10366 .99461 .I2IOO .99265 13831 99039 15557 .98782 17279 .98496 3 58 10395 99458 . I2I29 .99262 .13860 9935 15586 .98778 .17308 .98491 2 59 . 10424 99455 .12158 .99258 .13889 99031 15615 98773 17336 .98486 I 60 1453 99452 .12187 .99255 13917 .99027 15643 .98769 17365 .98481 o Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine i ' 8 4 83 82 81 80 NATURAL SINES AND COSINES. 79 10 i H 12 13 14 , Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine 17365 .98481 .19081 98163 .20791 97815 .22495 97437 .24192 .97030 60 I 17393 .98476 .19109 .98157 .20820 .97809 22523 9743 , 24220 .97023 59 2 .17422 .98471 .19138 .98152 .20848 97803 22552 .97424 .24249 97015 58 3 17451 .98466 .19167 .98146 .20877 97797 .22580 97417 24277 .97008 57 4 17479 .98461 19195 .98140 .20905 97791 .22608 .97411 24305 .97001 56 5 .17508 98455 .19224 98135 20933 .97784 22637 .97404 24333 .96994 55 6 17537 .98450 .19252 ' .98129 . 20962 97778 .22665 .97398 .24362 .96987 54 7 17565 .98445 . 19281 .98124 .20990 97772 .22693 97391 .24390 .96980 53 8 17594 .98440 .19309 .98118 .21019 97766 .22722 97384 .24418 96973 52 9 .17623 98435 .19338 .98112 .21047 9776o .22750 97378 . 24446 .96966 51 10 .17651 .98430 .19366 .98107 .21076 97754 .22778 9737 ! .24474 96959 5 ii .17680 .98425 19395 .98101 .21104 .97748 .22807 97365 24503 .96952 4Q 12 .17708 .98420 19423 .98096 .21132 97742 .22835 97358 2453 1 .96945 48 13 17737 .98414 .19452 .98090 .21161 97735 .22863 97351 24559 .96937 47 14 . 17766 .98409 .19481 .98084 .21189 97729 .22892 97345 24587 .96930 46 15 17794 .98404 19509 .98079 .21218 97723 .22920 97338 .24615 .96923 45 16 17823 .98399 19538 .98073 .21246 97717 .22948 97331 .24644 .96916 44 17 17852 .98394 19566 .98067 .21275 .97711 .22977 97325 .24672 .96909 43 18 .17880 .98389 19595 .98061 .21303 97705 23005 97318 .24700 .96902 42 19 .17909 98383 19623 .98056 21331 .97698 23033 973" .24728 .96894 20 17937 .98378 .19652 .98050 .21360 .97692 .23062 97304 24756 .96887 40 21 .17966 98373 . 19680 .98044 .21388 .97686 .23090 .97298 24784 .96880 39 22 17995 .98368 .19709 .98039 .21417 .97680 .23118 97291 .24813 96873 38 2 3 .18023 .98362 19737 ,98033 .21445 97673 23146 .97284 .24841 .96866 37 24 . 18052 98357 .19766 .98027 .21474 .97667 23175 .97278 .24869 .96858 36 25 .18081 98352 19794 .98021 .21502 .97661 23203 97271 .24897 .96851 35 26 . 18109 98347 19823 .98016 .21530 97655 .23231 97264 24925 .96844 34 27 . 18138 .98341 .19851 .98010 21559 .97648 .23260 97257 24954 ,96837 33 28 .18166 98336 .19880 .98004 .21587 97642 .23288 97251 .24982 .96829 3 2 29 18195 98331 . 19908 .97998 .21616 .97636 .23316 .97244 .25010 .96822 3 1 30 .18224 98325 19937 .97992 .21644 .97630 23345 97237 25038 .96815 30 31 .18252 .98320 !99 6 5 .97987 .21672 .97623 23373 .97230 . 25066 .96807 29 32 . 828 1 98315 19994 .97981 .21701 .97617 .23401 .97223 .25094 .96800 28 33 .18309 98310 .20022 97975 .21729 .97611 23429 .97217 .25122 .96793 27 34 18338 .98304 .20051 97969 .21758 .97604 23458 .97210 2515 1 .96786 26 35 .18367 .98299 .20079 97963 .21786 97598 .23486 97203 25179 .96778 25 36 18395 .98294 .20108 97958 .21814 97592 23514 .97196 .25207 .96771 24 37 .18424 .98288 .20136 97952 .21843 97585 23542 .97189 25235 .96764 23 38 .18452 .98283 .20165 .97946 .21871 97579 23571 .97182 .25263 .96756 22 39 .18481 .98277 .20193 .97940 .21899 97573 23599 .97176 25291 .96749 21 40 .18509 .98272 .20222 97934 .21928 .97566 23627 .97169 .25320 96742 2O 4i 18538 .98267 .20250 .97928 .21956 .97560 .23656 .97162 25348 .96734 Ig 42 .18567 .98261 .20279 .97922 .21985 97553 .23684 97155 25376 .96727 18 43 18595 .98256 .20307 979i6 .22013 97547 .23712 .97148 .25404 96719 17 44 .18624 .98250 20336 .97910 .22041 97541 23740 .97141 25432 .96712 16 45 . 18652 .98245 .20364 97905 . 22070 97534 23769 97134 .25460 . 96705 15 46 . 18681 .98240 -.20393 .97899 .22098 .97528 23797 .97127 .25488 .96697 14 47 .18710 .98234 .20421 97893 .22126 97521 .23825 .97120 25516 .96690 48 .18738 .98229 .20450 .97887 .22155 97515 23853 97"3 25545 .96682 12 49 .18767 .98223 .20478 .97881 .22183 .97508 .23882 .97106 25573 .96675 II 5 18795 .98218 .20507 97875 .22212 97502 .23910 .97100 .25601 ,96667 IO Si . 18824 .98212 20535 .97869 .2224O 97496 23938 .97093 25629 .96660 9 S 2 .18852 .98207 20563 .97863 .22268 .97489 .23966 .97086 25657 .96653 8 53 .18881 .98201 . 20592 97857 .22297 .97483 23995 97079 25685 .96645 7 54 .18910 .98196 . 20620 97851 22325 .97476 .24023 .97072 25713 .96638 6 55 .18938 .98190 .20649 97845 22353 .97470 .24051 97065 25741 .96630 5 56 .18967 .98185 .20677 97839 .22382 97463 .24079 .97058 .25769 .96623 4 57 .18995 .98179 . 20706 97833 .224IO 97457 .24108 97051 .25798 .96615 3 58 .19024 .98174 20734 .97827 22438 9745 .24136 97044 .25826 .96608 2 59 .19052 .98168 .20763 .97821 .22467 97444 .24164 .97037 25854 .96600 I 60 .19081 .98163 .20791 97815 .22495 97437 .24192 .97030 .25882 96593 o , Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine , 79 78 77 76 75 80 NATURAL SINES AND COSINES. '5 16 i? 18 19 Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine o I .25882 .25910 96593 96585 .27564 .27592 .96126 .96118 .29237 .29265 9563 .95622 .30902 30929 .95106 95097 32557 32584 94552 94542 60 59 2 25938 .96578 .27620 .96110 .29293 95613 30957 .95088 .32612 94533 58 3 .25966 9657 .27648 .96102 .29321 95605 30985 9579 .32639 94523 57 4 25994 .96562 .27676 .96094 .29348 95596 .31012 .95070 .32667 945M 56 5 .26022 96555 .27704 .96086 .29376 95588 .31040 .95061 32694 94504 5S 6 . 26050 96547 2773 1 .96078 . 29404 95579 .31068 95052 32722 94495 54 7 .26079 .96540 27759 .96070 29432 9557 1 31095 9543 32749 .94485 53 8 .26107 96532 .27787 .96062 . 29460 95562 3"2 3 95033 3 2 777 .94476 52 9 .26135 .96524 .27815 .96054 .29487 95554 3 II 5 I 95024 .32804 .94466 5i 10 .26163 965:7 27843 .96046 295*5 95545 .31178 95015 32832 94457 50. ii .26191 .96509 .27871 .96037 29543 95536 .31206 .95006 32859 94447 49 12 .26219 .96502 .27899 .96029 29571 95528 31233 94997 .32867 .94438 48 *3 .26247 .96494 .27927 .96021 .29599 955*<) .31261 .94988 329H .94428 47 M .26275 .96486 27955 96013 .29626 955" .31289 94979 32942 .94418 46 26303 .96479 27983 .96005 .29654 95502 31316 .94970 .32969 .94409 45 16 26331 .96471 .28011 95997 .29682 95493 3*344 .94961 .32997 94399 44 i? 26359 .96463 28039 .95989 .29710 95485 3 T 37 2 .94952 33024 .94390 43 18 .26387 96456 .28067 .95981 29737 95476 31399 94943 33051 .94380 42 T 9 .26415 .96448 28095 95972 29765 95467 3M27 94933 33079 9437 4i 20 26443 .96440 .28123 .95964 .29793 95459 3 T 454 .94924 .33106 .94361 40 21 .26471 96433 .28150 95956 .29821 95450 .31482 949!5 33134 9435 1 39 22 .26500 .96425 .28178 .95948 .29849 95441 .31510 .94906 33 l6 i .94342 38 23 .26528 .96417 .28206 .95940 .29876 95433 31537 .94897 33189 94332 37 24 .26556 .96410 .28234 9593 1 .29904 .95424 3 I 5 6 5 .94888 .33216 94322 36 2 5 .26584 .96402 .28262 95923 .29932 95415 3*593 .94878 33244 94313 35 26 .26612 . 963-^4 .28290 959!5 . 29960 9547 .31620 .94869 33271 9433 34 27 . 26640 .96386 .28318 95907 .29987 95398 .31648 .94860 33298 .94293 33 28 .26668 96379 .28346 .95898 30015 95389 3 l6 75 .94851 .33326 .94284 32 29 .26696 9637 1 28374 .95890 .30043 9538o 31703 .94842. 33353 94274 3i 3 .26724 .96363 . 28402 .95882 .30071 95372 3 1 73 .94832 333 Sl .94264 3 3 1 26752 9 6 355 .28429 95874 .30098 95363 .31758 94823 .33408 .94254 2 9 S 2 .26780 96347 28457 95865 .30126 95354 .31786 .94814 .33436 94245 28 33 .26808 .96340 .28485 95857 3 OI 54 95345 31813 .94805 33463 .94235 27 34 .26836 .96332 28513 95849 .30182 95337 .31841 94795 3349 .94225 26 35 .26864 .96324 .28541 .95841 .30209 95328 .31868 .94786 .335i8 .94215 25 36 .26892 .96316 .28569 .95832 3 237 953*9 .31896 94777 33545 .94206 24 37 .26923 .96308 28597 .95824 .30265 953!o 3 I 9 2 3 .94768 33573 .94196 23 38 .26948 .96301 .28625 .95816 .30292 95301 3*95* 94758 .33600 .94186 22 39 .26976 .96293 .28652 .95807 .30320 95203 3*979 94749 33627 .94176 21 40 .27004 .96285 .28680 95799 30348 95284 .32006 .94740 33655 .94167 2O 4 1 .27032 .96277 .28708 95791 30376 95275 3 2 034 9473 .33682 94 I 57 19 42 .27060 .96269 .28736 .95782 -30403 .95266 .32061 .94721 337 10 .94147 18 43 .27088 .96261 .28764 95774 3043 1 95257 .32089 .94712 33737 94137 17 44 .27116 .96253 .28792 .95766 30459 .95248 .32116 .94702 33764 .94127 16 45 .27144 .96246 .28820 95757 .30486 .95240 .32144 94693 33792 .94118 15 46 .27172 96238 .28847 95749 35 T 4 95231 .32171 .94684 33819 .94108 14 47 .27200 .96230 .28875 95740 3 542 95222 .32199 .94674 33846 .94098 *3 48 .27228 .96222 .28903 95732 3 570 95213 .32227 .94665 33874 .94088 12 49 .27256 .96214 .28931 95724 3597 .95204 32254 .94656 33901 .94078 11 50 .27284 .96206 .28959 95715 .30625 -95!95 .32282 .94646 33929 .94068 10 5* .27312 .96198 .28987 95707 30653 .95186 .32309 94637 33956 .94058 9 52 27340 .96190 .29015 .95698 .30680 95!77 32337 .94627 33983 .94049 8 53 27368 .96182 .29042 .95690 .30708 .95168 32364 .94618 .34011 94039 7 54 .27396 .96174 .29070 .95681 .30736 95^9 .32392 .94609 34038 . 94029 6 55 .27424 .96166 .29098 95673 30763 95150 32419 94599 34065 .94019 5 56 27452 .96158 .29126 .95664 .30791 .95142 32447 94590 34093 .94009 4 57 .27480 .96150 .29154 95656 .30819 95*33 32474 9458o .34120 93999 3 58 .27508 .96142 .29182 95647 .30846 95124 .32502 94571 34 J 47 .93989 2 59 .27536 .96134 .29209 95639 .30874 95"5 32529 .94561 34175 93979 I 60 27564 .96126 .29237 .95630 . 30002 .95106 32557 94552 .34202 93969 Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine r 1 74 73 72 71 7 NATURAL SINES AND COSINES. 81 2 2 . 2 o 2 2 3 i 2. 1 Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine o .34202 .34229 93969 93959 35837 .35864 93358 93348 3746* .37488 .92718 .92707 39073 .39100 .92050 .92039 .40674 .40700 9*355 9*343 60 59 2 34257 93949 3589* 93337 375'5 .92697 39*27 .92028 .40727 9*33* 58 3 34284 93939 .35918 93327 37542 .92686 39*53 .92016 40753 9*3*9 57 4 343** 93929 35945 933*6 37569 .92675 39*83 . 92005 .40780 9*307 56 5 34339 .93919 35973. 93306 37595 .92664 39207 .91994 .40806 .91295 55 6 .34366 93909 .36000 93295 37622 92653 39234 .91982 .40833 .91283 54 7 34393 93899 .36027 93285 37649 .92642 .39260 .91971 .40860 .91272 53 8 3442* .93889 36054 93274 37676 92631 39287 9*959 .40886 .91260 52 9 .34448 .93879 .36081 .93264 37703 .92620 .39314 .91948 .40913 .91248 5* 10 34475 .93869 .36108 93253 .37730 .92609 3934* 9*936 .40939 .91236 50 ii 34503 93859 36*35 93243 37757 92598 39367 .91925 .40966 .91224 49 12 34530 93849 .36162 93232 37784 .92587 39394 .91914 .40992 .91212 48 *3 34557 93839 .36190 .93222 .37811 92576 3942* .91902 .41019 .91200 47 34584 .93829 .36217 .93211 37838 92565 3944 s .91891 .41045 .91188 46 15 346*2 .93819 36244 93201 37865 92554 39474 9*879 .41072 .91176 45 16 34639 .93809 .36271 .93190 .37892 92543 3950* .91868 .41098 .91164 44 *7 .34666 93799 .3-6298 93*80 379*9 92532 39528 .91856 . 1125 .91152 43 18 34694 .93789 36325 93*69 37946 .92521 39555 .91845 **5* .91140 42 *9 .34721 93779 93*59 37973 .92510 3958* 9*833 . 1178 .91128 20 .34748 93769 36379 93*48 37999 92499 .39608 .91822 . 1204 .91116 40 21 34775 93759 .36406 93*37 .38026 .92488 39635 .91810 123* .91104 39 22 34803 93748 36434 93*27 38053 .92477 .39661 9*799 .91092 38 2 3 34830 9373 s .36461 .93116 .38080 .92466* .39688 .91787 . 1284 .91080 37 24 .34857 .93728 .36488 .93106 .38107 92455 397*5 9*775 . 1310 .91068 36 25 .34884 .93718 365*5 93095 38*34 .92444 3974* 9*764 *337 .91056 35 26 349*2 93708 .36542 93084 .38161 .92432 39768 9*752 *3 6 3 9*44 34 27 34939 .93698 .36569 9374 .38188 .92421 39795 .91741 *39 .91032 33 28 34966 .93688 .36596 .93063 382:5 .92410 .39822 9*729 . 1416 .91020 S 2 29 34993 93677 .36623 93052 .38241 92399 .39848 .91718 4*443 .91008 3 1 3 .35021 93667 .36650 93042 . 38268 .92388 39875 .91706 .41469 .90996 3 3i 35048 93657 .36677 933* 38295 92377 .39902 .91694 .41496 .90984 29 32 35075 .93647 36704 .93020 '38322 .92366 39928 9*683 4*522 .90972 28 33 35* 2 93637 3673* .93010 38349 92355 39955 .91671 4*549 .90960 27 34 35*30 35*57 .93626 .93616 36758 36785 92999 .92988 38376 .38403 92343 .92332 .39982 .40008 .91660 .91648 4*575 .41602 .90948 .90936 26 25 36 35*84 .93606 .36812 .92978 .38430 .92321 .40035 .91636 .41628 .90924 24 37 352*1 93596 36839 .92967 38456 923*0 .40062 .91625 4*655 23 38 35239 93585 .36867 92956 .38483 .92299 .40088 .91613 .41681 .90899 22 39 .35266 93575 .36894 92945 .38510 .92287 .40115 .91601 4*707 .90887 21 40 35293 93565 .36921 92935 .38537 .92276 .40141 .91590 4*734 90&75 20 4i 35320 93555 .36948 .92924 .38564 .92265 .40168 9*578 .41760 .90863 *9 42 35347 93544 36975 929*3 .3859* .92254 .40195 .9*566 .41787 90851 18 43 35375 93534 .37002 . 92902 .38617 92243 .40221 9*555 .41813 .90839 17 44 35402 93524 .37029 .92892 38644 .92231 .40248 9*543 . 1840 .90826 16 45 35429 935*4. 37056 .92881 -3867* .92220 .40275 9*53* . 1866 .90814 IS 46 35456 .93503 37083 .92870 .38698 .92209 .40301 9*5*9 . 1892 .90802 14 47 35484 93493 .37110 .92859 38725 .92198 .40328 .91508 *9*9 .90790 *3 48 355** 93483 37*37 .92849 38752 .92186 40355 .91496 *945 .90778 12 49 35538 .93472 .37164 .92838 .38778 92*75 .40381 .91484 1972 .90766 II 50 35565 .93462 37*9* .92827 .38805 .92164 .40408 9*472 . 1998- 9753 IO 5I 35592 93452 .37218 .92816 .38832 92*52 40434 .91461 .42024 .90741 9 52 .35619 .93441 37245 .92805 38859 .92141 .40461 9*449 .42051 .90729 8 53 35647 9343* .37272 .02794 .38886 .92130 .40488 9*437 .42677 .90717 7 54 35674 .93420 37299 .92784 389*2 .92119 .40514 9*425 .42104 . 90704 6 55 357* 934*0 37326 .92773 38939 .92107 .40541 9*4*4 .42*3 .90692 5 56 35728 934oo 37353 .92762 .38966 . 92096 .40567 .91402 .42156 .90680 4 57 35755 93389 3738o 9275* 38993 .92085 . -4594 .91390 .42183 .90668 3 58 35782 93379 37407 .92740 .39020 92073 .40621 9*378 .42209 90655 2 59 60 .358*0 .35837 93368 93358 37434 3746* .92729 .92718 .39046 39073 .92062 .92050 .40647 .40674 .91366 9*355 42235 .42262 .90643 . 9063 i I O , Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine , 6 9 6 3 6 7 6 6 6 5 NATURAL SINES AND COSINES. '5 26 27 28 29 Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine I .42262 .42288 .90631 .90618 43837 43863 .89879 .89867 45399 45425 .89101 .89087 .46947 46973 .88295 .88281 .48481 .48506 .87462 .87448 60 59 2 423*5 .90606 .43889 89854 4545* .89074 .46999 .88267 .48532 87434 58 3 .42341 .90594 439*6 .89841 45477 .89061 47024 .88254 48557 .87420 57 4 42367 .90582 .43942 . 89828 .45503 . 89048 .47050 . 88240 .48583 .87406 56 5 42394 .90569 .43968 .89816 .45529 89035 .47076 .88226 .48608 8739* 55 6 .42420 90557 43994 .89803 45554 .89021 .47101 .88213 .48634 87377 54 7 .42446 90545 .44020 .89790 4558o .89008 47*27 .88199 .48659 87363 53 8 42473 90532 .44046 .89777 .45606 .88995 47*53 .88185 .48684 87349 52 9 42499 .90520 .44072 .89764 45632 . 88o8l 47*78 .88172 .48710 87335 5* 10 42525 .90507 .44098 89752 45658 .88968 .47204 .88158 48735 .87321 50 ii 42552 .90495 44*24 89739 .45684 88955 47229 .88144 .48761 .87306 49 12 42578 .90483 44*5* .89726 457*0 .88942 47255 .88130 .48786 .87292 48 *3 .42604 .90470 44*77 897*3 45736 .88928 .47281 .88117 .48811 .87278 47 .42631 .90458 .44203 .89700 .45762 .88915 .47306 .88103 .48837 .87264 46 15 42657 . 90446 .44229 .89687 .45787 .88902 47332 .88089 .48862 .87250 45 16 .42683 90433 44255 .89674 458*3 .88888 47358 .88075 .48888 87235 44 *7 .42709 .90421 .44281 .89662 45839 88875 47383 .88062 .48913 .87221 43 18 .42736 .42762 . 90408 90396 44307 44333 .89649 .89636 .45865 .45891 .88862 .88848 47409 47434 .88048 .88034 .48938 .48964 .87207 87193 42 20 .42788 .90383 44359 .89623 459*7 .88835 .47460 .88020 .48989 .87178 40 21 .42815 9037* 44385 .89610 45942 .88822 .47486 .88006 .49014 .87164 39 22 .42841 .90358 .44411 89597 .45968 .88808 475** 87993 .49040 .87150 38 23 .42867 90346 44437 89584 45994 .88795 47537 .87979 .49065 .87136 37 24 .42894 90334 44464 8957* . 46020 .88782 .47562 .87965 .49090 .87121 36 25 .42920 .90321 .44490 89558 .46046 .88768 47588 8795* .49116 .87107 35 26 .42946 .90309 445*6 89545 .46072 88755 .47614 87937 .49141 .87093 34 27 42972 .90296 44542 .89532 .46097 .88741 47639 .87923 .49166 87079 33 28 42999 .90284 .44568 .89519 .46123 .88728 47665 .87909 .49192 .87064 32 29 43025 .90271 44594 .89506 .46149 88715 .47690 .87896 .49217 .87050 3 1 30 43051 .90259 .44620 89493 46175 .88701 477*6 .87882 .49242 .87036 30 3* 4377 .90246 .44646 .89480 .46201 .88688 4774* .87868 .49268 .87021 29 32 .43104 90233 .44672 .89467 .46226 .88674 47767 .87854 49293 .87007 28 33 43*3 .90221 .44698 .89454 .46252 .88661 47793 .87840 .49318 .86993 27 34 .43*56 . 90208 44724 .89441 .46278 .88647 .47818 .87826 49344 .86978 26 35 .43*82 .90196 4475 .89428 .46304 .88634 47844 .87812 .49369 .86964 25 36 43209 .90183 44776 894*5 4633 .88620 .47869 .87798 49394 .86949 24 37 43235 .90171 .44802 .89402 46355 .88607 47895 .87784 .49419 86935 23 38 43261 .90158 .44828 89389 .46381 88593 .47920 .87770 49445 .86921 22 39 .43287 .90146 .44854 .89376 .46407 .88580 .47946 .87756 .49470 .86906 21 40 433*3 90*33 .44880 .89363 46433 .88566 4797* 87743 49495 .86892 20 41 43340 .90120 .44906 .89350 .46458 88553 47997 .87729 4952* .86878 *9 42 43366 .90108 44932 89337 . 46484 88539 .48022 87715 .49546 .86863 18 43 43392 .90095 .44958 89324 465*0 .88526 .48048 .87701 4957* .86849 *7 44 .43418 .90082 44984 .89311 .46536 .88512 .48073 .87687 49596 .86834 16 45 43445 .90070 .45010 .89298 .46561- .88499 .48099 .87673 .49622 .86820 is 46 4347* .90057 45036 .89285 46587 .88485 .48124 .87659 .49647 .86805 14 47 43497 .90045 .45062 .89272 .46613 .88472 .48150 .87645 .49672 .86791 13 48 43523 .90032 .45088 .89259 .46639 .88458 .87631 .49697 86777 12 49 43549 .90019 45*14 .89245 .46664 88445 .48201 .87617 49723 .86762 II 5 43575 .90007 45*40 .89232 .46690 .88431 .48226 .87603 49748 .86748 IO 5I .43602 89994 .45*66 .89219 .46716 .88417 48252 .87589 49773 .86733 9 52 .43628 .89981 45*92 .89206 .46742 .88404 .48277 87575 .49798 .86719 8 53 43654 .89968 .45218 .89193 .46767 .88390 48303 .87561 .49824 .6704 7 54 .43680 .89956 45243 .89180 46793 88377 .48328 .87546 .49849 .86690 6 55 .43706 .89943 .45269 .89167 .46819 88363 48354 .87532 .49874 .86675 5 56 43733 .89930 45295 89*53 .46844 ..88349 48379 .87518 49899 .86661 4 57 43759 .89918 45321 .89140 .46870 .88336 .48405 .87504 49924 .86646 3 58 43785 89905 45347 .89127 .46896 .88322 .48430 .87490 .49950 .86632 2 59 .43811 .89892 45373 .89114 .46921 .88308 .48456 87476 49975 .86617 I 60 .89879 45399 .89101 .46947 .88295 .48481 .87462 .50000 .86603 |, Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine , < . 64 63 62 6l 60 NATURAL SINES AND COSINES. 83 f 3 . 31 32 33 34 Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine o I .50000 .50025 .86603 .86588 5*529 857*7 .85702 52992 530*7 .84805 .84789 .54464 .54488 83867 .8385* 559*9 55943 .82904 .82887 60 59 2 .50050 86573 .85687 5304* .84774 83835 55968 .82871 58 3 50076 .86559 51579 .85672 .53066 84759 54537 838*9 55992 82855 57 4 .50101 .86544 .51604 85657 .5309* 84743 5456i .83804 .56016 .82839 56 5 .50126 86530 .51628 .85642 53**5 .84728 54586 .83788 .56040 .82822 55 6 50*5* 86515 5*653 .85627 .8 47I2 .54610 .83772 .56064 .82806 54 7 .50176 .86501 .51678 .85612 53*64 .84697 54635 83756 .56088 .82790 53 8 .50201 .86486 5*73 85597 53*89 .84681 54659 83740 .56112 .82773 9 .50227 .86471 .51728 85582 532*4 .84666 54683 83724 56*36 82757 5* 10 .50252 .86457 5*753 85567 53238 .84650 .54708 .83708 .56160 .82741 5 ii 50277 .86442 5*778 8555* 53263 84635 54732 .83692 .56184 .82724 49 12 50302 .86427 5*803 .85536 53288 .84619 54756 83676 . 56208 .82708 48 T 3 .86413 .51828 8552* 533*2 .84604 5478i .83660 .56232 .82692 47 14 50352 .86398 .51852 .85506 53337 .84588 .54805 83645 .56256 .82675 46 15 50377 .86384 5*877 .85491 5336* 84573 .54829 .83629 . 56280 .82659 45 16 50403 .86369 5*902 85476 53386 84557 54854 .83613 56305 .82643 44 17 .50428 86354 .5*927 85461 534** .84542 54878 83597 .56329 .82626 43 18 50453 .86340 .5*952 .85446 53435 .84526 .83581 56353 .82610 42 *9 .50478 .86325 5*977 8543* .53460 .8451* 54927 83565 56377 .82593 20 50503 .86310 .52002 .85416 .53484 .84495 5495* 83549 .56401 82577 40 21 .50528 862Q5 .52026 .85401 53509 .84480 * 54975 83533 .56425 .82561 39 22 50553 .86281 .52051 85385 53534 .84464 54999 835*7 .56449 82544 38 2 3 50578 .86266 .52076 85370 53558 .84448 55024 .8350* .56473 .82528 37 24 . 50603 .86251 .52101 85355 53583 84433 .55048 .83485 56497 .82511 36 2 5 .50628 86237 .52126 8534 .84417 55072 .83469 5652* 82495 35 26 50654 .86222 .52151 85325 .53632 . 84402 .55097 83453 56545 .82478 34 27 .50679 .86207 52*75 853* 53656 .84386 .55121 83437 56569 .82462 33 28 .50704 .86192 .52200 .85294 .53681 84370 55*45 .83421 56593 .82446 3 2 29 . 50729 .86178 .52225 85279 53705 84355 55*69 83405 56617 .82429 3 1 5754 .86163 .85264 53730 84339 55*94 .83389 .56641 82413 30 3 1 5779 .86148 52275 .85249 53754 .84324 552*8 83373 .56665 .82396 29 3 2 .50804 .86133 52299 85234 53779 .84308 55242 83356 .56689 .82380 28 33 . 50829 .86119 52324 .85218 .53804 .84292 .55266 83340 567*3 .82363 27 34 .50854 .86104 52349 .85203 .53828 .84277 5529* .83324 56736 82347 26 35 .50879 .86089 52374 .85188 53853 .84261 553*5 83308 .56760 82330 25 36 .50904 .86074 52399 85*73 .53877 .84245 55339 .83292 .56784 .82314 24 37 . 50929 .86059 85157 .53902 .84230 55363 .83276 .56808 .82297 23 38 50954 .86045 .52448 85142 53926 .84214 55388 .83260 .56832 .82281 22 39 50979 .86030 52473 .85127 5395* .84198 554*2 .83244 .56856 .82264 21 40 5*004 .86015 .52498 .85112 53975 .84182 55436 .83228 .56880 .82248 20 4i 5*029 .86000 52522 .85096 .54000 .84167 5546o .83212 .56904 .82231 *9 42 5*54 .85985 52547 .85081 54024 .84151 55484 .83195 .56928 .82214 18 43 85970 52572 .85066 .54049 84*35 55509 .83179 .56952 .82198 *7 44 .51104 85956 52597 .85051 54073 .84120 55533 83163 56976 .82181 16 45 .51*29 .85941 .52621 85035 5497 .84104 55557 83*47 .57000 .82165 IS 46 5**54 .85926 .52646 .85020 54*22 .84088 5558* 83*3* .57024 .82148 *4 47 5**79 .85011 52671 85005 54*46 .84072 55605 83*15 57047 .82132 *3 48 .51204 .85896 .52696 .84989 54*7* .84057 5563 .83098 .57071 .82115 12 49 .51229 .85881 .52720 .84974 54*95 .84041 55654 .83082 57095 .82098 ii 50 5*254 .85866 .52745 .84959 .54220 .84025 55678 .83066 57**9 .82082 10 5, 5*279 .85851 52770 .84943 54244 .84009 55702 .83050 -57*43 .82065 9 52 5*304 .85836 .52794 .84928 .54269 83994 55726 83034 57*67 .82048 8 53 .85821 .52819 .84913 54293 .83978 5575 83017 57*9* .82032 7 54 5*354 .85806 .52844 .84897 543*7 .83962 55775 .83001 572*5 .82015 6 55 .85792 .52869 .84882 54342 .83946 55799 .82985 57238 .81999 5 56 5*404 85777 52893 .84866 .54366 .83930 55823 .82969 .57262 .81982 4 57 5*429 85762 529*8 .8485* 5439* 839*5 .55847 .82953 .57286 .81965 3 58 5*454 85747 52943 .84836 544*5 .83899 5587* .82936 573*o .81949 2 59 5*479 85732 52967 .84820 54440 .83883 .55895 .82920 57334 .81932 I 60 5*504 857*7 52092 .84805 .54464 .83867 559*9 .82904 57358 .81915 O , Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine , 59 58 57 56 55 84 NATURAL SINES AND COSINES. 35 36 37 38 39 Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine I 57358 57381 .81915 .81899 58779 .58802 .80902 .80885 .60182 .60205 .79864 .79846 .61566 61589 .78801 .78783 .62932 62955 777*5 .7/696 60 59 2 5745 .81882 . 58826 .80867 .60228 .79829 .61612 78765 .62977 77678 58 3 57429 .81865 58849 . 80850 .60251 .79811 61635 78747 .63000 .77660 57 4 57453 .81848 58873 80833 .60274 79793 .61658 .78729 .63022 .77641 56 5 57477 .81832 .58896 .80816 .60298 .79776 .61681 .78711 63045 .77623 55. 6 575 01 .81815 . 58920 .80799 .60321 79758 .61704 .78694 .63068 77605 54 7 57524 .81798 58943 .80782 .60344 7974* .61726 .78676 .63090 .77586 53 8 57548 .81782 58967 .80765 60367 79723 61749 .78658 63*13 .77568 S 2 9 57572 .81765 .58990 .80748 .60390 .79706 .61772 .78640 63*35 7755 5* 10 57596 .81748 .59014 .80730 .60414 .79688 6*795 . 78622 .63158 7753* 50 ii 57619 .81731 59037 .80713 .60437 .79671 .61818 .78604 .63180 775*3 49 12 57643 .81714 .59061 .80696 .60460 79653 .61841 .78586 .63203 77494 48 13 .57667 .81698 .59084 .80679 .60483 79635 .61864 78568 .63225 77476 47 14 57691 .81681 .59108 .80662 . 60506 .79618 .61887 78550 .63248 77458 46 15 577*5 .81664 .5913* .80644 .60529 . 79600 .61909 78532 .63271 77439 45 16 57738 .81647 59*54 .80627 60553 79583 .61932 .78514 .63293 .77421 44 17 57762 .81631 59*78 .80610 .60576 79565 61955 .78496 63316 .77402 43 18 .57786 .81614 .59201 .80593 .60599 79547 .61978 .78478 63338 .77384 42 ig .57810 8*597 59225 .80576 .60622 7953 .62001 .78^60 .63361 .77366 41 20 57833 .81580 .59248 .80558 .60645 795*2 . 62024 .78442 63383 77347 40 21 57857 81563 59272 .80541 .60668 .79494 . 62046 .78424 .63406 77329 39 22 57881 .81546 59295 .80524 .60691 79477 . 62069 .78405 .63428 773*0 38 2 3 5794 81530 593*8 80507 .60714 79459 .62092 78387 6345* .77292 37 24 .57928 8*5*3 .59342 .80489 60738 .79441 .62115 78369 63473 77273 36 25 57952 .81496 59365 .80472 .60761 .79424 .62138 7835* 63496 77255 35 26 57976 8*479 59389 80455 .60784 .79406 .62160 78333 .635*8 77236 34 27 57999 .81462 594*2 .80438 .60807 .79388 .62183 783*5 6354 .77218 33 28 .58023 8*445 59436 .80420 .60830 7937* .62206 78297 63563 77*99 32 29 .58047 .81428 59459 .80403 .60853 79353 .62229 .78279 63585 .77181 3 .58070 .81412 . 59482 .80386 .60876 79335 .62251 .78261 .63608 .77162 3 3 1 58094 8*395 595o6 .80368 .60899 793*8 .62274 .78243 .63630 77*44 29 S 2 .58118 .81378 59529 .80351 .60922 .79300 .62297 .78225 63653 77*25 28 33 58141 .81361 59552 80334 .60945 .79282 .62320 .78206 63675 77*07 27 34 .58165 8*344 59576 .80316 .60968 . 79264 62342 .78188 .63698 .77088 26 35 .58189 81327 59599 .80299 .60991 .79247 .62365 .78170 .63720 .77070 25 36 .58212 .81310 .59622 .80282 .61015 .79229 .62388 .78*52 63742 .77051 24 37 .58236 81293 . 59646 . 80264 .61038 .79211 .62411 78*34 63765 7733 23 38 .58260 .81276 .59669 .80247 .61061 79*93 62433 .78116 63787 .77014 22 39 .58283 .81259 59693 . 80230 .61084 79*76 .62456 .78098 .63810 . 76996 21 40 58307 .81242 597*6 .80212 .61107 .79^58 .62479 .78079 .63832 76977 2O 4i 58330 .81225 59739 .80195 .61130 .79140 .62502 .78061 .63854 76959 *9 42 58354 .81208 59763 .80178 6*153 .79122 .62524 .78043 .63877 .76940 18 43 58378 .81191 .59786 .80160 .61176 79*5 .62547 .78025 .63899 .76921 *7 44 .58401 .81174 39809 .80143 .61199 .79087 .62570 .78007 .63922 76903 16 45 58425 .81157 59832 .80125 .61222 .79069 .62592 .77988 63944 .76884 *5 46 .58449 .81140 59856 .80108 .61245 7905* .62615 77970 .63966 .76866 *4 47 .58472 .81123 .59879 .80091 .61268 79033 .62638 77952 63989 .76847 13 48 .58496 .81106 59902 .80073 .61291 .79016 .62660 77934 .64011 .76828 12 49 50 58519 58543 .81089 .81072 .59926 59949 .80056 .80038 .61314 6*337 .78998 .78980 .62683 .62706 779*6 .77897 64033 .64056 .76810 .76791 II 10 51 58567 .81055 59972 .80021 .61360 .78962 .62728 77879 .64078 .76772 9 5 2 58590 .81038 59995 .80003 61383 78944 .62751 .77861 .64100 76754 8 53 .58614 .81021 .60019 .79986 .61406 .78926 62774 77843 .64123 76735 7 54 58637 .81004 .60042 .79968 .61429 .78908 .62796 .77824 64*45 .76717 6 55 .58661 .80987 .60065 7995* 61451 .78891 .62819 .77806 .64167 .76698 5 56 .58684 .80970 .60089 79934 .61474 78873 .62842 .77788 .64190 .76679 4 57 58708 .80953 .60112 799*6 6i497 78855 .62864 .77769 .64212 .76661 3 58 59 58731 58755 .80936 .80919 60135 .60158 .79899 .79881 .61520 6*543 .78837 .78819 .62887 .62909 7775* 77733 .64234 .64256 .76642 .76623 2 I 60 58779 .80902 .60182 .79864 .61566 .78801 .62932 777*5 .64279 .76604 o , Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine , 54 53 1 52 51 50 NATURAL SINES' AND COSINES. 85 4 D 4 [ 4 2 4, 5 4^ n Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine o I .64279 64301 .76604 .76586 .65606 .65628 7547* 75452 .66913 66935 743*4 74295 .68200 .68221 73*35 73*i6 .69466 .69487 .71914 60 59 2 64323 76567 .65650 75433 .66956 74276 .68242 .73096 .69508 7*894 58 3 .64346 .76548 .65672 754*4 .66978 74256 .68264 .73076 .69529 7*873 57 4 .64368 7653 .65694 75395 .66999 74237 .68285 73056 69549 7*853 56 5 .64390 .76511 657*6 75375 .67021 .74217 .68306 73036 .69570 7*833 55 6 .64412 .76492 65738 75356 .67043 .74198 .68327 730*6 .69591 7*8*3 54 7 64435 76473 65759 75337 .67064 .74178 .68349 . 72996 .69612 7*792 53 8 64457 76455 .65781 .67086 74*59 .68370 .72976 69633 7*772 52 9 64479 76436 .65803 75299 .67107 74*39 .68391 72957 .69654 7*752 5* 10 .64501 .76417 .65825 75280 .67129 .74120 .68412 72937 .69675 7*732 So i 64524 .76398 .65847 .75261 .67151 .74100 .68434 729*7 .69696 .71711 49 2 64546 .76380 .65869 7524* .67172 . 74080 68455 72897 697*7 .71691 48 3 .64568 .76361 .65891 . 75222 67*94 .74061 .68476 .72877 69737 .71671 47 4 .64590 76342 659*3 .75203 .67215 .74041 .68497 72857 69758 7*650 46 .64612 76323 65935 75*84 67237 .74022 .68518 72837 .69779 7*630 45 6 64635 .76304 65956 75*65 67258 .74002 68539 .72817 . 69800 .71610 44 7 .64657 .76286 65978 75*46 .67280 73983 .68561 72797 .69821 7*590 43 8 .64679 .76267 .66000 75*26 .67301 73963 .68582 72777 .69842 7*569 42 9 .64701 .76248 .66022 75*7 67323 73944 .68603 72757 .69862 7*549 4* 64723 .76229 .66044 75088 67344 73924 .68624 72737 .69883 7*529 4 T .64746 .76210 .66066 .75069 67366 73904 .68645 .72717 .69904 7*508 39 2 3 .64768 .64790 .76192 76*73 .66088 .66109 75050 7503 67387 .67409 73885 73865 .68666 .68688 72697 72677 69925 .69946 .7*488 .71468 38 37 4 .64812 76*54 .66131 .75011 .67430 73846 .68709 72657 .69966 7*447 36 .64834 76*35 66153 74992 67452 .73826 68730 72637 .69987 7*427 35 6 .64856 .76116 66175 74973 67473 73806 .6875* .72617 .70008 7*407 34 27 .64878 .76097 .66197 74953 67495 73787 .68772 72597 .70029 .71386 33 28 .64901 .76078 .66218 74934 67516 73767 68793 72577 .70049 .7*366 3 2 29 .64923 76059 .66240 749*5 67538 73747 .68814 72557 .70070 7*345 3 1 3 .64945 .76041 .66262 .74896 67559 .73728 .68835 72537 .70091 7*325 30 31 .64967 .76022 .66284 .74876 .67580 737o8 68857 725*7 .70112 7*35 20 3 2 .64989 .76003 .66306 74857 .67602 .73688 .68878 .72497 .70132 .71284 28 33 .65011 75984 .66327 .74838 67623 73669 .68899 .72477 70*53 .71264 27 34 65033 75965 66349 .74818 .67645 73649 . 68920 72457 .70174 7*243 26 35 65055 75946 .66371 74799 .67666 73629 .68941 72437 70*95 .71223 25 36 65077 75927 66393 .74780 .67688 736*0 .68962 .724*7 .70215 -.71203 24 37 .65100 .75908 .66414 7476o .67709 73590 .68983 72397 . 70236 .71182 23 38 .65122 .75889 .66436 7474* 6773 73570 .69004 72377 .70257 .71162 22 39 65*44 7587 .66458 .74722 67752 7355* .69025 72357 .70277 .71141 21 40 .65166 7585* .66480 74703 67773 7353* .69046 72337 .70298 .71121 2O 4* .65188 75832 .66501 .74683 67795 735** .69067 723*7 703*9 .71100 19 42 .65210 758*3 .66523 . 74664 .67816 7349* .69088 .72297 70339 .71080 18 43 .65232 75794 66545 74644 67837 73472 .69109 .72277 .70360 7*059 *7 44 65254 75775 .66566 74625 .67859 73452 69130 72257 .70381 7*039 16 45 65276 75756 .66588 .74606 .67880 73432 .69151 .72236 .70401 .71019 15 46 .65298 75738' .66610 74586 .67901 734*3 .69172 .72216 .70422 .70998 *4 47 .65320 757*9 .66632 74567 67923 73393 .69193 .72196 7443 .70978 *3 48 65342 757 66653 .74548 67944 73373 .69214 .72176 70463 70957 12 49 65364 .75680 .66675 .74528 .67965 73353 69235 72*56 .70484 70937 II 50 .65386 .75661 .66697 74509 .67987 73333 69256 .72136 70505 .70916 10 51 .65408 75642 .66718 .74489 .68008 733*4 .69277 .72116 .70525 .70896 9 5 2 6543 75623 .66740 .74470 .68029 73294 .69298 72095 .70546 7-875 8 53 65452 .75604 .66762 7445* .68051 73274 .69319 72075 70567 70855 7 54 65474 75585 .66783 7443* .68072 73254 .69340 72055 .70587 .70834 6 55 65496 75566 .66805 .74412 .68093 73234 .69361 72035 .70608 .70813 5 56 .655*8 75547 .66827 74392 .68115 732*5 .69382 .72015 .70628 70793 4 57 6554 75528 .66848 74373 .68136 73*95 .69403 7*995 . 70649 70772 3 58 .65562 75509 .66870 74353 68157 73*75 .69424 7*974 .70670 .70752 2 59 65584 75490 .66891 74334 .68179 73*55 69445 7*954 .70690 773* I 60 .65606 7547* .66913 743*4 .68200 73*35 .69466 7*934 .70711 .70711 O , Cosine Sine Cosine Sine Cosine Sine Cosine Sine Cosine Sine -, 4 ? 4* 4 7 4 5 4. .0 3 86 NATURAL TANGENTS AND COTANGENTS. < > ] i 5 : , i ^ Tang Cotang Tang Cotang Tang Cotang i Tang Cotang Tang Cotang o .00000 Infin. .01746 57.2900 .03402 28.6363 .05241 19.0811 .06993 14.3007 60 I .00029 3437-75 01775 56.3506 .03521 28.3994 05270 i8.9755 .07022 14.2411 59 2 .00058 1718.87 .01804 55-4415 -0355 28.1664 05299 18.8711 .07051 14. 1821 58 3 .00087 1145.92 01833 03579 27.9372 .05328 18.7678 .07080 14-1235 57 4 .00116 859-436 .01862 53-7086 .03609 27.7117 05357 18.6656 .07110 14.0655 56 5 .00145 687.549 .01891 52.8821 -03638 27.4899 05387 18.5645 07139 14.0079 55 6 .00175 572-957 .01920 52.0807 -03667 27.2715 .05416 18.4645 .07168 13-9507 54 7 .00204 491.106 .01949 51.3032 .03696 27.0566 05445 18-3655 .07197 13.8940 53 8 .00233 429.718 .01978 50-5485 .03725 26.8450 05474 18.2677 .07227 13-8378 52 9 .00262 381.971 .02007 49.8157 03754 26.6367 .05503 18.1708 .07256 13.7821 Si 10 .00291 343-774 . 02036 49.1039 03783 26.4316 05533 18.0750 .07285 13.7267 50 ii .00320 312.521 .02066 48.4121 .03812 26.2296 .05562 17.9802 07314 13.6719 49 12 .00349 286.478 .02095 47-7395 .03842 26.0307 559i 17.8863 07344 13.6174 48 13 .00378 264.441 .02124 47.08,53 03871 25.8348 .05620 17-7934 7373 13-5634 47 15 .00407 .00436 245-552 229. 182 02153 .02182 46-4489 45.8294 .03900 .03929 25.6418 25-4517 .05649 .05678 17.7015 17.6106 .07402 .07431 13.5098 13-4566 46 45 16 .00465 214.858 .O22II 45.2261 .03958 25.2644 .05708 17-5205 .07461 13-4039 44 17 .00495 202.219 .O224O 44.6386 .03987 25-0798 05737 17-4314 .07490 i3-35 I 5 43 18 .00524 190.984 .02269 44.0661 .04016 24.8978 .05766 17-3432 07519 13.2996 4 2 19 oo553 180.932 .02298 43.5081 . 04046 24.7185 05795 17-2558 .07548 13.2480 20 .00582 171.885 .02328 42.9641 .04075 24.5418 .05824 17.1693 .07578 13-1969 40 21 .00611 163.700 02357 42.4335 .04104 24-3675 05854 17.0837 .07607 13.1461 39 22 .00640 156-259 .02386 41.9158 04133 24.1957 05883 16.9990 .07636 13.0958 38 2 3 .00669 149-465 .02415 41.4106 .04162 24.0263 .05912 16.9150 .07665 13.0458 37 24 .00698 143-237 .02444 40.9174 .04191 23-8593 .05941 16.8319 07695 12.9962 36 25 .00727 137-507 02473 40.4358 .04220 23.6945 0597 16.7496 07724 12.9469 35 26 00756 132.219 .O25O2 39-9655 .04250 23-5321 05999 16.6681 07753 12.8981 34 27 .00785 127.321 .02531 39.5059 04279 .06029 16.5874 .07782 12.8496 33 28 .00815 122.774 .02560 39.0568 .04308 23.2137 .06058 .07812 12.8014 S 2 29 .00844 118.540 .02589 38.6177 04337 23-0577 .06087 16.4283 .07841 I2-753 6 30 00873 114.589 .02619 38.1885 .04366 22 .9038 .06116 16-3499 .07870 I 2 . 7062 3 3 1 .00902 110.892 .02648 37-7686 04395 22.7519 .06145 16.2722 .07899 12 .6591 29 32 .00931 107.426 .02677 37-3579 .04424 22.6O2O .06175 16.1952 .07929 12.6l24 28 33 .00960 104.171 .02706 36-9560 04454 22.4541 .06204 16.1190 .07958 12.5660 27 34 .00989 101. 107 02735 36.5627 .04483 22.3081 .06233 16.0435 .07987 12.5199 26 35 .01018 98.2179 .02764 36.1776 04512 22. 1640 .06262 15-9687 .08017 12.4742 2 5 36 .01047' 95.4895 .02793 35-8006 .04541 2 .0217 .06291 15-8945 .08046 12.4288 24 37 .01076 92.9085 .O2822 35.4313 .04570 2 .8813 .06321 15.8211 .08075 12.3838 2 3 38 .01105 90.4633 .02851 35-o695 .04599 2 .7426 -06350 15-7483 .08104 12.3300 22 39 01135 88.1436 .02881 34.7151 .04628 2 .6056 -06379 15.6762 .08134 12.2946 21 40 .01164 85.9398 .02910 34-3678 .04658 2 .4704 .06408 15.6048 .08163 12.2505 2O 4i .01193 83.8435 .02939 34.0273 .04687 21.3369 06437 15-5340 .08192 12.2067 J 9 42 .01222 81.8470 .02968 33.6935 .04716 21.2049 .06467 15.4638 .08221 12.1632 18 43 .01251 79-9434 .02997 33.3662 04745 21.0747 .06496 15-3943 .08251 J2.I20I 17 44 .01280 78. 1263 . 03026 33-0452 04774 20.9460 .06525 15-3254 .08280 12.0772 16 45 .01309 76.3900 03055 32.7303 . 04803 20.8l88 -06554 15-2571 .08309 12.0346 15 46 .01338 74.7292 .03084 32.4213 .04833 20.6932 .06584 15-1893 08339 11.9923 M 47 .01367 73.!39o .03114 32.1181 .04862 20.5691 .06613 15.1222 .08368 11.9504 48 .01396 71.6151 .03143 31.8205 .04891 20.4465 .06642 15-0557 .08397 II.908 7 12 49 .01425 70.1533 .03172 31-5284 .04920 20.3253 .06671 14.9898 .08427 11.8673 II 50 -01455 68.7501 .03201 31.2416 .04949 2O.2O56 .06700 14-9244 .08456 11.8262 IO 5i .01484 67.4019 .03230 30.9599 .04978 20.0872 .06730 14-8596 08485 I1.7853 9 S 2 01513 66. 1055 .03259 30-6833 05007 19.9702 .06759 14-7954 .08514 11.7448 8 53 .01542 64.8580 .03288 30.4116 05037 19-8546 .06788 14-7317 .08544 11.7045 7 54 .01571 63-6567 .03317 30.1446 .05066 19.7403 .06817 14.6685 -08573 11.6645 6 55 .01600 62.4992 .03346 29.8823 05095 19.6273 .06847 14.6059 .08602 11.6248 5 56 .01629 61.3829 03376 29.6245 .05124 19.5156 .06876 14-5438 .08632 H.5853 4 57 .01658 60.3058 03405 29-3711 05153 19.4051 .06905 14.4823 .08661 11.5461 3 58 .01687 59-2659 03434 29.1220 .05182 19.2959 .06034 14.4212 .08690 11.5072 2 59 .01716 58.2612 03463 28.8771 .05212 19.1879 .06963 14.3607 .08720 11.4685 I 60 .01746 57.2900 .034Q2 28.6363 -05241 I9.o8ll .06993 14.3007 .08749 11.4301 O Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang ' 1 8< ) 8* B 8; 8^ ) 8c ,o NATURAL TANGENTS AND COTANGENTS. 87 5 6 7 1 g o 9 Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang o .08749 11.4301 . 10510 9- 51430 .12278 8.14435 .14054 7-11537 -15838 6-31375 60 I .08778 11.3919 . 10540 9.48781 .12308 8.12481 .14084 7. 10038 .15868 6.30189 59 2 ' .08807 "3540 . 10569 0.46141 . 12338 8. 10536 .14113 7.08546 .15898 6.29007 58 3 .08837 11-3163 10599 9-435I5 .12367 8.08600 14143 7.07059 -15928 6.27829 57 4 .08866 11.2789 . 10628 9.40904 .12397 8.06674 .14173 7-05579 15958 6.26655 56 5 .08895 11.2417 .10657 9 38307 .12426 8.04756 . 14202 7.04105 15988 6.25486 55 6 .08925 ii .2048 .10687 9-35724 .12456 8.02848 .14232 7.02637 . 16017 6.24321 54 7 .08954 n. 1681 .10716 9-33155 .12485 8.00948 . 14262 7.01174 .16047 6.23160 53 8 .08983 ii . 1316 . 10746 9-30599 12515 7.99058 .14291 6.99718 .16077 6.22003 S 2 9 .09013 11.0954 I0 775 9.28058 12544 7-97176 .14321 6.98268 . 16107 6 . 2085 i 51 10 .09042 11.0594 .10805 9-2553 .12574 7-95302 14351 6.96823 .16137 6.19703 So ii .09071 11.0237 .10834 9.23016 .12603 7-93438 .14381 6.95385 . 16167 6.18559 40 12 .09101 10.9882 .10863 9.20516 , 12633 7-91582 .14410 6.93952 .16196 6.17419 48 13 .09130 10.9529 . 10893 9.18028 .12662 7-89734 . 14440 6.92525 .16226 6.16283 47 14 .09159 10.9178 . 10922 9-15554 . 12692 7.87895 .14470 6.91104 .16256 6.15151 46 15 .09189 10.8829 .10952 9.13093 .12722 7.86064 14499 6 . 89688 .16286 6.14023 45 16 .09218 10.8483 .10981 9 . 10646 .12751 7.84242 14529 6.88278 .16316 6.12899 44 17 .09247 10.8139 . J'lOII 9.08211 . 12781 7.82428 14559 6.86874 16346 6.11779 43 18 .09277 10.7797* . 1 1040 9.05789 .12810 7.80622 .14588 6.85475 16376 6.10664 42 19 .09306 10.7457 . i 1070 9-03379 . 12840 7.78825 .14618 6.84082 .16405 6.09552 4i 20 09335 10.7119 .11099 9.00983 . 12869 7-77 35 .14648 6.82694 ^6435 6.08444 40 21 93 6 5 10.6783 .11128 8.98598 .12899 7-75254 .14678 6.81812 . 16465 6.07340 39 22 .09394 10.6450 .11158 8.96227 .12929 7.73480 .14707 6 . 79936 .16495 6.06240 38 2 3 .09423 10.6118 .11187 8.93867 .12958 7- 7*715 14737 6.78564 .16525 6.05143 37 24 09453 10.5789 .11217 8.91520 .12988 7.69957 .14767 6.77199 16555 6.04051 36 25 .09482 10.5462 .11246 8.89185 .13017 7.68208 14796 6-758JB -16585 6.02962 35 26 .09511 10.5136 . 11276 8.86862 13047 7.66466 .14826 6.74483 .16615 6.01878 34 27 .09541 10.4813 .11305 8.84551 .13076 7.64732 -14856 6- 73 '33 .16645 6.00797 33 28 .09570 10.4491 "335 8.82252 .13106 7.63005 .14886 6.71789 .16674 5-99720 S 2 29 .09600 10.4172 .11364 8.79964 13136 7.61287 149*5 6 . 7045 .16704 5-98646 3 1 3 .09629 10.3854 .11394 8.77689 13165 7-59575 .14945 6.69116 16734 5-97576 30 3 1 .09658 10.3538 .11423 8.75425 i3 J 95 7.57872 14975 6.67787 . 16764 5-9651 29 3 2 .09688 10.3224 .11452 8 73172 .13224 7.56176 .15005 6.66463 .16794 5-95448 28 33 .09717 10.2913 .11482 8.70931 i3 2 54 7-54487 15034 6.65144 .16824 5-9439 27 1 34 .09746 10.2602 11511 8.68701 .13284 7.52806 15064 6.63831 .16854 5-93335 26 35 .09776 10.2294 .11541 8.66482 I 33 I 3 7-5"32 .15094 6.62523 .16884 5.92283 25 36 .09805 10.1988 .11570 8.64275 I 3343 7 49465 .15124 6.61219 . 16914 5.91236 24 37 .09834 10.1683 . 11600 8.62078 J 3372 7.47806 15153 6.59921 1^944 5 90191 23 I 38 .09864 10.1381 . 11629 8.59893 13402 7-46i54 15183 6.58627 .16974 5.89151 22 39 .09893 10. 1080 11659 8.57718 13432 7-44509 .15213 6-57339 .17004 5.88114 21 40 .09923 10.0780 .11688 8-55555 .13461 7.42871 15243 6.56055 I 733 5.87080 2O 4 1 .09952 10.0483 .11718 8.53402 13491 7.41240 .15272 6-54777 .17063 5.86051 19 42 .09981 10.0187 .11747 8.51259 I.352I 7.39616 15302 6.5353 17093 5.85024 18 43 . OOII 9.98931 .11777 8.49128 1355 7-37999 15332 6.52234 .17123 5.84001 J 7 44 . 0040 9.96007 . 1806 8.47007 .13580 7.36389 15362 6.50970 I7I53 5.82982 16 45 . 0069 9.93101 . 1836 8.44896 .13609 7.34786 15391 6.49710 17183 5.81966 15 46 . 0099 9.90211 . 1865 8.42795 13639 7-33I9 .15421 6.48456 .17213 5-80953 14 47 . 0128 9-87338 1895 8.40705 .13669 7.31600 15451 6.47206 17243 5-79944 13 48 0158 9.84482 . 1924 8.38625 . 13698 7.30018 15481 6.45961 17273 5-78938 12 49 . 0187 9.81641 .11954 8.36555 .13728 7.28442 155" 6.44720 17303 5-77936 ii 5 . 0216 9.78817 . 1983 8.34496 '3758 7.26873 1554 6.43484 17333 5-76937 10 5i . 0246 9.76009 2013 8.32446 .13787 7.25310 15570 6.42253 .17363 5-75941 9 52 02 75 9.73217 . 2042 8 . 30406 .13817 7-23754 .15600 6.41026 17393 5 74949 8 53 35 9.70441 . 2072 8.28376 .13846 7.22204 1563 6.39804 17423 5.7396o 7 54 0334 9.67680 . 2IOI 8-26355 .13876 7.20661 .15660 6.38587 17453 5-72974 6 55 . 0363 9-64935 . 2131 8.24345 .13906 7.19125 .15689 6-37374 17483 5-71992 5 56 393 9.62205 . 2160 8.22344 13935 7-17594 15719 6.36165 17513 5-7IOI3 4 57 . 0422 9.59490 . 2190 8.20352 .13965 7.16071 15749 6.34961 17543 5-7037 3 58 0452 9.56791 . 2219 8.18370 '3995 7-14553 15779 6.33761 17573 5.69064 2 59 . 0481 9.54106 2249 8.16398 .14024 7.13042 .15809 6.32566 .17603 5-68094 I 60 0510 9.51436 . 2278 8.14435 .14054 7-II537 15838 6.31375 17633 5.67128 O / Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang / 8 4 8 3 8 2 8 1 8 D 88 NATURAL TANGENTS AND COTANGENTS. I I i I 2 I 3 i 4 Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang o i7 6 33 5.67128 19438 5- 1 4455 .21256 4.70463 23087 4-33148 -24933 4.01078 60 I .17663 5.66165 .19468 5-i365 .21286 .69791 .23117 4-32573 .24064 4.00582 59 2 17693 5-65205 . 19498 5.12862 .21316 .23148 .32001 24995 4.00086 58 3 17723 5.64248 19529 5.12069 21347 '. 1 68452 23179 3H30 .25026 3-99592 57 4 17753 5-63295 19559 5.11279 21377 .67786 23209 . 30860 .25056 3.99099 56 5 17783 5-62344 .19589 5.10490 .21408 .67121 .23240 .30291 .2=5087 3.98607 55 6 .17813 5.61307 .19619 5.09704 .21438 .66458 .23271 .29724 .25118 3.98117 54 7 17843 5.60452 . 19649 5.08921 .21469 65797 .23301 29*59 25149 3.97627 53 8 9 17873 17903 5-595" 5-58573 . 19680 .19710 5.08139 5.07360 .21499 21529 .65138 .64480 23332 23363 28595 .28032 .25180 25211 3-97139 3.96651 52 51 10 J 7933 5-57638 1974 5.06584 .21560 4.63825 23393 .27471 .25242 3.96165 So ii i79 6 3 5.56706 .19770 5.05809 .21590 4.63171 .23424 .26911 252.73 3.95680 49 12 17993 5-55777 . 19801 5-05037 .21621 4.62518 2 3455 .26352 -25304 3.95196 48 13 .18023 5-54851 .19831 5.04267 .21651 4.61868 23485 25795 -25335 3-94713 47 14 18053 5-53927 .19861 5-03499 .21682 4.61219 .23516 25239 .25366 3-94232 46 15 . 18083 5-537 .19891 5-02734 .21712 4.60572 23547 .24685 25397 3-93751 45 16 .18113 5-52090 .19921 5.01971 21743 4-59927 -23578 .24132 .25428 3-93271 44 17 .18143 S-Sii? 6 19952 5.01210 21773 4-59283 .23608 .23580 25459 3-92793 43 18 .18173 5.50264 .19982 5.00451 . 2 I 804 4.58641 23639 23030 25490 3.92316 42 19 .18203 5-49356 .20012 4-99^95 .21834 4.58001 .23670 .22481 25521 3-91839 4 1 20 18233 5-4845I .2OO42 4.98940 .21864 4-57363 .23700 4-21933 25552 3.91364 40 21 . 18263 5-47548 .20073 4.98188 .21895 4.56726 2373 1 4.21387 25583 3-90890 39 22 I&293 5-46648 .2OIO3 4-97438 .21925 4.56091 25762 4.20842 .25614 3.90417 38 23 .18323 5-4575 1 .20133 4.96690 .21956 4-55458 23793 4.20298 25645 3-89945 37 24 18353 5-44857 .20164 4-95945 .21986 4-54826 23823 4.19756 .25676 3-89474 36 25 18384 5-43966 1 90194 4.95201 .22017 4.54196 23854 4.19215 .25707 3 . 89004 35 26 .18414 5-43077 .20224 4.94460 .22047 4-53568 23885 4.18675 -25738 3-88536 34 2 7 .18444 5.42192 .20254 4-93721 .22078 4.52941 .23916 4.18137 .2,5769 3.88068 33 28 .18474 5.41309 .20285 4.92984 .22108 4.52316 .23946 4.17600 .25800 3.87601 32 29 .18504 5.40429 20315 4.92249 22139 4.51693 23977 4.17064 .25831 3-87136 3 1 30 1 .18534 5-39552 20345 4.91516 .22169 4.51071 .24008 4.16530 .25862 3.86671 30 31 .18564 5.38677 20376 4.90785 .222OO 4.50451 24039 4-15997 .25893 3 . 86208 29 32 18594 5-378o5 . 20406 4.90056 .22231 4-49832 .24069 4.15465 -25924 3-85745 28 33 .18624 5-36936 .20436 4.89330 .2226l 4.49215 .24100 4-14934 25955 3.85284 27 34 .18654 5.36070 .20466 4.88605 .22292 4.48600 .24131 4.14405 .25986 3.84824 26 35 .18684 5-35206 .20497 4.87882 .22322 4-47986 .24162 4-13877 .26017 3-84364 25 36 .18714 5-34345 .20527 4.87162 22353 4-47374 24193 4-I3350 .26048 3.83906 24 37 18745 5-33487 20557 4.86444 22383 4.46764 24223 4.12825 .26079 3-83449 23 138 18775 5-32631 .20588 4-85727 .22414 4-46i55 24254 4.12301 .26110 3.82992 22 39 .18805 5-3 r 778 .20618 4.85013 .22444 4-45548 -24285 4.11778 .26141 3-82537 21 40 18835 5.30928 .20648 4.84300 22475 4.44942 .24316 4-11256 .26172 3.82083 2O 4 1 .18865 5.30080 .20679 4-83590 22505 4-44338 24347 4.10736 . 26203 3.81630 iq 42 .18895 5-29235 .20709 4.82882 .22536 4-43735 24377 4.10216 26235 3-8ii77 18 43 18925 5-28393 .20739 4-82175 22567 4-43 T 34 .24408 4.09699 .26266 3.80726 17 44 18955 5-27553 .20770 4.81471 .22597 4-42534 2 4439 4.09182 .26297 3.80276 16 45 .18986 5-26715 .20800 4.80769 .22628 4.41936 .24470 4.08666 .26328 3.79827 15 46 . IgOl6 5.25880 .20830 4.80068 .22658 4.41340 24501 4.08152 .26359 3-79378 14 47 . 19046 5-25048 .20861 4-7937 .22689 4-40745 24532 4-07639 .26390 3-78931 i.3 48 . 19076 5.24218 .20891 4.78673 .22719 4.40152 .24562 4.07127 .26421 3-78485 12 49 . 19106 5-2339 1 .20921 4.77978 .22750 4-3956o 24593 4.06616 .26452 3.78040 11 5 .I9I3"6 5-22566 . 20952 4.77286 .22781 4.38969 .24624 4.06107 .26483 3-77595 IO 5i .19166 5-21744 .20982 4.76595 .22811 4-38381 24655 4-05599 -26515 3-77I52 9 52 .19197 5-20925 2 013 4-759o6 .22842 4-37793 .24686 4.05092 .26546 3.76709 8 53 .19227 5.20107 .2 043 4-752I9 .22872 4.37207 .24717 4.04586 26577 3.76268 7 54 19257 5- I 9293 .2 073 4-74534 .22903 4.36623 .24747 4 . 0408 i .26608 3.75828 6 55 .19287 5.18480 .2 104 4-7385I 22934 4.36040 .24778 4-03578 .26639 3.75388 5 56 I93I7 5.17671 .2 I 34 4-73I70 .22964 4-35459 . 24809 4-03076 . 26670 3- 7495 4 57 19347 5.16863 .2 l6 4 4.72490 .22995 4-34879 .24840 4.02574 .26701 3-74512 3 58 1 9378 5-16058 2 195 4.71813 .23026 4-34300 .24871 4.02074 26733 3 74075 2 59 .19408 5- 15256 .21225 4-7II37 . 23056 4-33723 .24902 4.01576 .26764 3 7364 I 60 .19438 5-14455 .21256 4.70463 .23087 4.33148 24933 4.01078 26795 3-73205 o , Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang ~| l 7 9 7 s. 7 1 7 6 7 5 NATURAL TANGENTS AND COTANGENTS. 89 15 16 17 18 '9 Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang .20795 3-73205 .28075 3.48741 3573 3.27085 32492 3.07768 34433 2 . 9042 I 60 I .26826 3.72771 .28706 3-48359 .30605 3-26745 32524 3.07464 34465 2.90147 59 2 .26857 3-72338 .28738 3-47977 30 6 37 3.26406 32556 3.07160 .34498 2.89873 58 3 .26888 3.71907 .28769 3-47596 .30669 3.26067 .32588 3.06057 3453 2.89600 57 4 .26920 3-7M76 .28800 3.47216 .30700 3.25729 .32621 3 - c6 554 34563 2.89327 56 5 .26951 3.71046 .28832 3-46837 373 2 3-25392 32653 3.06252 34596 2.89055 55 6 .26982 3.70616 .28864' 3-46458 .30764 3-25055 .32685 3- 595o .34628 2.88783 54 7 .27013 3.70188 .28895 3.46080 .30796 3.24719 32717 3-05649 .34661 2.88511 53 8 . 27044 3.69761 .28927 3.45703 .30828 3 24383 32749 3-05349 34693 2 . 88240 S 2 9 .27076 3-69335 28958 3-45327 .30860 3.24049 .32782 3-05049 .34726 2.87970 51 10 .27107 3.68909 .28990 3- 4495 i .30891 3-237I4 .32814 3-04749 34758 2.87700 5 n .27138 3.68485 .29021 3.44576 .30923 3-23381 .32846 3.04450 3479 1 2-87430 49 12 .27169 3.68061 29053 3.44202 3955 3.23048 .32878 3-o4!52 34824 2.87161 48 J 3 .27201 3.67638 .29084 3.43829 .30987 3.22715 .32911 3-03854 .34856 2.86892 47 J 4 27232 3.67217 .29116 3-43456 .31019 3.22384 3 2 943 3-03556 .34889 2.8662 4 46 15 .27263 3.66796 .29147 3-43084 3*5 l 3-22053 3 2 975 3.03260 34922 2.86356 45 16 .27294 3.66376 .29179 3-427*3 .31083 3.21722 33007 3.02963 34954 2.86089 44 J 7 .27326 3-65957 .29210 3-42343 Sins 3.21392 33040 3.02667 .34987 2.85822 43 18 27357 3-65538 .29242 3-4I973 3"47 3.21063 33072 3-02372 .35020 2-85555 42 '9 .27388 3.65121 29274 3.41604 .31178 3-20734 33 I0 4 3.02077 35 52 2.85289 4 1 20 .27419 3-64705 .29305 3.41236 .31210 3 . 20406 -33136 3-01783 35085 2.85023 40 21 27451 3.64289 29337 3.40869 .31242 3.20079 33169 3.01480, .35118 2.84758 39 22 .27482 3-63874 .29368 3.40502 31274 3- J 9752 33201 3.01196 35150 2 . 84494 38 2 3 275J3 3.63461 .29400 3.40136 .31306 3.19426 33233 3.00903 35i83 2.84229 37 24 27545 3.63048 29432 3-39771 31338 3.19100 .33266 3.00611 35216 2.83965 36 25 .27576 3.62636 .29463 3-39406 31370 3-18775 33298 3.00319 35248 2 . 83702 35 26 .27607 3.62224 29495 3-39042 .31402 3-18451 3333 3.00028 3528r 2.83439 34 27 .27638 3.61814 .29526 3-38679 31434 3.18127 33363 2.99738 353*4 2.83176 33 28 .27670 3.61405 29558 3-38317 .31466 3.17804 33395 2-99447 35346 2.82914 32 2Q .27701 3.60996 29590 3-37955 .31498 3.17481 334 2 7 2.99158 35379 2.82653 3 1 30 27732 3.60588 .29621 3-37594 31530 3-i7i59 3346o 2.98868 35412 2.82391 30 3 1 .27764 3.60181 29653 3-37234 .31562 3.16838 33492 2.98580 35445 2.82130 2 9 3 2 27795 3-59775 .29685 3-36875 3 I 594 3.16517 33524 2.98292 35477 2.81870 28 33 .27826 3 59373 .29716 3.36516 .31626 3.16197 33557 2 . 98004 35510 2. 8l6lO 2 7 34 .27858 3-58966 29748 3-36158 .31658 3-I5877 33589 2.97717 35543 2.81350 26 35 .27889 3-58562 .29780 3-358oo .31690 S-JSSS 8 .33621 2.97430 35576 2.81091 25 36 .27921 3.58160 .29811 3-35443 .31722 3.15240 33654 2.97144 .35608 2-80833 24 37 27952 3-57758 29843 3-35087 31754 3.14922 .33686 2.96858 35641 2.80574 23 38 27983 3-57357 29875 3-34732 .31786 3.14605 .33718 2.96573 .35674 2.80316 22 39 .28015 3 56957 . 29006 3-34377 .31818 3.14288 33751 2.96288 .35707 2 . 80059 21 40 .28046 3-56557 .29938 3-34023 .31850 3-13972 33783 2 . 96004 35740 2.79802 2O 4i .28077 3-5 6l 59 .29970 3-33 6 70 .31882 3.13656 .33816 2.95721 35772 2-79545 J 9 42 .28109 3-5576r .30001 3-333 I 7 3 I 9i4 3-13341 .33848 2-95437 35805 2.79289 18 43 .28140 3o5364 .30033 3-32965 .31946 3.13027 33881 2.95155 35838 2.79033 *7 44 .28172 3.54968 .30065 3.32614 3 T 978 3- I2 7 I 3 339*3 2.94872 35871 2.78 77 8 16 45 .28203 3-54573 .30097 3.32264 .32010 3.12400 33945 2.94591 35904 2 78523 15 46 .28234 3-54*79' .30128 3-3 I 9 I 4 .32042 3.12087 33978 2 . 94309 35937 2 . 78269 14 47 .28266 3-53785 .30160 3-3 I 565 .32074 3- I1! 775 .34010 2.94028 35969 2.78014 J 3 48 .28297 3-53393 .30192 3.31216 .32106 3.11464 34043 2.93748 .36002 2.77761 12 49 28329 3.53001 .30224 3.30868 32139 3- 1:[I 53 34075 2.93468 36035 a - 7757 II 5 .28360 3.52609 30255 3.30521 .32171 3.10842 .34108 2.93189 . 36068 2.77254 IO 5i .28391 3.52219 .30287 3-30I74 .32203 3-10532 .34140 2.92910 .36101 2.77002 9 5 2 .28423 3.51829 33 I 9 3.29829 32235 3.10223 34173 2.92632 36134 2.76750 8 53 28454 3-5 I 44 t 30351 3.29483 .32267 3.09914 34205 2.92354 .36167 2.76498 7 54 .28486 3-51053 .30382 3-29I39 .32299 3.09606 34238 2.92076 .36199 2.76247 6 55 .28517 3.50666 .30414 3-28795 32331 3.09298 34270 2.91799 .36232 2 . 75996 5 56 28549 3.50279 .30446 3' 28452 .32363 3.08991 .34303 2.91523 .36265 2.75746 4 57 .28580 3.49894 .30478 3.28109 .32396 3-08685 34335 2.91246 .36298 2.75496 3 58 .28612 3-49509 359 3.27767 .32428 3-08379 -34368 2.90971 3633 1 2.75246 2 59 .28643 3.49125 30541 3.27426 .32460 3.08073 .34400 2 . 90696 36364 2.74997 I 60 28675 3.48741 30573 3.27085 .32402 3.07768 34433 2.90421 36397 2.74748 O / Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang r 74 73 72 71 7 90 NATURAL TANGENTS AND COTANGENTS. 2 2 i 2 2 2 3 2 4 Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang o 3 6 397 2.74748 38386 2 . 60509 . 40403 2.47509 42447 2-35585 -44523 .24604 60 I 3 6 43 2-74499 .38420 2.60283 .40436 2.47302 .42482 2-35395 .44558 .24428 59 2 3 36463 . 36496 2.74251 2.74004 38453 .38487 2.60057 2.59831 .40470 . 40504 2.47095 2.46888 .42516 42551 2 - 35205 ^35oi5 44593 .44627 .24252 .24077 57 4 36529 2-73756 .38520 2.59606 .40538 2.46682 .42585 2.34825 .44662 23902 56 5 .36562 2.73509 38553 2.59381 .40572 2.46476 .42619 2 34636 .44697 .23727 55 6 36595 2.73263 38587 2.59156 .40606 2.46270 42654 2-34447 44732 23553 54 7 .36628 2.73017 .38620 2.58932 . 40640 2.46065 .42688 2.34258 .44767 23378 53 8 .36661 2.72771 38654 2.58708 .40674 2.45860 .42722 2 . 34069 .44802 .23204 52 9 36694 2.72526 .38687 2.58484 .40707 2-45655 42757 2.33881 .44837 .23030 5i 10 36727 2.72281 .38721 2.58261 .40741 2-4545 1 .42791 2.33693 .44872 .22857 50 ii 36760 2 . 72036 38754 2.58038 4775 2.45246 .42826 2.33505 .44907 .22683 49 12 36793 2.71792 .58787 2.578I5 .40809 2.45043 .42860 2- SSS 1 ? .44942 .22510 48 13 .36826 2.71548 .38821 2-57593 .40843 2.44839 .42804 2.33130 44977 22337. 47 J 4 36859 2.71305 38854 2-57371 .40877 2.44636 .42929 2.32943 .45012 .22164 46 15 .36892 2.71062 .38888 2-57150 .40911 2-44433 .42963 2.32756 4547 .21992 45 16 36925 2.70819 38921 2.56928 .40945 2.44230 .42998 2.32570 .45082 .21819 44 J 7 .36958 2.70577 38955 2 . 56707 .40979 2.44027 43 32 2.32383 45H7 .21647 43 18 .36991 2 70335 .38988 2.56487 .41013 2.43825 43067 2.32197 45152 .21475 42 JQ 37024 2.70094 . 39022 2.56266 .41047 2-43623 .43101 2.32012 45187 -21304 4i 20 37057 2.69853 39055 2.56046 .41081 2.43422 43 T 3 6 2.31826 .45222 .21132 40 21 .37090 2.69612 39089 2-55827 .41115 2.43220 .43170 2.3164! .45257 .20961 39 22 37 J 23 2.69371 .39122 2.55608 .41149 2.43019 43205 2- SHS 6 45292 .20700 38 23 37 J 57 2.69131 39156 2.55389 .41183 2 .42819 4323 2.31271 45327 .20619 37 24 37*9 2.68892 .39190 2-55170 .41217 2.42618 43274 2.31086 45362 .20449 36 25 37 2 23 2.68653 39223 2.5495 2 .41251 2.42418 43308 2 . 30902 45397 .20278 35 26 37256 2.68414 39257 2-54734 .41285 2.42218 43343 2.30718 45432 .20108 34 27 .37289 2.68175 .39290 2.54516 '^iSJQ 2.42019 43378 2-30534 45467 19938 33 28 37322 2.67937 39324 2.54299 4!353 2.41819 .43412 2.30351 45502 *97 6 9 32 29 37355 2.67700 39357 2 . 54082 41387 2.41620 43447 2.30167 45538 *9599 3 1 3 37388 2.67462 39391 2.53865 .41421 2.4I42I .43481 2.29984 45573 .19430 30 3i 37422 2.67225 394 2 5 2.53648 4*455 2.41223 435i6 2.29801 .45608 .19261 29 32 37455 2.66989 39458 2.53432 .41490 2. 4 I025 4355 2.29619 45643 .19092 28 33 .37488 2.66752 39492 2.53217 .41524 2.40827 43585 2.29437 .45678 . 18923 27 34 37521 2.66516 39526 2.53001 41558 2.40629 .43620 2.20254 45713 18755 26 35 37554 2.66281 39559 2.52786 .41592 2.40432 43 6 54 2.29073 45748 .18587 25 36 37588 2.660 4 6 39593 2.52571 .41626 2.40235 .43689 2.28891 45784 . 18419 24 37 .37621 2.65811 .39626 2-52357 .41660 2.40038 43724 2.28710 .45819 .18251 23 38 37654 2.65576 .39660 2.52142 .41694 2.39841 .43758 2.28528 45854 .18084 22 39 3768 7 2.65342 .39694 2.51929 .41728 2-39645 43793 2.28348 .45889 .17916 21 40 37720 2.65109 39727 2.5I7I5 .41763 2.39449 .43828 2.28167 .45924 17749 20 4 1 37754 2.64875 3976i 2.51502 41797 2.39253 .43862 2.27987 .45960 17582 19 42 37787 2 . 64642 39795 2.51289 .41831 2.39058 .43897 2 . 27806 45995 .17416 18 43 .37820 2.64410 .39829 2.51076 .41865 2.38863 43932 2 .27626 .46030 .17249 17 44 .37853 2.64177 .39862 2 . 50864 .41899 2.38668 43966 2.27447 .46065 .17083 16 45 .37887 2.63945 .39896 2.50652 4 J 933 2-38473 .44001 2.27267 .46101 .16917 15 46 .37920 2.63714 .39930 2.50440 .41968 2.38279 44036 2.27088 .46136 .16751 U 47 37953 2.63483 .39963 2.50229 .42002 2.38084 .44071 2.26909 .46171 .16585 13 48 2.63252 39997 2.500l8 .42036 2.37891 .44105 2 . 26730 .46206 . 16420 12 49 . 38020 2.63021 .40031 2.49807 .42070 2.37697 .44140 2.26552 .46242 *6255 II 5 38053 2.62791 .40065 2-49597 42105 2.37504 44175 2.26374 .46277 .16090 10 5i .38086 2.62561 .40098 2.49386 42139 2.37311 .44210 2.26196 .46312 15925 9 52 .38120 2.62332 .40132 2.49177 42173 2.37II8 44244 2.26ol8 .46348 .15760 8 53 38i53 2.62IO3 .40166 2.48967 .42207 2.36925 44279 2.25840 46383 I 559 6 7 54 .38186 2.6l8 7 4 .40200 2.48758 .42242 2-36733 .44314 2.25663 .46418 15432 6 55 . 38220 2.61646 .40234 2.48549 .42276 2.36541 44349 2.25486 .46454 .15268 5 56 38253 2.61418 .40267 2.48340 .42310 2 . 36349 44384 2.25309 .46489 .15104 4 57 .38286 2.6II90 .40301 2.48132 42345 2.36158 .44418 2.25132 .46525 .14940 3 58 59 .38320 38353 2.60736 40335 . 40369 2.47924 2.47716 42379 .42413 2.35967 2.35776 44453 .44488 2.24956 2.24780 .46560 46595 14777 .14614 2 I 60 .38386 2.60509 .40403 2.47509 42447 2.35585 44523 2.24604 .46631 M45I O / Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang t 6 ? 6, * 6 7 6( 5 6 5 NATURAL TANGENTS AND COTANGENTS. 91 * 2 5 2< 2 7 2 2< 71 ' i Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang o I .46631 .46666 2-1445* 2.14288 48773 .48809 05030 04879 50953 50989 96261 96120 53*7* 53208 88073 .87941 5543* 55469 .80405 .80281 ~6o~ 59 2 .46702 2.14125 48845 .04728 .51026 95979 53246 .87809 .55507 .80158 58 3 46737 2.13963 .48881 04577 .5*063 95838 53283 .87677 55545 .80034 57 4 .46772 2.13801 .48917 04426 .51099 95698 53320 87546 .55583 799** 56 5 .46808 2.13639 48953 04276 51*36 95557 53358 874*5 55621 .79788 55 6 .46843 2-13477 .48989 04*25 5**73 954*7 53395 .87283 55659 .79665 54 7 .46879 2.13316 .49026 03975 5*209 95277 53432 .87152 55697 79542 53 8 .46914 2.13*54 .49062 03825 5*246 95*37 53470 .87021 55736 794*9 5 2 9 .46950 2.12993 .49098 03675 5*283 94997 53507 .86891 55774 . 79296 5* 10 46985 2.12832 49*34 03526 5*3*9 94858 53545 .86760 .558*2 79*74 5 ii .47021 2. 12671 .49170 03376 5*356 .94718 .53582 .86630 55850 7905* 49 12 .47056 2.I25II .49206 03227 5*393 94579 .53620 .86499 .55888 . 78929 48 *3 .47092 2.12350 .49242 03078 5*430 94440 53657 .86369 55926 .78807 47 *4 .47128 2.I2I90 .49278 02929 5*467 .94301 55694 .86239 55964 78685 46 15 47*63 2.12030 493*5 02780 5*503 .94162 53732 .86109 .56003 78563 45 16 47*99 2.11871 4935* 02631 5*540 .94023 53769 85979 .56041 .78441 44 *7 47234 2. II7II 49387 02483 5*577 .93885 53807 .85850 56079 78319 43 18 .47270 2.H552 49423 02335 5*614 93746 53844 .85720 .56117 .78198 42 *9 47305 2.II392 49459 02187 5*5* .93608 .53882 8559* 56*56 .78077 4*, 20 4734* 2.II233 49495 .02039 .51688 93470 .53920 .85462 56i94 77955 40 | 21 47377 2.II075 49532 .01891 5*724 93332 53957 85333 .56232 77834 39 22 474*2 2 . 10916 .49568 01743 .51761 93*95 53995 85204 56270 777*3 38 23 .47448 2.10758 .49604 01596 .5*798 93057 54032 85075 56309 77592 37 24 47483 2. IO6OO .49640 .01449 .5*835 .92920 .54070 .84946 56347 7747* 36 2 5 475*9 2 . 10442 .49677 .01302 .51872 .92782 54*07 .84818 56385 7735* 35 26 47555 2.10284 497*3 01155 .51909 92645 54*45 .84689 56424 .77230 34 2 7 4759 2. IOI26 49749 .01008 .5*946 .92508 54*83 .84561 . 56462 .77110 33 28 .47626 2.09969 .49786 .00862 5*983 9237* .54220 84433 5650* .76990 3 2 29 .47662 2.09811 .49822 .00715 . 52020 92235 54258 .84305 56539 .76869 3* 30 .47698 2.09654 49858 .00569 52057 .92098 54296 .84177 56577 76749 3 3 1 47733 2.09498 .49894 .00423 .52094 .91962 54333 .84049 .56616 .76629 29 3 2 .47769 2.09341 4993* .00277 52*3* .91826 5437* .83922 56654 .76510 28 33 47805 2.09184 .49967 .00131 .52168 .91690 .54409 83794 .56693 .76390 27 34 .47840 2 .09028 .50004 .99986 52205 9*554 .54446 83667 5673* .76271 26 35 .47876 2.08872 .50040 .99841 .52242 .91418 .54484 .83540 56769 .76151 25 36 479*2 2.08716 .50076 .99695 .52279 .91282 54522 834*3 .56808 .76032 24 37 .47948 2.08560 9955 .523*6 9*147 .54560 .83286 .56846 759*3 23 38 .47984 2.08405 50*49 .99406 52353 .91012 54597 83*59 .56885 75794 22 39 .48019 2.08250 50*85 .99261 52390 .90876 54635 .83033 56923 75675 21 40 48055 2.08094 .50222 .99116 .52427 .90741 54673 .82906 .56962 75556 20 4* .48091 2.07939 .50258 .98972 52464 .90607 547*1 .82780 57000 75437 *9 42 .48127 2.07785 50295 .98828 .90472 54748 .82654 57039 753*9 18 43 .48163 2.07630 533* .98684 i 52538 .90337 .54786 .82528 .57078 .75200 *7 44 .48198 2.07476 50368 .9854 52575 . 90203 54824 .82402 .57116 75082 16 45 .48234 2.07321 .50404 98396 526*3 .90069 .54862 .82276 57*55 74964 *5 46 .48270 2.07167 .50441 98253 .52650 .89935 .54900 .82150 57*93 .74846 47 .48306 2.07014 50477 .98110 52687 .89801 54938 .82025 57232 .74728 *3 48 48342 2.06860 .97966 52724 .89667 54975 .81899 5727* .74610 12 49 .48378 2 . 06706 50550 97823 .52761 .89533 55013 .81774 5739 74492 II So .48414 2-06553 50587 .97681 .52798 .89400 5505 1 .81649 .57348 74375 10 51 .48450 2 . 06400 .50623 97538 52836 .89266 55089 .81524 57386 74257 9 5 2 .48486 2.06247 .50660 97395 52873 89*33 55*27 .81399 .57425 .74140 8 53 .48521 2.06094 .50696 97253 .52910 .89000 55*65 .81274 57464 .74022 7 54 48557 2.05942 50733 .97111 52947 .88867 55203 .81150 5753 73905 6 55 .48593 2.05790 .50769 .96969 .52985 .88734 5524* .81025 5754* .73788 5 56 .48629 2.05637 .50806 .96827 53022 .88602 55 2 79 .80901 5758o 7367* 4 57 .48665 2.05485 50843 .96685 5359 .88469 553*7 .80777 576*9 73555 3 58 .48701 2.05333 .50879 96544 53096 88337 55355 80653 57657 73438 2 59 48737 2.05182 .509*6 .96402 53*34 .88205 55393 .80529 57696 73321 I 60 48773 2.05030 50953 .96261 53*7* .88073 5543* .80405 57735 73205 i Cotang Tang Cotang Tang Cotang Tang Cotang Tang ,Cotang Tang \ I I ! ' j 6 4 6 3 6 2 6 1 6 ! i NATURAL TANGENTS AND COTANGENTS. 3 C > 33 3' > 3: 3^ f 1 Tang Cotang Tang Cotang Tang Cotang Tang ^otang Tang Cotang o 57735 .73205 .60086 .66428 .62487 .60033 .64941- .53986 07451 .4b256 ~6o~ I 57774 .73089 .60126 .66318 .62527 .64982 -53888 .67493 .48163 59 2 57813 72973 .60165 . 66209 .62568 .59826 .65024 53791 67536 .48070 58 3 57851 72857 .60205 .66099 .62608 59723 .65065 53693 67578 47977 57 4 .57890 .72741 .60245 .65990 .62649 .59620 .65106 53595 .67620 .47885 56 57929 .72625 .60284 .65881 .62689 59517 .65148 53497 .67663 .47792 55 6 .57968 .72509 . 60324 .65772 .62730 59414 .65189 .53400 .67705 .47699 54 7 .58007 72393 .60364 65663 .62770 593" .65231 .53302 67748 47607 53 8 .58046 .72278 .60403 65554 .6 2 8n .59208 .65272 53205 .67790 475*4 S 2 9 58085 .72163 .60443 65445 .62852 59105 .653J4 53 I0 7 .67832 47422 5i 10 .58124 72047 .60483 65337 .62892 .59002 65355 53 10 67875 4733 5 ii .58162 7*932 .60522 .65228 .62933 .58900 65397 5 2 9i3 .67917 .47238 49 12 .58201 .71817 . 60562 .65120 .62973 58797 .65438 .52816 .67960 .47146 48 J 3 .58240 .71702 .60602 .65011 .63014 58695 .65480 .52719 .68002 47053 47 14 58279 -71588 .60642 .64903 63055 58593 .65521 .52622 .68045 .46962 46 15 .58318 7*473 .60681 64795 63095 .58490 65563 52525 .68088 .46870 45 16 .58357 71358 .60721 .64687 .63136 58388 .65604 .52429 .68130 .46778 44 J 7 .58396 .71244 .60761 64579 63177 . 58286 .65646 52332 68173 .46686 43 18 58435 .71129 .60801 .64471 63217 .58184 .65688 52235 .68215 46595 42 iQ 58474 .71015 .60841 64363 .63258 .58083 .65729 52139 .68258 .46503 4i 20 58513 .70901 .60881 .64256 63299 5798i .65771 52043 .68301 .46411 40 21 58552 .70787 .60921 .64148 63340 57879 .65813 .51946 68343 .46320 39 22 5859 1 7 673 .60960 .64041 .63380 57778 65854 .51850 .68386 . 46229 38 23 .58631 .70560 .61000 63934 .63421 57676 .65896 5*754 .68429 46137 37 24 .58670 . 70446 .61040 .63826 .63462 57575 65938 .51658 .68471 .46046 36 2 5 .58709 .70332 .61080 63719 63503 -57474 .65980 5 I 562 68514 45955 35 26 .58748 .70219 .61120 .63612 63544 57372 .66021 -5M66 68557 45864 34 27 58787 .70106 .61160 63505 .63584 572?i .66063 5*37 .68600 45773 33 28 .58826 .69992 .61200 63398 .63625 57 I 7 .66105 5 I2 75 .68642 45682 32 20 .58865 .69879 .61240 .63292 .63666 .57069 .66147 .51179 .68685 45592 3i 30 .58905 .69766 .61280 .63185 .63707 56969 .66189 .51084 .68728 455i 3 3 1 .58944 .69653 .61320 .63079 .63748 .56868 .66230 . 50988 .68771 .45410 29 3 2 58983 .69541 .61360 . 62972 .63789 .56767 .66272 50893 .68814 45320 28 33 59022 .69428 .61400 .62866 63830 .56667 .66314 5797 68857 45229 27 34 .59061 69316 .61440 .62760 .63871 .56566 .66356 . 50702 .68900 45*39 26 35 .59101 .69203 .61480 .62654 63912 .56466 .66398 .50607 .68942 .45049 25 36 59^49 .69091 .61520 .62548 63953 .56366 . 66440 .50512 .68985 .44958 24 37 59179 .68979 .61561 .62442 .63994 .56265 .66482 504*7 .69028 .44868 23 38 .59218 .68866 .61601 .62336 64035 56165 .66524 50322 .69071 .44778 22 39 59258 .68754 .61641 .62230 .64076 .56065 .66566 . 50228 .69114 .44688 21 40 59297 .68643 .61681 .62125 .64117 55966 .66608 50133 69157 44598 2O 4i 59336 68531 .61721 .62019 .64158 .55866 .66650 .50038 .69200 .44508 10 42 59376 .68419 .61761 .61914 .64199 .55766 .66692 .49944 .69243 .44418 18 43 59415 .68308 .61801 .61808 . 64240 .55666 .66734 .49849 .69286 .44329 17 44 59454 .68196 .61842 .61703 .64281 55567 .66776 49755 .69329 44239 16 45 59494 .68085 .61882 61598 64322 55467 .66818 .49661 .69372 .44149 15 46 59533 67974 .61922 61493 .64363 55368 .66860 .49566 .69416 .44060 14 47 59573 67863 .61962 .61388 .64404 55269 .66902 .49472 69459 . 43970 13 48 .59612 .67752 .62003 .61283 .64446 55*7 .66044 49378 .69502 .43881 12 49 5965 1 .67641 .62043 .61179 .64487 557* .66986 .49284 69545 43792 11 5 .59691 6753 .62083 .61074 .64528 54972 .67028 .49190 .69588 43703 IO 5i 5973 .67419 .62124 .60970 .64569 54873 .67071 49097 .69631 .43614 g 52 5977 .67309 .62164 .60865 .64610 54774 .67113 .49003 .69675 43525 8 53 .59809 .67198 .62204 .60761 . 64652 54675 .67155 48909 .69718 43436 7 54 59849 .67088 .62245 60657 .64693 54576 .67197 .48816 .69761 43347 6 55 .59888 .66978 .62285 60553 64734 .54478 .67239 .48722 .69804 43258 5 56 .59928 .66867 .62325 .60449 64775 54379 .67282 .48629 .69847 .43160 4 57 59967 66757 .62366 .60345 .64817 54281 .67324 .48536 .69891 .43080 3 58 .60007 .66647 . 62406 .60241 .64858 54183 .67366 .48442 69934 .42992 2 59 .60046 .66538 .62446 .60137 .64899 .54085 .67409 .48349 69977 .42903 I 60 .60086 .66428 .62487 .60033 .64941 .53986 67451 .48256 .70021 .42815 O I Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang / 5 5 I 5' 1 5< 5 . 5. -0 ) NATURAL TANGENTS AND COTANGENTS. 93 f 3 5 3 6 3 7 3 8 3 9 Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang .70021 42815 72654 37638 75355 32704 .78129 .27994 .80978 2349 60 I .70064 .42726 .72699 37554 7540* .32624 78175 .27917 .81027 .23416 59 2 .70107 .42638 72743 3747 75447 32544 .78222 .27841 .81075 23343 58 3 .70151 4255 .72788 37386 75492 .32464 .78269 .27764 .81123 .23270 57 4 70194 .42462 72832 .37302 .75538 32384 .78316 .27688 .81171 .23196 56 5 .70238 42374 .72877 .37218 75584 32304 78363 .27611 .81220 23123 55 6 .70281 .42286 .72921 37*34 .75629 .32224 .78410 27535 .81268 .23050 54 7 70325 .42198 .72966 3705 .75675 32*44 78457 .27458 .81316 .22977 53 8 .70368 .42110 .73010 .36967 75721 .32064 .78504 .27382 81364 .22904 52 9 .70412 .42022 73055 36883 75767 3*984 7855* .27306 .81413 .22831 5* 10 .70455 41934 .73100 .36800 .758*2 .31904 .78598 .27230 .81461 22758 50 ii 7499 .41847 73*44 .36716 .75858 3*825 .78645 27*53 .81510 .22685 49 12 .70542 41759 73189 36633 75904 3*745 .78692 .27077 .81558 .22612 48 13 .70586 .41672 73234 36549 75950 .31666 .78739 .27001 .81606 22539 47 M .70629 .41584 73278 .36466 75996 .31586 .78786 .26925 81655 .22467 46 15 .70673 .41497 73323 36383 .76042 3*507 .78834 .26849 .81703 .22394 45 16 .70717 .41409 73368 .36300 .76088 .3*427 .78881 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378o 7 75264 .32865 78035 .28148 .80882 23637 .83811 .19316 2 59 .72610 37722 753* 32785 .78082 .28071 .80930 23563 .83860 .19246 i| 60 72654 37638 75355 .32704 .78129 .27994 .80978 .2349 .83910 19*75 o / Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang r 5' 5, 5 5- > 5 1 5 D 94 NATURAL TANGENTS AND COTANGENTS. 3 4 D 4 1 4 2 4 3 4 4 f Tang Cotang Tang Cotang Tang Cotang Tang Cotang Tang Cotang _J .83910 19175 .86929 15037 .90040 .11061 93252 .07237 .96569 03553 lol I .83960 i9 I0 5 .86980 .14969 .90093 .10996 .93306 .07174 .96625 03493 59 2 .8 4 OOQ 19035 .87031 .14902 .90146 .10931 .93360 .07112 .96681 3433 58 3 . 84059 .18964 .87082 .14834 .90199 . 10867 93415 .07049 96738 03372 57 4 .84108 . 18894 87133 .14767 .90251 . 10802 93469 .06987 .96794 03312 56 5 6 .84158 .84208 .18824 18754 .87184 .87236 .14699 .14632 .90304 .90357 10737 . 10672 93524 93578 .06925 .06862 .96850 .96907 .03252 03192 55 54 m .84258 .18684 .87287 14565 .90410 .10607 93633 .06800 .96963 .03132 53 8 .84307 .18614 87338 .14498 .90463 10543 .93688 .06738 .97020 .03072 52 9 .84357 .18544 87389 .14430 .90516 . 10478 93742 .06676 .97076 .03012 51 10 .84407 .18474 .87441 14363 .90569 .10414 93797 .06613 97133 .02952 50 ii .84457 . 18404 .87492 . 14296 . 9062 i 10349 .93852 .06551 .97189 .02892 49 12 .84507 .18334 87543 .14229 .90674 .10285 .93906 .06489 .97246 .02832 48 !3 .84556 . 18264 87595 .14162 .90727 . 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MATTER AOT> ITS PTtOPEHTIES. DEFINITIONS. 1. Natural and Physical Science. Science, which may be denned as a classified knowledge of nature, is divided into natural and physical science. Natural science concerns itself with the external form and internal structure of bodies. Physical science considers only the matter of which these bodies are composed. ILLUSTRATION. Geology, mineralogy, botany, and zoology, which investigate the form and structure of the earth, of minerals, of plants, and of animals, respectively, are natural sciences. Physics and chemistry, which consider the properties of matter itself, whether it is light or heavy, hard or soft, combustible or non- combustible, are physical sciences. 2. Matter is anything that possesses weight; that is, is acted on by gravitation. In studying matter, physical science considers: 1. The division of which matter is capable. 2. The attractions by which these particles are held together. 3. The motions these particles may have. 3. Division of Matter. Science recognizes three divi- sions of matter: masses, molecules, and atoms. 4. 2 PHYSICS. 4 A mass, or body, of matter is any portion of matter appre- ciable by the senses. A molecule is the smallest particle of matter that a body can be divided into without losing its identity. An atom is an indivisible portion of matter. Atoms unite to form molecules; a collection of molecules forms a mass, or a body. It has been calculated that the diameter of a molecule is larger than ^^rik^-^ of an inch, and smaller than ^^^^ of an inch. ILLUSTRATION. The sun and the grain of sand are both masses of matter ; the smallest particles of sugar or of salt that still show the properties of these substances, respectively, are molecules of sugar or of salt. The still more minute particles of carbon, of hydrogen, and of oxygen, which make up the molecule of sugar, or those of chlorine and of sodium, which compose the molecule of salt, are atoms. 4. Attraction of Matter. Three forms of attraction are recognized in science : 1. That form of attraction called gravitation, which is exerted between masses of matter. 2. That form which binds molecules together; called cohesion, when the molecules are alike, and adhesion, when the molecules are unlike. 3. That form of attraction called chemical attraction, or affinity, which draws atoms together. ILLUSTRATION. The planets are held to the sun, and stones are held to the earth by the attraction of gravitation. The molecules of silver or salt, being alike, are held together by cohesion ; those of granite or gunpowder, being unlike, are held together by adhesive attraction. The atoms that form a molecule of sugar or of salt are united by chem- ical attraction. Molecules united by cohesion, therefore, form homogeneous matter, while unlike molecules united by adhesion form heterogeneous matter. It is evident that there must be as many kinds of molecules as there are kinds of homogeneous matter. 5. Motions of Matter. The forms of motion admitted in science are three : 1. Mass motion, or visible mechanical motion. 2. Molecular motion, the motion of the molecule within 4 PHYSICS. 3 the mass, called heat, light, electricity, or magnetism, accord- ing to its character. 3. Atomic motion, which, though probable, is not yet fully established. 6. Province of Physics. Physics is that department of physical science which studies the results that flow from the molecular conditions of matter. ILLUSTRATION. Weight, which is a consequence of mass attraction; impact, which is a result of mass motion; tenacity, hardness, and elasticity, which depend on cohesion ; solution, capillarity, and diffusion, which result from adhesion; heat, light, electricity, and magnetism, which are molecular motions these are all objects of physical study and investigation. 7. Bodies are, as we have seen, collections of molecules. They exist in three forms or conditions : solid, liquid, and gaseotis. A solid body is one whose molecules change their relative positions with great difficulty; as, iron, wood, stone, etc. A liquid body is one whose molecules tend to change their relative position easily. Liquids readily adapt them- selves to the vessel that contains them, and their upper sur- face always tends to become level. Water, mercury, etc. are liquids. A gaseous body, or a gas, is one whose molecules tend to separate from one another; as, air, oxygen, etc. Gaseous bodies are sometimes called aeriform (air-like) bodies. They are divided into two classes : permanent gases and vapors. A permanent gas is one that remains a N gas at ordinary temperatures and pressures. A vapor is a body that at ordinary temperatures is a liquid or solid, but when heat is applied becomes a gas. One body may, however, exist in all three states; as, for example, mercury, which at ordinary temperatures is a liquid, becomes a solid (freezes) at 40 C., and a vapor at 350 C. Nearly all gases can be liqiiefied by great cold, and some can even be solidified. By means of heat nearly all solids may be either liquefied or vaporized. 4 PHYSICS. 4 8. Properties of Matter. All matter possesses Certain qualities that are, in scientific language, called properties. Some properties are common to all kinds of matter ; other properties are possessed only by certain kinds of matter. The former, termed general proper-ties, are: extension, impene- trability, weight, inertia, mobility, divisibility, porosity, com- pressibility, expansibility, elasticity, and indestructibility. Among the latter, which are known as special properties, are the following: hardness, tenacity, brittleness, malleability, and ductility. 9. Extension is the property of occupying space. Since all bodies must occupy space, it follows that extension is a general property. 10. Impenetrability is the property of matter that prevents two bodies from occupying the same portion of space at the same time. 1 1 . 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 cannot go, being held in suspension by the earth's attraction. Weight is measured by comparison with some standard. 12. Inertia means that a body can neither put itself in motion nor bring itself to rest. To do so, it must be acted on by some force. 13. Mobility means that a body can be changed in position by some force acting on it. 14. Divisibility is that property of matter which indi- cates that a body may be separated into parts. 15. Porosity is that property of matter which indicates that there is space between the molecules of a body. Mole- cules of bodies are supposed to be spherical, and hence there is space between them, as there would be between peaches in a basket. It is said that the molecules of water are larger 4 PHYSICS. 5 than those of salt; so that when salt is dissolved in water, its molecules wedge themselves between the molecules of the water, and unless too much salt is added, the water will occupy no more space than it did before. This does not prove that water is penetrable, for the molecules of salt occupy the space between the molecules of water. Water has been forced through iron by pressure, thus proving that iron is porous. 16. Compressibility is that property of matter which indicates that the molecules of a body may be crowded nearer together, so as to occupy a smaller space. 17. Expansibility is that property of matter which indi- cates that the molecules of a body may be forced apart, so as to occupy a greater space. 18. Elasticity is that property of matter which indi- cates that if a body be distorted within certain limits, it -will resume its original form when the distorting force is removed. Glass, ivory, and steel are more or less elastic. 19. Indestructibility indicates that matter can never be destroyed. A body may undergo thousands of changes; it may be resolved into its molecules, and its molecules may be resolved into atoms; these atoms may unite with other atoms to form other molecules and bodies, which may be entirely different from the original body; but through all these changes the number of atoms remains the same. The whole number of atoms in the universe is exactly the same now as it was millions of years ago, and will always remain the same. Matter is indestructible. 20. Hardness is that property of matter which indi- cates that some bodies may scratch other bodies. Fluids and gases do not possess hardness. The diamond is the hardest of all substances. 21. Tenacity is that property of matter which indicates that some bodies resist a force tending to pull them apart. Steel is very tenacious. 6 PHYSICS. 4 22. Brittleness is that property of matter which indicates that some bodies are easily broken; as, glass, crockery, etc. 23. Malleability is that property of matter which indi- cates that some bodies may be hammered or rolled into sheets. Gold is the most malleable of all substances. 24. Ductility is that property of matter which indicates that some bodies may be drawn into wire. Platinum is the most ductile of all substances. GRAVITATION. 25. Every body in the universe exerts a certain attract- ive force on every other body, which tends to draw the two bodies together. This attractive force is called gravitation. If a body is held in the hand, a downward pull is felt, and if released, it will fall to the ground."" This"piuTis commonly called weight, but it really is the attraction between the earth and the body. 26. Force of gravity is a term used to denote the attraction between the earth and bodies upon or near its surface. It always acts in a straight line between the center of the body and the center of the earth. The force of gravity varies at points on the earth's sur- face. It is slightly less on the top of a high mountain than at the level of the sea. For this reason, the weight of a body also varies. But if the weight of a body at any place be divided by the force of gravity at that place, the result is called the mass of the body. '27. The mass of a body is the measure of the actual amount of matter that it contains, and is ahvays the same. Let m mass of the body; W weight of the body; g force of gravity at the place where the body was weighed. weigfht of bodv W Then, mass = - $ ^Z O r m . (1.) force or gravity g 4 PHYSICS. 7 28. Law of Gravitation. The force of attraction by which one body tends to draw another body towards it is directly proportional to its mass, and inversely proportional to the square of the distance between the centers of the two bodies. 29. Laws of Weight. Bodies weigh most at the surface of the earth. Below the surface, the weight decreases as the distance to the center decreases. A bove the surface the weight decreases as the square of the distance increases. ILLUSTRATION. If the earth's radius is 4,000 miles, a body that weighs 100 pounds at the surface will weigh nothing at the center, since it is attracted in every direction with equal force. At 1,000 miles from the center it will weigh 25 pounds, since 4,000 : 1,000 = 100 : 25. At 2,000 miles from the center it will weigh 50 pounds, since 4,000 : 2,000 = 100 : 50. At 3,000 miles from the center it will weigh 75 pounds, and at the surface, or 4,000 miles from the center, it will weigh 100 pounds. If carried still higher, say 1,000 miles from the surface, or 5,000 miles from the center of the earth, it will weigh 64 pounds, since 5,000 2 : 4,000' 2 = 100 : 64. At 4,000 miles from the surface it will weigh 25 pounds, since 8,000 2 : 4,000' = 100 : 25. 30. Formulas for Gravity Problems. Let W = weight of a body at the surface; w = weight of a body at a given distance above or - below the surface; d distance between the center of the earth and the center of the body ; R radius of the earth = 4,000 miles. Formula for weight when the body is below the surface, wR-dW. (2.) Formula for weight when the body is above the surface, wd* = WR\ (3.) EXAMPLE 1. How far below the surface of the earth will a 25-pound ball weigh 9 pounds ? 8 PHYSICS. 4 SOLUTION. Use formula 3, w R = dW. Substituting the values of /?, W, and w, we have 9X4,000 = 0 Steel 7.84 490.0 Tin (cast) 7 29 455 6 Zinc (cast) 6 86 428 8 Antimony 6 71 419 4 Aluminum 2 60 162 5 LIQUIDS. Substance. Specific Gravity. Weight per Cubic Foot in Pounds. Ascetic acid 1 062 66 4 Nitric acid 1.420 88.8 Sulphuric acid 1.841 115.1 Muriatic acid 1.200 75.0 Alcohol 800 50 Turpentine 870 54 4 Sea -water (ordinary) 1 026 64 1 Milk 1 032 64 5 PHYSICS. WOODS. 11 Substance. Specific Gravity. Weight per Cubic Foot in Pounds. Ash .- 845 52 80 Beech 852 53 25 Cedar 561 35 06 Cork . . 240 15 00 Ebony (American) 1 331 83 19 Lignum vitse 1 333 83 30 Maple . . 750 46 88 Oak (old) 1.170 73 10 Spruce .500 31.25 Pine (yellow) .660 41 . 20 Pine (white) 554 34 60 Walnut 671 41 90 At GASES. F., and Under a Pressure of One Atmosphere. Substance. Specific Gravity. Weight per Cubic Foot in Pounds. Atmospheric air 1 0000 08073 Carbonic acid 1 5290 12344 Carbonic oxide 9674 07810 Chlorine 2 4400 19700 Oxvp^en 1 1056 08925 Nitrogen 9713 07841 Smoke (bituminous coal) 1020 00815 Smoke (wood) 0900 00727 *Steam at 212 F .4700 .03790 Hydrogen . . . .0692 00559 *The specific gravity of steam at any temperature and pressure, compared with air at the same temperature and pressure, is 0.622. PHYSICS. MISCELLANEOUS. Substance. Specific Gravity. Weight per Cubic Foot in Pounds. Kmery 4.00 250 Glass (average) 2 80 175 Chalk ... 2 78 174 Granite ... 2 65 166 Marble 2 70 169 Stone (common) 2 52 158 Salt (common) 2 13 133 Soil (common) 1 98 124 Clay 1 93 121 Brick 1 90 118 Plaster of Paris (average) 2 00 125 Sand 1.80 113 36. Determining Specific Gravity of Liquids. To find the specific gravity of any particular liquid, compared with that of water, it is only necessary to weigh equal bulks and divide the weight of the liquid by the weight of the water. To weigh eqtial bulks of liquid the simplest and most accurate way is to weigh them in succession in the same vessel, taking care that it is equally full on both occasions. A convenient vessel, commonly used when small quantities are to be weighed, is shown in Fig. 1. It is easily made by blowing a bulb on a glass tube. On that portion of the tube which is narrowed by drawing it out over a flame, a scratch is. made with a diamond or a file. The bulb is filled up to the scratch with the liquid, and is then weighed, emptied, cleaned, dried, filled with water, and again weighed. In these experiments care should always be taken that both liquids have the same temperature. This is easily accomplished by FIG. l. PHYSICvS. 13 immersing the bulb filled with liquid for some time in water, part of which is later used in the second weighing. Another form of apparatus for determining the specific gravity of liquids, devised by Dr. Sprengel, consists of an elongated U tube, Fig. 2, the ends of which terminate in the two capillary tubes A, B, bent at right angles in opposite directions. The shorter one A is narrower at the end than the longer one. The horizontal part of the wider tube is marked near the bend with a fine line. The U tube is filled by suction, the little bulb apparatus, Fig. 3, having been previously attached to the narrow capillary tube by a piece of rubber tubing. It is then detached from the bulb, placed FIG. 2. in water almost to the bends of the capillary tubes, left there until it has assumed the temperature of the water, and after careful adjustment of the volume of the liquid up to the mark in the wider capillary tube, it is taken out, dried, and weighed. 37. Determining the Specific Gravity of Solids. The determination of the specific gravity of a solid body is also made according to the principles explained, and may be made with the specific-gravity bottle. The bottle is first weighed full of water; the solid is then placed in the same pan of the balance and its weight determined; finally, the solid is put into the bottle, displacing an equal bulk of water, the weight of which is determined by the loss on again weigh- ing. In this way, the weights of the solid and that of an 14 PHYSICS. equal bulk of water are obtained. The former divided by the latter gives the specific gravity. EXAMPLE. A piece of lead weighs 73.68 grains. Glass bottle filled with water weighs 297.05 grains. Total 370.73 grains. After an equal volume of water was dis- placed by the lead, the weight was 864.15 grains. Hence, the displaced water weighed 6.58 grains. r/o ^*o Therefore, the specific gravity of lead is ~ ^- 11.2. Ans. o. oo 38. Another ingenious, but not quite as exact, method of finding the specific gravity of solids not easily soluble in water is based on the principle of Archimedes: When a body is im- mersed in a fluid, it loses in weight an amount equal to the weight of the fluid it displaces. , This principle is ap- plied as follows: Weigh the body first in the air, then weigh the body in water, suspending it by a string attached to a scale pan, as shown in Fig. 4. The difference between the t^vo weights will be the weight of an FlG - 4 - equal volume of water. The ratio of the weight in air to the difference thus found will be the specific gravity. Let Sp. Gr. = specific gravity ; W weight of the solid in air ; W = weight of the solid in water. Then the weight of a volume of water equal to the volume of the solid is W- W, and S - Gr - = 4 PHYSICS. 15 EXAMPLE. A body in air weighs 36.25 grains, and in water, 30 grains ; what is its specific gravity ? SOLUTION. Using formula 4, we obtain W 36.25 36.25 S P- Gr ' = W-W = 36.25-30 = -6725 = 5 ' 8 ' AnS ' 39. If the body is lighter than water, a piece of iron or other heavy substance must be attached to it, sufficiently heavy to sink both. Then weigh both bodies in air and both in water. Weigli both separately in air, and weigh the heavier body in water. Subtract the weight of the bodies in water from their iveight in air, and the result ivill be the weight of a volume of the water equal to the volume of the two bodies. Find the difference of the weights of the heavy body in air and in water, and the result will be the weight of a volume of water equal to the volume of the heavy body. Subtract this last result from the former, and the result will be the weight of a volume of water equal to the volume of the light body. The weight of the light body in air divided by the weight of its equal volume of water is the specific gravity of the light body. Let W weight of both bodies in air ; W = weight of both bodies in water; w = weight of light body in air ; W l = weight of heavy body in air ; W^ weight of heavy body in water. Then, the specific gravity of the light body is given by S P- Gr - = ' (5>) EXAMPLE. A piece of cork weighs 4.8 grains in air. A piece of cast iron weighs 36 grains in air and 31 grains in water. The weight of the iron and cork together in water is 15 grains; what is the specific gravity of the cork ? of the cast iron ? SOLUTION. Substituting in formula 5 the values given, q r _ w __ _ _ 4.8 _ - ( w- W'} -(Wi WJ) ~ (40.8 15) - (36 - 31) = 0- 23, the specific gravity of the cork. Ans. By formula 4, Sp. Gr. -yr^ rr^j- = o or = 7.2, specific gravity of the iron. Ans. - YV OO - OJ. 1G PHYSICS. 4 40. Determining Specific Gravity of Gases. The method of finding the specific gravity of a gas is of great importance to the chemist. The theory of the operation is as simple as if liquids are concerned, but as the process is more delicate and involves certain corrections for differences of temperature and pressure, it appears preferable to defer the consideration of this matter for the present. 4 1 . Hydrometers. Instruments called hydrometers are in general use for determining quickly and accurately the specific gravities of liquids and some forms of solids. There are two kinds ; viz. , 1. Hydrometers of constant weight, as Beaume's. 2. Hydrometers of constant volume, as Nicholson's. 42. A hydrometer of constant weight is shown in Fig. 5. It consists of a glass tube, near the bottom of which are two bulbs. The lower and smaller bulb is loaded with mercury or shot, so as to cause the instrument to remain in a vertical position when placed in the liquid. The upper bulb is filled with air, and its volume is such that the whole instrument is lighter than an equal volume of water. The point to which the hydrometer sinks when placed in water is usually marked, the tube being graduated above and below in such a manner that the specific gravity of the liquid can be read directly. It is customary to have two instruments : one with the zero point near the top of the stem, FIG. s. for use in liquids heavier than water, and the other with the zero point near the bulb, for use in liquids lighter than water. These instruments are more commonly used for determin- ing the degree of concentration or dilution of certain liquids, as acids, alcohol, milk, solutions of sugar, etc., rather than their actual specific gravities. They are then known as acidimeters, alcoholometers, lactometers, saccharimeters, etc., according to the use to which they are put. PHYSICS. 17 43. Nicholson's Hydrometer. This instrument is shown in Fig. 6. It consists of a hollow cylinder carrying at its lower end a basket d heavy enough to keep the apparatus upright when placed in water. At the top of the cylinder is a vertical rod, to which is attached a shallow pan a for holding weights, etc. The cylinder must be of such size that the apparatus may be so much lighter than water that a certain weight W must be placed FIG. 6. in the pan to sink it to a given point c on the rod. The body whose specific gravity it is desired to find must weigh less than W. It is placed in the pan a and enough weight w is added to sink the point c to the water level. It is evident that the weight of the given body is W w. The body is now removed from the pan a and placed in the basket d, an additional weight being added to sink the point c to the water level. Represent the weight now in the pan by W . The difference W w is the weight of a volume of water equal to the volume of the body. Hence, EXAMPLE. The weight necessary to sink the hydrometer to the point c is 16 ounces ; the weight necessary when the body is in the pan a is 7.3 ounces, and when the body is in the basket d, 10 ounces; what is the specific gravity of the body ? SOLUTION. Sp. Gr. = IV -w 16-7.3 W'-w ~ 10-7.3 _ 8.7 _ ~ - *' ' S< 44. Archimedes' principle gives a very easy and accurate method of finding the volume of an irregularly shaped body. Thus, subtract its weight in water from its weight in air ; 18 PHYSICS. divide by .03617 and the result will be the volume in cubic inches; or divide by 62.5 and the result will be the volume in cubic feet. If the specific gravity of the body is known, its cubical contents can be found by dividing" its weight by its specific gravity, and then dividing again by either .03617 or 62.5. EXAMPLE. A certain body has a specific gravity of 4.88 and weighs 76 pounds. How many cubic inches are there in the body ? 76 SOLUTION. = 479.72cu. in. Ans. 4. 38 X- 03617 45. "Since the weight of a cubic foot of water varies for different temperatures, and with the amount of impurities it contains, it is necessary to have some standard when getting the specific gravity. This standard is pure distilled water at its maximum density, which occurs at a temperature of 39. 2 F. At this temperature water weighs 62. 425 pounds per cubic foot ; but for ordinary calculations it is customary to take it as weighing 1,000 ounces, or 62.5 pounds per cubic foot. PROPERTIES OF AIR AND GASES. TENSION OF GASES. 46. The most striking feature concerning gases is that, no matter how small the quan- tity may be, they will always jill the vessels that contain them. If a bladder or football is 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. 7. The force that a gas always exerts when confined in a lim- ited space is called tension. FIG. 7. The word "tension" in this 4 PHYSICS. 19 case means pressure, and is only used in this sense in refer- ence to gases. 47. As water is the most common liquid, so is air the most common gas. It was supposed by the ancients that air was imponderable ; i.e., that it had no weight, and not until the year 1650 was the contrary proved. A cubic inch of air, under ordinary conditions, weighs nearly .31 grain. The ratio of the weight of air to water is about 1 : 774, which means that air is only T ^ as heavy as water. It is a well known fact that if a body is immersed in water and weighs less than the volume of water it displaces, this body will rise, extend partly but of the water, and float. The same is true, to a certain extent, of air. If a vessel made of light material is filled with a gas lighter than air, so that the total weight of both, vessel and gas, is less than the weight of the volume of air that they displace, the vessel will rise ; it is on this principle that balloons are made. 48. Since the air has weight, it is evident that the enormous quantity of air that constitutes the atmosphere must exert a considerable pressure upon the earth. We can easily notice this atmospheric pressure upon our persons. If we place the hand over the open mouth of the receiver of a powerful air pump, we find that the hand is pressed down- wards with great force. The total atmospheric pressure that the human body has to support amounts to several tons, but this pressure is not felt under ordinary circumstances because it is exerted equally in every direction. 49. This pressure of air may easily be proved by taking a long glass tube, closed at one end, and filling it with mer- cury. If the finger is placed over the open end, so as to keep the mercury from running out, and the tube is inverted and placed in a cup of mercury, as shown in Fig. 8, the mercury will fall, then rise, and after a few oscillations will come to rest at a height above the top of the mercury in the cup equal to about 30 inches. This height will always be constant under similar atmospheric conditions. Now, if the 20 PHYSICS. 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 occu- pied by the tube. According to Pascal's law, the pres- sure 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 sur- faces. Therefore the pressure of the atmosphere is transmitted in all directions. There being nothing in the tube except mercury to counter- balance the upward pressure of the air, the mercury falls in the tube until it exerts a downward pressure on the upper surface of the mercury in the cup sufficiently great to coun- terbalance the upward pressure pro- duced by the atmosphere. In order that there shall be equilibrium, the pressure of the air per unit of area on the upper surface of the mercury in the glass must equal the pressure (weight) exerted per unit of area by the mercury inside of the tube. Sup- pose the area of the inside of the tube is 1 square inch; then, since , mercury is 13.6 times as heavy as water, the weight of the mercurial columnis. 03617X13.6X30 = 14.7574 pounds. The actual height of the mercury is a little less than 30 inches, and the actual weight of a cubic inch of distilled water is a little less than .03617 pound. Taking these considerations into account, the average weight of the mercurial column at the level of the sea is 14.69 pounds, or, as it is usually expressed, 14.7 pounds. Since this weight, exerted upon 1 square inch of the liquid in the glass, just produced equilibrium, it is plain FIG. 8. PHYSICS. 21 that the pressure of the outside air is 14.7 pounds upon every square inch of surface. 5O. 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.it con- tained a gas of some kind, 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 ot gas present. The space is then called a partial vacuum. If the mercury falls 1 inch, so that the column is only 29 inches high, we say, in ordinary language, that there are 29 inches of vacuum. If it falls 8 inches, we say that there are 22 inches of vacuum; if it falls 16 inches, we say that there are 14 inches of vacuum ; etc. 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 QQ Og A -X14.7 = ^X14.7 = 1.96 Ib. per sq. in. oU oU In all cases where the mercurial column is used to measure a vacuum, the height of the column in inches gives the number of inches of vacuum. Thus, if the column were 5 inches high, or the vacuum gauge showed 5 inches, the vacuum would be 5 inches. 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 30x13.6 = 408 inches = 34 feet. This means that if a tube were filled with water, inverted, and placed in a dish of water in a man- ner similar to the experiment made with the mer- cury, the resulting height of the column of water FIG. 9. would be 34 feet. 22 PHYSICS. 4 51. The Barometer. The barometer is an instrument for measuring the pressure of the atmosphere. There are two kinds in general use the mercurial and the aneroid barometer. The mercurial barometer is shown in Fig. 9. The principle is the same as in the case of the inverted tube shown in Fig. 8. The tube and cup at the bottom are pro- tected by a brass or iron casing. At the top of the tube is a graduated scale that can be read to TT Vir of an inch by FIG. 10. means of a vernier. Attached to the casing is an accurate thermometer for 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 readings are reduced to this temperature. This standard 4 PHYSICS. 23 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. 52. In Fig. 10 is shown an illustration of an aneroid barometer. These instruments are made in various sizes, from the size of a large watch up to an 8 or 10 inch face. The barometer consists of a cylindrical box of metal, with a top of thin, elastic, corrugated metal. The air is removed from the box. When the atmospheric pressure increases, the top is pressed inwards, and when it is diminished, the top is pressed outwards by its own elasticity, aided by a spring beneath. These movements of the cover are trans- mitted and multiplied by a combination of delicate levers that 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 variations in tempera- ture. They are very portable, occupying but a small space, and are so delicate that they are said to show a difference in the atmospheric pressure when transferred from the table to the floor. They must be handled with care, as they are easily injured. The mercurial barometer is the standard. 53. With air, as with water, the lower we go, the greater the pressure, and the higher we go, 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, 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, it is 8.9 inches. 54. The density also varies with the altitude; that is, a cubic foot of air at an elevation of 5,000 feet above the sea level will not weigh as much as a cubic foot at sea level. This is proved conclusively by the fact that at a height of 3| miles the mercurial column measures but 15 inches, indi- cating that half the weight of the entire atmosphere is below 24 PHYSICS. 4 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 exceedingly rarefied state. It is by means of barometers that great heights are measured. The aneroid barometer has the heights marked on the dial, so that it can be read directly. With the mercurial barometer the heights must be calculated from the reading. 55. 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 is 8 X 5 X 14. 7 = 588 pounds ; but there is an equal pressure beneath the book to counteract the pressure on the top. It would now seem as though it would require a great force to open the book, since there are two pressures of 588 pounds each, acting in opposite directions, and tending to crush the book; so it would, but for the fact that there is a layer of air between each leaf, acting upwards and downwards with a pressure of 14. 7 pounds per square inch. If two metal plates are 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 the air thus excluded, it will take an immense force, compared with the weight 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 is laid upon a flat surface that has been previously moistened with water, it will require consid- erable force to separate them ; this is because the water helps to fill up the pores in the flat surface and in the glass, and thus creates a partial vacuum between the glass and the surface, thereby reducing the counterpressure beneath the glass. 4 PHYSICS. 25 EXPANSION OF GASES. 56. In Fig. 8 the space above the column of mercury was said to be a vacuum, and that if any gas or air was present, it would expand, its tension forcing the column of mercury downwards. If enough gas is admitted to cause the mercury to stand at 15 inches, the tension of the gas is 14 7 evidently - = 7.35 pounds per square inch, since the 2 pressure of the outside air of 14.7 pounds per square inch only balances 15 inches, 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 pressure (tension) of the gas in the tube is 7.35 potinds. If more gas is admitted, until the top of the mercurial column is just level with the mer- cury in the cup, the gas in the tube has then a tension equal to the outside pressure of the atmosphere. Suppose that the bottom of^the tube is fitted with a piston, and that the total length of the inside of the tube is 36 inches. If the piston is shoved upwards, so that the space occupied by the gas is 18 inches long, instead of 36 inches, the temperature remaining the same as before, it will be found that the ten- sion of the gas within the tube is 29.4 pounds per square inch. 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 14.7x2 = 29.4 pounds. If the piston is pushed farther up, so that the space occupied by the gas is only 9 inches, instead of 18 inches, the temperature still remaining the same, the pressure will be found to be 58.8 pounds per square inch. The volume has again been reduced one-half, and the pressure increased 2 times, since 29.4x2 = 58.8 pounds. The space now occu- pied 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 length, the volume is now -^g- = \ of its original volume, and its pressure is 58 8 _l_ = 4 times its original pressure. Moreover, if the tem- perature of the confined gas remains the same, the pressure 26 PHYvSICS. 4 and volume will always vary in a similar way. The law that states these effects is called Mariotte's /aw, and is as follows : 57. Marietta's Law. The temperature remaining the same, tJie volume of a given quantity of gas varies inversely as the pressure. The meaning of this is: If the volume of the gas is dimin- ished to i, i, |, etc. of its former volume, the tension will be increased 2, 3, 5, etc. times ; or, if the outside pressure is increased 2, 3, 5, etc. times, the volume of the gas will be diminished to ^, ^, J, etc. of its original volume, the tem- perature remaining constant. It also means that, if a gas is under a certain pressure, and the pressure is diminished to i, i, yL-, etc. of its original pressure, the volume of the con- fined gas will be increased 2, 3, 10, etc. times its tension decreasing at the same rate. Suppose 3 cubic feet of air to be under a pressure of 60 pounds per square inch in a cylinder fitted with a mov- able piston ; then, the product of the volume and pressure is 3 X 60 = 180. Let the volume be increased to 6 cubic feet, then the pressure will be 30 pounds per square inch, and 30x6 = 180, as before. Let the volume be increased to 24 cubic feet, it is then - 2 ^- 8 times its original volume, and the pressure is \ of its original pressure, or 60 X \ = 7^- pounds, and24x7|- = 180, as in the two preceding cases. It will now be noticed that if a gas is enclosed within a confined space, and allowed to expand without losing any heat, the product of the pressure and the corresponding volume for one position of tlie piston is the same as for any other position of the piston. If the piston were to compress the air, the same result would be obtained. Let p = pressiire for one position of the piston ; p l = pressure for any other position of the piston ; v = volume corresponding to the pressure/; 7'j = volume corresponding to the pressure p^. Then, /*> =/,',- (?) Knowing the volume and the pressure for any position of the piston and the volume for any other position, the pressure 4 PHYSICS. 27 may be calculated, or if the pressure is known for any other position, the volume may be calculated. EXAMPLE i. If 1.875 cubic feet of air be under a pressure of 12 pounds per square inch, what will be the pressure when the volume is increased (a) to 2 cubic feet ? (b] to 3 cubic feet ? (c] to 9 cubic feet ? SOLUTION. Solving formula 7 for/i, the unknown pressure, pv 72x1-875 (a) p l *- - ^-= -- = 67| Ib. per sq. in. Ans. 79 V 1 87^ (b) p, = - ^^ - = 45 Ib. per sq. in. Ans. (c) pi = 72x *' 875 = 15 ib. per sq. in. Ans. EXAMPLE 2. 10 cubic feet of air have a tension of 5.6 pounds per square inch ; what is the volume when the tension is (a) 4 pounds ? (b) 8 pounds ? (c) 25 pounds ? (d) 100 pounds ? SOLUTION. Solving formula 7 for v lt (a) y .= = 5 - 6 * 10 = 14cu.ft (b) Vi = 5 - 6 * 10 = 7 cu. ft. Ans. (c) Vi = 5 ' 6 * 10 = 2.24cu. ft. Ans. 25 (d) vt = 5 ' 6 1 ^ ) 1 =".66 cu. ft. Ans. 58. There are two ways of measuring the pressure of a gas: by means of an instrument called a manometer, and by means of a gauge. The manometer generally used is practi- cally the same as a mercurial barometer, except that the tube is much longer, so that pressures equal to several atmos- pheres may be measured, and is enlarged and bent into a U shape at the lower end ; both lower and upper ends are open, the lower end being connected to the vessel containing the gas, whose pressure it is desired to measure. The gauge is so common that no description of it will be given here. With both the manometer described above and the gauge, the pressures recorded are the amounts by which they exceed the atmospheric pressure, and are called the gauge pressures. To find the real pressure, called the absolute pressure, the 28 PHYSICS. 4 atmospheric pressure must be added to the gauge pressure. In all formulas in which the pressure of a gas or steam is used, the absolute pressure must be used, unless the gauge pressure is distinctly specified as being the proper pressure to use. For convenience, all pressures given in this Instruc- tion Paper and in the accompanying Question Paper will be absolute pressures, and the word absolute will be omitted to avoid its constant repetition. 59. As a necessary consequence of Mariotte's law, it may be stated that the density of a gas varies directly as the pressure, and inversely as the volume ; that is, the density increases as the pressure increases, and decreases as the vol- ume increases This is evident, since, if a gas has a tension of 2 atmos- pheres, or 14.7x2 = 29.4 pounds per square inch, it will weigh twice as much as the same volume would if the ten- sion were 1 atmosphere, or 14.7 pounds per square inch. For, let the volume be increased until it is twice as great as the original volume, the tension will then be 1 atmosphere. The total weight of the gas has not been changed, but there are now 2 cubic feet for every cubic foot of the original vol- ume, and the weight of 1 cubic foot now is only half as great as before. Thus, the density decreases as the volume increases, and as an increase of pressure causes a decrease of volume, the density increases as the pressure increases. Let D density corresponding to the pressure / and volume v; D 1 density corresponding to the pressure / 1 and volume z> r Then, / : D = p l : D^ or/>, = /, D, (8.) and v:D^ = v^.D, or v D = *;,/?,. (9.) 60. Since the weight is proportional to the density, the weights may be used in place of the densities in formulas 8 and 9. 4 PHYSICS. 29 Let W = weight of a cubic foot of air or other gas, whose volume is v and pressure is / ; W^ = weight of a cubic foot when the volume is v l and pressure is^. Then, pW l =p l W. . (1C.) vW = v^W v . (11.) EXAMPLE 1. The weight of 1 cubic foot of air at a temperature of 60 F., and under a pressure of 1 atmosphere (14.7 pounds per square inch), is .0763 pound; what will be the weight per cubic foot if the air is compressed until the tension is 5 atmospheres, the temperature still being 60 F. ? SOLUTION. Applying formula 1O, p Wi = pi W, or IX^i = 5 X- 0763. Hence, W l = .3815 Ib. per cu. ft. Ans. EXAMPLE 2. If in the last example the air had expanded until the tension was 5 pounds per square inch, what would have been its weight per cubic foot ? SOLUTION. Applying formula 1O, p Wi pi W. Here,/ = 14.7, /, = 5, and W = .0763. Hence, 14.7 X W^ = 5 X .0763, = .02595 Ib. per cu. ft. Ans. EXAMPLE 3. If 6.75 cubic feet of air at a temperature of 60 F., and a pressure of 1 atmosphere, are compressed to 2.25 cubic feet (the tem- perature still remaining 60 F.), what is the weight of a cubic foot of the compressed air ? SOLUTION.' Applying formula 11, v W V! W lt or 6.75 X .0763 = 2.25 X Wi. Hence, W, = 6 ' 75 X ' 768 = .2289 Ib. per cu. ft. Ans. 61. In all that has been said before, it has been stated that the temperature was constant; the reason for this will now be explained. Suppose 5 cubic feet of air to be confined in a cylinder placed in a vacuum, so that there will be no pressure due to the atmosphere, and suppose the cylinder to be fitted with a piston weighing, say, 100 pounds and having an area of 10 square inches. The tension of the gas will be iff. 10 pounds per square inch. Suppose that the tem- perature of the air is 32 F., and that it is heated until the 30 PHYSICS. 4 temperature is 33 F., i. e., until the temperature is increased 1, it will be found that the piston has risen a certain amount, and, consequently, the volume has increased, while the pres- sure is the same as before, or 10 pounds per square inch. If more heat is applied, until the temperature of the gas is 34 F., it will be found that the piston has again risen, and the volume again increased, while the pressure still remains the same. It will be found that for every increase of temperature there will be a corresponding increase of volume. The law that expresses this change is called Gay- Lussac's /aw, and is as follows: 62. Gay-Lussac's Law. If the pressure remains con- stant, every increase of temperature of 1 F. produces in a given quantity of gas an expansion of -^-^ of its volume at 32 F. If the pressure remains constant, it will also be found that every decrease of temperature of 1 F. will cause a decrease of ^-^ f the volume at 32 F. Let v = original volume of gas; v t final volume of gas ; t = f temperature corresponding to volume v\ t v = temperature corresponding to volume v^ <> That is, the volume of gas after heating (or cooling] is equal to the original volume multiplied by 460 plus the final temper- ature, divided by 1+60 plus the original temperature. .EXAMPLE. 5 cubic feet of air at a temperature of 45 are heated under constant pressure up to 177 ; what is its final volume ? SOLUTION. Applying formula 13, \ K /460 + 177\ ) = 5 ( 460T45 ) = 6 ' 307 CU " ft ' AnS ' 63. Suppose a certain volume of gas to be confined in a vessel so that it cannot expand ; in other words, suppose the piston of the cylinder before mentioned to be so fastened that it cannot move. Let a gauge be placed on the cylinder so that the tension of the confined gas can be registered. If 4 PHYSICS. 31 the gas is heated, it will be found that for every increase of temperature of 1 F. there will be a corresponding increase of ^3- of the tension. That is, the volume remaining con- stant, the tension increases ^^ f tne original tension for every degree rise of temperature. Let p = original tension; t = corresponding temperature; /j = final tension; /, = final temperature. That is, if a certain quantity of gas be heated (or cooled) from t to /,, the volume remaining constant, the resulting tension p l will be equal to the original tension multiplied by JfiO plus the final temperature, divided by 4.60 plus the original temperature. EXAMPLE. If a certain quantity of air is heated under constant vol- ume from 45 to 177, what is the resulting tension, the original tension being 14.7 pounds per square inch ? SOLUTION. Applying formula 13, 460T45 64. Absolute Temperature. According to the modern and now generally accepted theory of heat, the atoms and molecules of all bodies are in an incessant state of vibration. The vibratory movement in the liquids is faster than in the solids, and in the gases, faster than in either of the other two. Any increase of heat increases the vibrations, and a decrease of heat decreases them. From experiments and calculations, it has been concluded that at 460 below zero on the Fahren- heit scale all these vibrations cease. This temperature is called the absolute zero, and all temperatures reckoned from this point are called the absolute temperatures. The point of absolute zero has never been reached, the lowest recorded temperature being about 393 F. below zero, but, neverthe- less, it has a meaning, and is used in many formulas, being nearly always denoted by T. The ordinary temperatures 32 PHYSICS. 4 are denoted by /. When the word temperature alone is used, the meaning" is the same as ordinarily used, but when absolute temperature is specified, 460 F. must be added to the ordinary temperature. The absolute temperature cor- responding to 212 F. is 460 + 212 = 072 F. If the abso- lute temperature is given, the ordinary temperature may be found by subtracting 460 from the absolute temperature. For example, the absolute temperature being 520 F., the ordinary temperature is 520 - 460 = 60. 65. The relation between the pressure, volume, and temperature of air is given by the following formula: Let p = pressure of air in pounds per square inch ; v = volume of 1 pound of air in cubic feet; V = volume of W pounds of air in cubic feet ; T absolute temperature; W weight of air in pounds. Then, pv .37052 T. (14.) That is, the pressure in pounds per square inch, multiplied by the volume of a pound of air in cubic feet, is equal to .37052 times the absolute temperature corresponding to the pressure p and volume v. In this formula the weight of the air is 1 pound. EXAMPLE 1. The pressure upon 9 cubic feet of air weighing 1 pound is 20 pounds per square inch ; what is the temperature ? SOLUTION. Applying formula 14, p v = .37052 T, or 20 X 9 = .37052 T. 180 Hence, T = " = 485.8, nearly. . o i UOrO 485.8 460 = 25.8, the temperature. Ans. EXAMPLE 2. What is the volume of 1 pound of air whose tempera- ture is 60 F. under a pressure of 1 atmosphere ? SOLUTION. Applying formula 14, p v = .37052 T, and substituting, 14.7 X v = .37052 X (460 + 60) = .37052 X 520, .37052X520 ,. A or it = ^-~ = 13.107cu. ft. Ans. 66. If the weight of the air is greater or less than 1 pound, the following formula must be used: pV = .37052 WT. (15.) 4 PHYSICS. 33 That is, tJic pressure in pounds per square inch multiplied by tJie volume in eubic feet is equal to .3705% times the weight in pounds multiplied by the absolute temperature. EXAMPLE 1. 3 cubic feet of air weighing .35 pound are under a pressure of 48 pounds per square inch ; what is the temperature of the air ? SOLUTION. Applying formula 15, / V .37052 W T, and substi- 48X3 = . 37052 X. 35 X T, - = '"o- 4 - Then, 1,110.4 -460 = 650.4. Ans. EXAMPLE 2. What is the weight of 1 cubic foot of air at a tempera- ture of 32 , and under a pressure of 1 atmosphere ? SOLUTION. Applying formula 15, p V = .37052 W 7 7", and substi- 14.7 X 1 = -37052 X (460 + 32) X W, If .the pressure be taken as 14.69856 pounds per square inch, and the absolute zero as 459.4 instead of 460 below zero, and if .370514 be used instead of .37052 (more exact values), the weight of 1 cubic foot is =. 08073 ,.. nearly. EXAMPLE 3. What is the exact volume of 1 pound of air at a tem- perature of 32, and at a pressure of 1 atmosphere? Take absolute zero as 459.4, and the pressure as 14.69856 pounds per square inch. SOLUTION. p V = .370514 W T, or 14.69856 X V = .370514 X 1 X (459.4 + 32). 67. If in the formula p V . 37052 W T, both sides of the equation be divided by T (which, of course, does not alter pV the equality), there results the expression *-=- = . 37052 W. Let /,, Fj, and 7 1 , represent the pressure, volume, and temperature of the same weight of air in another state; then, /,F t = .37052 WT X . Dividing both sides by 7 1 ,, - = .37052 W. Therefore, since and are equal 34 PHYSICS. 4 to the same thing (i. e., . 37052 W), they are equal to each other, and , y p y f=f. (16.) This very important formula is the complete expression of Gay-Lussac's law, and is true for any of .the so called perma- nent gases. It was from this formula that formulas 13 and 13. were derived. Thus, let the pressure be constant; then, larly, letting the volume be constant, V = F,, and ^~- = OFT, or A = ^ = P (^^7)- So > also > b y letting the tem- perature be constant, T 7\ and ^~- = * ', or / V /, F,, which is the same as formula 7. In formulas 7, 16, 17, and 18, it matters not with what units the pressures and volumes are measured, except that they must be the same throughout the same example, and the pressures must always be absolute pressures. EXAMPLES FOR PRACTICE. 68. Solve the following: 1. A vessel contains 25 cubic feet of gas at a pressure of 18 pounds per square inch ; if 125 cubic feet of gas having the same pressure are forced into the vessel, what will be the resulting pressure ? Ans. 108 Ib. per sq. in. 2. A pound of air has a temperature of 126 and a pressure of 1 atmosphere; what volume does it occupy ? Ans. 14.77 cu. ft. . 3. A certain quantity of air has a volume of 26.7 cubic feet, a pres- sure of 19.3 pounds per square inch, and a temperature of 42 ; what is the weight ? Ans. 2.77 Ib. 4. A receiver contains 180 cubic feet of gas at a pressure of 20 pounds per square inch ; if a vessel holding 12 cubic feet be filled from the receiver until its pressure is 20 pounds per square inch, what will be the pressure in the receiver ? Ans. 18 Ib. per sq. in. 5. If 10 cubic feet of air having a pressure of 22 pounds per square inch and a temperature of 75 are heated until the temperature is 300, the volume remaining the same, what is the new pressure ? Ans. 31.251b. per sq. in. 4 PHYSICS. 35 MIXTURES OF GASES. 69. If two liquids that do not act chemically on each other are mixed together and allowed to stand, it will be found that after a time the two liquids have separated, and that the heavier has fallen to the bottom. If two vessels of equal capacities containing gases of different densities are put in communication with each other, the gases will be found to have mixed in equal proportions after a short time. If one vessel is higher than the other, and the heavier gas is in the lower vessel, the same result will occur. The greater the difference of the densities of the two gases, the quicker they will mix. It is assumed that no chemical action takes place between the two gases. When the two gases have the same temperature and pressure, the pressure of the mixture will be the same ; this is evident, since the total volume has not been changed, and unless the volume or temperature changes, the pressure cannot change. This property of the mixing of gases is a very valuable one, since, if they acted like liquids, carbonic-acid gas (the result of combustion), which is 1^ times as heavy as air, would remain next to the earth instead of dispersing into the atmosphere, the result being that no animal life could exist. 70. Mixture of Equal Volumes of Gases Having Unequal Pressures. If two gases Jiaving equal volumes and temperatures, but different pressures, are mixed in a vessel whose volume is equal to one of the equal volumes of the gas, the pressure of the mixture will be equal to the sum of the two pressures, provided that the temperature remains the same as before. EXAMPLE. Two vessels containing 3 cubic feet of gas, each at a temperature of 60 and subjected to pressures of 40 pounds and 25 pounds per square inch, respectively, are placed in communication with each other, and all the gas is compressed into one vessel. If the temperature of the mixture is also 60, what is the pressure ? SOLUTION. According to the rule just given, the pressure will be 40 + 25 = 65 pounds per square inch. This may be proved by appli- cations of Mariotte's law ; thus, compress the gas whose pressure is 25 pounds per square inch, until its pressure is 40 pounds ; its volume 36 PHYvSICS. 4 may be found thus: p v = p\Vi, or 25 X 3 = 40 X ^ ; whence, v = 1.875 cubic feet. Let communication be established between the two ves- sels. The pressure will evidently be 40 pounds, and the total volume 3 + 1.875 = 4.875 cubic feet. If this be compressed until the volume is 3 cubic feet, the temperature remaining at 60 throughout the whole operation, the final pressure may be found by formula 7, p v p\ VL Thus, 40 X 4.875 = p l X 3, and/! = - X4.875 = 65 pounds per square inch, as before. 71. Mixture of Two Gases HaAdng Unequal Vol- umes and Pressures. Let v and p = volume and pressure, respectively, of one of the gases; v l and p l = volume and pressure, respectively, of the other gas; Fand P = volume and pressure, respectively, of the mixture. Then, if the temperature remains the same, VP=vp + vJr (17.) That is, if the temperature is constant, the volume after mixture, multiplied by the resulting pressure, is equal to the volume of one gas before mixture multiplied by its pressure, plus the volume of the other gas multiplied by its pressure. EXAMPLE. Two gases of the same temperature, having volumes of 7 cubic feet and 4| cubic feet, and whose pressures are 27 pounds and 18 pounds per square inch, respectively, are mixed together in a vessel whose volume is 10 cubic feet. The temperature of the two gases and of the mixture being 60 F. , what is the resulting pressure ? SOLUTION. Applying formula 17, PV = Hence, P = = 27 Ib. per sq. in. Ans. 7 2 . Mixtu re of Two Vol times of Air Having Unequal Pressures, Volumes, and Temperatures. If a body of air having a temperature f lt a pressure /,, and a volume v^ is mixed with another volume of air having a temperature / 2 , a pressure / a , and a volume z> 2 , and the mixture has a volume V, pressure P, and a temperature t, then, either the new temperature /, the new volume V, or the new pressure P may be found, if the other two quantities are known, by the 4 PHYSICS. 37 following 1 formula, in which T^ T^ and T are the absolute temperatures corresponding to / J5 / 2 , and /: EXAMPLE. 5 cubic feet of air having a tension of 30 pounds per square inch, and a temperature of 80 F. , are required to be compressed together with 11 cubic feet of air having a tension of 21 pounds per square inch, and a temperature of 45 F., in a vessel whose cubical contents are 8 cubic feet. The new pressure is required to be 45 pounds per square inch. What is the temperature of the mixture ? SOLUTION. Substituting in formula 18, Hence, T = -_ = 489.66, nearly, and / = 29.66. Ans. EXAMPLES FOR PRACTICE. 73. Solve the following: 1. Two vessels contain air at pressures of 60 and 83 pounds per square inch, respectively. The volume of each vessel is 8.47 cubic feet. If all of the air in both vessels is removed to another vessel, and the new pressure is 100 pounds per square inch, what is the volume of the vessel, the temperature being the same throughout ? Ans. 12.11 cu. ft. 2. A vessel contains 11.83 cubic feet of air at a pressure of 33.3 pounds per square inch. It is desired to increase the pressure to 40 pounds per square inch by supplying air from a second vessel that contains 19.6 cubic feet of air at a pressure of 60 pounds per square inch. What will be the pressure in the second vessel after the pressure in the first has been raised to 40 pounds per square inch ? Ans. 55.96 Ib. per sq. in. 3. If 4.8 cubic feet of air having a tension of 52 pounds per square inch and a temperature of 170 are mixed with 13 cubic feet of air hav- ing a tension of 78 pounds per square inch and a temperature of 265, what must be the volume of the vessel containing the mixture, in order that the tension of the mixture may be 30 pounds per square inch and the temperature 80 ? Ans. 32.31 cu. ft. THE AIR PUMP. 74. The air pump is an instrument for removing air from an enclosed space. A section of the principal parts is shown in Fig. 11, and the complete instrument in Fig. 12. 38 PHYSICS. The closed vessel R is called the receiver, and the space enclosed by it is that from which it is desired to remove the air. FIG. 11. The receiver is usually made of glass, and the edges are ground so as to be perfectly air-tight. When made in the form shown, it is called a bell-jar receiver. The receiver rests upon a horizontal plate, in the center of which is an opening communica- ting with the pump cylinder C, by means of a bent tube /. The pump piston fits the cylinder accurately, and has a valve V opening upwards. At the junc- tion of the tube with the cylinder is another valve V also opening upwards. When the FIG. 12. 4 PHYSICS. 39 piston is raised the valve V closes, and, since no air can get into the cylinder from above, the piston leaves a vacuum behind it. The pressure on top of V being now removed, the tension of the air in the receiver R causes V to rise ; the air in the receiver then expands and occupies the space dis- placed by the piston, the space in the tube / and in the receiver R. The piston is now pushed down, the valve V closes, the valve V opens, and the air in C escapes. The lower valve Fis sometimes supported, as shown in Fig. 11, by a metal rod passing through the piston and fitting it somewhat tightly. When the piston is raised or lowered, this rod moves with it. A button near the upper end of the rod confines its motion to within very narrow limits, the piston sliding upon the rod during the greater part of the journey. 75. Degrees and limits of Exhaustion. Suppose that the volume of R and / together is four times that of C, and that there are, say, 200 grains of air in R and t, 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 spaces R, t, and C. The ratio between the sum of the spaces R and /, and the total space R + tC is f; hence, 200 Xf = 160 grains = the weight of air in R and / after the first stroke. After the second stroke, the weight of the air in R and / would be 200 X f X | = 200 X (4 ) 2 = 200 X if = 128 grains. At the end of the third stroke the weight would be [200 X (|) 2 ] X f- = 200 X (i) 3 = 200 X Jfo = 102. 4 grains. At the end of n strokes the weight would be 200 X (I)". It is evident that it is impossible to remove all of the air that is contained in R and t by this method. It requires an exceed- ingly good air pump to reduce the tension of the air in R to ^5- of an inch of mercury. When the air has become rarefied to this extent, the valve V will not lift, and, consequently, no more air can be exhausted. 76. Sprengel's Air Pump. In Fig. 13, c d is a glass tube longer than 30 inches, open at both ends, and connected, 40 PHYSICS. by means of india-rubber tubing, with a funnel A filled with mercury and supported by a stand. Mercury is allowed to fall into this tube at a rate regulated by a clamp at c. The lower end of the tube c d fits in the flask B, which has a spout at the side a little higher than the lower end of c d ; the upper part has a branch at x to which a receiver R can be tightly fixed. When the clamp at c is opened, the first por- tions of the mercury that run out close the tube and prevent air from entering from below. These drops of mercury act like little pistons, carrying the air in front of them and forcing it out through the bottom of the tube. The air in R expands to fill the tube every time that a drop of mercury falls, thus crea- ting a partial vacuum at R, which becomes more nearly complete as the process goes on. The escaping mercury falls into the dish //, from which it can be FIG - 13 - poured back into the funnel from time to time. As the exhaustion from R goes on, the mercury rises in the tube c d until, when the exhaustion is complete, it forms a continuous column 30 inches high; in other words, it is a barometer whose Torricellian vacuum is NOTE. A theoretically perfect vacuum is sometimes called a Torri- cellian vacuum. 4 PHYSICvS. 41 the receiver R. This instrument necessarily requires a great deal of time for its operation, but the results are very com- plete, a vacuum of ^--J-o-g- of an inch of mercury being sometimes obtained. By the use of chemicals in addition to the above, a vacuum of -g^-oVon f an i nc ^ f mercury has been obtained. HEAT. NATURE OF HEAT. 77. As to the exact nature of heat, scientists differ, but all modern thinkers and investigators agree that heat is a form of energy, and that it is a kind of motion. It is not our pur- pose to enter into the different theories regarding heat, but as much of the generally accepted theory will be given as will be necessary to make clear the principles that are to follow. 78. In Art. 3 it was stated that bodies are composed of molecules. Notwithstanding the extreme minuteness of the molecules, they play a very important part in the modern theory of heat. Each molecule attracts the molecules sur- rounding it in a manner similar to the attraction between earth and bodies near its surface, only with an immensely greater force in proportion to their sizes. Without going into any theory regarding the precise nature of heat, it will be taken for granted that each and every molecule has a rapid vibratory motion to and fro, and that the molecules are kept from getting beyond a certain distance from one another by the attractive force between them. This attractive force is called cohesion; without it, everything throughout the universe would instantly crumble into the finest dust. In Art. 15 it was stated that the molecules are supposed to be round ; it is likewise supposed that they are at a con- siderable distance apart, compared with their diameters. When heat is applied to a body, the number of the vibrations 42 PHYSICS. 4 is increased proportionally to the amount of heat supplied. In consequence of this increase, the distance through which a molecule moves is increased, and the force of cohesion that binds the molecules together is lessened. If enough heat is added to a solid, the force of cohesion is so far overcome that the body melts. If heat is supplied in sufficient quantity, the melted body becomes a vapor, and so long as it is kept at this temperature the force of cohesion has no effect, in con- sequence of the number of vibrations having been so far increased and the distance between any two molecules hav- ing become too great for the force of cohesion to act. If the vapor is cooled, the number of vibrations and also the dis- tance between any two molecules will decrease ; the force of cohesion begins to act and the body becomes a liquid. If cooled further and a sufficient quantity of heat is removed (in other words, if the number of vibrations is so far decreased that the molecules are comparatively near together), the body becomes a solid and remains so until the temperature is again increased to the melting point. 79. If a body is heated and brought near the hand, the sensation of warmth is felt; if heat is removed from this same body and it is again brought near the hand, the sensa- tion of cold is felt. The heat that thus manifests itself is called sensible heat, because any change from any state to a hotter or colder state is indicated at once by the sense of feeling. The more sensible heat a body possesses, the hotter it is ; the more sensible heat is taken away from it, the colder it is. TEMPERATURE. 8O. The most common heat measure we have is that which we gain by means of the sensation of warmth it produces. According to the character of this sensation a body is said to be cold, warm, or hot. These terms all refer to the power that one body has of communicating heat to other bodies. The measure of this power is termed 4 PHYSICS. 43 temperature, which may be more exactly embodied in the following" definition: The temperature of a body is a measure of tJie intensity of its licat, and is further defined as the thermal state of a body considered with reference to its power of communicating heat to other bodies. 8 1 . For scientific purposes, the sensations are not suffi- ciently accurate methods of measuring temperature ; accord- ingly, temperature is usually measured by certain of the effects produced by heat. Among these, one of the most convenient is, that most bodies expand with a rise of tem- perature. This expansion is distinctly perceptible in solids; it occurs to a greater extent in liquids, and most of all with gases. For the general purposes of temperature measure- ment, mercury and alcohol are the most convenient sub- stances, the former because it boils only at a very high temperature, and the latter because it does not solidify at the greatest known cold produced by ordinary means. One of these liquids, enclosed in a suitable vessel, constitutes the temperature-measuring instrument termed a thermometer. 82. In constructing 1 a thermometer, a bulb is blown at one end of a glass tube of a very narrow bore ; the bulb and a portion of this tube are next filled with carefully purified mercury; this is boiled, and thus all air and moisture is driven out of the tube; the open end is then hermetically sealed by fusing" the glass itself. The bulb and a portion of the tube are thus filled with mercury and the remainder of the tube is a vacuum, save for the presence of a minute quantity of mercury vapor. On heating the bulb of this instrument the mercury expands and rises considerably in the stem. Heat has a tendency to so distribute itself throughout any body or series of bodies as to make them of the same temperature; consequently, if the thermometer is placed in contact with the body whose temperature it is desired to measure, a redistribution of heat occurs until the two are at the same temperature. That is to say, if the body is the colder it receives heat from the thermometer, and if 44 PHYSICS. it is hotter it yields heat to the thermometer, until the tem- perature of the two is the same. The two being in efficient contact, this stage is indicated by the mercury becoming stationary in the thermometer tube. Now the mercury is constant for any one temperature; therefore, to register ^==^ temperature, it is only necessary to have, further, a scale or series of graduations attached to the stem of the instrument, by which the temperature may always be read. 83. In Fig. 14 is shown a mercurial thermom- eter with two sets of graduations on it. The one on the left, marked F, is the Fahrenheit ther- mometer, so named after its inventor, and is the one commonly in use in this country and in Eng- land; the one on the right, marked C, is the centigrade thermometer, and is used by scientists throughout the world on account of the graduations being better adapted for calculations. 84. In graduating thermometers two fixed points of temperature are almost universally em- ployed. These are the temperatures of melting ice, and of the steam from boiling water. Certain simple precautions being taken, these temperatures are always constant. It is, in addition, necessary to graduate the thermometers, so as to register temperatures intermediate between these two points, and also below and above them. The most convenient system of graduation is that of Celsius, known as the centigrade scale. In this scale the distance between the melting point of ice (or the freezing point of water, as it is commonly called) and the temperature of steam at atmospheric pressure is divided into 100 equal graduations or degrees. (This latter tem- perature is more commonly described as being that of the boiling point of water. ) The freezing point is called zero, or 0, and the boiling point 100. Degrees of the same FIG. 14. 4 PHYSICS. 45 value are carried above and below the boiling and freezing- points, respectively; the temperature below is considered negative, and is counted downwards from zero; thus, 10 below freezing point is 10, and so on. Degrees above the boiling point are simply counted upwards from zero; thus, 10 above the boiling point is 110, and so on. 85. In graduating Fahrenheit thermometers, the distance between the freezing and boiling points is divided into 180 equal parts, and degrees of the same size are carried above and below the boiling and freezing points. Fahrenheit assumed that the greatest cold attainable was 32 below the freezing point, and accordingly took that point as his zero and reckoned from it upwards. The freezing point thus became 32 F. and the boiling point 32 -f 180 = 212. Degrees below the Fahrenheit zero are reckoned downwards from zero and are considered as negative. In Germany and Russia temperature is reckoned on the Reaumur scale, in which the freezing point is and the boiling point is 80. 86. Of these three thermometers, the centigrade is used the most, but since the Fahrenheit instrument is the one in general use in this country, all temperatures in this section will be understood to be in Fahrenheit degrees, unless other- wise stated. 87. It is frequently necessary to be able to compare these scales, and to translate temperatures from any one into another. For example, what would 80 C. be on the Fah- renheit scale ? Since the number of degrees between the freezing point and the boiling point on the centigrade scale is 100, and on the Fahrenheit 180, it is evident that, if F = number of degrees Fahrenheit, and C number of degrees centigrade, F : C :: 180 : 100, or F = } C = f C. C 100/T 5 P u -- r -- r. 46 PHYSICS. 4 Therefore, to change centigrade temperatures into their corresponding Fahrenheit values: Rule. Multiply the temperature, centigrade, by \ and add 32 ; the result ivill be the temperature, Fahrenheit. To change Fahrenheit temperatures into their correspond- ing centigrade values: Rule. Subtract 32 from the temperature, Fahrenheit, multiply by f, and the result ivill be the temperature, centi- grade. Expressing these two rules by means of formulas, Let /,. = temperature centigrade; t f = temperature Fahrenheit. Then, t f = /, + 32, (19.) and 4. = -(/>- 32). (2O.) EXAMPLE 1. Change (a) 100' C., (b) 4 C., and (c) -40 C. into Fah- renheit temperatures. SOLUTION. (a) t f = f 4 + 32 = f XlOO + 32 = 212 F. Ans. (b) t f = |x4 + 32 = 39.2 F. Ans. (c) t f = f X -40 + 32 = -40 F. Ans. EXAMPLE 2. Change (a) 60 F., (b) 32 F., and (c) 20 F. into their corresponding centigrade temperatures. SOLUTION. (a) t c = (/ r -32)| = (60-32)f = 15| C. Ans. (b) 4 = (32-32)f = C. Ans. (c) 4 = (- 20 - 32)f = - 28f C. Ans. 88. The absolute temperature is the temperature meas- ured above the point of absolute zero (see Art. 64). Hence, on the Fahrenheit scale the absolute temperature T is 460 -{- 1 when t the ordinary temperature, and is above zero. If t is below zero, its value is negative, and the absolute temperature Tis 460 -j-( 1) = 460 t. Throughout this section, where temperatures are men- tioned, t will denote the ordinary temperature indicated by the thermometer, and T, the absolute temperature. EXAMPLE. What are the absolute temperatures corresponding to the Fahrenheit temperatures 212, 32, and 39.2? 4 PHYSICS. 47 SOLUTION. 460 + 212 = T = 672 F. Ans. 460 + 32 = T = 492 F. Ans. 460 -39.2 = T = 420.8 F. Ans. The absolute temperature on the centigrade scale is T = 273i + / when t is above zero, or T = 273^ - 1 when t is below zero. EXAMPLE. What are the absolute temperatures corresponding to 100 C., 4 C., and 40 C.? SOLUTION. 2731 +100 = T =-. 373i C. Ans. 2731 + 4 = T = 2771 C. Ans. 2731 - 40 = T = 2331 C. Ans. EXAMPLES FOR PRACTICE. 89. 1. What are the absolute temperatures corresponding to (a) 120 R.? (b) 120 C.? (c) 120 F.? f (a} 338 j- R> Ans. I (6) 393|C. [ (c) 580 F. 2. Change 10 R. to the corresponding Fahrenheit and centigrade readings. Ans. 91 F. ; - 12| C. OF BODIES BY HEAT. 90. The volume of any body solid, liquid, or gaseous is always changed if the temperature is changed; nearly all bodies expand whin heated, and contract when cooled. In solids having definite figures, the expansion may be consid- ered in three ways, according to the conditions. 1. The expansion in one direction, as the elongation of an iron bar; this is called linear expansion. 2. Surface expansion, where the area is increased. 3. Cubical expansion, where the increase in the whole volume is considered. 91. In Fig. 15 is shown an apparatus for exhibiting the linear expansion of a solid body. A metal rod A is fixed at one end by a screw B, the other end passing freely through 48 PHYSICS. the eye 6", held in the post, and pressing against the short arm of the indicator F. The rod is heated as shown, and its elongation causes the indicator to move along the arc D E. An illustration of surface expansion is afforded nearly every day in machine shops, particularly in locomotive FIG. 15. shops, where piston rods, crankpins, etc. are shrunk in, and tires are shrunk on their centers. In shrinking a tire on, it is bored a little smaller than the wheel center. The tire is then heated until the area of its circumference is expanded enough to allow it to slide over the wheel center. On being FIG. 16. cooled with water, the tire contracts, and tends to regain its original area, but is prevented by reason of the wheel center being a trifle larger. This causes the tire to hug the center with immense force, and prevents it from coming off. 4 PHYSICS. 49 Cubical expansion may be illustrated by means of a Grave- sandes' ring. This consists of a brass ball , Fig. 16, which at ordinary temperatures passes freely through the ring m, of very nearly the same diameter. When the ball is heated it expands so much that it will no longer pass through the ring. The expansion of liquids is clearly shown in the mercurial and alcohol thermometers. The expansion of gases was treated to some extent in Arts. 56 to 72. 92. Coefficient of Expansion. Suppose that the tem- perature of the metal rod shown in Fig. 15 was 32 F. before heating, and that its length at that temperature is exactly 10 feet. It is found that when the temperature is raised 1, or to 33, the bar is 10 f eet -|- TYOT mcn l n g- The linear expansion is then (10 feet + TTTO- inch) 10 feet = y^V o" inch, and the ratio between this expansion and the original length of the bar is : 10X12 = .. * .,_ = .000006944. 1,200 1,200X120 For every increase of temperature of 1, this rod elongates .000006944 of its length. This number .000006944, which is equal to the expansion of the rod for 1 rise of temperature divided by the original length, is called the coefficient of linear expansion. Had the temperature of the rod been increased 100 instead of 1, the amount of elongation would have been .000006944x100 = .0006944 of its length, or .0006944X120 = .083328 inch, or T V inch. 93. The table following contains the coefficients of expansion for a number of solids, mercury, and alcohol, and the average cubical expansion of gases. No liquids are given except mercury and alcohol, for the reason that the coeffi- cient of expansion for liquids is different at different tem- peratures. 50 PHYSICS. TABLE 2. TABLE OF COEFFICIENTS OF EXPANSION. Name of Substance. Linear Expansion. Surface Expansion. Cubical Expansion. Cast iron .00000617 .00001234 00001 8.50 Copper 00000955 00001910 00002864 Brass 00001037 00002074 00003112 Silver 00000690 00001390 00002070 Bar iron .00000686 00001372 00002058 Steel (tin tempered) . . . Steel (tempered) .... .00000599 . 00000702 .00001198 .00001404 .00001798 .00002106 Zinc 00001634 . 00003268 .00004903 Tin 00001410 00002820 00003229 Mercury Alcohol .00003334 .00019259 .00006668 .00038518 .00010010 .00057778 Gases .00203252 94. Let L = length of any body ; / = amount of expansion or contraction due to heating or cooling the body ; A area of any section of the body; a increase or decrease of area of the same section after heating or cooling the body ; V volume of the body ; v increase or decrease in volume due to heating or cooling the body; C\ coefficient of expansion taken from col- umn 1, Table 2; C 3 coefficient taken from column 2, Table 2 ; 6" 3 = coefficient taken from column 3, Table 2 ; / difference in degrees of temperature between the original temperature and the temperature of the body after it has been heated or cooled. 4 PHYSICS. 51 Then, / = LC,t. a = A C t. v = VC\t. EXAMPLE. How much will a bar of untempered steel 14 feet long expand, if its temperature is raised 80 ? SOLUTION. Since only one dimension is given, that of length, linear expansion only can be considered. From Table 2, the coefficient of linear expansion per unit of length for a rise in temperature, of 1 is found to be .00000599 for untempered steel. Hence, using formula 21, /= L dt, and substituting, 14 X . 00000599 X 80 = .0067088 foot, or .0067088 X 12 = .0805056 inch. Ans. 95. This expansion seems very small, but in engineer- ing 1 constructions, when long pieces are rigidly connected, it must be taken into account. If the cross-section of the above bar was "Z inches square, and the bar was fitted tightly between two supports, an expansion of the above amount would exert a pressure against the supports of about 58,000 pounds. Suppose that an iron rod 1^ inches in diameter and 100 feet long were used as a tie-rod in constructing a bridge ; that it was put in place and securely fastened to two rigid supports diiring a warm day in summer when the temperature in the sunlight was, say, 110. On a cold day in winter when the thermometer registered zero, the amount that the bar would tend to shorten owing to this change in temperature would be, substituting these values in formula 21, .00000686X100X110 = .07546 foot = .90552 inch. If this rod were rigidly secured so that it could neither stretch nor shorten, it would then exert a pull on the sup- ports of about 33, 400 pounds. EXAMPLE 1. The wheel center of a locomotive driver is turned to exactly 50 inches in diameter. If the steel tire be bored 49.94 inches in diameter, to what temperature must the tire be raised in order that it may be easily shoved over the center ? Assume that the diameter of the tire is expanded to y-gVu f an inch larger than the center, and that the original temperature is 60. 52 PHYSICS. 4 SOLUTION. For this case, formula 22 may be used. The original diameter of the tire is 49.94 inches, and it is to be increased to 50.001 inches. The area of a circle 49.94 inches in diameter is 1,958.79 square inches. Area of circle 50.001 inches in diameter is 1,963.58 square inches. The difference between them is 1,963.58-1,958.79 = 4.79 square inches = a in formula 22. Hence, since C 2 = .00001198, and A = 1,958.79, substitute these values in a = A C 2 /, and 4.79 = 1,958.79 4. 7Q X. 00001198 x/ = .023466 /. Therefore, / = ^^ = 204.125, and 204. 125 + 60 = 264.125. Ans. NOTE. Owing to the form of the equation here denoted by formula 22, and to the manner in which the coefficients C 2 were determined, this example may be more easily solved by means of formula 21. Thus, regard the diameter as a linear dimension and apply formula SI. Increase in diameter = / = 50.001 49.94 = .061 inches. L = 49.94 and Ci = .00000599. Substituting, .061 = 49.94 X .00000599 X /, or, 49.94 The slight difference in the two answers is immaterial, and is to be expected. EXAMPLE 2. What is the decrease in volume of a copper cylinder 30 inches long and 22 inches in diameter, if cooled from 212 to 0, the measurement being taken at a temperature of 70 ? SOLUTION. 212 70 = 142 = the increase in temperature above 70. Use formula 23, v = V C 3 1. V = 22 2 X .7854 X 30 = 11,404 cu. in. v = 11,404 X .00002864 X 142 = 46.38 cu. in. 11,404 + 46.38 = 11,450.38 cu. in. = the volume at 212. 70 = 70 = the difference in temperature. V = 11,404 X .00002864 X 70 = 22.86 cu. in., nearly. 46.38 + 22.86 = 69.24 cu. in. Ans. 96. The bars of a furnace must not be fitted tightly at their extremities, but must be free at one end, otherwise in expanding they will split the masonry. In laying the rails on railroads, a small space is left between the successive rails; for, if they touched, the force of expan- sion would cause them to curve. Water pipes are fitted to one another by means of telescope joints, which allow room for expansion; so, also, are steam pipes, by means of the so called expansion joints. If a glass vessel is heated or cooled too rapidly, it cracks, especially if it is thick; the reason for this is that since glass is a poor conductor of heat, 4 PHYSICS. 53 the sides become unequally heated, and, consequently, unequally expanded, which causes a fracture. 97. It will be found on trial that the three preceding formulas will not work back ; i. e. , if the length of a bar after it has been heated is found by formula 21, and an attempt is made to reduce the bar to its original length by again applying formula 21 and substituting for t the same value as in the first case, the value obtained for / will be slightly different in the two cases. The difference, however, is so slight that it is neglected in practice. If, however, the student desires to obtain exactly the same result in both cases, he must use the following more cumbersome formula, in which t^ /. 2 , / 15 / 2 are, respectively, the original and final temperatiires, the original and final lengths, and C l has the same value as in formula 21 : This formula is always used when calculating the expan- sion of gases by substituting v^ z/ a , and -^^ for /,, / 2 , and C lt respectively. The result obtained will be exactly the same as those obtained by formula 12. For, substituting the values as directed, the formula becomes 492 / - 32 NX 7; 492 + /,-32 * 492 98. Although, as stated previously, the expansion of solids and liquids is very nearly uniform throughout all ranges of temperature, water is a marked exception to the general rule. If water is cooled down from its boiling point, it continually contracts until it reaches 39.2 F., when it begins to expand, until it freezes at 32 F. On the other hand, if water at 32 F. is heated, it contracts until it reaches 39.2 F., when it commences to expand. Therefore, the 54 PHYSICS. density of water is greatest where this change occurs. The importance of this exception is seen in the fact that ice forms on the surface of the water, since it is lighter than the warmer body of water lying at varying depths below it. Were it not for this fact, all the large bodies of water would freeze solid and would so affect the earth's climate that it would be unin- habitable. t The coefficient of expansion of water is such a very changeable quantity (varying with the temperature) that a special table is necessary. 99. The effect of heat on the expansion of gases was treated in Arts. 61 to 67, and will not be repeated here. It should be stated, however, that the constant .37052 used in formulas 14 and 15 has that value for air only. For other gases it varies. If the value of this constant for any gas be represented by R, formula 15 becomes pV = RWT. (25.) 1 GO. The value R for several gases is given in the fol- lowing table: TABLE 3. Gas. Volume of 1 Lb. at 32 F. and a Tension of 1 At- mosphere (14.7 Lb. per Sq. In.). Cubic Feet. Weight of 1 Cu. Ft. at 32 F. and a Tension of 1 At- mosphere (14.7 Lb. per Sq. In.). Pounds. R. Air 12 3880 08073 . 37052 Oxvtren 11 2056 .08925 .33552 Nitrogen 12 7226 07860 .38143 Hydrogen 178.8910 .00559 5.34946 EXAMPLE 1. What is. the volume of 3 ounces of hydrogen gas having a tension of 20 pounds per square inch and a temperature of 80 ? SOLUTION. 3 ounces = T 8 of a pound. Since / = 80 , T 460 + 80 = 540, R = 5.34946 from Table 3. Hence, by formula 25, p V = R W T, or 20 V = 5.34946 X A X 540 = 541.6328, and V - o7 r = 27.08164 cu. ft. ; say, 27.082 c.u. ft. Ans. 4 PHYSICS. 55 EXAMPLE 2. What is the weight of 10 cubic feet of oxygen having a tension of 1 atmosphere and a temperature of 60 ? SOLUTION. By formula 25, p V = A' W T, or 10 X 14.7 = .33553 X H'X 520. 147 Hence, 147 = 174.4704 W, and W = ". ' .- .84255 Ib. Ans. 1 74. 4 4 U4 In Table 2 the coefficient of expansion for gases was given as . 00203252 ; this is the fraction ^i^ reduced to a deci- mal. This value of the coefficient of expansion is very nearly the same for all gases, particularly for those that are very difficult to liquefy. EXAMPLES FOll PRACTICE. 1O1. Solve the following: 1. (a) How much will an iron tie-rod 60 feet long expand when the temperature is raised from 40 to. 110 ? (b) Calculate also by formula 24. (c) What is the difference in the two results ? ( t a ^ 345744 in Ans. { (b) .845725 in. [ (0 .000019 in. 2. To what temperature must a steel tire of 59.93 inches internal diameter be raised in order that its diameter may be 60.0015 inches ? Original temperature = 71. Ans. 270. 3. What is the volume of .68 pound of nitrogen gas having a ten- sion of 20 pounds per square inch and a temperature of 345 ? Ans. 10.44 cu. ft. HEAT PROPAGATION. 1O2. Conduction of Heat. Conduction is the slow progress of the vibratory motion from places of higher to places of lower temperature in the same body. Different bodies possess very different conducting powers, the good conductors being those in which conduction is most rapid, and the bad conductors those in which it is very slow. Experiments have established a numerical comparison of the conducting powers of many bodies. Representing the con- ducting power of silver by 100, the following table shows 56 PHYSICS. the conducting 1 power of a number of other metals as deter- mined by Wiedemann and Franz : TABLE 4. Silver 100.0 Copper 73 . 6 Gold 53.2 Brass 23.1 Zinc 19.0 Tin., 14.5 Iron 11.9 Steel 11.6 Lead 8.5 Platinum 8.4 Rose's alloy 2.8 Bismuth. . 1.8 1O3. As a class, the metals are the best conductors; the fluids, both liquid and gaseous, are very poor conductors, their conducting" power being hardly appreciable. Water, for example, can be made to boil at the top of a vessel, while a cake of ice is fastened within a few inches of the surface. If thermometers are placed at different depths, while water boils at the top, it is found that the conduction of heat downwards is very slight. Organic substances conduct heat poorly. This enables trees to withstand great and sudden changes in the atmos- phere without injury. The bark is a poorer conductor than the wood beneath it. Cotton, wool, straw, bran, etc. are all poor conductors. Rocks and earth are poorer conductors, the less dense and homogeneous the mass ; hence, the length of time required for the sun's rays to penetrate the earth. The mean highest temperature of the air near the ground in Central Europe is in the month of July, but at a depth of from 25 to 28 feet in the earth the highest temperature occurs in the month -of December. 1O4. Dynamical Theory of Heat. Before going any further, it will be convenient to explain here the view now generally taken as to the mode in which heat is propagated. On this subject, it is stated in Ganot's Physics, * ' A hot body is one whose molecules are in a state of vibration. The higher the temperature of a body, the more rapid are these 4 PHYSICS. 57 vibrations, and a diminution in temperature is but a dimin- ished rapidity of the vibration of the molecules. The propagation of heat through a bar is due to a gradual com- munication of this vibratory motion from the heated part to the rest of the bar. A good conductor is one that readily takes up and transmits the vibratory motion from molecule to molecule, while a bad conductor is one that takes up and transmits the motion with difficulty. But even through the best of the conductors, the propagation of this motion is comparatively slow. How, then, can be explained the instantaneous perception of heat when a screen is removed from a fire, or when a cloud drifts from the face of the sun ? In this case, the heat passes from one body to another with- out affecting the temperature of the medium that transmits it. In order to explain these phenomena, it is imagined that all space, the space between the planets and the stars, as well as the interstices in the hardest crystal and the heaviest metal in short, matter of any kind is permeated by a medium having the properties of matter of infinite tenuity, called ether. The molecules of a heated body being in a state of intensely rapid vibration, communicate their motion to the ether around them, throwing it into a system of waves, which travel through space and pass from one body to another with the velocity of light. When the undulations of the ether reach a given body, the motion is given up to the molecules of that body, which, in their turn, begin to vibrate ; that is, the body becomes heated. The process of this motion through the ether is termed radiation, and what is called a ray of heat is merely one series of waves moving in a given direction. HEAT MEASUREMENT. THERMAL UNITS. 1O5. Quantity of Heat. Temperature is not a measure of quantity of heat, for a thermometer would indicate the same temperature both in a vessel containing a pint, and one containing a gallon, of boiling water, although it is 58 PHYSICS. 4 evident that the one must contain eight times as much heat as the other; further, to raise the gallon of water to the boiling point, eight times the amount of heat necessary to similarly raise the pint is required. This leads us to the mode of measuring and indicating quantity of heat. Quantity of heat is measured by the amount necessary to raise a certain weight of some body from one temper atiire to another fixed temperature. 1O6. There are three units in use for measuring the quantity of heat: 1. The British, thermal unit, which represents the amount of heat required to raise 1 pound of water 1 Fahren- heit. Instead of writing out the words "British thermal unit " in full, it is customary to abbreviate them to B. T. U. Thus, 7 pounds of water raised 15 F. , would equal 7x15 = 105 B. T. U. The unit of heat used in this section will be the British thermal unit. 2. The thermal unit, which represents the amount of heat that is necessary to raise 1 pound of water 1 centigrade. Since 1 C. = f X 1 F., it follows that the thermal unit is f times as large as a B. T. U. Hence, to change B. T. U. into thermal units, multiply the number of B. T. U. by f. To change thermal units into B. T. U., multiply the number of thermal units by f. 3. The calorie, which represents the amount of heat necessary to raise 1 kilogram of water 1 centigrade. One kilogram = 2.2 pounds and 1 C. = |Xl F. ; hence, a calorie = 2.2 Xf 3.96 B. T. U. The calorie is used in Germany, France, and in other countries using the metric system of weights and measures. SPECIFIC HEAT. 1O7. When equal weights of two different substances having the same temperature are placed in similar vessels and subjected for the same length of time to the heat of the same lamp, or are placed at the same distance in front of the same fire, it is found that their final temperatures will differ g 4 PHYSICS. 59 considerably ; thus, mercury will be much hotter than water. But, as, from the conditions of the experiment, they have each been receiving the same amount of heat, it is clear that the quantity of heat sufficient to raise the temperature of mercury through a certain number of degrees will raise the same weight of water through a less number of degrees; in other words, it requires more heat to raise a certain weight of water 1 than it does to raise the same weight of mercury 1. Conversely, if the same quantities of water and of mer- cury at 200 are allowed to cool down to the temperature of the room, the water will require a much longer time for the purpose than the mercury; hence, in cooling through the same number of degrees, water gives up more heat than mercury. The number of heat units or parts of a heat unit required to raise the temperature of a given quantity of any substance 1 is called the specific heat of that substance. It will be seen from the above definition, that tJic specific Jieat of a substance is the ratio between the amount of heat required to raise the temperature of the substance 1 and the amount of the heat required to raise the temperature of the same weight of water 1. 1O8. If the specific heat of lead were given as .0314, it would mean that the amount of heat required to raise a cer- tain weight of lead 1 would raise the same weight of water only .0314 of 1; or it would mean that .0314 B. T. U. would raise the temperature of 1 pound of lead 1 F. EXAMPLE. The specific heat of copper is .0951 ; how many B. T. U. will it take to raise the temperature of 75 pounds 180? SOLUTION. Since it takes .0951 B.T. U. to raise 1 pound of copper 1, it will take .0951 X 75 X 180 to raise 75 pounds 180. Hence, .0951 X 75 X 180 = 1,283.85 B.T.U. Ans. In the example just given, if it had been required to raise 75 pounds of water 180 (that is, from the freezing point to the boiling point), it would have taken 75x180 = 13,500 B. T. U.,and 1>283 ' 85 = .0951 = the specific heat of copper. 13,500 60 PHYSICS. 109. The following" is the formula for finding the num- ber of B. T. U. required to raise the temperature of a substance a given number of degrees, or for finding the number of B. T. U. given up by a body in cooling a given number of degrees. Let W = weight of body in pounds ; s specific heat of the substance composing the body; / = original temperature of body; /j = final temperature of body; n = number of B. T. U. required, or given up, in changing the temperature of the body from t to ff. Then, n = Wfa-t)*. (26.) EXAMPLE. A piece of wrought iron weighing 31. 3 pounds and having a temperature of 900 is cooled to a temperature of 60 ; how many units of heat does it give up ? The specific heat of wrought iron is . 1 1 38. SOLUTION. Apply formula 26, /?= lV(tit)s. Substituting, n = 31.3 X (900-60) X .1138 = 2,992.03 B. T. U. Ans. If a body be cooled from a temperature / clown to a tem- perature /,, the value of ;/ will be negative, the minus sign indicating that the body was cooled. 110. In the following table are given the specific heats of a number of substances under constant pressure; TABLE 5. SOLIDS. Copper 0.0951 Gold 0.0324 Wrought iron 0.1138 Steel (soft) 0.1165 Steel (hard) 0.1175 Zinc. 0.0956 Brass 0.0939 Glass . . 0.1937 Cast iron 0. 1298 Lead 0.0314 Platinum 0.0324 Silver .... 0.0570 Tin 0.0562 Ice 0.5040 Sulphur 0.2026 Charcoal. . . 0.2410 PHYSICS. LIQUIDS. 61 Water . . . Alcohol . . Mercury . Benzine . . Glycerine 1.0000 0.7000 0.0333 0.4500 0.5550 Lead (melted) 0. 0402 Sulphur (melted) 0.2340 Tin (melted) 0.0637 Sulphuric acid 0.3350 Oil of turpentine 0.4260 GASES. Gases. Constant Pressure. Constant Volume. Air 23751 16847 Oxvsren . 21751 15507 Nitrogen 0.24380 17273 Hydrogen 3 40900 2 41226 Superheated steam .... Carbonic oxide . . . 0.48050 24790 0.34600 17580 Carbonic acid 0.40400 15350 111. The reason that there are two values for the spe- cific heat of gases is that it takes less heat to raise the tem- perature of a gas when the volume is constant than when the pressure is constant, but the volume varies. Thus, con- sider a closed cylinder filled with gas. If heat is applied, the pressure and temperature will both increase. Denoting the specific heat at constant pressure by s p and at constant volume by s v , the number of heat units required to heat the gas from / to t will be s v W(T T). If, however, the cyl- inder is imagined to be fitted with a frictionless piston, free to move up or down, and heat be applied, the gas will expand, overcoming a resistance equal to the weight of the piston plus the pressure of the atmosphere. Hence, in addi- tion to the heat required to increase the vibratory movement of the molecules, heat is also required to overcome the outer pressure, which remains constant in this case. The number of heat units necessary will then be s p 62 PHYSICS. 4 Mixing Two Bodies of Unequal Temperatures. If a certain quantity of water having a temperature of 40 is mixed with a like quantity having a temperature of 100, it is evident that the temperature after mixing will be - & = 70. But if 5 pounds of water having a temperature of 40 is mixed with 5 pounds of copper having a temperature of 100, the temperature after mixing will not be 70. The resulting temperature may be found by the following formula, provided there is no change of the state in the body (ice melting into water, etc. ) : ^V, + **.*,+ **>.*, + etc. in which / is the final temperature of the mixture ; W l9 s^ and / the weight, specific heat, and temperature, respect- ively, of one body; l}\, J 2 , and / 2 , the same for second body; and W^, s 3 , and / 3 , the same for a third body, etc. Remembering that the specific heat of water is 1, and getting the specific heat of copper from Table 5, the tem- perature / will be 5X1X40 + 5X. 0951X100 5X1 + 5X.0951 : 4o - 21 EXAMPLE 1. If 21 pounds of water at a temperature of 52 are mixed with 40 pounds of water at a temperature of 160, what is the tempera- ture of the mixture ? SOLUTION. Since the specific heat of water is 1, it maybe left out in applying formula 27, and the temperature is found to be 21X58 + 40X160 21+40 EXAMPLE 2. A copper vessel weighing 2 pounds is partly filled with water having a temperature of 80 and weighing 7.8 pounds. A piece of wrought iron weighing 3 pounds 4 ounces and having a temperature of 780 is dropped into this water. What is the final temperature of the mixture ? SOLUTION. Substituting the values given in formula 27, and remem- bering that the original temperatures of the copper vessel and the water that it contains are the same (3 Ib. 4 oz. = 3.25 lb.), we have _ 2 X. 0951 X 80 + 7.8 X 80 + 3.25 X . 1138X780 _ , 2X.095T+7.8 + 3.25X.H38 Ans. 4 PHYSICS. 63 EXAMPLE 3. A wrought-iron ball weighing 1 pound is placed in a reheating furnace : when it has attained the temperature of the furnace, it is taken out and placed in a copper vessel weighing \ pound and contain- ing exactly 2 pounds of water at a temperature of 75". Assuming that no water escapes as steam, and that the temperature of the ball, water, and vessel after mixing is 156', what is the temperature of the furnace ? SOLUTION. Substituting the values given in formula 27, 1 X .1138 X A + 2 X 75 + .5 X .0951 X 75 1 X- 1138 + 2 + . 5 X- 0951 .1138 /, +153.56625 156 = 2.16135 -- ' or 156 X 2.16135 = .1138 /, + 153.56625. Hence, .1138/, = 183.60435, 183.60435 , ;, A or /, = 1138 = 1,613.4 . Ans. 113. By means of formula 27, the specific heat of a substance may be obtained. Thus, in ^V.+ r^V.+ rr ^^-h/F 2 5 2 +/f suppose that the specific heat s 3 is required and that all of the other quantities, including" /, are known. Then, solving the above equation for s,^ or EXAMPLE. A silver vessel weighing 13 ounces is suspended by a string; 1 pound 4 ounces of water having a temperature of 120 is poured into it, and in this is placed a piece of metal weighing 14 ounces and having a temperature of 100. If the temperature of'the vessel was 72-, and the temperature of the mixture is f 17, what is the specific heat of the piece of metal ? SOLUTION. Using formula 28, and letting IV { , Si, and A represent, respectively, the weight, specific heat, and temperature of the silver vessel; \V^, j a , and / 2 , the same for the water; and ^F 3 , j 3 , and / 3 , the same for the piece of metal, 13X.057(72-117) + 20X1(120-117) -33.345 + 60 ... . 14(117-100) ~238 All weights must be reduced to either pounds or ounces before substituting. ' 64 PHYSICS. 4 EXAMPLES FOR PRACTICE. 114. Solve the following : 1. How many units of heat are required to raise the temperature of 10 ounces of platinum from 80 to 2,000 ? Ans. 38.88 B. T. U. 2. In order to determine the specific heat of a certain alloy, a piece weighing 12 ounces was heated to a temperature of 320, and was then immersed in 2 pounds 6 ounces of water contained in a lead vessel weighing 4 pounds 7 ounces. The temperature of the water and of the vessel being 70, what was the specific heat of the alloy if the tempera- ture of the mixture was 79 ? Ans. .1202. 3. In order to determine the temperature of the interior of a chim- ney, a silver bar weighing 20 ounces is placed in it until it has attained the same temperature. It is then immersed in 1 pound of water con- tained in a brass vessel weighing 10 ounces. The temperature of the vessel and water being 65, and of the mixture 981, what was the tem- perature within the chimney ? Ans. 596. 4. (a) An iron casting weighing 3 tons is cooled from 2,100 to 100 ; how many units of heat does it give up ? (b) If all this heat could be utilized, how many pounds of coal would it be equivalent to, assuming that 1 pound of coal gives out 14,500 B. T. U. during its combustion ? . ( (a) 1,557,600 B. T. U. S ' ( (b) 107. 4'2 Ib. iLATENT HEAT. 115. In all that has been said in the preceding pages only the phenomena relating to sensible heat have been con- sidered. If a quantity of pounded ice at a temperature of 32 be put in a vessel and held over the flame of a spirit lamp, heat passes rapidly into the ice and melts it; but a thermometer resting in this mixture of ice and water shows no tendency to rise ; it will remain at 32 until all of the ice has been melted. Where has the heat gone that was sup- plied to the ice? This question was first investigated by Dr. Black, of Edinburgh, in 1760, and is easily explained by the modern dynamical theory of heat. Dr. Black took a pound of water and a pound of ice, both having a temperature of 32, and placed them in two vessels suspended in a chamber that was kept at as nearly uniform a temperature as possible. At the end of half an hour the temperature of the water was 39. 2, but the ice did not reach 4 PHYSICS. 65 that temperature till 10| hours had passed, being- melted, of course, in the meantime. Dr. Black reasonably assumed that the ice received the same quantity of heat that the water did in each half hour, because it was placed in exactly the same position in regard to the surrounding air; that is to say, it received 39. 2 32 = 7.2 units of heat every half hour, or 14.4 units every hour, and 14.4 X 10J = 151.2 units in 10 hours; hence, it took 151.2 7.2 = 144 units of heat to change the pound of ice at 32 into water at 32. More accurate determinations have fixed this number as 142.65, and this value will be used in this section whenever the occa- sion arises for using it. If a pound of water having a temperature of 212 is mixed with a pound of water having a temperate of 32, the temperature of the mixture will be 21 2 I 9 IL = 122 ; the boiling water giving up 90 and the cold A water receiving 90, thus bringing both to a common tem- perature. If a pound of ice. at a temperature of 32 is mixed with a pound of water at a temperature of 212, the temper- ature of the mixture will be only 50.675, instead of 122, as in the previous case. Here the water has given up 212 50.675 = 161. 325 units of heat in order to bring both bodies to a common temperature. Since the temperature of the ice was raised from 32 to 50.675, it follows that 50. 675 32 = 18. 675 units of heat were used to raise the tem- perature of the ice after it had been melted into water, and 161.325 18.675 = 142.65 units of heat were necessary to convert the. ice at 32 into water of the same temperature. This extra number of units of heat, which is necessary to convert a solid into a liquid of the same temperature without raising the temperature of the solid, is called the latent 'heat of fusion, and the temperature at which this change of state in the body takes place is called the melting point, or temperature of fusion. All solids probably have a latent heat of fusion, the word probably being used because some solids have never been melted, except at such high temper- atures that accurate measurements are not possible, but its value varies greatly for different substances, being greater 66 PHYSICS. 4 for ice than for any other known solid, while for frozen mer- cury its value is only 5. 09 ; that is, to change 1 pound of frozen merctiry at its temperature of fusion (37.8 F.) into liquid mercury of the same temperature requires only 5.09 units of heat. Now, it is reasonable to suppose that if it requires 142.65 units of heat to convert a pound of ice at 32 into water at 32, then the same number of heat units would be given up when water at 32 is changed into ice at 32; experiment has shown that this is true. 116. If water is heated to its boiling point of 212 under a constant pressure of 14. 09 pounds per square inch, it has been found by experiment that it will require about 966 units of heat per pound of water to change it into steam at 212. This extra number of units of heat necessary to convert a liquid into a gas, or rather vapor, of the same temperature and pressure, is called the latent heat of vaporization, and the temperature at which this change of state takes place is called the temperature of vaporization. 117. According to the modern theory of heat, the extra quantity of heat necessary for a change of state of a body is used in forcing the molecules of a body farther apart, and in overcoming the force of cohesion. This latent heat is not lost, but performs work in giving additional energy to the molecules of a body, and it always reappears when the body resumes its former state. Thus, for instance, a pound of steam under a pressure of 1 atmosphere contains 966 -|- 180 = 1,146 units of heat more than a pound of water at 32. Hence, if 1 pound of steam at 212 is mixed with -f--fj- = 5.37 pounds of water at 32, the temperature of the mixture will be exactly 212, or the boiling point of water; in other words, the steam raised 5.37 pounds of water from the freez- ing point to the boiling point without lowering its own tem- perature, by merely changing from steam into water. If a pound of water at a temperature of 32 is changed into ice of the same temperature, 142.65 units of heat will be given up during this change of state. PHYSICvS. 118. In the following table are given the temperatures of fusion and of vaporization, and the latent heats of fusion and vaporization whenever they have been determined with sufficient accuracy: TABLE 6. Substance. Tempera- ture of Fusion. Tempera- ture of Vaporiza- tion. Latent Heat of Fusion. Latent Heat of Vaporiza- tion. Water 32 212 142 65 966 6 Mercury Sulphur -37.8 228 3 662 824 5.09 13 26 157 Tin 446 25 65 Lead 626 9 67 Zinc 680 1,900 50.63 493 Alcohol Unknown 173 372 Oil of turpentine . Linseed oil .... 14 313 600 124 Aluminum 1 400 Copper . 2 100 Cast iron 2 192 3 300 Wrought iron .... Steel 2,912 2 520 5,000 Platinum 3 632 Iridium 4 892 119. The following examples will show the purpose of Tables 5 and 6: EXAMPLE 1. How many units of heat are required to change 12 pounds of ice at a temperature of 20 C. into steam at 212 F. ? SOLUTION. By formula 19, t f (f X 20)-f-32 = 4 F. This is equivalent to 32 + 4 = 36 F. below the freezing point. In Table 5, the specific heat of ice was given as .504; hence, it will take 12 X 36 X -504 = 217.728 B. T. U. to raise the temperature of 12 pounds of ice from 4 to 32. To convert this ice into water of 32 will require 142.65 X 12 = 1,711.8 B. T. U. To raise this water from 32 to a tem- perature of 212 will require 12 X 180 = 2,160 B. T. U. To convert it into steam of 212 will require 966.6 X 12 = 11,599.2 B. T. U. The total number of units of heat required will be 217.728 + 1,711.8 + 2,160 + 11,599.2 = 15,688.728 B. T. U. Ans. 68 PHYSICS. 4 EXAMPLE 2. How many units of heat will it take to evaporate 25 pounds of mercury from a temperature of 70 ? SOLUTION. The temperature of vaporization of mercury is 662, and the specific heat is .0333; the increase in temperature from 70 will be 662 70 = 592. The number of units of heat required will be 25 X 592 X -0333 = 492.84 heat units. The latent heat of vaporization is 157; hence, 492. 84 + 25 X 157 = 4,417. 84 B.T.U will be required. Ans. 12O. A solid may be changed into a liquid, not only by melting it, but also by dissolving it, as salt or sugar is dis- solved in water. Since the particles of the solid body must be torn asunder in opposition to the forces that hold them together, it is reasonable to suppose that a certain amount of heat will be required to do this. That such is a fact may be easily proved by any one having a thermometer. Place a thermometer in a vessel of water and leave it there until it indicates the temperature of the water; then put in the water some salt or sugar and stir so as to make it dissolve more quickly, and it will be found that the mercury has fallen several degrees. In fact, if any solid is dissolved in a liquid that does not act chemically upon it, the temperature of the mixture will be lower than if the solid did not dis- solve. It is this principle that is taken advantage of in the so called freezing mixtures. A mixture of 1 part of nitrate of ammonia and 1 part of water will reduce the temperature from 50 to 4, a fall of 46. The effects are still more striking when both bodies are solids, one of which is already at the freezing point. Thus, a mixture of 2 parts of snow or finely powdered ice and 1 part of common salt will reduce the temperature from 50 to 0, a range of 50 ; while a mix- ture of 4 parts of potash and 3 parts of snow or powdered ice will lower the temperature from 32 to 51, a range of 83. Latent heat plays an important part in every-day life. It takes a long time and severe cold to freeze the water of a river to any depth, even though the thermometer goes far below the freezing point. This is because 142.65 units of heat must be given up by every pound of water, after being brought to the freezing point, before the ice can form. 4 PHYSICS. 69 If it were not for this, the rivers, lakes, and other bodies of water would be frozen solid as soon as the water reached the freezing point and would be melted as soon as the tempera- ture went above that point. In the spring, all of the snow on the hills would be melted during a warm day, and great floods would be the consequence. As it is, 142.65 units of heat must be supplied to every pound of snow at 32 to con- vert it into water at 32. EXAMPLES FOB PRACTICE. 1.22. Solve the following : 1. If a pound of steam at 212 and 7 pounds of ice at 32 are mixed, what will be the resulting temperature ? Ans. 50.5. 2. How many units of heat are required to vaporize 10 pounds of mercury from a temperature of 100 ? Ans. 1,757.146 B. T. U. 3. How many pounds of oil of turpentine at 60 can be vaporized by 1 pound of coal, if the coal gives out 14 ,500 B. T.U. during combustion? Ans. 62.56 Ib. 4. How many pounds of water at 32 can be vaporized by 1 pound of coal? Ans. 12.646 Ib. 5. How many pounds of coal are required to raise 100 pounds of wrought iron from 85 to its melting point ? Ans. 2.219 Ib. SOURCES OF HEAT. 123. Heat is derived from the following sources: 1. Physical sources, which include the radiation of heat from the sun, terrestrial heat, change of state in bodies, and electricity. 2. Chemical sources, resulting from chemical combina- tions, such as combustion, oxidation, etc. 3. Mechanical sources, comprising friction, percussion, and pressure. 124. Physical Sources. 1. The most intense of all of the sources of heat is the sun. The majority of scientists are of the opinion that all of the heat received or given up 70 PHYSICS. 4 by the earth has, or has had, its source in the sun. It would be out of place here to enter into this theory fully. It is the amount of heat radiated from the sun and received by the earth that causes the change of seasons, that causes the waters in the rivers, lakes, and seas to evaporate and form the clouds, to be again precipitated as rain or snow. With- out it, no living thing, animal or vegetable, could exist. 2. The earth possesses a heat peculiar to itself, called terrestrial heat. When a descent is made below the surface, the temperature is found to gradually increase. This is not caused by the heat radiated from the sun, for the material comprising the earth is such a poor conductor that the heat of the sun's rays penetrates only a very short distance below the surface. The explanation usually given for this phe- nomenon is that the interior of the earth is in a molten con- dition. The terrestrial, heat exerts but a slight effect, not raising the temperature of the surface more than ^V f a degree. 3. If a liquid is poured iipon a finely divided solid, as a sponge, flour, starch, roots, etc., the temperature will be increased from 1 to 10, according to conditions. This phe- nomenon may be called heat produced by capillarity. 4. The heat produced by a change of state has already been described ; it is the heat given off when a body is con- verted from a gas or liquid to a liquid or solid. 5. Extremely high temperatures may be produced by the electric current. By means of it, quicklime, firebrick, osmium, porcelain, and several other substances, which until very recently have resisted every attempt to melt them, may be made to run like water. 125. Chemical Sources. Whenever two or more sub- stances that act chemically upon one another are brought together and allowed to combine, heat is evolved ; and when- ever the heat caused by this chemical union is sufficiently intense to raise the resultant substances to a temperature at which they emit light, this act of union is termed com- bustion. PHYSICS. 71 This subject will not be treated here, but will be considered fully by itself in Theoretical Chemistry. Mechanical Sources. 1. The friction between any two bodies rubbed together produces heat. Rubbing one hand briskly against the other will soon make the hands too warm for comfort. The friction between a journal and its bearing causes heat; the heat causes the journal and bearing to expand, the journal expanding more rapidly on account of being smaller and being heated more quickly; the expansion causes a greater pressure on the bearing, pro- ducing more friction and heat. If the bearing is not properly oiled, the heat will become so intense in a short time that the soft metal in it will melt. When shooting stars strike the earth's atmosphere, their velocity is so great (sometimes as high as 150 miles a second) that the friction of the atmosphere causes them to take fire almost instantly. Wherever there is fric- tion, there is heat. 2. Heat is also generated by percussion. The repeated blows of a hammer upon a piece of iron, lead, or other metal will soon make it quite hot. 3. The generation of heat by pressure was spoken of in connection with gases; that is, the .temperature rises when a gas is compressed. This is also true of solids and liquids, but the results are not so marked in their cases. The production of heat by the compression of gases is easily shown by means of the pneumatic syringe shown in Fig. 17. This consists of a glass tube with thick sides, hermetically closed with a leather piston. At the bottom is a small cavity in which a piece of cotton, moistened with ether or carbon disulphide, is placed. The tube being filled with air, the piston is suddenly 72 PHYSICS. 4 plunged downwards. Thus compressed, the air generates so much heat that the cotton is ignited, and can be seen burning when the piston is suddenly withdrawn. The ignition of the cotton in this experiment indicates *a tem- perature of at least 570, since it will not ignite at a lower temperature. MOHT. PROPAGATION OF LIGHT. 127. When the nerves of the retina of the eye are stimulated, the sensation is commonly described by saying that light is seen. Light is, then, a pure sensation; and it may be caused in several ways. Ordinarily it is produced as the result of some action that is taking place at some dis- tance from* the observer such as the combustion of gas in a lamp flame, or the chemical processes of the sun. 128. Theory of Jjight. Sir Isaac Newton assumed that luminous bodies emit infinitely small particles in straight lines, which, by penetrating the transparent parts of the eye and falling upon the nervous tissue, produce light. Modern philosophers, however, in developing the theory of sound, came to the conclusion that light might be the effect of undulations, or little waves, propagated with inconceivable velocity through the highly elastic medium of excessive tenuity called ether, which fills all space and lies between the particles of material substances. 129. Transparent, Translucent, and Opaque Bodies. Bodies that allow light to pass, so that objects can be clearly distinguished through them, are called transparent; the air, water, and clear glass are common examples. Oiled paper, roughened glass, and porcelain allow light to pass, but objects cannot be clearly distinguished through them ; they are trans- lucent bodies. Opaque bodies, such as wood or stone, do not 4 PHYSICS. 73 allow light to pass through them. Opaque bodies, when very thin, are translucent, and transparent bodies, when very thick, become translucent and even opaque. 130. Propagation of right. A medium is a sub- stance, as air and glass, through which light passes. If the density and composition of the medium is the same in all parts, it is called homogeneous ; water, carefully prepared glass, and the atmosphere, if we consider only a small por- tion of it, are homogeneous mediums. In a homogeneous medium light is propagated in straight lines ; this is our every-day experience. The rays of the sun, as they trace their way through our room, are straight ; in order that light may pass through a tube, the tube must be straight. 131. Ray and Pencil of Ijight. A light proceeding from a luminous body is supposed to be made up of straight lines, called rays of light. A number of rays form a pencil of light. Velocity of light. A ray of light emitted from a luminous body proceeds in a straight line with extreme velocity. Through a series of astronomical observations, an approximate knowledge of this velocity has been obtained. As the moon revolves about the earth, so revolve the satel- lites of Jupiter about that planet, and the time of revolution of each satellite has been ascertained from its periodical entry into or exit from the shadow of the planet. The time required by -one satellite is only 42 hours. The Danish astronomer Romer observed that this period appeared longer when the earth in its passage around the sun increased the distance between Jupiter and itself, and again that the periodic time became shorter when the earth moved towards Jupiter. Although the difference is rather small for a single revolution of the satellite, it increases by the addition of many revolutions during the passage of the earth from its nearest to its greatest distance from Jupiter, a passage that occupies about 6 months, till it amounts to 16 minutes and 16 seconds. From this, Romer concluded that the light of 7.4 PHYSICS. the sun reflected from the satellite required that time to pass through a distance equal to the diameter of the orbit of -the earth, and, since this space is, approximately, 190,000,000 miles, the velocity of light must be about 190,000 miles per second. REFLECTION AND REFRACTION. 133. Let it be assumed that a ray of light is deflected by a mirror J/, Fig. 18, so that it strikes point A^ of a hori- zontal mirror L, and is from there reflected in the direction A*. If we then draw a line from point N to O at right angles to the mirror and measure the angles M N O and O N R, we find that these two angles are equal, and we further notice that the two rays lie in a plane which is at right angles to the mirror N. The line A" O perpendicular to the mirror is called the normal. The ray M N is called the incident ray and the ray N R is the reflected ray. The angle that the incident ray makes with the normal is the angle of incidence, and the angle made by the normal and the reflected ray is the angle of reflection. 134. The general laws of reflection are as follows: 1. The angle of incidence is equal to the angle of reflec- tion. 2. In whatever position the incident ray meets the reflect- ing surface, a plane can always be so placed that it passes through the incident ray> 'the reflected ray, and the normal. 4 PHYSICS. - 75 135. The same laws also hold good if a mirror or any other reflecting surface be curved, as a portion of a sphere, since a curve may be considered as made up of a multitude of minute planes. Parallel rays, however, cease to be parallel when reflected from such a surface ; they become conver- gent or divergent according as the reflecting surface is con- cave or convex. Suppose that A B, Fig. 19, is a section of a concave mirror. Point C, the center of the sphere, is known as the center of FIG. 19. curvature, point O as the apex, and CO as the principal axis. A very small portion of the mirror at any point A will be a plane, of which A C, the radius, will be the normal; and if any incident ray, as S A, strikes the mirror at A, the reflected ray will be A F (angle SA C = angle FA C)\ the same is .true for any small portion of the mirror, as, for instance, L, O, and B. If the arc A B, the section of the reflecting surface, is small compared to the radius CO, it is found, both by measurements and experiments, that all incident rays, as SA, TL, X B, parallel to the principal axis, after reflection, pass through the same point Fin the prin- cipal axis. This point, which is midway between C and the mirror, is called \}^Q principal focus, and the distance between F and O is known as the focal length. 136. In Art. 13O it was stated that the rays of light proceed in straight lines; this, however, is true only as long 76 PHYSICS. as the medium through which these rays pass is homogene- ous. Should, however, the density or chemical nature of the medium through which the ray travels vary, the ray of light is bent from its original course into a new one ; or it is said to be refracted. Fig. 20 shows a thick plate of glass through which a ray of light is sup- posed to travel, B being the point of the surface where the ray A enters. This ray, instead of proceeding onwards in the direction C, will be bent downwards towards point D, and on leaving the plate and entering into the air on the other side, will again be bent, but in an opposite direction, so as to travel parallel to the continuation of the original path, provided there is the same medium on the upper as on the lower side of the plate. The general law of refraction may, then, be expressed as follows: When a ray of light passes from a rarer to a denser medium it is usually refracted towards a line perpendicular to the surface of the latter, and, conversely, when it leaves a dense, medium for a rarer one, it is refracted away from a line per- pendicular to the surface of the denser substance. In the former case the angle of incidence is greater than that of refraction; in the latter it is less. In Fig. 20, angle ABE is the angle of incidence, and D B E' is the angle of refrac- tion. 137. The amount of refraction, for the same medium, varies with the obliquity with which the ray strikes the sur- face. When perpendicular to the latter, the ray passes with- out change of direction ; and in other positions the refraction increases with the obliquity. 138. Let A, Fig. 21, represent a ray of light falling upon the surface of a heavy plate glass at the point B. From this point let a perpendicular fall and be continued into the new medium, and around B as a center let a circle be drawn. Then, PHYSICS. 77 according to the general law of refraction, stated in Art. 136, the refraction must be towards the perpendicular, in the direction B C, for example. If now the lines a a and a' a', which are the sines of the angles of incidence and refraction, respectively, are drawn at right angles to the perpendiculars, and their lengths compared by means of a scale, these lengths will be found, in this case, to be in- the ratio of 3 : 2. Or, if another ray such as E is taken, which is A refracted in the same manner towards F y the bending being natu- rally greater from the increased obliquity of the ray, and if the sines of the new angles of incidence and re- fraction are then com- pared, they will be found to bear to each other the same ratio; i expressed by the rule : FIG. 21. e., 3 : 2. This fact maybe, then, So long as the light passes from one to the other of tivo media, the ratio of the sines of the angles of incidence and refraction remains constant. 139. Different bodies possess different refractive power, and, as a rule, the densest substances refract most. The method adopted for describing the comparative refractive power of different bodies is to state the ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium ; this ratio is known as the index of refraction of the two substances. It is greater or less than unity, according as the second medium is denser or rarer than the first. If the index of refraction of any particular substance is once known, the effect of the latter upon a ray of light enter- ing it at any angle can be easily calculated by the law of sines. PHYSICS. 14O. In Table 7 the indices of refraction of several substances are given. It is assumed that the ray of light passes into the substances in question from the air. TABKE 7. INDICES OF REFRACTION. Ice 1.30 Water 1.34 Fluorspar 1.40 Plate glass 1.50 Rock crystal 1.60 Chrysolite 1.69 Carbon bisulphide 1.70 Garnet 1.80 Phosphorus 2. 20 Diamond 2.50 Lead chromate 3.00 Cinnabar . . 3.20 141. Some Results of Refraction. A stick partly sub- merged in water appears to be bent at the surface ; the tip of the stick and all that is under water seems to be raised. The atmosphere is less dense the higher we ascend, and we can assume it to be made up of layers of air of various densities. The rays of light from the sun or a star, instead of descending in straight lines, are re- fracted at every layer. Thus, the light from S t in. Fig. 22, reaches the eye as if it came from S f ; that is, a star appears higher in FIG - ** the heavens than it really is. For the same reason, the sun is seen before it is above, and after it is below, the horizon. 142. Irregular Refraction. The rays of light are straight in air and straight in water, provided the density is uniform. This is not the case if from any cause the density is not uniform. The quivering of objects, when seen over heated coke or on a hot day, is due to unequal refraction, the density being constantly changing. PHYSICS. 79 1 43. Total Reflection. When a ray of light passes from one medium into another that is less refracting, as from water into air, it has been seen in Art. 136 that the emerging ray is again bent from its course. Hence, when light is propagated in a mass of water from CtoA, Fig. 23, there is always a value of the angle of incidence CAB such that the angle of refrac- tion fiADisa right angle, in which case, the refracted ray emerges parallel to the surface of the water. This angle CAB is called the critical angle, since, for any greater E A B, for instance the incident ray can no longer emerge, but undergoes reflection. This is called total reflection, because the incident light is entirely reflected. The critical angle for water and air is 48^-. If you raise a glass of water above your head you can easily find a position in which the rays from the eye to every part of the under surface make a greater angle than 48^ with the normal. A brilliant mirror due to the total reflec- tion of light will be observed. 144. Transparent glass when finely ground loses its transparency; light is diffused at the surface, and the few rays that pass through the upper layer of the fine particles are refracted in passing from the glass to the air between the surfaces; the result of this combined irregular reflection and refraction is that the glass appears opaque. If a liquid, whose refractive index is the same as that of glass, is poured over the glass, the result will be the same, in respect to the action of light, as if the surface were made smooth again and all the spaces were filled with glass; that is, powdered glass becomes in its effects a glass plate and. is again transparent. 80 PHYSICS. 4 145. Prisms. A prism may be defined as a portion of a transparent medium bounded by plane inclined surfaces. The line along which these surfaces meet is called the refracting edge, and the angle between them, the refracting angle of the prism. The side opposite the refracting edge is called the base of the prism, and any section through it, perpendicular to this edge, is called a principal section. If a ray of light falls upon one of the inclined surfaces of a prism, the change in the direction of the ray that is, its deviation instead of being neutralized at the second surface, is still further increased. Let ray A, Fig. 24, be incident to the surface PQof a prism at O* At O the refraction will be towards the normal N FIG. 24. and the ray will be refracted to O'. Draw normals at O and O'. At O' the second refraction takes place, but since the face PS is inclined to the face P Q, the second deviation is in the same direction as the first; that is, towards the base of the prism. Evidently the deviation of the ray at the first surface is A ON t O O' or d = i r, and at the second is A'O'N'tO'Oora" = i'-r'. The total deviation is the sum of the partial deviations, or D d-\-d f = (i r)-\-(t' r'). That is, the total deviation in the figure, or the angle A I />, or D, between the incident and emergent rays is equal to the sum of the partial deviations d-\-d', or I O O' and I O' O, since the exterior angle of the triangle is equal to the sum of the two interior angles. Moreover, the deviation may be PHYSICS. 81 obtained in terms of the index (;/) and the prism angle. Since D (ir)-\- (i' r') or (i + i') (r + r'), and as the prism angle is equal to r-\-r f , the deviation D = (i-^-i') a. If the prism be thin and the incident ray nearly normal, we may write / = nr and i' = nr', whence i-\-i' = n(r-\-r r ) or // <7, and D n a a or (// 1) a\ which means that the deviation is equal to the prism angle multiplied by the index of refraction less 1. 146. When the deviation produced by a prism is a minimum, the angles formed by the ray with the normals within the prism are equal, as are also the angles without it; that is, r = r' and / = i'. Consequently, we can simplify the formula for deviation, D writing it D = 2z 2r. In Fig. 25 we see that the angle at t formed by the two normals is for that reason equal FIG. 25. to #, the refractive angle of the prism. But, being exterior to the triangle O O' t, it is equal to the sum of the two interior angles, that is, 7'-(-r / or2r; therefore, a = 2r and r \a. Hence, D 2/ POLARIZATION. 158. Double Refraction. A large number of crystals possess the property of double refraction, which means that if a ray of light passes through such a crystal, it becomes split or divided into two rays; one, which is called the ordi- nary ray, follows the general law of refraction, while the other, which is known as the extraordinary ray, takes an entirely different, new course, which depends on the position of the crystal. This remarkable property is found in nearly all bodies that crystallize in any other than the cubical form. The phenomenon is explained by assuming that the ether in these doubly refracting bodies is not equally elastic in all directions, and that, consequently, the vibrations, in certain directions at right angles to each other, are transmitted with unequal velocity. A transparent variety of.calcite found in Iceland, and hence known as Iceland spar possesses this property in a marked degree. If a piece of this substance is placed upon a sheet of white paper on which an ink spot or any other mark has previously been made, two images of the mark will be seen. 90 PHYSICS. 4 159. Polarization. If a ray of light CD is thrown upon a plate of glass A A, as shown in Fig. 33, at an angle of 56 45', that portion of the ray which is reflected will be found to have acquired properties it did not before possess ; for, on throwing it at the same angle upon a second glass plate B B, it will be observed that there are two particular positions of the latter (indicated by B f B') ; namely, those in which the planes of incidence are at right angles to each other, when the ray of light is no longer reflected, but entirely refracted. Light that has undergone this change is said to be polarized. The other portion of the ray that passes through the first, or polarizing, plate A A is also to a certain extent in this peculiar condition i.e., polarized and by using a number of similar plates, placing one behind the other, this effect may be greatly increased. It must, however, be remarked that the light polarized by transmission is in an opposite state to that polarized by reflection; that is, when examined by the second, or analyzing, plate, held at the angle before mentioned, it will be seen to be reflected when the other is transmitted, and to be dispersed when the first is reflected. 16O. Angle of Polarization. Not every transparent substance is capable of polarizing light. Glass, water, and a number of other substances, however, produce this change, 4 PHYSICS. 91 each having a particular angle at which the effect is the greatest. The angle of polarization of a substance is, then, the angle that the incident ray must make with the perpen- dicular to a plane surface of that substance in order that the polarization be complete. For glass this angle is 56 45'; for water, 52 45'; for quartz, 57 32'; for diamond, 68; etc. . If in the preceding experiment the plane had been inclined at a larger or smaller angle, the ray of light would not have been completely polarized, which would be shown by the ray being partially reflected from the analyzing plate B B in all positions. 161. Polarization in Opposite Directions. The ordi- nary and extraordinary rays produced by sending a ray of light through a substance possessing the property of double refraction are found on examination to be completely polar- ized, but in a very peculiar manner. The one is capable of reflection when the other is transmitted or refracted. This fact is expressed by saying that the two rays are polarized in opposite directions. With a good sized piece of transparent Iceland spar, the two oppositely polarized rays maybe widely separated and examined separately by an analyzing plate. 162. Some of the doubly refracting substances absorb one of these rays, but not the other. Tourmaline, a mineral, is the best known medium possessing this property. When a ray of light is sent through a plate of tourmaline cut at right angles to the axis, an ordinary and an extraordinary ray are formed, which are polarized in planes at right angles to each other. The mineral, however, rapidly absorbs the ordinary ray, and if the emer- ging polarized extraordinary ray is observed through a second similar plate, held exactly in the same position as the first one and as shown in Fig. 34 (a), little change will be noticed; but when the second plate is turned around as shown in Fig. 34 (#), so as to make the axes cross at right angles, it will be 92 PHYSICS. 4 found that the light is almost wholly stopped. A plate of this mineral in this way is a valuable test for detecting polarized light, and light that has not undergone the change. Instead of tourmaline plates, which are, owing to the thick- ness required for complete polarization, not very transpar- ent, and are mostly more or less colored, two Nicol's prisms, or the cheaper and quite as effective Foucault's prisms, are used, as they possess the same properties as tourmaline. Two of these prisms, placed in exactly the same position, one behind the other, allow light to pass through them ; but by turning the second prism a cloudiness is produced, until, by continuing to turn the prism, a point is reached that causes perfect darkness. This phenomenon occurs whether the light is colored or not. 163. INTicol's prism consists of a bar of Iceland spar, about 1 inch in height and | inch in breadth ; the prism is cut in the plane that passes through the obtuse angles, as shown in Fig. 35 (a). The two halves are joined by a layer (*> FIG. 35. of Canada balsam. The principal involved in the construc- tion of this prism is the following: The refractive index of Canada balsam, being 1. 549, is less than the index of the ordi- nary ray of Iceland spar, 1. 654, and greater than the index of its extraordinary ray, 1.483; hence, if a ray of light CD, Fig. 35 (/;), enters, the prism, the ordinary ray is totally reflected on the surface A B and takes the direction D E F, by which it is refracted out of the prism, while the extraor- dinary ray D G emerges alone. The prism is used as an analyzer and also as a polarizer, possessing the same prop- erties as tourmaline, but, as has been already mentioned, is preferable because it gives a colorless field of view. 4 PHYSICS. 93 In Foucatilt's prism, which is constructed on exactly the same principle, the layer of Canada balsam is replaced by a layer of air, the prism being held together by a brass or aluminum mounting. 164. Circular Polarization. A ray of light colored by passing, for example, through a plate of red glass, which has passed through the first of two Nicol's prisms, and which is then obstructed in consequence of the position of the second prism, regains its original property of penetrating the second prism if a plate of rock crystal is placed between the two prisms in the path of the ray. The passage of the ray through the analyzer (second prism) can now only be interrupted by a further turn of the analyzer. The rotation required depends on the thickness of the plate and the color of the ray. It increases from red in the following order: yellow, green, blue, and violet. This kind of polarization is known as circular polarization, and for a long time quartz or rock crystal was the only solid substance known to exhibit this peculiar property of produ- cing it. Scientific investigations and experiments, however, have brought to light since then a number of substances that possess this property in a still higher degree; thus, for instance, a plate of cinnabar acts nearly fifteen times more powerfully than a plate of equal thickness of rock crystal. The direction of the rotation that is required to produce this complete opaqueness of the analyzer depends on the composition of the interposed substance. A substance is said to be dcxtro- or levo-rotary, according as it is necessary to turn the analyzer around in the direction of the hands of a watch, or in the opposite direction. The angle through which the polarizer has to be turned is called the angle of rotation. 165. It has been observed that solutions of many organic substances show the property of circular polarization, thoiigh, as a rule, to a less extent than rock crystal. Thus, solutions of cane sugar and tartaric acid possess right-handed polariza- tion, while oil of turpentine and albumin are left-handed. In all of these solutions the angle of rotation varies directly 94 PHYSICS. 4 with the strength of the solutions, and the thickness of the column through which the light passes; hence, circular polarization has become an important auxiliary in chemical analysis. This power of circular polarization is found most com- monly in vegetable and animal products and their immediate derivatives, as will be seen in the description of such bodies in Organic Chemistry. It is assumed that this power depends on some peculiarity in the structure of the molecules. 166. In order to determine the amount of polarization any liquid may exhibit, it is poured into a glass tube and closed with glass plates. This is then placed between two Nicol's prisms, which have previously been so arranged that the analyzer appears opaque. Through the polarizing power of the liquid, the analyzer becomes transparent, and from the magnitude of the angle through which the analyzer has to be turned to appear opaque again, the polarizing power of the liquid is calculated. 167. An apparatus of this kind is widely used and is of the greatest importance in the sugar industry. It is called the saccliarimeter, and is used for determining the con- centration of solutions of cane sugar. The instrument was devised by Biot, and through certain additions perfected by Soleil. Though a full description of the saccharimeter, its uses, and instructions in regard to calculations, etc. cannot be given in this section, a short explanation of this apparatus will enable the student to comprehend more fully the impor- tance of circular polarization in connection with chemistry, and .will give him some idea as to the way in which this physical phenomenon is titilized. This apparatus is shown in Fig. 3G. The two Nicol's prisms are in the 'fastenings A and /?, respectively. After the prisms are arranged so that the analyzer A appears opaque, the tube, filled with a solution of sugar, is placed as marked by dotted lines in the figure. The light will now PHYSICS. 95 be again enabled to pass through the analyzer, and prism A has to be turned through a certain angle till entire darkness again appears. The magnitude of this angle is read on the circular disk C, which is divided into degrees, and upon which moves an index D connected with the prism A. If we assume that the tube used is exactly 10 inches long and the solution to be analyzed contains exactly 10 per cent. FIG. 36. of cane sugar, the angle of rotation will be 19.6. Now, the magnitude of this angle is directly proportional to the length of the liquid column and also to the percentage of sugar in Solution. If, therefore, a solution containing x per cent, of sugar in a tube y inches long produces an angle of rotation of z degrees, the percentage of sugar will be given by the equation : y x " : io x io' or? * = 100-s- (29.) The apparatus shown in Fig. 36 is of the simplest kind, and can only be used conveniently when pure cane sugar is to be analyzed; if the solution contains cane sugar and uncrystallizable sugar, the latter possessing left-handed 96 PHYSICS. 4 rotation and the former right-handed, only the difference of these two actions is obtained. The whole quantity has to be changed into uncrystallizable sugar, the experiment repeated, and from the results of the two observations the quantity of each kind of sugar calculated. Soleil has added to this form of saccharimeter a number of improvements that facilitate observations and make measurements more exact. MAGNETISM. 168. Magnets are substances that Have the property of attracting pieces of iron or steel, and the term magnetism is applied to the cause of this attraction. Magnetism exists in a natural state in an ore of iron, which is known in chemistry as magnetic oxide of iron or magnetite. This magnetic ore was first found by the ancients in Magnesia, a city in Asia Minor; hence, the name magnet. It was also discovered that when a small bar of this ore is suspended in a horizontal position by a thread, it has the property of pointing in a north-and- south direction. From this fact the name lode stone, "leading stone," was given to the ore. When a bar or needle of hardened steel is rubbed with a piece of lodestone, it acquires magnetic properties similar to those of the lodestone, without the latter losing any of its own force. Such bars are called artificial magnets. Artificial magnets that retain their magnet- ism for a long time are called permanent mag- nets. The common form of artificial or permanent magnet, Fig. 37, is a bar of steel bent into the shape of a horseshoe and then hardened and magnetized. A piece of soft iron called an armature, or a keeper, is placed across the two FIG. 37. f ree ends and helps to prevent the magnet from losing its magnetic force. PHYSICS. 1 69. If a bar magnet is dipped into iron filings, the filings are attracted towards the two ends and adhere there in tufts ; while towards the center of the bar, half-way between the two ends, there is no such tendency (see Fig. 38). That part of the magnet where there is no apparent magnetic attrac- FlG. 38. tion is called the neutral line ; and the parts around the ends where the attraction is greatest are called poles. An imaginary line drawn through the center of the magnet from end to end, connecting the two poles together, is the axis of 'magnetism. 17O. A compass consists of a magnetized steel needle, Fig. 39, resting upon a fine point and capable of turning freely in a horizontal plane. When not in the vicinity of other magnets or mag- netized iron, the needle will always come to rest with one end pointing towards the north and the other towards the south. The end pointing northward is the north- seeking pole, or, simply, the north pole, and the opposite end is the south- sec king pole, or south pole. This polarity applies as well to all mag- nets. If the north pole of one magnet is brought near the south pole of another magnet, attraction takes place; but if two north poles or two south poles are brought together, they repel each other. In general, like magnetic poles repel eacJi other; unlike poles attract each other. 98 PHYSICS. 171. The earth is a great magnet whose magnetic poles coincide nearly, but not quite, with the true geographical north and south poles. A freely suspended magnet, there- fore, will always point in a north-and-south direction. 173. It is impossible to produce a magnet with only one pole. If a long bar magnet is broken into any number of parts, each part will still be a magnet and have two poles one north and one south pole. 173. Substances that are not in themselves magnets that is, do not possess poles and neutral lines, but are capable of being attracted by a magnet are called magnetic substances. In addition to iron and its alloys, the following metals are magnetic substances: nickel, cobalt, manganese, cerium, and chromium. These metals, however, possess magnetic properties in a very inferior degree compared with iron and its alloys. All other known substances are called non-magnetic substances. 174. The space surrounding a magnet is called a mag- netic field; or, in other words, a magnetic field is a place where a freely suspended magnetic needle will always come to rest, pointing in the same direction. ^\\^)\\;Y,U! 1 1 \Hfu r '/'W'''% FIG. 40. FIG. 41. 175. Magnetic attractions and repulsions are assumed to act in a definite direction and along imaginary lines called lines of magnetic force, or, simply, lines of force, and every magnetic field is assumed to be traversed by such lines of PHYSICS. 99 force. Their position in any plane may be shown by placing a sheet of paper over a magnet and sprinkling fine iron filings over the paper. In the case of a bar magnet lying on its side, the iron filings will arrange themselves in curved lines extending from the north" to the south pole, as shown in Fig. 40. A view of. the magnetic field looking towards either pole of a bar magnet would exhibit merely radial lines, as shown by the filings in Fig. 41. FIG. 42. Every line of force is assumed to pass out from the north pole, make a complete circuit through the surrounding medium, and into the south pole, thence passing through the magnet to the north pole again, as shown in Fig. 42. This NOTE. In all diagrams, the direction of the lines of force will be represented by arrowheads upon dotted lines. 100 PHYSICS. is called the direction of tJic lines of force, and the path they take is called the magnetic circuit. The direction of the lines of force in any magnetic field can be traced by a small, freely suspended magnetic needle or a small compass, such as indicated by m in Fig. 42. The north pole of the needle will always point in the direction of the line of force, the length of the needle lying tangent to the lines of force at that point. If the needle is moved bodily in the direction towards which the north pole points, its center or pivot will describe a path coinciding with the direction of the lines of force along that part of the magnetic field. Lines of force can never intersect ; when two opposing mag- netic fields are brought together, as indicated by the iron filings in Figs. 43 and 44, the lines of force from each will ymm^ y/im FIG. 43. FIG. 44. be crowded and distorted from their original direction until they coincide in direction with' those opposing and form a resultant field in which the direction of the lines of force will depend on the relative strengths of the two opposing nega- tive fields. The resulting poles thus formed are called con- sequent poles. In every magnetic field there are certain stresses that produce a tension along the lines of force and a pressure across them ; that is, they tend to shorten themselves from end to end, and repel one another as they lie side by side. 4 PHYSICS. 101 17(>. When a magnetic substance is brought into a mag- netic field, the lines of force in that vicinity crowd together and all tend to pass through the substance. If the substance is free to move on an axis, but not bodily towards the mag- net pole, it will always come to rest with its greatest extent or length in the direction of the lines of force. The body will then become a temporary magnet, its south pole being situated where the lines of force enter it, and its north pole where they pass out. The production of magnetism in a magnetic substance in this manner is called magnetic induc- tion. The production of artificial magnetism in a hardened- steel needle or bar by contact with lodestone is only a special case of magnetic induction. 177. The amount, or quantity, of magnetism is expressed by the total number of lines of force contained in a magnetic circuit. 178. Magnetic density is the number of lines of force passing through a unit area measured perpendicularly to the direction of the lines. ELECTRICITY. INTRODUCTION. 179. Electricity is the name given to that which directly causes all electrical phenomena. The word is derived from the Greek word vJAe/crpov, meaning "amber." Although electrical science has made great advances in the last few years, the exact nature of electricity is unknown. Recent researches tend to demonstrate that all electrical phenomena are due to a peculiar state or stress of a medium called ether ; that, when in this condition, the *et1ier pos- sesses potential energy, or capacity for doing ^vork, as is 102 PHYSICS. 4 manifested by attractions and repulsions, by chemical decom- position, and by luminous heating, and various other effects. 180. In all probability, electricity is not a form of mat- ter, for the only physical properties it possesses in common with material substances are indestructibility and elasticity ; it does not possess 'cv eight, extension, nor any of the other physical properties of matter. 181. Electrical science is founded upon the effects pro- duced by the action of certain forces upon matter, and all knowledge of the science is deduced from these effects. The study of the fundamental principles of the science is an analysis of a series of experiments and the classification of the results, under laws and rules. It is not necessary to keep in mind any hypothesis as to the exact nature of elec- tricity; its effects and the laws that govern them are quite similar to those of well known mechanical and natural phenomena, and will be best understood by comparison. Electricity may either appear to reside upon the surfaces of bodies as a charge, under high pressure, or flow through their substance as a current, under comparatively low pressure. That branch of science which treats of charges upon the surfaces of bodies is termed electrostatics, and the charges are said to be static charges. Electrodynamics is that branch which treats of the action of electric currents. ELECTROSTATICS. 183. When a glass rod or a piece of amber is rubbed with a piece of silk or fur, the parts rubbed will be found to have the property of attracting light bodies, such as pieces of silk, wool, feathers, gold leaf, pith, etc., which, after momentary contact, are repelled. These attractions and repulsions are caused by a static charge of electicity, residing upon the surfaces of those bodies. A body in this condition is said to be electrified. PHYSICS. 103 FIG. 45. A better experiment for demonstrating this action is to suspend a small pith ball by a silk thread from a support or bracket, as shown in Fig. 45. Such an apparatus is spoken of as an electric pendulum. If a static charge of electricity is developed on a glass rod by rubbing it with silk, and the rod brought near the pendulum, the ball will be attracted to the rod ; but, after momentary contact, it will be repelled. By this contact the ball becomes electrified, and, so long as the two bodies retain their charges, mutual repulsion will take place whenever they are brought near each other. If a stick of sealing wax, electrified by being rubbed with fur, is brought near to another pendu- lum, the same results will be produced the ball will fly towards the wax, and, after contact, will again be repelled. But the charges respectively developed in the preceding cases are not in the same condition. For if, after the pith ball has been touched with the glass rod and repelled, the electrified sealing wax is brought in the vicinity, attraction takes place between the ball and the sealing wax. Similarly, if the pendulum is charged with the electrified sealing wax, the ball will be repelled by the wax and attracted by the glass rod. An electric charge excited upon glass by rubbing it with silk has been . termed a positive charge (+) ; and that devel- oped on resinous bodies by friction with flannel or fur, a negative charge (). 184:. Neither charge is produced alone, for there is always an equal quantity of both charges produced, one charge appearing on the body rubbed, and an equal amount of the opposite charge upon the person or thing rubbing. 185. The intensity of the charge developed by rubbing the two substances together is independent of the actual 104 PHYSICS. 4 amount of friction that takes place between the bodies. For, in order to obtain the highest possible degree of electrifica- tion, it is only necessary to bring every portion of one surface into intimate contact with every particle or every portion of the other surface; when this is done, no extra amount of rubbing can develop any greater charge upon either sub- stance. 186. From these experiments are derived the following laws : 1. When two dissimilar substances are placed in contact, one of them ahvays assumes the positive and the other the negative condition, altJiough the amount may sometimes be so small as to render its detection very difficult. 2. Electrified bodies with similar charges are mutually repellent, ivliile electrified bodies with dissimilar charges are mutually attractive. 187. In the following list, called the electric series, the substances are arranged in such order that each receives a positive charge when rubbed with any of the bodies follow- ing, and a negative charge when rubbed with any of those preceding it. 1. Fur G. Cotton 11. Sealing wax 2. Flannel 7. Silk 12. Resin 3. Ivory 8. The body 13. Sulphur 4. Crystals 9. Wood 14. Gutta percha 5. Glass 10. Metals 15. Guncotton For example, glass, when rubbed with fur, receives a neg- ative charge ; but when rubbed with silk it receives a positive charge. 188. An electroscope is an instrument for detecting static charges of electricity, and for determining whether they are positive or negative ; it does not measure the intensity of the charges. The pith ball suspended by a silk thread acts as a simple elec- PHYSICS. 105 troscope. A more sensitive electroscope is showrrin Fig. 40, and consists of two gold leaves suspended with- in a glass jar J, which serves to protect them from drafts of air and to support them from con- tact with the earth. The gold leaves a are sup- ported side by side in the jar by a brass rod or wire b, which passes through a cork in the mouth of the jar. The upper end of the brass rod is furnished with a flat metallic plate or ball c. An electrified body, such as the rod d, brought into the vicinity of the electroscope will cause the leaves to repel each other, and they will remain in that position for some time. To determine tJie condition of a charge by tlie electroscope: First, charge the gold leaves with a known charge, such as that developed upon glass when rubbed with silk. The leaves will spread apart, being electrified with a positive charge. When they are thus charged, the approach of a "body positively charged will cause them to open still more widely; while on the approach of one negatively charged, they will close together. FIG. 46. 189. The torsion balance is an instrument used to measure the force exerted between two electrified bodies. It consists of an arm or lever of some light insulating material, such as a straw or piece of wood, provided at one end with a gilt pith ball ?/, Fig. 47, and sus- pended in a glass jar by a fine silver wire. The wire passes FIG. 47 106 PHYSICS. 4 up through a glass tube and is fastened to a brass stopper b, called the torsion Jiead. 'The torsion head is graduated in degrees and is capable of being revolved around upon the glass tube. Another gilt pith ball m is fastened to the end of the vertical glass rod a, which is inserted through an open- ing in the top of the jar. A narrow strip of paper, also divided into degrees, encircles the glass jar at the level of the two pith balls. To use the torsion balance : Turn the torsion head around until the two pith balls m and n just touch each other. Remove the glass rod a, and communicate the charge to be measured to the gilt ball m. Replace the glass rod in the jar. The two gilt balls will touch each other momentarily and half of the charge will pass from m to n. As both balls possess similar charges, they will immediately repel each other; the ball n being driven around, twists up the wire to a certain extent. The resistance of the wire to torsion will eventually balance the force of repulsion, and the ball n will come to rest, the balls being separated by a certain distance. In any wire, the force of torsion is proportional to the amount of twist, or, in this case, to the angle of torsion ; hence, the force exerted between the two balls can be measured by the angle described by the ball n. 19O. By means of the torsion balance, it is proved that the force exerted between two bodies, statically charged witJi electricity, varies inversely as the square of the distance between them. Thus, suppose two electrified bodies \ inch apart repel each other with a certain force ; at a distance of 1 inch, the force would only be one-sixteenth as great. This law is equally true for the force of attraction between two bodies with dissimilar charges. A unit quantity of electricity is that charge which, when placed in air at a distance of 1 centimeter from another equal and similar charge, will be repelled with the force of 1 dyne. NOTE. (a) The centimeter, or the unit of length, represents of the distance from the pole to the equator on the surface 1,000,000,000 4 PHYSICS. 107 of the earth, and is equal to .3937 inch, or 1 inch equals 2.54 centi- meters, nearly. A square centimeter is the area contained in a square each of whose sides is 1 centimeter in length ; 1 square centimeter equals. 155 square inch, or 1 inch equals 6.45 square centimeters, nearly. A cubic centimeter is the volume contained in a cube each of whose edges is 1 centimeter in length; 1 cubic centimeter equals .06102 cubic inch, or 1 cubic inch equals 16.387 cubic centimeters. (b) The gram, or the unit of mass or quantity of matter, represents the quantity of matter contained in a cubic centimeter of pure water at the temperature of its maximum density, which is 4 C., or 39.2 F., and is equal in weight to 15.432 grains. (c] The second, or the unit of time, represents ^ of an ordinary day of 24 hours. This system of units is designated as the absolute or C. G. S. system, to distinguish it from other systems based upon other fundamental units. The secondary units, derived from ^hese fundamental units, are the unit of velocity and the dyne. The unit of velocity, or the rate at which a body changes its relative position, is determined by dividing the distance in centimeters through which a body travels by the time in seconds required to travel that distance. The unit of velocity is, therefore, 1 centimeter per second. To compute the velocity in centimeters per second, divide the num- ber of centimeters by the number of seconds. The dyne, or the unit of force, is that force which, by acting upon a mass of 1 gram for 1 second, can give to it a velocity of 1 centimeter per second. 191. In either case, whether of attraction or repulsion, the force at any given distance is equal to the product of the two quantities of electricity on the bodies. That is, if a cer- tain body were charged with 4 unit quantities of electricity, and another with 3 unit quantities, then the force exerted between them would be 12 times as great as if each had con- tained a charge of 1 unit. 192. Only that part of a dry glass rod which has been rubbed will be electrified; the other parts will produce neither attraction nor repulsion when brought near an elec- troscope. The same is true of a piece of sealing wax or resin. These bodies do not readily conduct electricity, that is, they oppose or resist the passage of electricity through them. Therefore, electricity can reside only as a charge upon that part of their surfaces where it is developed. Experi- ments show that when a metal receives a charge at any point, the electricity immediately passes or flows through its substance to all parts. Metals, therefore, are said to be 108 PHYSICS. 4 good conductors of electricity. Bodies have, accordingly, been divided into two classes; namely, non-conductors or insulators, those bodies that offer an infinitely high resistance to the passage of electricity ; and conductors, or those that offer a comparatively low resistance to its passage. This distinction is not absolute, for all bodies conduct electricity to some extent, while there is no known substance, that does not offer some resistance to its flow. 11)3. Rlcctrical resistance maybe defined as a general property of matter, varying with different substances, by virtue of which matter opposes or resists the passage of elec- tricity. 1 94. Conductivity is the facility with which a body trans- mits electricity, and is the opposite of resistance. For instance, copper is of low resistance and high conductivity; wood is of high resistance and low conductivity. 195. In giving the following list and dividing the differ- ent substances into two classes, it should be understood that it is done only as a guide for the student. Between, these classes are many substances that might be included in either, and no hard-and-fast line can be drawn. The list is arranged in order of the conductivity of the different substances, begin- ning with silver, which is the best known conductor. f Paper . Oils Porcelain Wood Silver NON-CON- Copper DUCTORS CON- Other metals OR DUCTORS i INSULA- Charcoal Ordinary water TORS Silk Resins Gutta percha Shellac Ebonite Paraffin Glass Dry air 4 PHYSICS. 109 196. Arr electric charge will be induced in a conductor when that conductor is brought into the vicinity of an elec- trified body. This effect is termed electrostatic induction, and the range of space in which it can take place is an elec- trostatic field. If the conductor^ B, Fig. 48, is supported from contact with the earth by insulators, and is then brought into the electrostatic field of the conductor C] but not touching (7, which is electrified with ^.positive charge, the following facts will be observed: 1. A charge will be produced on A B, as is shown by the pith balls ... J FIG. 48. spreading apart. 2. This charge will be negative at the end A nearest C, and positive at the end B farthest from C, as can be shown by an electroscope. 3. The charges at A and B are equal to each other; for, if the conductor A B is removed from the vicinity of the conductor C without having touched C, the opposite charges immediately neutralize each other; that is, no electrifica- tion will be indicated by the pith balls. 4. Again, as C is brought nearer and nearer A, the charges on opposite signs on the approaching surfaces attract each other 'more and more strongly until C is brought very near A, when a spark darts across the intervening space. The two charges, rushing together, neutralize each other, leaving the induced positive charge, which was formerly repelled to the end B of the conductor, as a permanent charge over all the surface of A B. 5. Or, if the conductor A B is touched by a conductor connected to the earth when it is under the influence of C, the positive charge will be neutralized by the earth, and the negative charge will remain when A B is removed from the field of C. The charge that passes to the earth from A B is 110 PHYSICS. 4 called a free charge, while that charge which is held by the inductive influence of C is a bound cliarge. Both free and bound charges can be negative or positive, depending upon the sign of the charge on C. When two conducting bodies, both electrified with equal dissimilar charges, are touched together momentarily, the two charges will neutralize each other, no trace of either remaining; but if they are unequal, the smaller charge will neutralize an equal amount from the larger and leave a charge that is equal to the difference between the two original charges, the sign of the remaining charge being the same as that of the larger one. Before the bodies can be separated, the remaining charge will divide equally between the two bodies. For example, two gilt balls A and B are charged respectively with -j- 20 and 4 units of electricity. When the balls are placed in contact, the 4 charge on B will neutralize a +4 charge on A and leave a -f 16 charge, which immediately divides equally between the two balls; that is, a charge of -j- 8 units remains on each ball when they are separated. It is found that the effect of this electrostatic induction is greatly increased by placing some other substance, such as glass or paper, between the two bodies. 197. The facility with -which a body allows electrostatic induction to act across it is called its inductiiw capacity. The inductive capacity varies with different substances, but almost all non-conductors are better than air. 198. The electrophorus, Fig. 40, is an instrument devised for the purpose of obtaining an almost unlimited number of static charges of electricity from one single charge, and is based upon the principle of electrostatic induction. It consists of two main parts: a thin cake of resinous material cast in a round metal dish or pan B, about 1 foot in diameter; and a round disk A of slightly smaller diameter, made of metal and provided with a glass handle. In using PHYSICS. Ill the electrophorus, the resinous cake must first be beaten or rubbed with a warm piece of woolen cloth or fur. The disk, or cover, is then placed on the cake, touched momen- tarily with the finger to liberate the free charge, then removed by taking up by the handle. It is now found to be powerfully electrified with a positive charge; so much so, indeed, as to yield a considerable spark when the hand is FIG. 49. brought near it. The cover may be replaced, touched, and again removed, and will thus yield any number of sparks; the original charge on the resinous plate meanwhile remain- ing practically as strong as ever. 199. A static charge of electricity is not usually distrib- uted uniformly over the surface of conducting bodies. Experiments show that there is more electricity on the edges and corners than upon their flatter parts. The term electric density is used to signify the quantity of electricity residing in a small area at any part of a body, the distribution being supposed to be imiform over that small part of the surface. The electric density is the quotient arising from dividing the total charge of electricity in units of quantity residing upon the surface of a body, by the area of the surface in square inches. For example, a charge of 240 units of electricity is 112 PHYSICS. imparted to a sphere, the surface area of which is 40 square inches; then, the electric density over the surface of the sphere is - 2 T \- = 6 units of electricity per square inch. 200. Electrostatic machines have been devised for the purpose of obtaining larger static charges than can be developed by rubbing the glass rod or by the electrophorus. They consist, mainly, of two parts, one for producing and the other for collecting the charges. There are three important kinds of electrostatic machines : the cylinder, the plate ', and the induction machines. 20 1. The cylinder machine, as usually constructed, con- sists of three principal parts: a cylinder of glass, revolving upon a horizontal axis ; a rubber, or cushion of horsehair, to which is attached a long silk flap; and an insulating metallic cylinder called a prime conductor. In Fig. 50, the FIG. 50. cushion of horsehair a, covered with a coating of amalgam of zinc, presses against the glass cylinder b from behind, allowing the silk flap s to rest upon the upper half of the glass. The prime conductor C is provided at one end with a row of fine metallic spikes and is placed in front of the machine with the row of spikes projecting towards the glass cylinder. When the glass cylinder is rotated, a positive charge is produced upon the glass and a negative charge upon the rubber. The positive charge is carried around upon the glass cylinder, and, just before reaching a position opposite the row of spikes, it acts inductively upon the prime conductor, 4 PHYSICS. 113 attracting a negative charge to the near end and repell- ing a positive charge to the far end. When the positive charge arrives in front of the row of spikes, it will be neu- tralized by the attracting negative charge from the con- ductor, leaving the glass in a neutral condition, ready to be excited again. A positive charge now remains upon the prime conductor and can be utilized for other experiments. 202. The plate machine is similar in all respects to the cylinder machine with the exception that a glass or ebonite plate is used instead of the glass cylinder, and there, are usually two sets of rubbers or cushions, instead of c>ne. Each set of cushions is double; that is, it is made in two parts, with the plate revolving between them. One set of cushions is placed at the top of the machine, and the other at the bottom, with silk flaps extending from each over a quadrant of the plate. The charge is collected on two prime conductors connected by a metal rod, and each is provided with a row of fine spikes at one end. They are placed in such a position that the two rows of fine spikes project towards the glass plate at opposite sides of its horizontal diameter. The electrostatic action of the machine is in all respects the same as that of the cylinder machine. 203. The induction machine differs widely in its action from the two machines previously described. It requires an initial charge from some exterior source to start its action. The initial charge acts inductively across a revolving glass plate, and produces other charges; these charges in turn are conveyed by the moving parts to some other point, where they increase the initial charge, or furnish a supply of electricity to a prime conductor. The two principal machines of this class are the Holtz and the Wimshurst. 204. It has been shown that opposite charges attract and hold one another; that electricity cannot flow through glass, and yet can act across it by induction. If a piece of tin foil is pasted upon the middle of each face of a thin plate 114 PHYSICS. 4 of glass, and one of the pieces is electrified with a positive charge and the other with a negative charge, the two charges will attract each other ; or, in other words, they are held, or bound, by each other. It will be found that these two pieces of tin foil may be charged a great deal stronger in this manner than either of them could possibly be if they were stuck to the glass alone and then electrified. This property of retaining and accumulating a large quantity of static charges, which two conductors possess when placed side by side and separated from each other by a non-con- ductor, is termed capacity. 2O5. A condenser is an apparatus for condensing or accumulating a large quantity of static charges of electricity on a comparatively small surface. It consists of two con- ductors separated by a thin layer of some non-conducting material; one of the plates is entirely insulated from the earth, and the other is connected to it by a conductor. The capacity of a condenser depends upon the size and form of the condensing plates, the thinness of the insula- ting material between them, and the inductive capacity of the insulating material. A convenient form of condenser is called the Ley- den jar, Fig. 51. It consists of a glass jar J coated up to FIG. 51. a certain height on the inside and outside with tin foil. A brass knob a is fixed on the end of a stout brass wire, which passes downwards through a lid or stopper of dry, well 4 PHYSICS. 115 varnished wood, and connected by a loose piece of brass chain with the inner coating of the jar. To charge the jar, the knob is held to the prime conductor C of an electrical machine, the jar being either held in the hand by the outer tin foil coating or connected to the earth by a wire or chain. When a positive charge is thus imparted to the inner coating, it acts inductively on the outer coating, attracting a negative charge in the face of the outer coating nearest the glass, and repelling a positive charge to the out- side of the outer coating. This outer charge then passes through the hand or any conductor to the earth. 2O 7. An electrostatic battery consists of a number of Leyden jars whose inside coatings are all connected together and whose outside coatings are all connected to the earth. ELECTRODYNAMICS. 208. In dealing with electric currents the word potential will be substituted for the general and vague phrase electrical condition. The term potential, as used in electrical science, is analo- gous to pressure in gases, head in liquids, and temperature in heat. When an electrified body, positively charged, is connected to the earth by a conductor, electricity is said to flow from the body to the earth; and, conversely, when an electrified body negatively charged is connected to the earth, electricity is said to flow from the earth to that body. That which determines the direction of floiv is the relative electrical potential or pressure of the two charges with respect to the earth. 209. It is impossible to say with certainty in which direction electricity really flows, or, in other words, to declare which of two points has the higher and which the 116 PHYSICS. 4 lower electrical potential, or pressure. All that can be said with certainty is, that when there is a difference of electrical potential, or pressure, an electric current tends to flow/r centi (TTTO)> anc ^ milli (roVo)- These prefixes form the key to the entire system. In the following tables, the abbreviations of the principal units of these sub- multiples begin with a small letter, while those of the mul- tiples begin with a capital letter. The abbreviations are those adopted by the International Congress of Metric Weights and Measures. Chemists commonly use c. c. instead of cm. 3 for cubic centimeter. TABLE 1. MEASURES OF LENGTH. 10 millimeters (mm.) = 1 centimeter cm. 10 centimeters = 1 decimeter dm. 10 decimeters = 1 meter . . m. 10 meters = 1 dekameter Urn. 10 dekameters. = 1 hektometer Hm. 10 hektometers = 1 kilometer Km. MEASURES OF SURFACE (NOT LAND). 100 square millimeters (mm 2 .) = 1 square centimeter cm 2 . 100 square centimeters 1 square decimeter dm 2 . 100 square decimeters = 1 square meter m 2 . MEASURES OF VOLUME. 1,000 cubic millimeters (mm 3 .).. r . = 1 cubic centimeter cm 3 . 1,000 cubic centimeters = 1 cubic decimeter dm 3 . 1,000 cubic decimeters =1 cubic meter m 3 . 10 THEORETICAL CHEMISTRY. 10 milliliters (ml.) 10 centiliters 10 deciliters 10 liters 10 dekaliters 10 hektoliters. . , MEASURES OF CAPACITY. = 1 centiliter . . . = 1 deciliter = 1 liter = 1 dekaliter . . . = 1 hektoliter. . . = 1 kiloliter... cl. dl. 1. Dl. HI. Kl. NOTE. The liter is equal to the volume occupied by 1 cubic decimeter of pure distilled water at its temperature of maximum density, or 4 C. MEASURES OF WEIGHT. 10 milligrams (mg.). . 10 centigrams 10 decigrams 10 grams 10 dekagrams 10 hektograms 1,000 kilograms 1 centigram eg. 1 decigram dg. 1 gram g. 1 dekagram Dg. 1 hektogram Hg. 1 kilogram Kg. Iton.. T. NOTE. The gram is the weight of 1 cubic centimeter of pure dis- tilled water at a temperature of 4 C. ; the kilogram is the weight of 1 liter of water ; the ton is the weight of 1 cubic meter of water. TABLE 2. CONVERSION TABLE: ENGLISH MEASURES INTO METRIC. English. Metric. Metric. Metric. Metric. Inches to Meters. Feet to Meters. Pounds to Kilograms. Gallons to Liters. 1 0.0253998 0.3047973 0.4535925 3.7853122 2 0.0507996 0.6095946 0.9071850 7.5706244 3 0.0761993 0.9143919 1.3607775 11.3559366 4 0.1015991 1.2191892 1.8143700 15.1412488 5 0.1269989 1.5239865 2.2679625 18.9265610 6 0.1523987 1.8287838 2.7215550 22.7118732 w ^ 0.1777984 2.1335811 3.1751475 26.4971854 8 0.2031982 2.4383784 3.6287400 30.2824976 9 0.2285980 2.7431757 4.0823325 34.0678098 10 0.2539978 3.0479730 4.5359250 37.8531225 THEORETICAL CHEMISTRY. 11 TABLE III. CONVERSION TABLE: METRIC MEASURES INTO ENGLISH. Metric. English.. English. English. English. Meters to Inches. Meters to Feet. Kilograms to Pounds. Liters to Gallons. 1 39.37043*2 3.2808693 2,2046223 0.2641790 2 78.740864 6.5617386 4.4092447 0.5283580 3 118.111296 9.8426079 6.6138670 0.7925371 4 157.481728 13.1234772 8.8184894 1.0567161 5 196.852160 16.4043465 11.0231117 1.3208951 6 236.222592 19.6852158 13.2277340 1.5850741 7 275.593024 22.9660851 15.4323564 1.8492531 8 314.963456 26.2469544 17.6369787 2.1134322 9 354.333888 29.5278237 19.8416011 2.3776112 10 393.704320 32.8086930 22.0462234 2.6417902 15. By using- these tables, one measure can be converted into another by simple addition. It should be remarked that it is not necessary to use all the figures of the different decimals, four decimal places being all that are commonly used; it is only in very exact calculations that all seven decimal places are- required. ILLUSTRATION. Change 6,471.8 feet into meters. Any number, as 6,471.8, may be regarded as 6,000 + 400 + 70 + 1 + .8; also 6,000 = 1,000 X 6; 400 = 100 X 4, etc. Hence, looking in the left-hand column of Table 2 for figure 6 (the first figure of the given num- ber) we find opposite in the third column, which is headed "Feet to Meters," the number 1.8287838. Now, using but four decimal places and increasing the figure in the fourth decimal place by 1 (since the next figure is greater than 5), we get 1.8288. In other words, 6 feet = 1.8288 meters; hence, simply moving the decimal point three places to the right, 6,000 feet = 1, 000 X 1.8288 = 1,828.8. In like manner, 400 feet = 121.92 meters; 70 feet = 21. 336 meters; 1 foot = .3048 meter: 12 THEORETICAL CHEMISTRY. 5 and .8 foot = .2438 meter. Adding, as shown below, the result is 1,972.6046 meters. 1 828.8000 1 2 1.9200 21.3360 .3048 .2438 1 9 7 2.6 4 6 Again, convert 19.635 Kg. into pounds. The work should be perfectly clear from the explanation given below. The result is 43.2877 pounds. 2 2.0 4 6 2 19.8416 1.3228 .0661 .0110 4 3.2 8 7 7 The only difficulty in applying these tables lies in locating the decimal point. Its position may always be determined by applying the following statement : If the figure consid- ered lies to the left of the decimal point, count each figure in order, beginning with units (but calling units place zero) until the desired figure is reached ; then move the decimal point to the right as many places as the figure being con- sidered is to the left of the decimal point. Thus, in the number 6,471.8 above, the 6 lies three places to the left of the 1, which is in units place; hence, the decimal point is moved three places to the rig/it. By exchanging the words "right" and "left," the statement will also apply to deci- mals. Thus, in the second case above, the 5 of the number 19. 635 lies three places to the rig/it of units place; hence, the decimal point in the number taken from the table is moved three places to the left. ENGLISH WEIGHTS AND MEASURES. 10. Familiarity with English weights and measures is assumed ; still, the following particulars may prove useful : One United States gallon has a volume of 231 cubic inches, and contains 4 quarts, or 8 pints. A gallon of pure water weighs 128 ounces, or 58,318 grains; hence, a pint of pure 5 THEORETICAL CHEMISTRY. 13 water weighs 1C ounces, or 1 pound, and from this has come the saying "a pint is a pound," although, of course, a pint of any liquid having a specific gravity differing from that of water would have a different weight. The measure termed ^ fluid ounce is a measure of volume, and not of weight, and is equal to y 1 ^ part of a pint. In other words, it is the vol- ume of 1 ounce of pure water. The fluid ounce bears^the same relation to the avoirdupois ounce as does the cubic centimeter to the gram. When directions are given to dissolve 1 part by weight of a solid in 10 parts (or any other number of parts) by weight of water, either ounces in fluid ounces or grams in cubic centi- meters may be taken. As it is the general custom to use the centigrade ther- mometer in all chemical calculations and investigations, the temperatures given in this and the following Instruction Papers will always be in degrees centigrade, unless other- wise stated. The student should bear in mind that the absolute tem- perature of a body is its temperature in degrees C. -f- 273-1, but in all practical work the fraction is dropped and 273 is used. THE BALANCE. 17. Though the student has, at this early stage, no need of a balance and a set of weights, a full knowledge of how to manipulate the- former is of such vast importance to chemists that a full description will not be out of place at this point. A most useful form of balance, having a weighing capacity of from 1 milligram to 500 grams, is shown in Fig. 1. The balance is enclosed in a glass case with sliding doors, and rests on a solid wooden box A A, in which is a drawer that may be used for weights, etc. A metallic pillar B is screwed to the top of the box, and supports, in a cradle C at the top, the central knife edges of the beam D. On the ends of the beam, at E, are also fitted knife edges, on which hang the hooks F, and from these are suspended the bows G which 14 THEORETICAL CHEMISTRY. carry the pans H. In the center of the front of the base of the balances is fixed a handle, or knob /. In the position shown in the figure, the pans are at rest on the balance base. But, on turning this knob to the right, the whole of the -bal- ance is raised by means of an eccentric, which actuates a sliding rod within the pillar B, and raises the cradle C. The FIG. i. pans are then free to swing, and- their motion is shown by the index finger K, which is fastened to the center of the beam. The end of this index moves in front of a graduated scale, which has a zero mark in the center. Before using the balance, see that it is level and that the index is exactly in front of the 0-ma.rk. On raising the pans by turning the handle /, the index should vibrate an equal number of degrees each side of the zero. Should it not do so, one of the pans is slightly heavier than the other, probably through not being quite clean ; if so, turn the handle back and set the balance to rest and then dust the whole instrument very THEORETICAL CHEMISTRY. 15 carefully, and test again. If still out, the balance may require adjusting, which is done by means of a screw at the end of the beam. If the left-hand pan appears to be too heavy, the screw must be screwed outwards from the center of the beam, and in the opposite direction if the right pan is too heavy. However, do not be too quick in adjusting; try repeatedly to get the scales to balance before touching the adjustment screw; many a good balance has been spoiled and accurate weighing has been made impossible by too frequent adjustment. Every chemist should keep his balance and weights sacred, and should treat them with tJie greatest care and consideration. The weights are usually kept in a box, as shown in Fig. 2. The larger weights are gener- ally made of brass, while the smaller ones are made of aluminum; each weight rests in a separate compartment under a glass lid A B. For the purpose of lifting the FIG. 3. weights, a forceps C is used, which has its place in the box. Accurate weights must tinder no consideration be touched with the fingers, but always lifted with the forceps. 16 THEORETICAL CHEMISTRY. 5 The arrangement of the weights varies with different makes. Fig. 3 shows the usual and most convenient arrangement for a set of weights of from 500 grams to .0001 gram. Before attempting to weigh, the student must learn not only the denomination of each weight, but also its place in the box. He must be quite as well able to re&d tJie weights he has placed in the balance pan from the empty space in the box as from the weights themselves. 18. Method of Weighing. Let us suppose that it becomes necessary to weigh a glass flask. This, in the first place, must be thoroughly cleaned and dried, then placed on the left-hand pan of the balance. Let us assume the weight of the flask to be 238. 847 grams and let us see how these figures are obtained. First, with the balance at rest, place the apparatus in position. Then take the 200-gram weight from the box by means of the forceps and place it on the right-hand pan ; lift the balance by turning the knob / to the right, and notice which pan is heavier ; lower the balance by returning the knob. In this case the weight will not be heavy enough, and the index will swing to the right. Put in addition the 100-gram weight in the pan and lift the balance the weight is too heavy and the index will swing to the left. Do not forget that every time a weight (even the smallest) is added or removed, the balance must be set to the position of rest; i. e. , lowered by turning the handle to the left, and that this must- be done very gently and care- fully. Take off the 100-gram weight and try 50 too much ; remove the 50 and try the 20 not enough ; add another 20 too much; remove the 20 and add 10 not enough; add 5 not enough; add 2 not enough;- add another 2 too much; remove the 2 and add 1 not enough. As the weights get nearer and nearer to the weight of the flask the beam begins to swing more and more slowly. Having now got to a point at which another gram more than turns the scale, commence in the same way with the fractions of a gram. Try first .5 gram not enough; add .2 not enough; 5 THEORETICAL CHEMISTRY. 17 add another .2 too much; remove the .2 and try .1 not enough. As at this stage the beam swings very slowly, it is advisable to start it by wafting a current of air down on one of the pans with the hand, until the index finger swings very nearly to the extremity of the scale. Try next the . 05-gram weight ; the index finger swings perhaps just 6 divisions to the left and 5 to the right the weight is too much ; remove the .05 and try .02 not enough; add another .02 not enough. Next add .005 not enough; but more likely than not- the index finger swings only a fraction of a degree farther to the right than to the left. Add .002 the index finger swings the same number of degrees each side of zero. The weights now exactly balance the flask. Now we must read the weights; this must first be done from the box, reading the empty spaces. In this case we have 200 + 20 + 10 + 5 + 2 4-1 = 238. Against the words "weight of the flask " write this number in your note book. Next read off the decigram weight spaces ; there are empty .5 + .2 + .1 = .8. Write "8" after 238. The centigrams come next; there are .02 + .02 = .04. Write '"4" after 238. 8. The milligrams are read in the same manner, and are .005 +.002 = .007. Write "7" after 238.84. The whole figure will then read, "Weight of flask = 238.847 grams." Having thus read the weight from the empty spaces in the box, next take out the weights and check off your reading as the weights are returned to their proper places. This double reading greatly reduces the chances of error in recording the weight of a substance. This operation has been described in full, because it is the foundation of exact chemical work; these operations are much shorter in practice than they appear on paper, and the student should in no way become discouraged. Remember in every case to use the greatest care when working with a balance; never touch scales or weights with the fingers, nor leave them about out of their proper place in the box; on the contrary, take every precaution to see that they are not soiled or other- wise injured. 18 THEORETICAL CHEMISTRY. 5 LAWS OF CHEMICAL COMBINATION. MOI^ECUXES AND ATOMS. 19. Chemical Definition of Molecule. A molecule is a group of two or more atoms that are united by their affinity. It is the smallest part of any substance that can be obtained by physical division and still exist in a free or uncombined state. 20. Classification of Molecules. Molecules are divided into two classes: 1. Elemental molecules, which are formed by like atoms. 2. Compound molecules, which are formed by unlike atoms. Matter composed of molecules containing- like atoms is termed simple, or elementary, matter ; matter whose molecules are composed of dissimilar atoms is called compound matter. ILLUSTRATION. A mass of gold, of silver, of charcoal, of sulphur, etc., is formed of molecules whose atoms are alike ; therefore, these sub- stances are simple, or elementary, matter. The molecules that com- pose a mass of marble, of sulphuric acid, or of water are composed of unlike atoms ; these substances are examples of compound matter. 2 1 . The simplest way to distinguish elemental molecules from compound molecules is to cause a rearrangement of the atoms between two similar molecules. Elemental molecules do not yield any new kind of matter, whereas compound molecules produce elemental molecules. Let us, for instance, assume that a a and a a are two mole- cules, each composed of two atoms a and a ; then we never can obtain, by any rearrangement, any other molecules than a a and a a ; but should the molecules be a b and a b, that is, compound molecules, and each be composed of the dissimilar atoms a and b, then, by a rearrangement of the atoms, the elemental molecules a a and b b would result. 22. dumber of Elemental Molecules. Although the number of substances we observe around us seems practi- cally unlimited, yet there are comparatively few elemental 5 THEORETICAL CHEMISTRY. 19 molecules. The number of these that has been positively ascertained so far is approximately seventy, and, as every elemental molecule is composed of atoms that are similar to the molecule, it is self-evident that the number of elemental atoms is also approximately seventy. 23. Nomenclature of Elemental Molecules and Atoms. Elemental molecules and their atoms always pos- sess the same name, which, in some instances, is the one by which these substances are known in every-day life, as, for instance, gold, silver, iron, copper, etc. ; in other cases the name is chosen on account of certain striking properties, etc. that the body may exhibit. So, for instance, chlorine obtained its name from the Greek name of its color; cczsium, which is the Latin name for "sky-blue," from the blue lines which characterize this element's spectra, etc. 24. Avogadro's Law and the Conclusions There- from. The Italian physicist Avogadro, in 1811, and, inde- pendently, Ampere, a French chemist, in 1814, as a result of various investigations and experiments, established the fol- lowing law, which by right of precedence is generally known as the law of Avogadro. Equal volumes of all substances, cither elemental or com- pound, in the gaseous state, at the same temperature and pressure, contain an equal number of molecules. From this law it obviously follows : 1. That the molecules of all gaseous bodies must be of equal size. 2. That the weight of any molecule compared with that of a molecule of hydrogen is proportional to the weight of any given volume also compared with an equal volume of hydrogen. If, for instance, 1 liter of chlorine weighs 35.5 times as much as 1 liter of hydrogen, 1 molecule of chlorine must weigh 35.5 times as much as 1 molecule of hydrogen, if the above law is true. 25. If we mix 1 volume of hydrogen and 1 volume of chlorine and expose the mixture to the light, we obtain 20 THEORETICAL CHEMISTRY. 5 2 volumes of hydrochloric acid; and if we assume that 1 volume of these gases contained 500 molecules of hydrogen and chlorine, respectively, we will have 1,000 molecules of the compound. Submitting hydrochloric acid to an analysis, we find that each of its molecules is composed of 1 atom of hydrogen and 1 atom of chlorine; and since the 1,000 mole- cules of hydrochloric acid were formed from 500 molecules of hydrogen and 500 molecules of chlorine, it is evident that each of these molecules must have furnished 2 atoms. From this fact we can state that a molecule of hydrogen is composed of 2 atoms. If, then, we further assume that the weight of an atom of hydrogen is 1, so as to serve as a unit, the weight of a mole- cule of hydrogen, i. e. , its molecular weight, will be 2. 26. By density of a body is meant its mass or quantity of matter, compared with the mass or quantity of matter of an equal volume of some standard body arbitrarily chosen; as hydrogen is chosen as this standard, we may obtain the molecular weight of any substance, elemental or compound, by multiplying its density by 2. ILLUSTRATION. : The density of oxygen is 16; that is, a given volume of it weighs 16 times as much as the same volume of hydrogen ; its molecular weight must, then, evidently be 32, or 16 times greater than that of hydrogen. 27. Atomicity. The number of atoms that compose any elemental molecule is obtained by dividing its molecular weight by its atomic weight. We call an elemental molecule monatomic, diatomic, triatomic, tetratomic, or hexatomic according as its atomicity is 1, 2, 3, 4, or 6. Most elemental molecules are diatomic, and those ele- ments that are not volatile, and that, consequently, cannot be weighed in the gaseous state, are arbitrarily classed among the diatomic molecules. ILLUSTRATION. The molecular weight of chlorine is 71, its atomic weight is 35.5; hence, the chlorine molecule contains 2 atoms, or is said to be diatomic. Arsenic has a molecular weight of 300 and an atomic weight of 75 ; hence, a molecule of arsenic has an atomicity of 4, or is said to be tetratomic. THEORETICAL CHEMISTRY. 28. Table 4 contains the elemental molecules the atom- icity of which has been experimentally determined. TABLE 4. ATOMICITY. Monatomic Mercury Cadmium Zinc Barium Iodine (at 1,500) Bromine (at 1,800) Diatomic Hydrogen Oxygen Fluorine Chlorine Bromine Iodine (below 1,000) Nitrogen Sulphur (above 800) Selenium (above 1,200) Tellurium Tr iatomic Ozone Selenium (below 800) Tetratomic Phosphorus Arsenic Hexatomic Sulphur (about 500) ATOMIC WEIGHT': VALENCE. 29. For chemical purposes an atom may be defined as the unit quantity of an element that may enter into, or be expelled from, a chemical combination. The atomic, or combining, weight is the relative weight of the atom of each element compared with that of hydrogen, which, being the lightest, is taken as unity. 30. Quantity of Combining Power. Assuming the combining power of an atom of hydrogen to be 1, the com- bining power of other atoms will be 1, 2, 3, 4, 5, 0, or 7. This means that some atoms possess a combining power equal to that of hydrogen and unite with one atom of it, while other atoms possess a higher combining power and unite with 2, 3, 4, 5, 6, or 7 atoms of hydrogen. This quantity of combining power is signified by valence; it expresses the number of hydrogen atoms an atom can com- bine with or be exchanged for; since, however, an atom is 22 THEORETICAL CHEMISTRY. 5 not limited to only one combination with another substance, but can form several compounds with it, its valence neces- sarily varies. When the valence of a substance changes, it usually increases or decreases by 2 ; so that an atom of an element may have a valence of 2, 4, or 6, or of 1, 3, 5, or 7. Atoms with even valences are termed artiads, those with odd valences, perissads. According as their valence is 1, 2, 3, 4, 5, 6, or 7, atoms are called monads, dyads, triads, tetrads, pentads, Jicxads, or Jieptads, which names are derived from the Greek numerals. When using the adjective term, the Latin numerals are em- ployed, and we say an atom is univalent, bivalent, trivalent, quadrivalent, quinquivalent, sextivalent, or scptivalent. 31. To understand valence fully, we must remember that atoms combine through their affinity. This affinity is not equally strong in the atoms of the different kinds of ele- mental matter. Taking the affinity of hydrogen as unity, any other atom whose affinity is completely satisfied by uni- ting with 1 atom of hydrogen has the same valence. Other atoms exist whose affinity requires for saturation 2, 3, 4, 5, 6, or 7 atoms; hence their valence which is the same for all similar atoms is said to be 2, 3, 4, etc. The valence of oxygen is 2, because 1 atom of oxygen requires 2 atoms of hydrogen to satisfy its affinity. An atom of phosphorus demands 5 atoms of hydrogen to become fully saturated; hence its valence is 5. 32. Graphic Description of the Valence of Atoms. In order to have a tangible idea of valence, we assume that atoms possess a number of bonds, or links. This is graphically O- O Monad. Dyad. Triad. Tetrad. Pentad. Hexad. Heptad. FIG. 4. expressed by a circle with short lines radiating from it, the number of which indicates the valence. The graphic repre- sentation given in Fig. 4 shows the seven classes of atoms. The number of bonds only, not their direction, is significant. THEORETICAL CHEMISTRY. 23 The valences of the different elements are shown in Table 5. Each element that possesses more than one valence is placed under that heading which signifies the most frequently occurring valence of that element. TABLE 5. VALENCE OF ATOMS.* PERISSADS ARTIADS . Monads Dyads Hydrogen Lithium Oxygen Magnesium Rubidium Zinc Caesium Cadmium Potassium Beryllium Sodium Mercury Silver Copper Fluorine, I, III Calcium Thallium, I, III Strontium Chlorine, I, III, V, VII Barium Bromine, I, III, V, VII Cobalt, II, IV Iodine, I, III, V, VII Nickel, II, IV Sulphur, II, IV, VI Selenium, II, IV, VI Triads Boron Tellurium, II, IV, VI Manganese, II, IV, VI Aluminum Tetrads Gallium Germanium Indium Silicon Yttrium Thorium Scandium Zirconium Ytterbium Carbon Cerium Titanium, II, IV Lanthanum Tin, II, IV Erbium Platinum, II, IV Gold, I, III Palladium, II, IV Antimony, III, V Lead, II, IV Bismuth, III, 'V Iron, II, IV, VI Nitrogen, III, V Chromium, II, IV, VI Phosphorus, III, V Arsenic, III, V Hexads Tungsten, IV, VI Molybdenum, II, IV, VI Ruthenium, II, IV, VI Pentads Rhodium, II, IV, VI Columbium Iridium, II, IV, VI Tantalum Osmium, II, IV, VI Vanadium, III, V Uranium, II, IV, VI * The valence of many of the elements has not been determined with certainty, and consequently it is not claimed that this table is abso- lutely correct. We have aimed to make it as correct, and at the same time as simple, as possible. 24 THEORETICAL CHEMISTRY. 5 SYMBOLS A^D FORMULAS. 33. For convenience of description, each element has an abbreviation of its full name, called its symbol. These abbreviations consist, where practicable, of the initial letter of the Latin name of the element. As, however, there are about seventy elements, and only twenty-six letters in the alphabet, a large number of the symbols are composed of the initial and another distinctive letter selected from the name. Thus, the three elements carbon, chlorine, and copper (cuprum) all have names commencing with U C"; carbon being the most important has the letter C for its symbol, while 67 and Cu stand for chlorine and copper, respectively. As all compound bodies are the result of the combination of elements, they may conveniently be expressed symbolically by placing side by side the symbols of the constituent ele- ments. The symbol of a compound is termed -a, formula. Thus, common salt consists of sodium, whose symbol is Na, and chlorine, whose symbol is 67; accordingly, NaClvs, writ- ten for its formula. The multiplication of atoms is expressed by placing an Arabic numeral to the right of and below the symbol. Thus, S 9 stands for 2 atoms of sulphur; N 3 for 3 atoms of hydrogen. The multiplication of a molecule (formula) is expressed either by enclosing the formula in brackets and placing the multiplier outside to the right of and below the formula, or by placing the multiplier to the left of and in front of the formula. Either (H^O)^ or 4// 2 6> stands for 4 molecules of water ; either (6> 2 ) 4 or 46> 2 stands for 4 molecules of oxygen, each molecule consisting of 2 atoms. The valence of an atom is expressed by placing a Roman numeral or a corresponding number of dashes to the right of and above it. 7 iv , or C'" : ', stands, for instance, for the quadrivalent atom of carbon and 0", or O" , stands for the bivalent atom of oxygen. 5 THEORETICAL CHEMISTRY. 25 34. Combination in Definite Proportions. The symbols, however, signify not only the names of elements, but according to the law the proportions by weight, according to which bodies combine, are invariable for each combination these symbols also indicate the mini- mum proportion in which the different atoms combine. For instance, O not only signifies oxygen, but it also always stands for 16 parts by weight of it; Cl, which stands for chlorine, always signifies 35.5 parts by weight. In this way each of the elements has its own symbol, representing the name and the definite proportion, by weight, in which it combines. From this it is evident that we not only express by a for- mula the qualitative, but also the quantitative, composition of a substance. ILLUSTRATION. H^SO* stands for sulphuric acid, consisting of //a = 2 parts by weight of hydrogen ; S = 32 parts by weight of sulphur ; (9 4 = 64 parts by weight of oxygen. 35. Combination in Multiple Proportions. The combinations of oxygen and hydrogen, H 9 O (water) and Hfli (hydrogen peroxide), afford examples of two elements combining in more than one proportion. It will be observed from the two formulas that in the latter compound exactly twice' the quantity of oxygen is combined with the same quantity of hydrogen as in the former. Taking, also, other series of compounds into consideration, it will be seen that a simple relation exists between the propor- tions of each element dn the various compounds. In fact, not only does chemical combination invariably occur in definite proportion, but when tivo elements combine in more than one proportion they unite in multiple propor- tions. The reason why combination occurs in multiple proportion is, that when two elements combine, 1 atom of the first may combine with either 1, 2, or 3 atoms of the second, but combination with the fraction of an 26 THEORETICAL CHEMISTRY. 5 atom would necessarily be a contradiction of the atomic theory. ILLUSTRATION. Nitrogen forms five distinct compounds with oxygen ; if such quantities of the compound be taken as contain the same weight of nitrogen, the weights of oxygen will be proportional to the numbers 1, 2, 3, 4, and 5. Hyponitrous oxide A^O, 28 parts of nitrogen and 16 parts of oxygen. Nitrogen dioxide A r 2 <9 2 , 28 parts of nitrogen and 32 parts of oxygen. Nitrous oxide A^Og, 28 parts of nitrogen and 48 parts of oxygen. Nitrogen tetroxide N 9 O 4 , 28 parts of nitrogen and 64 parts of oxygen. Nitric oxide N 2 O 5 , 28 parts of nitrogen and 80 parts of oxygen. THE CHEMICAL ELEMENTS. 36. The names of the elements, their symbols, and their combining weight are given in Table 6. The combining weights given are those derived from experiments. For most calculations the nearest whole number or whole num- ber with .5 may be employed. Thus, chlorine maybe taken as 35.5, oxygen as 16, and so on. 37. Metals and !Non-Metals. The elements have been divided into two groups, according to whether they are metallic or non-metallic in their nature. The metallic prop- erties are very decided in a metal such as either iron, copper, or gold; and it is equally obvious that such bodies as sulphur and phosphorus are not metals. But, although the proper- ties of these are very definite, a number of other bodies are much less marked in character ; in fact, there is no well denned line of division between the two classes, as the one vseries gradually overlaps the other. Thus, the element arsenic, which occupies an intermediate position, is placed by some chemists among the metals, and by others among the non-metals or metalloids. In physical character the metals are, as a rule, opaque bodies with a peculiar luster, known as metallic luster, and they are comparatively good conductors of heat and electricity. These 5 THEORETICAL CHEMISTRY. 27 properties, however, do not belong exclusively to the metals, for carbon in the form of graphite has, as is well known, a very decided metallic luster, and conducts electricity well. Chemically, the metals, as a whole, form oxides that act as bases, while the non-metallic oxides form acids, but even in this respect the series overlap each other. (The meaning of the terms oxide, base, and acid is fully explained in subse- quent articles.) The non-metals in Table 6 are indicated by being printed in heavy type, while the metals are printed in ordinary type. As hydrogen is the lightest known element, it has been customary to call its atomic weight 1, and state the atomic weights of other elements in terms of hydrogen as 1. That is, the atomic weight of an element has meant that the ele- ment was that many times as heavy as hydrogen. As it somewhat simplifies accurate chemical calculations to call the atomic weight of oxygen 16, there is a tendency at present to arbitrarily assign this value to oxygen, and to assign atomic weights to the other elements that will agree with this arbitrarily chosen standard. In the accompanying table, taken from the atomic weights published early in the year 1899 by the American Chemical Society, weights based on each of these standards are given in separate columns. As it seems more rational to call the lightest element 1, and use this as the measure for the others, this standard has been retained in this and the following sections. These atomic weights are the results of the latest deter- minations, and are probably more accurate than any pre- viously published; but as they are not absolutely correct, and will probably be changed from time to time, it is thought best to use the atomic weights most generally used, in this and the succeeding papers, but if the student prefers he can easily substitute the values here given in his later work. THEORETICAL CHEMISTRY. TABLE 6. NAMES OF ELEMENTS, THEIR SYMBOLS, AND ATOMIC WEIGHTS. Name. Symbol. Atomic Weight. H = 1. O 16. Aluminum Al Sb As Ba Bi B Br Cd Cs Ca C Cc Cl Cr Co Cb Cu E F G Ge Au H In I Ir Fe La Pb Li Mg Mn 26.91 119.52 74.44 136.39 206 . 54 10.86 79.34 111.54 131.89 39.76 11.91 138.30 35.18 51.74 58.55 93.02 63.12 165.06 18.91 69.38 71.93 195.74 1.00 112.99 125.89 191.66 55.60 137.59 205.36 6.97 24.10 54.57 27.11 120.43 75.01 137.43 208.11 10.95 79.95 112.38 132.89 40.07 12.00 139.35 35.45 52.14 58.99 93.73 63.60 166.32 19.06 69.91 72.48 197.23 1.008 113.85 126.85 193.12 56.02 138.64 206.92 7.03 24.28 54.99 Antimony (stibium) . . . Arsenic Barium Bismuth .... .... Boron Bromine? Cadmium Caesium Calcium Cartoon. Cerium Chlorine Chromium Cobalt Columbium Copper (cuprum) Erbium Fluorine Gallium Germanium Gold (aurum) Hydro r en Indium Iodine Iridium Iron (ferrum) Lanthanum Lead (plumbum) . . Lithium Magnesium Manganese . THEORETICAL CHEMISTRY. 29 TABL.E 6 Continued. Atomic Weight. H = 1. O = 16. Mercury (hydrargyrum) Molybdenum Hg Mo 198.49 95 26 200.00 95 99 Nickel Ni 58 24 58.69 N 13 93 14 04 Osmium Os 189 55 190 99 o 15 88 16 00 Palladium Pd 105 56 106 36 Phosphorus . P 30 79 31 02 Platinum Pt 193 41 194 89 Potassium K 38 82 39.11 Rhodium Rh 102 23 103 01 Rubidium Rb 84.78 85 . 43 Ruthenium Rn 100 91 101 68 Samarium Sin 149 13 150 26 Scandium Sc 43 78 44.12 Selenium St? 78 58 79.17 Silicon Si 28.18 28.40 Silver (arp'entum) 107 11 107.92 Sodium (natrium) Na 22.88 23.05 Strontium Sr 86.95 87.61 Sulphur s 31 83 32.07 Tantalum Ta 181.45 182.84 Tellurium . ... Te 126 52 127.49 Thallium ....-...:- Tl 202.61 204.16 Thorium Th 230.87 232 . 63 Tin (stannum) Sn 118 15 119.05 Titanium Ti 47.79 48.15 Tungsten (wolfram) . . . Uranium W u 183.43 237 77 184.83 239 59 Vanadium V 50.99 51 . 38 Ytterbium Yb 171.88 173.19 Yttrium Y 88 35 89 02 Zinc Zn 64.91 65.41 Zirconium Zr 89 72 90 . 40 30 THEORETICAL CHEMISTRY. 5 At ordinary temperatures, two of the elements mercury and bromine are liquids; five ox)^gen, hydrogen, chlorine, fluorine, and nitrogen are gases. The remaining elements are solids ; but at varying temperatures all have been lique- fied, with the exception of carbon, which so far has only been slightly softened with the highest temperature obtain- able. 38. Electropositive and Electronegative Elements. A more scientific, but not generally accepted, division is to distinguish the elements according to the quality of their combining power. This means that they are divided into: 1. Positive elements, or those that are attracted to the negative electrode in electrolysis. 2. Negative elements, or those that are attracted to the positive electrode. ILLUSTRATION. If salt is decomposed through the influence of elec- tricity, it will be noticed that the chlorine atoms gather at the positive electrode, and are, consequently, called negative, while the sodium atoms collect at the negative electrode and are considered positive atoms. In Table 7 the principal elements are arranged according to their electrochemical character, each atom being positive to any atom placed above it, and negative to any one placed below it. It should be remarked here, that it is absolutely necessary that the student should make himself thoroughly familiar with the names of the elements, their symbols, atomic weights, and electrochemical characters; in fact, he should try to learn these things by heart ; nothing looks so bad as to see a chemist referring to his books to find the symbol, com- bining weight, etc. of an element. 39. Meta-Elements. In Art. 22 the number of ele- ments is given as " approximately seventy"; the reason for not giving a more positive figure is that the standard adopted for elements has become doubtful through recent researches. It has, for instance, been observed that the spectrum of THEORETICAL CHEMISTRY. 31 TABLE 7. ELECTROCHEMICAL CHARACTER OF ELEMENTS. Negative End Oxygen Sulphur Nitrogen Fluorine Chlorine Bromine Iodine Selenium Phosphorus Arsenic Chromium Vanadium Molybdenum Tungsten Boron Carbon Antimony Tellurium Tantalum Columbium Titanium Silicon Tin Hydrogen Gold Osmium Iridium Platinum Rhodium Ruthenium Palladium Mercury Silver Copper Uranium Bismuth Gallium Indium Germanium Lead Cadmium Thallium Cobalt Nickel Iron Zinc Manganese Lanthanum Cerium Thorium Zirconium. Aluminum Scandium Erbium Yttrium Ytterbium Beryllium Magnesium Calcium Strontium Barium Lithium Sodium Potassium Rubidium Caesium Positive End 32 THEORETICAL CHEMISTRY. 5 didymium is capable of resolution into nine separate constit- uents, and that yttrium may be made to yield five different components, each giving a different phosphorescent spec- trum. From this the conclusion has been drawn that such elements as didymium, yttrium, samarium, decipium, philip- pium, holmium, thulium, dyprosium, terbium, gnomium, etc. are not elements in the true sense of the word, but that they are formed of certain simpler substances, which are at present unknown to us, and which are provisionally called meta-elements. 4O. Determination of Atomic Weights. The atomic weight has been defined in Art. 29 as the relative weight of any atom referred to hydrogen as unity. In order to deter- mine the atomic weight of an atom, we have to know: 1. The quantity, by weight, of the element that combines with 1 atom of hydrogen. 2. The molecular weight of the hydrogen compound. As the absolute weight of each constituent of a certain quantity, usually 100 parts, of a hydrogen compound may be obtained by analysis ; the quantity, by weight, of a body that unites with 1 part of hydrogen may be obtained by simple proportion, and as the molecular weight of any substance represents the sum of the weights of its constituent atoms, the combining weights obtained by analysis, when added together, must either give the molecular weight, or a num- ber of which the molecular weight is a multiple. EXAMPLE 1. A quantity of hydrogen iodide contains, according to analysis, 99.22 per cent, of iodine and .78 per cent, of hydrogen; the molecular weight of hydrogen iodide is 128. What is the atomic weight of iodine ? SOLUTION. The proportion .78 : 99.22 = 1 : 127 shows that in this compound the smallest quantity of iodine that can combine with 1 part of hydrogen weighs 127 times as much as hydrogen, and as the sum of the combining weights (127 + 1) is 128, which is also the molecular weight of hydrogen iodide, the atomic weight of iodine is 127. Ans. EXAMPLE 2. Hydrogen arsenide, or arsine, is shown by analysis to contain 3.85 per cent, of hydrogen and 96.15 per cent, of arsenic; its molecular weight is 78. What is the atomic weight of arsenic ? 5 THEORETICAL CHEMISTRY. 33 SOLUTION. From the proportion 3.85 : 96.15 = 1 : 25, it is seen that 25 parts of arsenic combine with 1 part of hydrogen. If there is only 1 atom of hydrogen in 1 molecule of arsine, then 25, being the smallest quantity, by weight, in which arsenic combines, will be its atomic weight ; but adding the two combining weights (25 + 1) together we obtain only 26, which, however, is exactly one-third of the given molecular weight of arsine. Hence, a molecule of arsine is composed of 3 parts of hydrogen and 25 X 3 = 75 parts of arsenic. 75 is, therefore, the atomic weight of arsenic. Ans. 41. Indirect Method for Determination of Atomic Weights. Some elements do not unite directly with hydro- gen ; the comparison with hydrogen must, then, be made by means of another element, for which purpose chlorine is generally chosen. EXAMPLE. There is no known hydrogen compound of silver, but there exists a well known combination of silver and chlorine, called horn-silver. On analysis, horn-silver yields 75.26 per cent, of silver and 24.74 per cent, of chlorine ; its molecular weight is 143.5. What is the atomic weight of silver ? SOLUTION. As 35. 5 parts of chlorine combine with 1 part of hydrogen, the quantity of silver that combines with 35.5 parts of chlorine is its atomic weight'; 24.74 : 75.26 = 35.5 : 108. Since 108 + 35.5 = 143.5, the molecular weight, 108, thus obtained must be the atomic weight of silver. Ans. EXAMPLES FOB PRACTICE. 42. Solve the following : 1. The density of arsenic vapor is 150; what is its molecular weight? Ans. 300. 2. The density of phosphorus is 62 ; what is its molecular weight ? Ans. 124. 3. The density of phosphorus is 62, and its atomic weight is 31 ; how many atoms compose a phosphorus molecule ? Ans. 4 atoms. 4. The density of bromine is 80, its atomic weight is 80 ; how many atoms compose a bromine molecule ? Ans. 2 atoms. 5. Water, a compound of hydrogen and oxygen, has a molecular weight of 18. On analysis it yields 88.89 per cent, of oxygen and 11.11 per cent, of hydrogen ; what is the atomic weight of oxygen ? Ans. 16. 6. Marsh gas is a compound containing 75 per cent, of carbon and 25 per cent, of hydrogen. Its density is 8 ; what is the atomic weight of carbon ? Ans. 12. 34 THEORETICAL CHEMISTRY. 5 7. The analysis of aurous chloride gives 84.7 per cent, of gold and 15.3 per cent, of chlorine. The density of the compound is 116 ; what is the atomic weight of gold ? Ans. 196.5. 8. The analysis of stannous chloride gives 62.48 per cent, of tin and 37.57 per cent, of chlorine. The molecular weight of the compound is 189; what is the atomic weight of tin ? Ans. 118. 43. Relation of Atomic Weight to Specific Heat. When equal weights of different bodies are raised through the same number of degrees of temperature, they take up different amounts of heat ; or, in other words, different bodies possess different capacities for heat. Thus, 1 kilogram of water requires thirty times as much heat as 1 kilogram of mercury to raise its temperature 1 degree ; that is, if the quantity of heat required to raise the temperature of 1 kilo- gram of water 1 degree is represented by 1, the quantity of heat required to raise the same weight of mercury 1 degree will be represented by -$. This fraction expresses what is called the specific heat of mercury. The specific heat of a solid or liquid body is, then, the amount of heat required to raise the temperature of a certain weight of this body 1 degree, the amount of heat required to raise the temperature of an equal weight of water 1 degree being taken as unity. The specific heat of a substance varies according as the substance is in the solid, liquid, or gaseous state, but the specific heat of all solid elements exhibits a remarkable rela- tion to their atomic weight. In 1819 Dulong and Petit observed that if, instead of calculating the specific heats for equal weights, such weights be taken as represent the atomic weights of the solid elements, the numbers expressing the capacity for heat of the different atoms are sensibly equal ; from which it follows that all solid elements possess the same atomic heat, and that the specific heats of solid elements are inversely as their atomic weights. Hence, we have in the determination of specific heat a means of checking the atomic weight of a solid element, or of ascertaining it in doubtful cases. This is done by dividing 5 THEORETICAL CHEMISTRY. 35 the atomic heat of a solid element by its specific heat, the quotient thus obtained being the atomic weight. The num- ber 6.4 represents, approximately, the atomic heat of the solid elements. SPECIFIC HEAT. ATOMIC HEAT. ATOMIC WEIGHT. ILLUSTRATION. Lead .031 6.41 206.77 Platinum .032 6.33 197.81 Silver... .059 6.37 107.97 Sulphur. .2026 6.48 31.98 Tin 054 6.37 117.96 Copper.. .0951 6.04 63.51 COMPOUND MOLECULES. 44. A compound molecule may be defined as a molecule composed of dissimilar atoms, which are united according to the law of valence. The number of atoms that take part in this union is apparently unlimited; while a molecule of hydrochloric acid contains but two atoms, and a molecule of sulphuric acid 7 atoms, certain organic compounds contain a much higher number. Protagon, for instance, a compound of lecithin and cerebrin, both obtained from the substance of the human brain, contains 384 atoms. The molecular weight of a compound molecule is equal to the sum of the weights of its component atoms, and also to twice its density in the gaseous state. Thus, an atom of chlorine weighing 35.5 unites with an atom of hydrogen weighing 1, to form a molecule of hydrogen chloride (hydro- chloric acid), whose molecular weight is 36.5. Or, 2 atoms of hydrogen weighing 2, 1 atom of sulphur weighing 32, and 4 atoms of oxygen weighing 64 unite to form a sulphuric-acid molecule weighing OS. 45. Compound molecules are classified as: 1. Binary molecules; that is, molecules which, whatever the number of atoms maybe, contain only two different kinds of atoms. 2. Ternary molecules; that is, molecules whose dissimilar atoms are united by the aid of some third atom. 36 THEORETICAL CHEMISTRY. 5 BINARY COMPOUNDS. 46. Nomenclature of Binary Molecules. The names of chemical compounds are fixed by rule, but in the case of the more well known compounds the old, every-day names are commonly used; hardly any one, for instance, would think of saying ' ' hydrogen monoxide " for water. The name of a compound should, as far as possible, indicate its compo- sition; this end is attained by making the name of a binary compound consist of derivatives of the names of its compo- nents. There are, unfortunately, several modifications of each name, and English-speaking chemists have not always used a uniform method of nomenclature ; the student, there- fore, must make himself familiar with all of them, as, in the course of reading various books and manuals on chemistry, he is sure to meet with the same body under different names. Very little practice is sufficient, however, to overcome this little difficulty. The names of the binary molecules are derived' from their constituent atoms. The most frequently occurring binaries are those composed of a metal and a metalloid (non-metal) ; the name of the metal is first written, and then that of the metalloid, one or more syllables being removed from the latter and the termination ide added. ILLUSTRATION. Copper and oxygen yield copper oxide. Magnesium and oxygen yield magnesium oxide. Silver and sulphur yield silver sulphide. Zinc and phosphorus yield zinc phosphide. Calcium and iodine yield calcium iodide. Aluminum and bromine yield aluminum bromide. Sodium and chlorine yield sodium chloride. Potassium and nitrogen yield potassium nitride. Barium and fluorine yield barium fluoride. - Cadmium and selenium yield cadmium selenide. The termination ide is always characteristic of a binary compound. 47. Modification of This Rule. As it frequently happens that more than one compound of the same elements 5 THEORETICAL CHEMISTRY. 37 is known, it becomes necessary to distinguish these com- pounds. Whenever the metal atom enters into combination with more than one valence, this fact is indicated in the com- pound by changing the termination of the name of this atom into ic or ous. Should the metal act with only two valences, then in the higher one its name takes the termination ic, and in the lower one the termination o us. ILLUSTRATION. Bivalent mercury and oxygen yield murcuric oxide. Univalent mercury and oxygen yield mercurous oxide. Quadrivalent tin and chlorine yield stannic chloride. Bivalent tin and chlorine yield stannous chloride. 48. Use of Prefixes. If the metallic constituent acts with more than two valences, the termination ic being given on the discovery of the compound is generally arbi- trarily assigned, and a further discrimination becomes again necessary; this distinction is marked by the use of a prefix. A compound in which the valence of the first, or metallic, constituent is less than in the ous, takes the prefix Jiypo, which means "under." When the valence is above ic, the prefix per is used. The termination of the second con- stituent, however, always remains ide. ILLUSTRATION. Quinquivalent chlorine and oxygen form chloric oxide. Trivalent chlorine and oxygen form chlorous oxide. Univalent chlorine and oxygen form hypochlorous oxide. Septivalent chlorine and oxygen form perchloric oxide. Sexivalent sulphur and oxygen form sulphuric oxide. Quadrivalent sulphur and oxygen form sulphurous oxide. Bivalent sulphur and oxygen form hyposulphurous oxide. Quinquivalent nitrogen and oxygen form nitric oxide. Trivalent nitrogen and oxygen form nitrous oxide. Univalent nitrogen and oxygen form hyponitrous oxide. 38 THEORETICAL CHEMISTRY. 5 49. Use of ^Numeral Prefixes. In order to indicate the number of atoms of each constituent, the Greek numer- als are prefixed to the name of the constituent. ILLUSTRATION. 1 atom of Cand 2 atoms of O form carbon dioxide. 1 atom of P and 5 atoms of Br form phosphorous pentabromide. 2 atoms of Fe and 3 atoms of O form diferric trioxide. 8 atoms of Ti and 4 atoms of TV form trititanic tetra- nitride. 50. Formation of Binaries. A binary molecule con- sists of atoms with either equal or different valences. Atoms having the same valence unite in the proportion of 1 to 1 ; that is, the atoms mutually saturate each other and their chemism is satisfied. ILLUSTRATION. K' and C/', both monads, form K' Cl' or K-CL S" and O", both dyads, form S"O" or S=O. Pt and Sz iv , both tetrads, form /V iv S/ iv or Pf^Si. When atoms with different valences unite, each atom has to furnish the same number of bonds to satisfy their chem- ism. This number is the least common multiple of the two different valences. The absolute number of atoms of each constituent is obtained by dividing" the least common multi- ple by the valence of each atom. ILLUSTRATION. When a tetrad and a hexad combine, each must furnish 12 bonds, 12 being the least common multiple of 4 and 6. To do this, 3 tetrads and 2 hexads are required. The following formulas will further illustrate this law : //', a monad, and O", a dyad, form H'f>" . N 1 ", a triad, and O", a dyad, form A'0' 3 '. Sn iv , a tetrad, and S", a dyad, form Sn iv S". 7 V , a pentad, and O", a dyad, form / 2 V <9' 5 '. C/ vii , a heptad, and O", a dyad, form CY<9'/. 6V iv , a tetrad, and F , a monad, form Si ' F'^. Whenever a perissad and an artiad atom form a molecule, the required number of atoms is inversely as the valence of each. 51. It was previously stated that atoms do not exist in a free state; consequently, the formation of compound 5 THEORETICAL CHEMISTRY. 39 molecules may be considered as a process in which an exchange of atoms takes place between a number of mole- cules that are brought in close proximity ; a fact that may easily be explained by the theory that dissimilar atoms possess a far greater attraction for each other than similar atoms. ILLUSTRATION. A molecule of sulphur S-S is brought under suitable conditions in contact with a molecule of oxygen O-O. The stronger attraction of the atoms of oxygen for the atoms of sulphur causes them to form a new combination, and the result, in this case, is SO + SO, or 2 molecules of hyposulphurous oxide. Graphically, it may be repre- sented in the following way : After Combination. Before Combination. 52. Saturated .and Uiisa titrated Molecules. The groups of atoms now being considered are called saturated molecules, because the bonds of their constituent atoms are all mutually engaged. We distinguish between these and u nsatnr at ed groups of atoms, which, possessing free bonds, may enter into combination like single atoms. These unsat- urated groups of atoms are known as compound radicals (or radicles). They do not exist free in nature, although, like atoms, -by combining with another similar group they may form a saturated molecule. Their valence is equal to the number of their unsatisfied bonds, which is the difference of the valence of their components. ILLUSTRATION. Water consists of H-O-H '; supposing 1 atom of hydrogen to be removed, the unsaturated group H-O- remains, which, though consisting of 2 atoms, is able to enter into the formation of a molecule like any single atom. It possesses one free bond and acts therefore as a monad. It is expressed in formulaic form thus: (//'(?")'. Combining with another similar group it forms a saturated molecule (ff'O"}' + (N'O")' = H-O-O-H, or H*O*. Or the pentad P v may be partly saturated by the dyad O" , forming the trivalent compound 40 THEORETICAL CHEMISTRY. 5 radical (P v O")" f , which may combine like a trivalent atom, say with 67, forming (P v O")'"C/ 3 . 53. Names of Compound Radicals. The names of compound radicals terminate in yl. The root of the name comes either from constituents of the radical or from some compound into which it enters. ILLUSTRATION. The compound radical (H'O") r is called hydroxyl, (P V O")'" is termQ&phosphoryl, (C iv 7/ 3 )' is termed methyl, from methyl alcohol, of which it is a constituent. The compound radicals (H^N}' amidogen, (CN)' cyanogen, and (H^N}' ammonium are the only three exceptions to this rule. Molecules containing a compound radical, united to an atom, are regarded as binaries and named accordingly. ILLUSTRATION. (NO*)' and Cl form nitryl chloride, (CO)" and 5 yield carbonyl sulphide, etc. TERNARY COMPOUNDS. 54. The name ternary is applied to those compounds containing three elements. The third atom, which performs a linking function, must be a polyad, an atom whose valence is greater than 1, since no monad can join, other atoms together. Among the compounds higher than the binary series are those formed by the union of water with oxides, thus forming acids and hydrates (hydroxides) and derivatives of these bodies. 55. Acids. The name acid is a familiar one, because it is continually applied in every-day parlance to anything that is sour. A number of bodies possess this distinction in com- mon ; to the chemist the sourness of an acid is but an acci- dental property, as, according to his definition of these bodies, substances are included as acids that are not sour to the taste. An acid may be defined as a body containing hydrogen, which hydrogen may be replaced by a metal, or group of elements equivalent to a metal. It may also be defined as a compound that will unite with a base, forming a salt and water. As a class the acids arc sour; they are also 5 THEORETICAL CHEMISTRY. 41 active chemical agents. Most acids are characterized by the property of changing the color of a solution of litmus, a naturally blue body, to a red tint. Oxygen is a constituent of most acids, the members of this group being distinguished as oxy acids. The oxides that by union with water form acids are termed anhydrides, or anhydrous acids. They are in most cases non-metallic oxides, but sometimes consist of metals combined with a compara- tively large number of atoms of oxygen. There are a few acids in which oxygen is absent ; these are termed hydr acids; hydrochloric acid HCl is an example of this class. Accord- ing to the -definition given, hydrogen is an essential constitu- ent of all acids. It should, however, be mentioned that some chemists apply the term "acid" to what are here termed "anhydrides"; consequently, CO 3 (carbon dioxide or carbonic anhydride) is sometimes described as carbonic- acid gas. This name is now, however, being replaced by the one more in accordance with the theories of modern chemical science. 56. Bases and Alkalies. The oxides that, when dis- solved in water, restore the blue color to reddened litmus are called alkalies; they form a subdivision of a larger class of oxides, the whole of which combine readily with acids, and are known as bases. A base is a compound, usually an oxide or a hydrate, of a metal (or group of elements equivalent to a metal), and this metal (or group of elements] is capable of replacing the hydro- gen of an acid, when the two are placed in contact. The greater number of metallic oxides are bases. Bases, as well as acids, differ considerably in their chemical activity. As already mentioned, certain bases that dissolve in water are termed alkalies. An alkali is a base of a specially active character. It can easily be recognized by its solubility in water, to which it imparts a soapy taste and feeling, and also by its ability to restore the blue color to vegetable blues, such as litmus, that have been previously reddened by an acid. The principal 42 THEORETICAL CHEMISTRY. 5 alkalies are sodium Jiydrate NaHO and potassium Jiydrate KHO. A solution of ammonia gas in water is also alkaline, and is often termed ammonium Jiydrate. Paper tinted yellow with turmeric is also used as a test for alkalies, which give the paper a reddish-brown color. The hydrates are mostly compounds of metallic oxides with water; they are also sometimes termed hydroxides. (The word hydroxide, however, is objectionable, because it infringes on the restriction of ide to the names of binary compounds. ) Their formation is represented by the follow- ing equations: NajO + H 9 O = 2 NaHO sodium oxide water sodium hydrate CaO 4- H t O = Ca(OH\ calcium oxide water calcium hydrate 57. Salts. When an acid and a base react on each other, the substance or compound produ'ced by the replacement of the hydrogen of the acid by the metal of the base is termed a salt. Water is also formed during the reaction. The action of the stronger acids and bases on each other is very violent; the resulting salts are usually without action on litmus. This is, however, not always the case, for when a strong acid combines with a weak base, the salt is acid to litmus, of which nitrate of mercury is an example. When the base is strong and the acid is weak, as in sodium carbonate, the salt has an alkaline reaction. The following are instances of the formation of salts by the union of acids and bases : HCl + NaHO = NaCl + H^O hydrochloric sodium sodium , acid hydrate chloride HCl + NH^HO = NHfl + H t O hydrochloric ammonium ammonium acid hydrate chloride T T r* /") I / r "Vv /") /^"yv C /") I T T S~) Jj o (_/ ( C-C/ ^^ O#>jC/ r- tt {s sulphuric calcium calcium acid oxide sulphate wai 5 THEORETICAL CHEMISTRY. 43 .Litmus, which has been repeatedly spoken of in connec- tion with the ternary compounds, is itself a salt of a vegetable acid and a vegetable base possessing a blue color. An acid, when added, displaces the weak vegetable acid and with the remaining vegetable base forms a salt; the liberated litmus acid, being red, gives the red tint to the solution. On adding to this solution another base, this latter combines with the stronger acid, displacing in turn the vegetable base. The displaced vegetable acid and base again unite, and the original blue color is also restored. 58. Nomenclature of Acids and Salts. The names of acids are derived from those of their principal constituents, changing them into adjectives ending in ic. From sulphur we have sulpJiuric acid ; from nitrogen, nitric acid ; etc. Hydr acids are distinguished by the prefix liydro; for instance, hydrochloric acid. The names of the corresponding salts are derived from the same root by adding ate. Thus, the salts of sulphuric and nitric acids are called sulphates and nitrates, respectively. If an element should form two oxides, both of which combine with water to form acids, then the acid contain- ing the higher proportion of oxygen receives the name ending in ic, while for the termination of the other acid ous is chosen. ILLUSTRATION. Two oxides of sulphur are known, S<9 2 , or sulphurous oxide, and SO 3 , or sulphuric oxide. Each combines with water to form acids. ' SO, + sulphurous at AT sulphurous oxide acid SO 3 + H^O = H^SO, The salts of acids whose names end in ous have the ter- mination ite. H^SO^ + ZNaHO = Na^SO^ + 2ff t O sulphurous sodium sodium . acid hydrate "" sulphite 44 THEORETICAL CHEMISTRY. 5 The same termination that is applied to the base is also applied to the salt. Thus, two different sulphates of mercury are obtained, namely, when sulphuric acid is caused to react with mercurous oxide and mercuric oxide, respectively. H,SO t + Hg,0 = J/g.SO t + Hft sulphuric mercurous mercurous acid oxide sulphate H^SO^ + HgO = HgSO^ + H,0 sulphuric mercuric mercuric acid oxide sulphate The salts of hydracids, being necessarily binaries, have names ending in ide. 1HCI + HgO = HgCl, + H^O hydrochloric mercuric mercuric acid oxide chloride 59. Basicity of Acids. In the formation of salts, the hydrogen of an acid is replaced by a metal. Hydrogen, being univalent, is replaced by different metals in propor- tions varying according to their valence. The valence of the displacing metals must be equal to that of the total number of hydrogen atoms displaced. Thus, a monad will replace 1 atom of hydrogen; a dyad, 2 atoms, and so on. The quantity of a base required to form a salt with an acid depends on the number of atoms of hydrogen present capable of replacement by the base; the measure of this quantity is termed basicity of the acid. Those acids containing but 1 atom of replaceable hydrogen are termed monobasic, those with 2 or 3 atoms of replaceable hydrogen are called dibasic and tribasic, respectively. A monobasic acid requires 1 atom of a monad metal to form a salt, while with a bivalent metal 2 molecules of the acid are necessary. In every case such a number of mole- cules of acid and atoms of metal must be taken as will give the same total valence of replaced hydrogen atoms as that 5 THEORETICAL CHEMISTRY. 45 of the replacing metal. The following typical examples will illustrate this principle: MONOBASIC ACID. HNO* + Na'OH = Na' NO* + nitric acid nitric acid nitric acid sodium hydrate calcium hydrate bismuth hydrate sodium nitrate calcium nitrate bismuth nitrate sulphuric acid sulphuric acid 3// 2 S<9 4 sulphuric acid DIBASIC ACID. ZNa'OH = Na^SO, + sodium sodium hydrate sulphate a"(OH^ = CaSO< + calcium calcium hydrate sulphate bismuth bismuth hydrate sulphate TRIBASIC ACID. phosphoric acid phosphoric acid phosphoric acid sodium hydrate sodium phosphate calcium hydrate bismuth hydrate calcium phosphate = Bi'"PO, bismuth phosphate 6O. Acid, Normal, and Basic Salts. Salts such as those whose formulas are given in the preceding examples are known as normal salts. A normal salt may be defined as a salt produced by the replacement of the hydrogen of the acid by its valence equivalent of a metal, or by a group of elements that acts as a metal, in a base. In the case of those acids containing more than 1 replace- able atom of hydrogen, if a portion only of such hydrogen is 46 THEORETICAL CHEMISTRY. 5 replaced, the result is a salt that still possesses an acid char- acter. An acid salt may, then, be defined as a salt produced by the partial replacement of the hydrogen atoms ofapolybasic acid by the metal of a base ; as, for instance: H^SO^ + NaHO = NaHSO^ + HJ) sulphuric sodium hydrosodium acid hydrate sulphate Certain oxyacids exist that possess the power of combining with more of the base than is necessary to produce a normal salt, thereby forming what are termed basic salts. A basic salt may, then, be defined as a salt produced by the combina- tion of an acid with a higher proportion of a base than is necessary for the formation of a normal salt. Yellow basic mercuric sulphate, known as turpeth mineral, is an example of a basic salt, the additional quantity of base being distin- guished here by being placed after the comma in the formula Its formula is generally written thus: 61. Classification of Acids. Oxyacids are sometimes divided into ortho- and meta-acids, but this distinction is not important and these terms are often used in a rather loose way. In its strictest sense, an ortho-acid is one containing as many hydroxyl groups as the maximum valence of the principal element. Thus, phosphorus, having a maxi- mum valence of 5 (as is shown by the formation of PCl^), might be expected to form an acid having the formula P(OH) b , and this would be called ortho-phosphoric acid. As a matter of fact, this acid is not known, and only a very few acids formed in this way do exist ; hence, the term is often used in a broader sense and is applied to the normal tri- basic acids. Thus, PO(OH) Z is frequently called ortho- phosphoric acid. A meta-acid is one formed by extracting water from an ortho-acid. Hence, if we use the term ortho in its strictest sense in the case just mentioned, and call P(OH\ ortho- phosphoric acid, the acid usually called ortho- phosphoric PO(OH)^, which would be formed by removing one molecule 5 THEORETICAL CHEMISTRY. 47 of water from P(OH}^ would be a meta-acid; and by remov- ing another molecule of water we would have another meta-acid PO^OH), which is usually known as meta-plios- phoric acid. Thus, we would have two meta-acids of phos- phorus. To avoid confusion, when the term ortJio is used in its strictest sense, the term anhydro-acids is sometimes applied to those obtained by the removal of water from the ortho-acids, and the term meta is reserved for a particular acid. This distinction is only important in studying 1 the theoretical formation of acids, but the terms ortho and meta are important in organic chemistry where they are applied in a different manner. VOLUME RELATIONS OF MOLECULES. 62. Molecular Volume. The fact that all gases, whether elementary or compound, expand and contract at the same rate when subjected to variations of temperature and pressure, has an important bearing on their probable molecular conditions. Their similarity in this respect has led to the assumption expressed in Avogadro's law (see Art. 24), from which it follows that, at the same temperature and under the same pressure, the volume of any gaseous molecule is the same, whatever may be the nature and com- position of the gas. Every molecular formula, therefore, not only expresses the weight but also the volume of the molecule. The molecule of hydrogen is chosen as the standard of molecular volume ; but as it is sometimes con- venient to speak of atomic volume which, however, is inconsistent since atoms do not exist free the volume occupied by an atom of hydrogen is taken as unity ; the vol- ume occupied by a molecule of hydrogen will be 2, and, since all molecules occupy the same volume, the volume of a molecule of any gaseous body is assumed to be 2, also, when this system is used. 63. Density and Molecular Weight. The density of a gas has already been defined as the weight of any volume 48 THEORETICAL CHEMISTRY. 5 compared with that of the same volume of hydrogen (see Art. 26). As the molecule of hydrogen contains two atoms, its molecular weight, expressed in terms of atomic weight, is, consequently, 2. The molecular weight of any body in the gaseous state is the weight of that volume which occu- pies the same space as do 2 parts by weight of hydrogen. Conversely, as the molecular weight is the sum of the weights of the constituent atoms, the density of a gas may be determined from its formula. As the molecule of hydro- gen weighs 2, the density of any gas, whether elementary or compound, is identical with the half of its molecular weight. For instance, the molecule of steam weighs 1 8, and occupies the same space as the molecule of hydrogen, which weighs 2 ; the quantity of steam occupying the same space as 1 vol- ume -of hydrogen must, therefore, weigh - = 9. 64. Absolute Weight of Hydrogen. Hydrogen being taken as the unit of density, its absolute weight is of great importance. As a result of most careful weighing, it has been found that 1 liter of hydrogen at the normal tempera- ture and pressure weighs, approximately, .0896 gram, or 11.16 liters weigh 1 gram. It is absolutely necessary for the student to remember that the weight of 1 liter of hydro- gen at and 760 millimeters pressure is .0896 gram, and he should therefore thoroughly impress it on his mind. Know- ing this, there is absolutely no difficulty in calculating the weight of any other gas whose composition is known. The weight in grams of a liter of any gas is its density multiplied by .0896; further, the weight in grams of 11.16 liters of any gas is identical with the number representing its density. The density of oxygen, for instance, being 16, the weight of 1 liter at N. T. P.* is .0896x16 = 1.4336 grams. The weight of a compound gas is found just as easily. The den- sity of hydrochloric acid is 18.18; therefore, the weight of 1 liter of this acid is .0896 X 18.18 1.6289 grams, or 11.16 liters weigh 18.18 grams. * N. T. P. is used as abbreviation of normal temperature and pres- sure. THEORETICAL CHEMISTRY. 40 This constant, .0896, is of such frequent occurrence in chemical calculations that it has been proposed to give it a distinct -name, crith ; the weight of a liter of any other gas is then expressed as so many criths. 65. Diffusion of Gases. Diffusion is the intermixture of molecules brought about by their power of moving against one another. This power is possessed in the highest degree by gaseous molecules, all gases being capable of perfect and comparatively rapid intermixture. Some liquids, such as alcohol and water, also intermix perfectly, although com- paratively slowly ; while other liquids, such as oil and water, diffuse into each other only to a very limited extent. When alcohol is poured on to water it forms a separate layer on the surface of the water, because it is specifically the lighter of the two. 'After a time, however, the two layers will no longer be discernible, and the liquid will be a homogeneous mixture of alcohol and water. In the same way, hydrogen will float on air, but only for a very short time, since the rate of diffusion of this gas is ex- tremely rapid. A homogeneous mixture of hydrogen and air will be speedily formed. The molecules of all gases are constantly moving, but they do not move with the same velocity, so that some gases diffuse more rapidly than others. This has been discovered by confining a gas in a vessel closed by some material having very .minute pores, and immersing the vessel in an atmosphere of some other gas. The pas- sage of the molecules through the pores of the material closing the vessel is sufficiently slow to allow of a comparison between the velocities of passage or rates of diffusion of the two gases. For this purpose, a diffusion tube (see Fig. 5) is employed. It consists of a glass tube A, closed at one end by a plate of piaster of Paris B. If this tube is filled with hydrogen and FIG. 5. 50 THEORETICAL CHEMISTRY. 5 its open end immersed in colored water, the water will be observed to rise rapidly in the tube on account of the rapid escape of the hydrogen through the pores of the plaster of Paris. The external air, of course, passes into the tube through the pores at the same time, but much less rapidly than the hydrogen passes out, so that the ascent of the column of water C marks the difference between the volume of hydrogen that passes out and the volume of air. that passes into the tube in a given time, and allows a measurement to be made of the rate of diffusion, that is, of the velocity with which the gas issues as compared with the velocity with which the air enters, this velocity being always taken as unity. To determine the rate of diffusion it is, of course, necessary to maintain the water at the same level within and without the diffusion tube, so as to exclude the influence of pressure. This method has disclosed the law of diffusion, known as Graham's laiv, namely, that the rate of diffusion of gases is inversely as the square root of tlieir density. Thus, hydro- gen and oxygen have, respectively, densities of 1 and 10; therefore, hydrogen diffuses four times as rapidly as oxygen. 66. Combination by Volume. Gay-Lussac, who investigated the proportions in which gaseous volumes enter into combination, established the following laws: 1. The ratio in which gases combine by volume is always a simple one. 2. The volume of the resulting gaseous product bears a simple ratio to the volume of its constituents. The laws of combination by volume, which at the time of Gay-Lussac's investigation (between 1805 and 1808) were simply experimental, have recently been investigated by Clausius, who has shown that they are really nothing but a very simple deduction from the law of Avogadro. According to Avogadro's law, equal volumes of all gaseous bodies, contain the same number of molecules. If, there- fore, the number of molecules is in some way decreased, the 5 THEORETICAL CHEMISTRY. 51 volume must necessarily be decreased also. If we assume that in a given volume of any gas each molecule is diatomic, and that by some means the molecule can be made tetratomic the number of atoms remaining the same the number of molecules will be reduced one-half, since each molecule con- tains twice the number of atoms; and this reduction of molecules is naturally accompanied by a corresponding decrease of the volume ; that is, the volume of the gas will also be reduced one-half. Or, if we assume that the diatomic molecule could be made triatomic, the number of molecules, and, consequently, the volume, would be reduced one-third. To apply this argument to the facts of volume combination, we will consider separately the combination of hydrogen with monads, dyads, triads, and tetrads, assuming all their mole- cules to be diatomic. 1. Combination of a Monad ivitli Hydrogen. Each atom of a monad combines with an atom of hydrogen; before they combine, their molecules are each .diatomic, and each molecule of the compound is also diatomic; in accordance with the principles of Art. 24, if there was 1 volume of each before the combination, the number of compound molecules after the combination takes place is exactly equal to the sum of the molecules of the monad and hydrogen before they combined, and the result is 2 volumes of the compound. Monads, then, combine with one another, volume to vol- ume, yielding 2 voluirfes of the product. ILLUSTRATION. The monad atom chlorine Cl and the monad atom hydrogen //"combine to form hydrogen chloride HCL Their molecules being diatomic, they unite, molecule to molecule, volume to volume, and, consequently, the number of molecules and the volume they occupy will remain exactly the same after combination has taken place as it was prevfous to it. We may represent this molecularly thus : H Cl H- Cl | + | yield H Cl H-Cl, 52 THEORETICAL CHEMISTRY. and volumetrically thus: and C1 9 form (HCl) 9 2. Combination of a Dyad with Hydrogen. A dyad atom will require 2 atoms of hydrogen ; 1 molecule will require 2 molecules of hydrogen ; and 1 volume, 2 volumes of hydro- gen. Or, to express this fact in other words, dyads combine with monads in the ratio of 1 to 2. As the resulting molecule is triatomic, it is evident that 3 diatomic molecules have combined to form 2 triatomic mole- cules, and that the decrease in the number of molecules also causes a corresponding decrease in volume. Thus we see that 3 volumes of simple gas give 2 volumes of compound gas, a condensation of 3 volumes to 2 volumes taking place during combination. Dyads, then, combine with monads in the ratio of 1 to 2, and yield 2 volumes of the product. ILLUSTRATION. An atom of oxygen is bivalent, and its molecule dia- tomic. One atom of oxygen requires 2 atoms of hydrogen to form a molecule of water gas, and 1 molecule of oxygen requires 2 molecules of hydrogen, and 1 volume requires 2 volumes. This may be repre- sented molecularly thus : Before combination. After combination. HOH H-O-H I II I H-O-H HOH 3. Combination of a Triad with* Hydrogen. One triad atom requires 3 monad atoms for saturation, 1 molecule requires 3 molecules, and 1 volume requires 3 volumes; since the newly produced molecule is tetratomic, the number of molecules and the corresponding volume they occupy must be reduced one-half. Triads, then, unite with monads in the ratio of 1 to 3, yielding 2 volumes of the product. ILLUSTRATION. The trivalent atom of nitrogen TV requires 3 atoms of hydrogen to form 1 molecule of ammonia. THEORETICAL CHEMISTRY N HHH HHH III + III \1X + N HHH N 53 or JB, + B, As 4 diatomic molecules yield 2 tetratomic molecules, 1 volume of nitrogen and 3 volumes of hydrogen will yield 2 volumes of ammonia gas. 4. Combination of a Tetrad with Hydrogen. One tetrad atom requires 4 monad atoms for saturation ; 1 tetrad mole- cule will require 4 monad molecules, and 1 volume of any tetrad will require 4 volumes of any monad; as the resulting molecule consists of 5 atoms, the 5 original volumes will be condensed to two. Tetrads, then, unite with monads in the ratio of 1 to 4 and yield 2 volumes of the product. ILLUSTRATION. The tetrad atom of carbon C combines with 4 monad atoms of hydrogen to form marsh gas. That is, 5 diatomic molecules unite to form 2 pentatomic molecules, or, as in this case, 1 volume of carbon and 4 volumes of hydrogen yield 2 volumes of marsh gas. HHHH | | | | HHHH H | C \ H H | H-C-H | H The preceding combinations may be graphically repre- sented thus: (2) THEORETICAL CHEMISTRY. (3) (1) 1 monad molecule and 1 monad molecule yield 2 diatomic mole- cules. (2) 1 dyad molecule and 2 monad molecules yield 2 triatomic mole- cules. (3) 1 triad molecule and 8 monad molecules yield 2 tetratomic molecules. (4) 1 tetrad molecule and 4 monad molecules yield 2 pentatomic molecules. 67. There are two more cases to be considered, namely, where the molecules are either monatomic or tetratomic. 1. As all known monatomic molecules are dyads, the combination with monad atoms would be : 1 atom and 2 atoms give 3 atoms ; 1 molecule and 1 molecule give 1 molecule; 1 volume and 1 volume give 1 volume ; or, in other words, monatomic dyad molecules combine with diatomic monad molecules in equal volume and yield 1 vol- ume of the product. ILLUSTRATION. The monatomic dyad zinc Zn combines with the diatomic monad chlorine Cl ; that is, 1 atom of zinc combines with THEORETICAL CHEMISTRY. 55 2 atoms of chlorine, or 1 molecule of zinc combines with 1 molecule of chlorine, or 1 volume of zinc with 1 volume of chlorine. The product is the triatomic molecule zinc chloride Z.nCli, a condensation of half the original volume taking place. 2. As all known tetratomic molecules are triads, the atomic combination is: 1 atom and 3 atoms yield 4 atoms ; 1 tetratomic molecule and 6 diatomic molecules yield 4 tetratomic molecules ; 1 volume and 6 volumes yield 4 volumes; which means that tetratomic triad molecules combine with diatomic monad molecules in the ratio of 1 to 6 volumes, producing 4 volumes of the product. ILLUSTRATION. Phosphorus is a triad, possessing a tetratomic mole- cule. When it unites with chlorine, we have, atomically, P and C/ 3 giving PC/a, and molecularly, /* 4 and 6C/ 2 giving (/*C/ 3 ) 4 , or by vol- ci a C1 2 Cl, Cl g Cl. C1 9 or, as it is also sometimes graphically expressed, Cl, Cl, CHEMICAL CALCULATIONS. REACTIONS. 68. Molecular Stability. All molecules are more or less liable to chemical change. Their atoms may be altered in number, in kind, or in relative position by various exter- nal influences. A molecule is the more stable in proportion as it resists this tendency to change. 5G THEORETICAL CHEMISTRY. 5 69. Chemical Keactioiis and Reagents. The term chemical reaction is applied to any change that takes place in the atoms composing a molecule. The .substance applied to produce the change is called a reagent. 70. Chemical reaction always occurs within the mole- cule. Hence, when two substances react on each other, the changes that result maybe considered as taking place between single molecules ; and as all molecules in homogeneous matter are alike, and what is true of one molecule must be true for any mass of them, it follows that a molecular change repre- sents a mass change, also. 7 1 . Chemical Equations. Since every chemical change or reaction is simply an alteration in the association and position of atoms within the molecule^ and, consequently, a change of the constitution of the molecule, this reaction may be expressed by an equation. The substances entering into the reaction are called the factors; those issuing from the reaction are called the products. The equation representing the reaction is written accord- ing to the following rule : Place the formulas of the factors connected by the sign of plus as t lie first member of the equation, and the formulas of the products also connected by the sign of plus as the second. ILLUSTRATION. The reaction of the 2 molecules A B and CD would then be expressed as follows : A B + CD = A D + B C. 72. A chemical equation, however, not only expresses the fact of reaction, but it also indicates the quantities, by weight concerned in it, and since the atoms remain exactly the same after the reaction as they were before, being only differently associated, it is evident that a loss of weight cannot occur through chemical reaction, and the weight of the bodies resulting from a chemical change must be the 5 THEORETICAL CHEMISTRY. 57 same as that of the bodies before the change, whatever it may be, had occurred. 73. Chemists classify chemical reactions in the following manner: Analytical reactions represent the separation of a complex molecule into more simple ones. ILLUSTRATION. 2// 2 <9 = 2// 2 + O 2 water hydrogen oxygen Synthetical reactions represent the union of simple mole- cules to form one or more complex ones. ILLUSTRATION. H* + C/ 2 = 1HCI hydrogen chlorine hydrochloric Metathetical reactions represent an exchange or transpo- sition of atoms between molecules. ILLUSTRATION. Pb(NO-^ + Na^SC* = Pb(SO*) + 2NaNO 3 lead sodium lead sodium nitrate sulphate sulphate nitrate 74. Facility of Chemical Reaction. Since a chemical reaction is the result of the reciprocal action of atoms and has for its effect a change in the composition of the molecule, it is evident that it can only take place when these atoms, and, consequently, their molecules, are brought into intimate rela- tion, or, more exactly, when the molecules of one of the bodies enter within the sphere of action of other bodies. This sphere is rather limited, as the affinity of the atoms is only exercised at infinitely small distances. In consequence, affinity is often retarded by cohesion, which maintains the relations between molecules of solid bodies. These forces are frequently in opposition, and in order that the first may obtain supremacy it is necessary that the other shall yield. To increase the affinity between two or more bodies it becomes necessary to diminish their cohesion. On this condition the molecules can enter within the sphere of their reciprocal attraction, and the atoms of one body can attract those of the other. 58 THEORETICAL CHEMISTRY. 5 Experience teaches us that this cohesion can to a certain extent be overcome and chemical reaction facilitated when the acting substances are brought into the liquid or gaseous state. Hence, either heat by which bodies are vaporized, or solution, by which bodies are liqtiened, is an important aid in producing chemical reaction. 75. Precipitation. If we place some solution of mer- curic chloride in a test tube and add to it, drop by drop, an iodide of potassium solution, a red powder is formed, which gradually settles down to the bottom of the tube. This pow- der is mercuric iodide, while potassium chloride remains in solution. Whenever a substance separates from a solution through the addition of another, as the mercuric iodide has done in the example cited, the term precipitate is applied to the sep- arated substance, and the substance that will produce a pre- cipitate is known as a precipitant. The process of producing a precipitate is known as precipitation. Those conditions of chemical change depending upon solu- bility are stated in Berthollet's /aw, which may be expressed as follows : When, on mixing two substances in solution, a new com- pound can be formed that is insoluble in tJie solvent employed, such compound will be formed and will appear as a precipitate. 76. Another law, also established by Berthollet, relates to the products of reactions that are volatile instead of being insoluble solids ; it may be stated as follows : When, on mixing different substances, a new substance that is volatile can be produced by the rearrangement of the atoms of the partaking- substances, such new substance ivill be pro- duced and will appear as a gas. ILLUSTRATION. If we mix sodium nitrate NaNO 3 and sulphuric acid fJ- 2 SO* together, through a rearrangement of the atoms of these two 5 THEORETICAL CHEMISTRY. 59 substances, hydrosodium sulphate and nitric acid are produced, accord- ing to the equation : NaNO 3 + HtSOt = HNO 3 + HNaSO* sodium sulphuric nitric hydrosodium nitrate acid acid sulphate Nitric acid, being volatile, escapes as a gas when heat is applied. By means of Table 8, which shows the solubility of various compounds, the resulting chemical change of these compounds, covered by the first-mentioned law of Berthollet, may be predicted, and as the student progresses in his study and experimenting he will soon acquire sufficient knowledge to predict also the changes covered by the second-mentioned law of Berthollet, which requires some familiarity with the volatility of the various compounds. ILLUSTRATION. The reactions obtained by mixing calcium chloride and sodium carbonate may be expressed by the following equation : Cad* + Na^CO s = CaCO 3 + ZNaCl calcium sodium calcium sodium chloride carbonate carbonate chloride Will there be a precipitate ? Consulting Table 8, we find that calcium carbonate is insoluble, and we can predict that under the given conditions calcium carbonate will be obtained in the solid form as a precipitate. The reactions obtained by mixing calcium hydrate and carbon dioxide may be expressed by the equation : Ca(OH}* + CO* = CaCO 3 + H*O calcium carbonic calcium hydrate dioxide carbonate Will there be a precipitate ? Consulting the table, we find that calcium carbonate is insoluble in water, and, therefore, we can predict that under the given conditions calcium carbonate will be obtained as a precipitate. 77. Modes of Chemical Action. Although chemical actions vary considerably in their character, they can all be classified under the following five heads : 1. Direct Union or Synthesis. EXAMPLE. Zn + C/ 2 = ZnCl* _: zinc chlorine fiO THEORETICAL CHEMISTRY. OIU9S.IV ouny v is o d W S 6 Case L To find, therefore, the percentage composition of a molecule when the molecular weight, number of atoms, and atomic weights are known, use the following rule : Rule. Multiply tJie atomic weight by the number of atoms, and the product by 100; divide by the molecular ivcigJit, and the quotient will be the percentage amount of that con- stituent. EXAMPLE. The formula for potassium chlorate is KCIO^\ assigning to each atom its weight, we have the equation K + Cl + O 3 = KC/0, 39 35.5 (16x3) 122.5 which means that 122.5, being the sum of the constituent atoms, repre- sents the molecular weight of potassium chlorate, and that 122.5 parts of KCIO-A contain 48 parts of oxygen, 39 parts of potassium, and 35.5 parts of chlorine. What quantities of these constituents will be found in 100 parts of KCIOJ. SOLUTION. (1) Oxygen; x= = 39. 18g of oxygen. Ans. \&&. o (2) Chlorine; x = 10 * 8 ' 5 ; 5 X 1 = 28.98^ of chlorine. Ans. I**, o (3) Potassium; x= 10 * 3 *? X * = 31. 84g of potassium. Ans. \ii. 5 . - Total, 100. 00. 82. Other Problems by This Formula. In formula 1 the four quantities , ;/, /;/, and x are employed. Any three of these being known, the fourth can, of course, be easily found. There are, consequently, three more cases to be considered. Case II. To find the number of atoms of the constituents of a compound, that is, the chemical formula of the molecule, having given the percentage amount of the constituents, their 6(5 THEORETICAL CHEMISTRY. 5 atomic weights and the molecular weight of the compound, formula 1, by transposition, becomes n = mx , (2.) from which we derive the following rule: Rule. Multiply the molecular weight by the percentage of the given constituent, and divide the product by the atomic iv eight of the constituent, multiplied by 100; the quotient is the number of atoms of that constituent in the molecule. Ex AMPLE. ^Sulphuric acid has a molecular weight of 98; its percent- age composition is: hydrogen 2.04 per cent., sulphur 32.65 per cent. , and oxygen 65.31 per cent. The atomic weight of hydrogen is 1, that of sulphur 32, and that of oxygen 16. What is the formula of sulphuric acid? SOLUTION. The atomic weights of hydrogen, sulphur, and oxygen, are, respectively, 1, 32, and 16; the number of atoms of each element is obtained by formula 2. 98X2.04 IX 100 98x32.65 32 X 100 98X65.31 = 2, number of atoms of hydrogen 1, number of atoms of sulphur; = 4, number of atoms of oxygen. 16 X 100 The molecular formula of sulphuric acid is, therefore, H^SO*. Ans. Case ///. Having the molecular weight, the percentage composition, and the number of atoms of any constituent in the molecule, to find the atomic weight of that constituent, we obtain from formula 1, by transposition, mx , Q \ = looT? < 8< > from which we have the following rule : Rule. Multiply the molecular weiglit by the percentage of the constituent whose weight is desired, and divide the prod- uct by the number of atoms multiplied by 100; the quotient is the atomic weight required. EXAMPLE. Mercurous chloride contains 2 atoms of mercury, consti- tuting 84.96 per cent, of the whole; the molecular weight of mercurous chloride being 470.34, what is the atomic weight of mercury? 5 THEORETICAL CHEMISTRY. 07 SOLUTION. Substituting these values in formula 3, we have a - 2 V 100 = ^^' the atom i c weight of mercury. Ans. Case IV. Having given the atomic weight of any con- stituent, its percentage, and the number of atoms of it in the molecule, to find the molecular weight of the compound, we obtain, by a final transposition of formula 1, WOan m = . (4.) Whence, the following rule : Rule. Multiply the atomic weight of the given constituent by the number of its atoms, and this product by 100; divide the final product by the percentage of that constituent, and the quotient is the molecular weight. EXAMPLE. One molecule of sulphuric acid contains 1 atom of sul- phur, which is 32.65 per cent, of it; the atomic weight of sulphur being 32, what is the molecular weight of sulphuric acid ? SOLUTION. Substituting these values in formula 4, m = ^ - = 98, the molecular weight of sulphuric acid. Ans. o^. bo 83. Calculation of an Atomic Group. It some- times becomes necessary to calculate the percentage of a group of atoms in a molecule. Formula 1 enables us to do this, using a to indicate the weight of the group, and n the number of such groups in the molecule. EXAMPLE. The mineral magnesite MgCO* is decomposed by heat into MgO and.Q9 2 ; how much magnesium oxide MgO is there in 100 parts of magnesite ? SOLUTION. The molecular weight of MgO is 40, and of MgCO* is 84. a n X 100 By formula 1, - , we have m 40 X = 4 7.62# of magnesium oxide. Ans. 84. Other Than Percentage lumbers. In some cases it is necessary to calculate the quantity of a constituent in more or less than 100 parts. The answers to such problems can, of course, be obtained by stating the proportion for 68 THEORETICAL CHEMISTRY. 5 each problem ; but they may also be derived from formula 1 by putting y, the quantity of the constituent, in place of x, and s, the quantity of the compound, in place of 100. The formula then becomes from which we obtain the following rule : Rule. Multiply the weight of the constituent contained in one molecule by the weight of the compound given in "the prob- lem, and divide this product by the molecular weight; the quotient will be the weight of the required constituent. EXAMPLE. How much iodine may be obtained from 311 grams of potassium iodide KI, the atomic weight of iodine being 127, and the molecular weight of potassium iodide being 166 ? SOLUTION. Substituting the proper values in formula 5, we obtain ' Hence, 311 grams of potassium iodide yield 237.9 grams of iodine. Ans. Finally, if we wish to know the quantity of a compound necessary to yield a certain weight of the constituent, we obtain by transposition the following formula : (6.) a n EXAMPLE. How much potassium iodide would be required to yield 59 grams of iodine ? SOLUTION. Substituting the proper values in formula 6, 2 = = = 77.12 grams of potassium iodide. Ans. 85. Calculations From Equations. The same simple arithmetical principles are applied in calculating the weights of substances that enter into or are produced by chemical changes. The change should be expressed by an equation. The molecular weights of all the participating compounds are calculated from a table of atomic weights and are placed below their formulas. Having got thus far, the problems may be solved by making the following proportion : o THEORETICAL CHEMISTRY. 69 The molecular weight of the given substance is to the abso- lute weight of it given in the problem, as the molecular weight of the required substance is to its required quantity. Let M = molecular weight of the given substance; ;;/ molecular weight of the required substance ; W absolute weight of the given substance ; w = absolute weight of the required substance. We have the proportion M : W = ;;/ : w, whence we obtain the following formulas : M ---. ^. (7.) (8.) m Mw m = ~W~' ^ ') =. (10.) - Hence, if we know any three of the above values (M, W, m, and w), we can find the fourth by using one of the above formulas. The application of these formulas to the four possible cases is shown by the following examples : Case I. Knowing the absolute and molecular weights of the given substance, and also the absolute weight of the required substance, to find the molecular weight of the required substance. > EXAMPLE. A certain chemical reaction is represented by the follow- ing equation: Zn(NO) 3 + A" 2 C<9 3 = ZnCO 3 + (KNO& zinc potassium zinc potassium nitrate carbonate carbonate nitrate 189 138 125 202 If 156 grams of zinc nitrate yield 103. 17 grams of zinc carbonate, and the molecular weight of zinc carbonate is 125, what is the molecular weight of zinc nitrate ? SOLUTION. Substituting the values just given in formula 7, we obtain M = * ' = 189, the molecular weight of zinc nitrate. Ans. 10o. 17 70 THEORETICAL CHEMISTRY. 5 Case II. Knowing the molecular weight of both the given and also the required substances, to find the absolute weight of the given substance necessary to yield a certain weight of the required substance. EXAMPLE. Sodium nitrate is prepared by the action of hydro- disodium phosphate upon strontium nitrate, according to the following equation : Oi = HSrPO, strontium hydro-di- hydrostrontium sodium nitrate sodium phosphate phosphate nitrate 211 142 183 170 How much hydro-disodium phosphate will be required in order to yield 809 grams of sodium nitrate ? SOLUTION. Substituting the proper values in formula 8, we obtain 142 V 309 W = ^=Q - = 258. 1 grams of hydro-disodium phosphate. Ans. Case III. Knowing the absolute and molecular weights of the given substance and also the absolute weight of the required substance, to find the molecular weight of the latter. EXAMPLE. Sodium carbonate is produced by the action of sodium sulphide upon calcium carbonate according to the following equation : Na^S + CaCO 3 = Na^CO 3 + CaS sodium calcium sodium calcium sulphide carbonate carbonate sulphide 78 100 106 72 103.17 grams of sodium carbonate are obtained from 97.33 grams of calcium carbonate ; the molecular weight of calcium carbonate being 100, what is the molecular weight of sodium carbonate ? SOLUTION. Substituting these values in formula 9, we obtain 100X103.17 97.33 = 106, the molecular weight of sodium carbonate. Ans. Case IV. Knowing the molecular weights of both the given and the required substance, and also the absolute weight of the given substance, to find the absolute weight of the required substance. EXAMPLE. Hydropotassium sulphate is produced by the action of sulphuric acid upon potassium nitrate, according to the following equation : + H*SO< = H(NO,} + HKSO, potassium sulphuric nitric hydropotassium nitrate acid acid sulphate 101 98 63 136 5 THEORETICAL CHEMISTRY. 71 How much hydropotassium sulphate is yielded by the decomposition of 500 grams of potassium nitrate by sulphuric acid ? SOLUTION. Substituting the proper values in formula 1O, we obtain 500 V 136 w = - rrr: = 673.27 grams of hydropotassium sulphate. Ans. In all the above problems it has been assumed that each molecule of the factor yielded one of the product. If in any reaction this is not true, then J/and m must represent the sum of the molecular weights expressed in the equation. 86. As all gaseous molecules have the same volume, we are able to read every equation representing a reaction between gaseous bodies, not only quantitatively but also volume trically. Thus, the equation 2// 2 + <9 2 = 2// 2 6> 4 32 36 shows that 2 molecules of hydrogen and 1 molecule of oxygen yield 2 molecules of water. It may also be read: 2 volumes of hydrogen and 1 volume of oxygen yield 2 volumes of water, and calculations may be made from this as well as from the weights. EXAMPLE. How much water would be yielded by the explosion of 10 cubic centimeters of hydrogen ? SOLUTION. Since 2 volumes of hydrogen yield 2 volumes of water, 10 cubic centimeters of hydrogen will yield 10 cubic centimeters of water in the gaseous state steam. 8 7 . Relation of the Hydrogen Unit to the Air Unit. Density has .already been denned as the weight of any volume of gas compared with that of the same volume of hydrogen, measured at the same temperature and pressure. (These conditions are always understood in speaking of comparative weights of gases.) The term specific gravity may likewise be defined as the weight of any volume of gas compared with that of the same volume of air. The density of hydrogen is assumed to be 1, and its specific gravity is .0693. It is then obvious that by multiplying the density of any gas by the specific gravity of hydrogen, i. e., .0693, the specific gravity of the gas in question may be obtained, and 72 THEORETICAL CHEMISTRY. 5 that, conversely, by dividing the specific gravity of any gas by .0693 its density may be found. ILLUSTRATION. The density of oxygen is 16; its specific gravity, therefore, will be 16 X -0693 = 1.108. The specific gravity of chlorine is 2.46; its density, therefore, is 2.46-T-.0693 = 35.5. PRESSURE, VOLUME, AND TEMPERATURE OF GASES. 88. If the temperature of a confined gas remains the same, the pressure and volume will always vary according to Mariotte's law, which is as follows : The temperature remaining the same, the quantity .of gas varies inversely as the pressure. The meaning of this is : If the volume of the gas is dimin- ished to \, \, \, etc. of its former volume, the pressure will be increased 2, 3, 5, etc. times, or if the outside pressure is increased 2, 3, 5, etc. times, the volume of the gas will be diminished to \, \, \, etc. of its original volume, the temper- ature remaining constant. As a necessary consequence of Mariotte's law, it may be stated that the density of a gas varies directly as the pressure, and inversely as the volume ; that is, the density increases as the pressure increases, and deer eases, as the volume increases. Hence, the volume of gases changes with the variation of atmospheric pressure as measured by the barometer; that is, the volume is increased when the barometer falls, and decreased as the barometer rises. The normal pressure to which it is usual to refer gaseous volumes is 760 millimeters of mercury. If we represent the volume of a gas under a height H of the barometer by V, and under any other height h by v, then, according to the above law, we obtain the proportion : V : v rr h : H, which is expressed by the formula : Rule. Multiply the given volume of the gas by the given 5 THEORETICAL CHEMISTRY. 73 height of the barometer ; divide t/te product by !t>0, and the quotient tlius obtained will be the normal volume. 89. Mariotte's law holds when the temperature of the gas remains constant; if the temperature varies, for every increase of temperature there will be a corresponding increase of volume. The law that expresses this change is called Gay-Lussac's law, and may be stated as follows : If the pressure remains constant, every increase of temper- ature of 1 produces in a given quantity of gas an expansion of ^\^ of its volume at 0. If the pressure remains constant, it will also be found that every decrease of temperature of 1 will cause a decrease of 24 of the volume at 0. Let v = original volume of gas; i\ = final volume of gas ; / temperature corresponding to volume v; t l = temperature corresponding x to volume v v . Then, Rule. The volume of gas after heating (or cooling) equals the original volume multiplied by 273 plus the final tempera- ture, divided by 273 plus the original temperature. EXAMPLE 1. A gas measures 30 cubic centimeters atO; what will it measure at 72? SOLUTION. Substituting in formula 12 the proper values, we obtain .^i EXAMPLES. A certain quantity of gas measures 36.6 cubic centi-. meters at 96 ; what will it measure at ? SOLUTION. Substituting in formula 12 the proper values, we obtain (O*7Q i 7O\ g 7 3^ Q ) - 37.912 c. c. Ans. *" = 36 ' 6 = 27 ' 08 c ' c ' Ans ' EXAMPLE 3. A certain quantity of oxygen measures 560 cubic centi- meters at 15 ; what will the same quantity measure at 95 ? SOLUTION. Substituting in formula 12 the proper values, we obtain = 715.55 c. c. Ans. 74 THEORETICAL CHEMISTRY. 5 EXAMPLES FOR PRACTICE. 90. Solve the following : 1. The formula of calcium chloride is CVzC/ 2 ; its molecular weight is 111. What percentage quantities of the two constituents will be found in 100 parts of calcium chloride ? , j 36$ of calcium. ( 64$ of chlorine. 2. Potassium chlorate has a molecular weight of 122.5, and is com- posed of 89.18 per cent, of oxygen, 28.98 per cent, of chlorine, and 31.84 per cent, of potassium. What is the formula of potassium chlo- rate ? Ans. KCIO*. 3. Salt NaCl contains 39.32 per cent, of sodium, whose atomic weight is 23. In a molecule of salt there is but 1 atom of sodium. What is the molecular weight of salt? Ans. 58.5. 4. A quantity of ammonia gas had a volume of 350 cubic centimeters when measured at 74 ; what volume would it have at 0? Ans. 275.3 c. c. 5. The molecular weight of silver nitrate AgNO* is 170 ; it contains 63. 53 per cent, of silver, and has but 1 atom of silver in a molecule. What is the atomic weight of silver? Ans. 108. 6. Nitric acid is prepared by the action of sulphuric acid on potas- sium nitrate, according to the equation : KNO* + H^SO, = HNO* + HKSO, potassium sulphuric nitric hydropotassium nitrate acid acid sulphate 101 98 63 136 (a) How much potassium nitrate is necessary to yield 36 grams of nitric acid? Ans. 57.71 grams. (b) How much nitric acid may be produced from 500 grams of potassium nitrate ? Ans. 311.88 grams. (c} How much sulphuric acid will be required to yield 36 grams of nitric acid ? Ans. 56 grams. 7. The density of marsh gas C// 4 is 8; how many grams does a liter of it weigh ? Ans. .7168 gram. 8. The specific gravity of hydrogen iodide is 4,41; what is its molecular weight ? Ans. 127.26. CRYSTALLOGRAPHY. 91. Most chemical substances, when they pass from the liquid or gaseous into the solid stato, assume some definite geometric form, or are said to crystallize. Crystals are pro- duced when a substance, such as nitrate of potash, is dis- solved in water and the solution is allowed to gradually 5 THEORETICAL CHEMISTRY. 75 evaporate; when a body, such as sulphur, is melted and allowed to solidify by cooling-; or when a volatile substance, such as iodine, is vaporized, and the vapor condensed on a cool surface. One of the most common methods of crystallizing- a solid substance consists in dissolving it in hot water and allowing the solution to cool slowly. The more slowly it cools, the larger and more symmetrical are the crystals. A hot saturated solution is not generally the best for crys- tallizing, because it deposits the dissolved body too rapidly. Thus, the hot solution of saltpeter prepared as above would solidify to a mass of minute crystals on cooling, but if 100 grams of saltpeter are dissolved in 120 cubic centimeters or 4 measured ounces of boiling water, the solution will form crystals 2 or 3 inches long when slowly cooled in a covered vessel. If the solution is stirred while cooling, the crystals will be very minute, having the appearance of a white powder. Some solids, however, refuse to crystallize, even from a hot saturated solution, if it is kept absolutely undisturbed. Sodium vSulphate affords a good example of this. If to boil- ing water in a flask the sulphate is added until the water refuses any longer to take it up, it will be found that the water has dissolved more than twice its weight of the salt, and that a solution results which boils at 104.5. If this solution is allowed to cool in the open flask, an abundant crys- tallization will take place, for cold water will dissolve only about one-third of its weight of sulphate. But if the flask is tightly corked while the solution is boiling, it may be kept for several days without crystallizing. In this condition we say the solution is supersaturated. On withdrawing the cork, the air entering the partly vacuous space above the liquid will be seen to disturb the surface slightly, and from that point beautiful prisms will shoot through the liquid, until the whole has become a nearly solid crystalline mass. A considerable elevation of temperature is observed, conse- quent on the passage from the liquid to the solid form. If the solution of sodium sulphate is somewhat weaker, con- taining exactly two-thirds of its weight of the sulphate, it 76 THEORETICAL CHEMISTRY. 5 may be cooled without crystallizing 1 , but a touch with a glass rod will start the crystallization immediately. The crystallization of a supersaturated solution is caused by contact with a crystal of the salt itself. Minute crystals of sodium sulphate are present in the floating dust of the air and cause the crystallization w r hen they fall into the supersaturated solution. A perfectly clean glass rod may be dipped into the solution without causing crystalliza- tion, but a rod that has been exposed to air will have some particles of sodium sulphate on it and will start crystalliza- tion ; if the rod is heated so as to render the sodium sulphate from the dust anhydrous, it will no longer cause crystalliza- tion unless it is drawn through the hand. Air filtered through cotton wool does not cause super- saturated solutions to crystallize. If the solution contain- ing- two-thirds of its w r eight of sodium sulphate is allowed to cool in a flask closed by a cork furnished with two tubes plugged with cotton wool, it will be found that on with- drawing the plugs and blowing through one of the tubes dipping into the solution, crystallization does not take place ; but if air is blown by a pair of bellows into the same solution it will crystallize at once. A striking illustration of the power of unfiltered air to start crystallization is afforded by a solution of alum pre- pared by saturating a volume of water at 90 and allowing it to cool in a flask, the mouth of which is closed by a plug of cotton wool. In this state it may be kept for weeks with- out crystallizing, but on withdrawing the plug, crystallization will be seen to commence at once at a few points on the sur- face immediately under the opening of the neck, and will spread slowly from these, octahedrons of alum of half an inch or more in diameter being built up in a few seconds, the temperature, at the same time, rising very considerably. In the laboratory, stirring is always resorted to in order to induce crystallization, if it does not take place immediately. Thms, it is usual to test for potassium in a solution by adding tartaric acid, which should cause the formation of minute 5 THEORETICAL CHEMLSTRY. 77 crystals of hydropotassium tartrate (cream of tartar), but the test seldom succeeds unless the solutions are stirred briskly with a glass rod. 93. The crystals of sodium sulphate produced in the experiment described above contain, in a state of combina- tion with the salt, more than half their weight of water. Their composition is anhydrous sodium sulphate JVa 9 SO t , 142 parts, or 1 molecule, water . ... . . H^O, 180 parts, or 10 molecules, as expressed by the formula Na^SO^Hfl, or Na^SO^ Wag (aq being the abbreviation of aqua, the Latin name for water). If some of the crystals are pressed between blotting paper to remove adhering water, and left exposed to air, they will gradually effloresce, or become covered with a white, opaque powder. This powder is the anhydrous sodium sulphate into which the entire crystals ultimately become converted by exposure to air, since most crystals containing water have their crystalline forms destroyed by the loss of the water, which is commonly spoken of as water of crystallization. Colored salts containing water of crystallization generally change color when the water is removed. The sulphate of copper affords a very good example to this. The beautiful blue prismatic crystals of this salt contain anhydrous sulphate of copper CuSO, 159.5 parts, or 1 molecule, water . . . . . . . //aO, 90 parts, or 5 molecules, which is expressed by the formula CuSOjbHjD. When these are exposed to the air at the ordinary temper- ature they remain unchanged ; but if heated to the boiling point of water they become opaque and are easily crumbled into a grayish-white powder. This powder consists of anhydrous sulphate of copper CuSO 4 , 159.5 parts, or 1 molecule, water H $34 ' 689 ' " \ (b) $37,935. (18) Solve the following by cancelation: 72X48X28X5 ? 80x60x50x16x14 W 96X15X7X6 : W 70X50X24X20 A J W 8 ' Ans ' ( (b) 32. 1 ARITHMETIC. 3 (19) If a mechanic earns $1,500 a year for his labor, and his expenses are $908 per year, in what time can he save enough to buy 28 acres of land at $133 an acre ? Ans. 7 yr. (20) A freight train ran 365 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 mi. (21) If the driving wheel of a locomotive is 16 feet in circumference, how many revolutions will it make in going from Philadelphia to Pittsburg, the distance between which is 354 miles, there being 5,280 feet in one mile ? Ans. 116,820 rev. (22) What is the quotient of: (*) 589, 824 -T- 576? (b) 369,730,620-^-43,911? (c) 2,527,- 525-4-505? (d) 4, 961, 794, 302 -^ 1,234? ( 1,024. 8,420. 4,020,903. (23) A man paid $444 for a horse, wagon, and harness. If the horse cost $264 and the wagon $153, how much did the harness cost ? Ans. $27. (24) What is the product of: (a) 1,024X576? (b) 5,005x505? (c) 43,911x8,420? ( (a) 589,824. AnsJ () 2,527,525. ( (c) 369,730,620. (25) If a man receives 30 cents an hour for his wages, how much will he earn in a year, working 10 hours a day and averaging 25 days per month ? Ans. $900. (26) What is a fraction ? (27) What are the terms of a fraction ? (28) What does the denominator show ? (29) What does the numerator show ? (30) How do you find the value of a fraction ? 4 ARITHMETIC. 1 (31) Is -^- a proper or an improper fraction, and why ? (32) Write three mixed numbers. (33) Reduce the following- fractions to their lowest terms : t> A* A, H- Ans. $, i, i, |. (34) Reduce 6 to an improper fraction whose denomina- tor is 4. Ans. - 2 ^. (35) Reduce 7J, 13 T \, and lOf to improper fractions. Ans. *, (36) What is the value of each of the following: - 1 /, -L 7 -, > > H ? . Ans. (37) Solve the following: (a) 35- I5|~4f. Ans. (c) (38) i + | + f = ? Ans. (39) i + f+A = ? Ans. i (40) 424-31f + 9^ = ? Ans. (41) An iron plate is divided into four sections; the first contains 29f square inches; the second, 50f square inches; the third, 41 square inches; and the fourth, 69^- square inches. How many square inches are in the plate ? Ans. 190 T 9 sq. in. (42) Find the value of each of the following : 16 4 + 3 . . Ans - w *. 16 8 *W A- (43) The numerator of a fraction is 28, and the value of the fraction J ; what is the denominator ? Ans. 32, 1 ARITHMETIC. 5 (44) What is the difference between (a) J- and T ^ ? (b) 13 and 7^ ? ( c ) 312^ and 229 \ ? (() TV Ans. -{ (*) 5 T V ((c) 83H- (45) If a man travels 85 T \ miles in one day, 78^ miles in another day, and 125|~J miles in another day, how far did he travel in the three days ? Ans. 289ffJ- mi. (46) From 573f tons take 216f tons. Ans. 357^ T. (47) At f of a dollar a yard, what will be the cost of 9 yards of cloth ? Ans. 3^-f dollars. (48) Multiply f of f of ^ of |f of 11 by of of 45. Ans. (49) How many times are f contained in f of 16 ? Ans. 18 times. (50) Bought 211^ pounds of old lead for 1 cents per pound. Sold a part of it for 2^ cents per pound, receiving for it the same amount as I paid for the whole. How many pounds did I have left ? Ans. 52 Ib. (51) Write out in words the following numbers: .08, .131, .0001, .000027, .0108, and 93.0101. (52) How do you place decimals for addition and sub- traction ? (53) Give a rule for multiplication of decimals. (54) Give a rule for division of decimals. (55) State the difference between a fraction and a decimal. (56) State how to reduce a fraction to a decimal. 1-32 ARITHMETIC. 1 (57) Reduce the following fractions to equivalent deci- mals: |, J, A, T and Ans. .5. .875. .15625. .65. [ .125. (58) Solve the following: 32.5 +.29 +1.5 , 1.283X8+1* Ans. - 589 + 27X163-8. 40.6 + 7.1 X (3.029 -1.874) 25 + 39 6.27 + 8.53-8.01 i 2.5029. 6.3418. 1,491.875. ) 8.1139. (59) How many inches in .875 of a foot ? Ans. 10^- in. (60) What decimal part of a foot is T 3 ^ of an inch ? Ans. .015625. (61) A cubic inch of water weighs .03617 of a pound. What is the weight of a body of water whose volume is 1,500 cubic inches ? Ans. 54.255 Ib. (62) 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 feet of fencing at $. 50 a foot, how many tons of coal were in the car? Ans. 22 T. (63) Divide 17,892 by 231, and carry the result to four decimal places. Ans. 77.4545+. (64) What is the value of the following expression car- ried to three decimal places : 74. 26 X 24 X 3. 1416 X 19 X 19 X 350 33,000X12X4 = ? Ans. 446. 619- . (65) Express: (a) .7928 in 64ths; (b) .1416 in 32ds; (c) .47915 in 16ths. f (a) fi- Ans. (b) A. IW A- 1 ARITHMETIC. (66) Work out the following examples: (a) 709. 63 -.8514; (b) 81.963-1.7; (c) 1-.001; (e) 872.1 -(.8721 + .008); (/) -(6.704-2.38). Ans. - 18-. 18; (d) (5. 028 + .0073) (a) 708.7786. (b) 80.263. 17.82. .999. 871.2199. .7113. w (d) w I (/) (67) Work out the following: (a) -.807; (b) .875 -f; (<:) (A + -435) - (AV--07); (^/) What is the difference between the sum of 33 millionths and 17 thousandths, and the sum of 53 hundredths and 274 thousandths ? Ans. (a) .068. (b) .5. (c) .45125. (d) .786967. (68) What is the sum of .125, .7, .089, .4005, .9, and .000027? Ans. 2.214527. (69) 927.416 + 8.274 + 372.6 + 62.07938 = ? Ans. 1,370.36938. (70) Add 17 thousandths, 2 tenths, and 47 millionths. Ans. .217047. (71) Find the products of the following expressions : (a) .013 X. 107; (b) 203x2.03 X. 203; (c) 2.7x31.85x (3.16 -.316); (d) (107.8 + 6.541- 31.96) X 1.742. (a) Ans. (b) (c) (d) .001391. 83.65427. 244.56978. 143.507702. (72) Solve the following (a) (A - 13) X 1625+1; ( + .013- 2.17) X13J-7A- (X.21) -(.02X A); (*) f () -384375. Ans. J (b) .1209375. I (c) 6.4896875. ARITHMETIC. 1 (73) Solve the following : (a) .875-4-i; (*) |^.5; M^^Jr T ^ .1^5 r (#) 1.75, Ans. J (^) 1.75. 1(0 -5. (74) Find the value of the following" expression : 1.25X20X3 87 + (11X8)' 459 + 32 Ans. 210f (75) From 1 plus .001 take .01 plus .000001. Ans. .990999. ARITHMETIC. (ARTS. 1-168. SEC. 2.) (1) What is 25 per cent, of 8,428 Ib. ? Ans. 2,107 Ib. (2) What is 1 per cent, of $100 ? Ans. $1. (3) What is 1 per cent, of $35,000 ? Ans. $175. (4) What per cent, of 50 is 2 ? Ans. 4$. (5) What per cent, of 10 is 10 ? Ans. 100$. (6) Solve the following: (a) Base = $2,522 and percentage = $176.54. What is the rate ? (b) Percentage = 16.96 and rate = 8 per cent. What is the base? (c) Amount = 2 16. 7025 and base = 213.5. What is the rate? (d) Difference = 201.825 and base = 207. What is the rate ? (b) 212. (c) (d) Ans. (7) A farmer gained 15$ on his farm by selling it for $5,500. What did it cost him ? Ans. $4,782.61. (8) A man 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 in a year ? Ans. $441. 75. (9) If 37^ per cent, of a number is 961.38, what is the number? Ans. 2,563.68. 2 ARITHMETIC. 2 (10) A man owns of a property. 30$ of his share is worth $1,125. What is the whole property worth ? Ans. $5,000. (11) What sum diminished by 35$ of itself equals $4,810 ? Ans. $7,400. (12) A merchant's sales amounted to $197.55 on Monday, and this sum was 12|$ of his sales for the week. How much were his sales for the week ? Ans. $1,580.40. (13) The distance between two stations on a certain rail- road is 16.5 miles, which is 12-j-$ of the entire length of the road. What is the length of the road ? Ans. 132 mi. (14) After paying 60$ of my debts I find that I still owe $35. What was my whole indebtedness ? Ans. $87.50. (15) Reduce 28 rd. 4 yd. 2 ft. 10 in. to inches. Ans. 5, 722 in. (16) Reduce 5,722 in. to higher denominations. Ans. 28 rd. 4 yd. 2 ft. 10 in. (17) How many seconds in 5 weeks and 3.5 days ? Ans. 3,326,400 sec. (18) How many pounds, ounces, pennyweights, and grains are contained in 13,750 gr. ? Ans. 2 Ib. 4 oz. 12 pwt. 22 gr. (19) Reduce 4,763,254 links to miles. Ans. 595 mi. 32 ch. 54 li. (20) Reduce 764,325 cu.in. to cu.yd. Ans. 16 cu.yd. 10 cu.ft. 549 cu.in. (21) What is the sum of 2 rd. 2 yd. 2 ft. 3 in. ; 4 yd. 1 ft. 9 in. ; 2 ft. 7 in.? Ans. 3 rd. 2 yd. 2 ft. 1 in. (22) What is the sum of 3 gal. 3 qt. 1 pt. 3 gi. ; 6 gal. 1 pt 2 gi. ; 4 gal. 1 gi. ; 8 qt. 5 pt. ? Ans. 16 gal. 3 qt. 2 gi. (23) What is the sum of 240 gr. 125 pwt. 50 oz. and 3 Ib. ? Ans. 7 Ib. 8 oz. 15 pwt. 2 ARITHMETIC. 3 (24) What is the sum of 11 16' 12"; 13 19' 30"; 20 25"; 26' 29"; 10 17' 11" ? Ans. 55 19' 47". (25) What is the sum of 130 rd. 5 yd. 1 ft. 6 in. ; 215 rd. 2 ft. 8 in. ; 304 rd. 4 yd. 11 in. ? Ans. 2 mi. 10 rd. 5 yd. 7 in. (26) What is the sum of 21 A. 67 sq.ch. 3 sq.rd. 21 sq.li. ; 28 A. 78 sq.ch. 2 sq.rd. 23 sq.li.; 47 A. 6 sq.ch. 2 sq.rd. 18 sq.li. ; 56 A. 59 sq.ch. 2 sq.rd. 16 sq.li. ; 25 A. 38 sq.ch. 3 sq.rd. 23 sq.li. ; 46 A. 75 sq.ch. 2 sq.rd. 21 sq.li.? Ans. 255 A. 3 sq.ch. 14 sq.rd. 122 sq.li. (27) From 20 rd. 2 yd. 2 ft. 9 in. take 300 ft. Ans. 2 rd. 1 yd. 2 ft. 9 in. (28) From a farm containing 114 A. 80 sq.rd. 25 sq.yd., 75 A. 70 sq.rd. 30 sq.yd. are sold. How much remains ? Ans. 39 A. 9 sq.rd. 25 sq.yd. (29) From a hogshead of molasses, 10 gal. 2 qt. 1 pt. are sold at one time, and 26 gal. 3 qt. at another time. How much remains ? Ans. 25 gal. 2 qt. 1 pt. (30) If a person were born June 19, 1850, how old would he be August 3, 1892 ? Ans. 42 yr. 1 mo. 14 da. (31) A note was given August 5, 1890, and was paid June 3, 1892. What length of time did it run ? Ans. 1 yr. 9 mo. 28 da. (32) What length of time elapsed from 16 min. past 10 o'clock A. M., July 4, 1883, to 22 min. before 8 o'clock p. M., Dec. 12, 1888 ? Ans. 5 yr. 5 mo. 8 da. 9 hr..22 min. (33) If 1 iron rail is 17 ft. 3 in. long, how long would 51 rails be, if placed end to end ? Ans. 53 rd. 1| yd. 9 in. (34) Multiply 3 qt. 1 pt. 3 gi. by 4.7. Ans. 4 gal. 2 qt. 1.7 gi. (35) Multiply 3 Ib. 10 oz. 13 pwt. 12 gr. by 1.5. Ans. 5 Ib. 10 oz. 6 gr. 4 ARITHMETIC. 2 (36) How many bushels of apples are contained in 9 bbl., if each barrel contains 2 bu. 3 pk. 6 qt. ? Ans. 26 bu. 1 pk. 6 qt. (37) Multiply 7 T. 15 cwt. 10.5 Ib. by 1.7. Ans. 13 T. 3 cwt. 67.85 Ib. (38) Divide 358 A. 57 sq.rd. 6 sq.yd. 2 sq.ft. by 7. Ans. 51 A. 31 sq.rd. 8 sq.ft. (39) Divide 282 bu. 3 pk. 1 qt. 1 pt by 12. Ans. 23 bu. 2 pk. 2 qt. i pt. (40) How many iron rails, each 30 ft. long, are required to lay a railroad track 23 mi. long ? Ans. 8,096 rails. (41) How many boxes, each holding 1 bu. 1 pk. 7 qt., can be filled from 356 bu. 3 pk. 5 qt. of cranberries ? Ans. 243 boxes. (42) If 16 square miles are equally divided into 62 farms, how much land will each contain ? Ans. 165 A. 25 sq.rd. 24 sq.yd. 3 sq.ft. 80+ sq.in. (43) What is the square of 108 ? Ans. 11,664. (44) What is the cube of 181.25 ? Ans. 5,954,345.703125. (45) What is the fourth power of 27.61 ? Ans. 581,119.73780641. (46) Solve the following: (#)106 2 ; (b) (182i) 2 ; (c) .005 2 ; (d) .0063 2 ; (e) 10. 06 2 . Ans. -i (a) 11,236. (b) 33,169.515625. (c) .000025. (d) .00003969. (e) 101.2036. (47) Solve the following: (a) 753 3 ; (b) 987. 4 3 ; (c) .005 3 ; (d) .4044 3 . Ans. - (a) 426,957,777. (&) 962,674,279.624. (c) .000000125. (d) .066135317184. 2 ARITHMETIC. 5 (48) What is the fifth power of 2 ? Ans. 32. (49) What is the fourth power of 3 ? Ans. 81. (50) What are the values of: (a) 67.85 2 ? (b) 967,845 2 ? (a) 4,603.6225. Ans. - (b) 936,723,944,025. W A- (51) What is (a) the tenth power of 5 ? (b) the fifth power of 9? \(a) 9,765,625. ' ( (b) 59,049. (52) Solve the following: (a) 1.2 4 ; (b) II 6 ; (c) I 7 ; (d} .Ol 4 ; W - 1& - [ (a) 2.0736. (b) 1,771,561. Ans. - (V) 1. (d) .00000001. . (e) .00001. (53) Find the values of the following: (a) .0133 3 ; (b) 301.011 3 ; (c) (i) 3 ; (d) (3f) 3 . Ans. (a) .000002352637. (b) 27, 273, 890. 942264331. (d) 52, or 52.734375. (54) In what respect does evolution differ from involution ? NOTE. In the answers to the following examples, a minus sign after a number indicates that the last digit is not quite as large as the num- ber printed. Thus, 12.497 indicates that the number really is 12.496+ , and that the 6 has been made a 7 because the next succeeding figure was 5 or greater. For example, had it been desired to use but three decimal places in example 46 (<), the answer would have been written 33,169.516-. (55) Find the square root of the following: (a) 3,486,- 784.401; (b) 9,000,099.4009; (c) .001225. [(a) 1,867.29+. Ans. I (b) 3,000.017-. I (c) .035. G ARITHMETIC. (56) Extract the square root of (a) 10, 795. 21; (b) 73,008.04; (c) 90; (d) .09. ( (a) 103.9. b 270.2. (d) .3. (57) Extract the cube root of (d) .32768; () 74,088; (c) 92,416; (rf) .373248. ( (a ) .6894+. (,) 45.212-. () fl!7,649; (c) .-000064; (d) ff. f a} 9 . ( + &try. (<:) From 14* + 4 - Qc - 3d take 110 - 2 + 4^ - 4*/. Ans. i (b) x 3 %x*y -f- xy I (c) 3 (22) Find the numerical values of the following expres- sions when a = 16, b = 10, and x = 5 : (#)i (ab^x -f- 614,400. Ans. (*) i ' (<:) 23,400. (23) Find the sum of the following : (a) ^xyz Sxyz 5xyz, Qxyz xyz + Zxyz-, (b) 3 2 + lab + 4 -a* + 5ab-b\ 18 2 - 20^ - 19 2 , and (<;) 4ww + 3^ 4) 5a. I r 3 3 TRIGONOMETRIC FUNCTIONS. 5 (31) In a right triangle A B C, right-angled at C, side A C = 17.5, side B C = 21.3; find the other three parts. f 39 24' 23". Ans. I 50' 35' 37". I 27. 57, nearly. (32) In the formula, t = I " *, it is required rr ,, -f- Vr^S^ to transform the formula so that / 2 will stand alone in the first member. In other words, solve for / 2 . Ans , .. (33) A can do a piece of work in 5 days, B in 6 days, and ( C in 7-J days; in what time will they do it, working together ? Ans. 2 days. (34) Simplify : (a) ; + ab (') x+- Ans. < Z-x X. a b. b 4 (35) () Arrange a*t>* + 2^ + 3 - 7^ 2 ^ 3 + 6# 4 4 according to the decreasing powers of a ; (b) according to the increas- ing powers of b. (c) With a? -\- 1 + 2a 3 -\- ax arranged accord- ing to the increasing powers of a, should the 1 be placed first or last, and why ? (36) (a) Express with radical signs : .**; 3,r*j-i; (b) Clear a~ l fa-\ -- -?-{-(m n)~ l _ 3 of negative expo- nents. (c) Express with fractional exponents: 6 ELEMENTARY ALGEBRA AND 3 (37) Divide : (. &CIX J X ., X ,,. \)TJl ft ofl i oil d] - ! by ; (b) 4 a - a by 4 a ; () ^ ;tr - Ans. -< (^) 2w'w + 1. (38) Multiply: \ / I * ^ ' \ / and (<:) a 3 -f- 3# 2 $ 2^ 3 by 5^ 2 f() 24r 4 - Ans. j (b) 2x* [(c) -5tf 5 +6# 4 by (39) Perform the indicated additions: X . X V X* X X x-y y-x + nr-- Ans. - (a) (*) 84 (40) Factor the following: (a) 9;r 4 + VZx^f + 4j (41) Divide : 9^ 3 -l by 3^-1; 3. f (a) 3x* Ans. (42) Why are letters used in algebra, and in what ways do they differ from figures ? 3 TRIGONOMETRIC FUNCTIONS. 7 (43) (a) Reduce 1 + 2,r ^ to a fractional form. 3 f 2 I 2jtr I 1 () Change - - to a mixed quantity. r io.M^+4 A O^V Ans. 56 lb ' Sulphur, 7 lb. Charcoal, 6 lb. (52) Find the values of the following: v ; (b) m, (a) -1 ^ M ; W ' (a) -5^/V. (b) lOaW Ans. 4 (c) 3;V. (53) In a triangle ABC, side A B = 70 feet, side B C = 42 feet, and angle A = 36 10'. Find the angles B and C and the side A C. f Angle B = 64 14'. Ans. \ Angle C = 79 36'. I Side A C 64.1 ft. (54) (a) What is the supplement of 72 C (b) What is the complement of 22 34' 17" ? ir 36"? L07 48 r 24". 67 25' 43". (55) In a triangle ^ B C, angle ^ = 57 34. 5', angle C = 44 22.5' and side A B = 344 feet. Find angle B and the sides A C and C. f Angle ^ = 78 3'. Ans. -I Side^ C" = 481.22ft. ) How is a molecule constituted ? (22) (a) Define absolute zero, (b) Define absolute tem- perature. (23) A compass C is placed between the north and south poles of two magnets, as shown in Fig. 1. Towards which pole will the north pole of the compass needle tend to point, and why ? FIG. i. (24) The current in a horizontal conductor is flowing from the north towards the south. In what direction will the north pole of a compass needle point if the compass is placed under the conductor ? (25) What do you understand by a liquid body ? (26) If a body weighs 1 pound at a distance 100 miles from the center of the earth, what will it weigh (a) at the surface ? (b) at 100 miles above the surface ? Take the earth's radius as 4,000 miles. ( (a) 40 Ib. Ans. ' v ' (b) 38.0731b. (27) Define (a) electrostatic; (b) electrodynamic. (28) How far above the surface of the earth will a 2-pound ball weigh 3. ounces ? Ans. 9,064 mi. (29) Referring to electricity, what is meant by a con- tinuous current ? (30) If 4 cubic feet of copper alloy weigh 2,000 pounds, (a) what is the specific gravity ? (b) What is the weight of a cubic inch ? . ( (a) Sp. Gr. = 8. *'\(b) .28941b. (31) Two gilt balls A and B are statically charged with + 4 and 24 units of electricity, respectively; if the balls are allowed to touch momentarily and are then placed at a 4 PHYSICS. 4 distance of 2 centimeters from each other, will they attract or repel each other, and what will be the force in dynes exerted between them ? Ans. Repelled with a force of 25 dynes. (32) State what you understand by inertia. (33) Under what conditions does electricity appear ? (34) A piece of wood weighs 11^ ounces in the air. It is attached to a piece of marble weighing 5 pounds in the air and 3 pounds 2 ounces in water. Both together weigh 2 pounds 9 ounces in water. What is the specific gravity (a) of the wood ? (b) of the marble ? ( (a) .555. AnS '|() 2.667. (35) What will be the sign of the static charge left on the cover of an electrophorus, if the cake is made of sulphur instead of resin and is rubbed with a piece of silk ? (36) A bottle weighs 2 pounds in air and 10 ounces in water. A pound of sugar is put into the bottle, and the bottle then weighs 16 ounces in water. What is the specific gravity of the sugar? Ans. '1.6. (37) What will be the sign of the static charge developed (a) on a glass rod when rubbed with fur ? (b) on a piece of hard rubber when rubbed with silk ? (c) on a piece of flannel when rubbed against a piece of amber ? (38) The weight necessary to sink a Nicholson's hydrom- eter to a fixed point on the rod is 2 pounds 8J ounces. The weight necessary to sink the hydrometer to this same point when a piece of slate is in the basket is 1 pound 11 ounces, and when the slate is in the upper pan, 12 ounces, (a) What is the specific gravity of the slate ? (b) What is its volume ? (39) For what purposes is an electroscope used ? (40) What is a vacuum ? Illustrate it. (41) A closed vessel fitted with a piston contains oxygen under a pressure of 3 atmospheres. If the piston is so moved that the volume is 2J times its. former volume, what is the 4 PHYSICS. 5 tension of the gas in pounds per square inch, the tempera- ture remaining the same ? Ans. 17. G4 Ib. per sq. in. (42) For what is the torsion balance used in electrical measurements ? (43) If you are told that the vacuum gauge of a condenser shows 23 inches vacuum, what do you understand by it ? What, in that case, is the pressure in the condenser ? (44) Two gilt balls are statically charged with -f- 20 and 5 units of electricity, respectively ; with what force in dynes will they attract each other when placed (a) at a distance of 1 centimeter from each other ? (b) at a distance of 5 centimeters ? Ans. \ W 10 dynes ' ( (b) 4 dynes. (45) What is the weight of 100 cubic feet of hydrogen having a temperature of 80 and a tension of 18 pounds per square inch? Ans. .6231 Ib. (46) Four cubic feet of air are heated under a constant pressure from 40 to 115. What is the resulting volume? Ans. 4.6 cu. ft. (47) From what is the word electricity derived ? (48) The temperature of the discharged air of an air compressor is 120 and the pressure is 40 pounds per square inch ; when it has cooled to the temperature of the surround- ing air, which is 55, what is its pressure ? Ans. 35.517 Ib. per sq. in. (49) State the principal sources of heat. (50) Describe Nichol's prism. (51) What is the weight of a cubic foot of air at 60 under a pressure of 1 atmosphere ? Ans. .0763 Ib. (52) State the law of attraction and repulsion of statically charged bodies. (53) State, in your own words, what you understand by the " temperature of a body." 6 PHYSICS. 4 (54) The temperature of a wrought-iron bar 4 inches square and 20 inches long is raised 1,200. What is (a) its cubical expansion ? (b) its linear expansion ? (c) How much larger will the area of a cross-section be ? ((a) 7. 903 cu. in. Ans. I (b) .16464 in. I (c) .2634 sq. in. (55) What do you understand (a) by a plano-convex lens ? (b) by a concavo-convex diverging lens ? (c) How are lenses constructed ? (56) W T hat are the physical properties of electricity ? (57) Two vessels, the volumes of which are each 7^ cubic feet, are filled with air; the temperature is the same in both, but the tension in one is 2 atmospheres and in the other 40 pounds per square inch. If all the air in one vessel is compressed into the other, what is the. tension of the mix- ture after it has cooled down to the original temperature ? Ans. 69.4 Ib. per sq. in. (58) Three gases oxygen, hydrogen, and nitrogen are mixed together in an iron retort containing 40 cubic feet. The volume and the tension of the oxygen are 12 cubic feet and 1 atmosphere, respectively; of the hydrogen, 10 cubic feet and 2 atmospheres; of the nitrogen, 8 cubic feet and 3 atmospheres. The temperature of the separate gases and of the mixture remaining the same throughout, what is the tension of the mixture ? Ans. 20.58 Ib. per sq. in. (59) How much would a steel wire rope 900 feet long shorten, if cooled from 90 to 28 ? Ans. 4.01 in. (60) Define magnetic density. (61) (a) Name the kind of thermometer in general use. Reduce 44 R. (b) to the corresponding centigrade tempera- ture ; (c) to the corresponding Fahrenheit temperature. ( (b) 55 C. AnS -)(,) 131 F. (62) Define (a) good conductors; (b) bad conductors. 4 PHYSICS. 7 (63) State what is meant by terrestrial heat. (64) Show in a rough sketch (a) the refracting edge, (b) the refracting angle, and (c) the base, of a prism. (65) A compass C is placed alongside of a bar magnet opposite the neutral line, as shown in Fig. 2. Towards which pole of the magnet will the south pole of the compass needle tend to point, and why ? (66) A vessel containing 13 cubic feet of air, having a tem- perature of 73 and a tension of 1 atmosphere, is placed in communication with another vessel containing 18 cubic feet of air, at a temperature of 53, and a tension of 30 pounds per square inch. What is the new temperature, if the ten- sion of the mixture is 20 pounds per square inch ? Ans. 20.65. (67) When you rub a piece of sulphur with a piece of flannel, what charge does the sulphur receive ? (68) Forty cubic feet of air having a temperature of 100 and a tension of 90 pounds per square inch, are mixed with 57 cubic feet of air having a temperature of 130 and a tension of 80 pounds per square inch. The tension of the mixture is 120 pounds per square inch and the temperature is 110. What is the volume ? Ans. 67.248 cu. ft. (69) Change (a) -10 R, (b) 25 F., and (c) 2,200 F. into the corresponding centigrade readings. Ha) -23iC. Ans.() -3fC. [(c) 1,204|C. (70) Give the names of the known magnetic substances. (71) Twelve cubic feet of gas are heated from 65 to 390 ; what is the increase in volume, the pressure remaining constant? Ans. 7.4286 cu. ft. (72) Define (a) electrical resistance ; (b) conductivity. 8 PHYSICS. 4 (73) (a) Change 798 B. T. U. to calories, (b) Change 40 calories to B. T. U. j ( a ) 201.515 calories. *'\(b) 158.4 B.T.U. (74) A piston rod is 4 inches in diameter and a steel piston is bored 3.9985 inches in diameter to receive it. To what temperature must the piston be heated, assuming its original temperature to be 80 and the diameter of its bore after heating to be 4.001 inches, to allow the rod to enter freely ? Ans. 184.4. (75) If, in the last example, the whole piston were raised to the same temperature as the bore, and its weight were 360 pounds, how many units of heat would be required, assuming a loss of 12$ from various causes ? Ans. 4,975.6 B. T. U. (76) In order to find the temperature of a stove fire, a piece of cast iron weighing 1 pound is placed in it. A copper vessel weighing 1| pounds is partly filled with 3^ pounds of water, the temperature of both being 85. After the cast iron piece has been placed in the vessel of water, the tem- perature of the mixture is found to be 128. What was the temperature of the fire ? Ans. 1,252. (77) A solution of cane sugar placed in a tube 12 inches long, and as observed through the saccharimeter, requires an angle of rotation of 29 to reproduce entire darkness. What per cent, of sugar is contained in this solution ? Ans. 12.33$. (78) What is supposed to be the cause of all electrical phenomena ? (79) What do you understand by "specific heat " ? (80) A vessel has a volume of 6.7 cubic feet. A vacuum gauge attached to it shows 17J inches. How much air at atmospheric pressure will it be necessary to admit to have the vacuum gauge show (a) 5 inches ? (b) inches ? ((a) 2.791 cu. ft. " { (b) 3.908 cu. ft. (81) If 4,516 cubic inches of gas having a temperature of 4 PHYSICS. 9 260 are cooled down to a temperature of 80, the pressure remaining the same, what is its new volume ? Ans. 1.06 cu. ft. (82) (a) What do you understand by dispersion of light ? (b) Into what colors is light decomposed ? (c) What infer- ence has been drawn from the presence of Fraunhofer's lines in the solar spectrum ? (83) What is sensible heat ? (84) How many units of heat would be required to vapor- ize 12 pounds of solid mercury having a temperature of 50 ? (Use the same value for the specific heat of solid mercury as given in the table for liquid mercury.) Ans. 2,229.595 B. T. U. (85) What is the absolute temperature corresponding to (a) 96 F. ? (b) 32 C. ? (c) 180 C. ? (d) 650 F. ? (e) - 40 C. ? (86) (a) Name the rays produced by double refraction. (b) Name a mineral which possesses the property of double refraction. (87) What is meant by the term quantity of heat ? (88) State what you know of the theory of light. (89) State what you know about a magnetic field. (90) (a) What do you understand .by a translucent body ? (b) What is a homogeneous medium ? (91) (a) What is meant by polarization in opposite direc- tions ? (b) Which media possess the property of polarization ? (92) How can you explain the quivering of objects seen over very hot bodies ? (93) (a) What do you understand by circular polarization ? What does it mean if a substance is said to be levo-rotary ? (94) What is the approximate velocity of light ? (95) What do you understand by the index of refraction ? (96) State some experiments for detecting the presence of static charges of electricity. THEORETICAL CHEMISTRY. (1) Calcium carbonate CaCO 3 may be precipitated by heat, the reaction being represented by the following equa- tion : CaHJ^CO^ = CaCO, + Hft + CO, calcium calcium carbon bicarbonate carbonate dioxide 'How much calcium carbonate would be precipitated by heating 296.3 grams of calcium bicarbonate ? Ans. 182.9 grams. (2) Explain what you understand by chemical combina- tion in definite and multiple proportion. (3) Give a short account of how you would proceed to weigh a test tube on a chemical balance. (4) Define the terms inflammable body and supporter of combustion ; explain why these terms are purely relative. (5) How many atoms are contained (a) in a molecule of mercury ? (b) in a molecule of ozone ? (c) in a molecule of arsenic ? (6) A white powder is shaken up with distilled water; how would you ascertain whether any of it is dissolved ? (7) State Avogadro's law and the deductions therefrom. (8) Define meta-acids. (9) What is meant by the term allotropy ? 2 THEORETICAL CHEMISTRY. 5 (10) The analysis of a compound of chlorine and antimony gives 46. 61$ of chlorine and 53. 39$ of antimony; the density of this compound, is 114.25. What is the atomic weight of antimony ? (11) Define matter. Describe its general properties and various physical states. (12) Define combustion, and state what you understand by that term. (13) Explain the meaning of the terms analysis and syn- tJicsis. (14) In which way does methyl-orange differ from phenol- phthalein ? (15) Which of the following elements are metalloids and which are metals: Cl, K, P, S, N, Na, As, H, Fe ? (16) (a] What do you understand by a compound radical? (b) How are compound radicals named ? (17) Give examples of oxidizing and reducing agents. (18) Define force, and state why chemical action may be properly regarded as a force. (19) What is molecular volume, and how is it expressed ? (20) What is the percentage composition of chromic acid H^CrO^ ? f 1. 69$ of hydrogen. Ans. -j 44.30$ of chromium. I 54.01$ of oxygen. (21) A certain volume of oxygen, under a barometric pressure of 781 millimeters, measures 542 cubic centimeters. What is its true volume, measured at 760 millimeters ? Ans. 556.97 c. c. (22) What is chemistry? (23) A glassful of dirty water is given to you. Describe the plan you would adopt, first, for rendering the water clear, and, secondly, for ascertaining whether or not the clear water contains any dissolved matter. 5 THEORETICAL CHEMISTRY. 3 (24) State the difference between a mechanical mixture and a chemical combination. (25) Mention the elements that at ordinary temperatures are, respectively, gases and liquids. (26) State what you know of chemical and physical changes. (27) How would you prepare salt from sea-water ? (28) An analysis of a body having a molecular weight of 84 gave the following results : C= 14.290 O = 5 7.1 40 H= 1.190 Na = 2 7.3 80 1 00.000 What is its chemical formula ? Ans. HNaCO^. (29) State what you know about atomicity. (30) What relation, if any, exists between the atomic weight and the specific heat of bodies ? (31) (a) Define the terms precipitation, precipitate, and precipitant, (b) Give Berthollet's law concerning precip- itation. (32) How many volumes of carbon dioxide CO^ will be formed when a mixture of 4 volumes of carbon monoxide CO and 4 volumes of oxygen O is burned ? Show the reac- tion through an equation ; state what volume of oxygen, if any, remains unburned. (33) Explain the meaning of the terms sublimation, vol- atile, fixed, and ignition. (34) What are the effects on matter of gravitation, cohe- sion, and chemical attraction, respectively ? (35) Define the terms acid, base, and salt. (36) Explain the meaning of the term nascent. 4 THEORETICAL CHEMISTRY. 5 (37) Give a short account of the metric system of weights and measures. (38) By what properties is a metal distinguished from a non-metal ? (39) Define the terms monobasic and dibasic as applied to acids. (40) What is the percentage composition of borax Na t B 4 O, ? f 22. 77$ of sodium. Ans. j 21.78$ of boron. 155.45$ of oxygen. (41) (a) What is expressed by a chemical equation ? (b) How is such an equation constructed ? (42) What do you understand by a fluid ounce ? (43) State what you know about the nomenclature of acids and salts. (44) What are stoichiometrical calculations ? (45) Define density. (46) (a) How many atoms may a compound molecule contain ? (b) How are compound molecules classified ? (47) (a) State Gay-Lussac's two laws concerning the proportions in which gaseous volumes combine, (b) How may these laws be derived from Avogadro's law ? (48) (a) In what way may chemical changes occur ? (b) Illustrate these changes by using the letters A, B, etc. instead of symbols. (49) Ammonium iodide may be prepared according to the equation : NH + HI = NH I 3 4 ammonia hy ^a dlC ^SuSe" 111 If 56. 1 grams of ammonia yield 478. 5 grams of ammonium iodide, and the molecular weight of ammonium iodide is 145, what is the molecular weight of ammonia ? Ans. 17. 5 THEORETICAL CHEMISTRY. 5 (50) Explain the terms crystal, dimorphous, amorphous, water of crystallization, and isomorphous. (51) (a) Define molecule, (b) State how molecules are classified. (52) What do you understand (a) by a symbol ? (b) by a formula ? (53) (a) How do atoms of different valence combine ? (b) Suppose a pentad and a dyad combine, how many atoms of each are necessary to form a saturated compound ? (54) In what respects does a mixture of copper and sulphur differ after being" heated from its condition previous to the application of heat ? (55) A compound, hydrodisodium phosphate, is found on analysis to contain hydrogen = 0. 7 % sodium = 3 2.4$ phosphorus = 2 1.8$ oxygen = 4 5.1 % 1 00.0$ Its molecular weight is 142. What is its formula ? Ans. (56) What is the simplest way to distinguish between elemental and compound bodies ? (57) Why does chemical combination always occur in either definite or multiple proportion ? (58) Alumina is composed as follows : aluminum = 5 3.4$ oxygen = 4 G. 6 $ 1 00.0$ Its molecular weight is 103. What is its formula ? Ans. Al 9 O f (59) Potassium sulphate may be obtained according to the following equation : THEORETICAL CHEMISTRY. 2KNO, + potassium sulphuric potassium nitrous nitrite acid sulphate acid How much potassium nitrite will be required to yield 392 grams of potassium sulphate ? Ans. 383 grams. (60) Define (a) a monad ; (b} a heptad ; (c) a pentad. (61) Give the names of SrO, NaCl, P 2 <9 5 , 2 <9 3 , SnCl^ (62) A molecule of potassium nitrate contains 1 atom of potassium, which is by weight 38. 7$ of it. The molecular weight of potassium nitrate is 101. What is the atomic weight of potassium ? Ans. 39. (63) Describe the process of gaseous diffusion, and state the law governing its rate. (64) What do you understand by an electronegative element ? (65) Define the meaning of the term perissad^ (66) State how acids are classified. (67) A molecule of zinc carbonate contains 1 atom of zinc, which is by weight 52$ of it ; the molecular weight of zinc carbonate is 125. What is the atomic weight of zinc ? Ans. 65. (68) What will a gas measure at 0, which at 100 measured 40. 1 cubic centimeters ? Ans. 29. 349 c. c. (69) State how you would symbolically express the valence of an atom. (70) (a) State the principles that underlie the nomencla- ture of binary molecules, (b} Which termination is always characteristic of a binary molecule ? (71) Why may mass reactions be represented by molecular formulas ? (72) Zinc sulphide ZnS contains 67$ of zinc, or 1 atom of zinc in each molecule. The atomic weight of zinc is 65 ; what is the molecular weight of zinc sulphide ? Ans. 97. 5 THEORETICAL CHEMISTRY. 7 (73) Phosphonium iodide PHJ may be obtained accord- ing to the following equation : PH 3 + HI = PHJ hydrogen hydriodic phosphonium phosphide acid iodide If 883.2 grams of hydriodic acid yield 1,117.8 grams of phosphonium iodide, whose molecular weight is 162, what is the molecular weight of hydriodic acid ? Ans. 128. (74) What volume would 350 cubic centimeters of ammonia gas, measured at 74, have at 94 ? Ans. 370.17 c. c. (75) Define atomic weight. (76) Write in words the meaning of each of the following symbols: H, O, 6> 2 , 3H f (77) How many liters of hydrogen weigh 1 gram ? (78) Ammonium nitrate NH^NO^ decomposes under the influence of heat into one molecule of hyponitrous oxide N t O and 2 molecules of water H^O. How many parts of hyponi- trous oxide N^O are there in 100 parts of ammonium nitrate ? Ans. 55 parts. (79) Antimonous sulphide may be precipitated according to the following equation : KSb'Cl^ + 3Sff t = KHCl + Sb^ antimonous hydrogen hydrochloric antimonous chloride sulphide acid sulphide 78-57 grams of antimonous sulphide are obtained from 105. 6 grams of antimonous chloride, the molecular weight of which is 228. 5. What is the molecular weight of antimonous sulphide ? Ans. 340. (80) You are required to neutralize with caustic soda a quantity of sulphuric acid diluted with water. Describe how you would do this, and how you would ascertain that the neutralization has been effected. (81) State what the valence of an atom means. (82) What data are necessary to determine the atomic weight of an element ? 8 THEORETICAL CHEMISTRY. 5 (83) Salt NaCl contains 39.32$ of sodium, the atomic weight of which is 23 ; what is the molecular weight of salt ? Ans. 58.5. (84) Sodium sulphate may be formed according to the following equation: ff,S0 4 + NafO, = Na^SO, + CO, + H^O sulphuric sodium sodium carbon acid carbonate sulphate dioxide 98 106 142 44 18 How much sodium sulphate may be yielded by the decom- position of 396.11 grams of sodium carbonate ? Ans. 530. 64 grams. (85) (a) What do you understand by the density of a gas ? (b) The volume of a given weight of a gas has to be calcu- lated at the standard of a temperature and pressure ; what does this mean ? (86) What is (a) a chemical reaction ? (b) a chemical reagent ? (87) How much calcium phosphate is required to yield 501 kilograms of phosphorus ? The atomic weight of phos- phorus is 31, and the molecular weight of calcium phosphate is 310. Ans. 5,010 Kg. (88) What volume would 420 cubic centimeters of hydro- gen, measured at 739 millimeters pressure, occupy at 760 millimeters pressure ? Ans. 408.4 c. c. (89) (a) What are the properties common to acids ? (b) In what respect does sulphuric acid differ from sulphurous acid ? (90) (a) How are chemical reactions classified ? (b) Illus- trate by using the letters A, B, etc. instead of symbols. (91) How much does 1 liter of hydrogen weigh ? (92) What do you understand by molecular stability ? 5 THEORETICAL CHEMISTRY. 9 (93) Give the formulas of potassium iodide, lead sulphide, calcium chloride, and cupric oxide. (94) Read the following equation by weight : potassium . sulphuric potassium nitric nitrate acid sulphate acid (95) How much lead maybe obtained from 932 kilograms of lead sulphide ? The atomic weight of lead is 207 and the molecular weight of lead sulphide is 239. Ans. 807.21 Kg. INDEX. NOTE. All items in this index refer first to the section (see Preface, Vol. I) and then to the page of the section. Thus, "Combustion 5 86" means that combustion will be found on page 86 of section 5. A. Sec. Absolute pressure ................ 4 " temperature ........... 4 Abstract numbers ............... 1 Acid, Definition of ................ 5 " Normal, and basic salts... 5 " sulphuric, Determination of ........... ............... 5 Acidimetry and alkalimetry ---- 5 Acids and salts, Nomenclature of 5 " Basicity of .................. 5 " Classification of ............ 5 Action, chemical, Modes of ...... 5 Addition .......................... 1 " in algebra ............... 3 " "decimals ............. 1 " of denominate numbers 2 " "fractions .............. 1 " " " in algebra.. 3 " " like quantities ....... 3 " "monomials ........... 3 " "polynomials .......... 3 Rule for ................. 1 " Sign of...- ............... 1 table ................ .... 1 Agents, Oxidizing ................ 5 " Reducing ................ 5 Aggregation, Symbols of ........ 1 " ........ 3 Air, Properties of ................ 4 " pumps ........................ 4 " unit, relation to hydrogen unit ........................ 5 Algebra, Definition of ............ 3 " Notation used in ....... 3 " Signs used in ........... 3 " Use of letters in ........ 3 Algebraic expressions, Reading 3 Page. Sec. Page. 27 Algebraic expressions, Terms of 3 5 31 Alkalies and bases 5 41 1 Allotropy 5 81 40 Amount (Percentage) 2 2 45 Analysis and synthesis 5 7 Aneroid barometer 4 23 94 Angle, Complement of 3 62 91 " Functions of 3 63 43 " Supplement of 3 62 44 " To find, from function... 3 64 45 Angles or arcs, Measurement of 2 11 59 Antecedent of a ratio 2 43 4 Apparatus, List of 5 97 10 Arabic notation 1 2 38 Arithmetic, Definition of 1 1 15 Fundamental proc- 27 esses of 1 4 39 Atom, Definition of 4 2 12 " " " 5 3 12 Atomic group, Calculation of 5 67 15 " weight, Definition of 5 21 8 " " Determination 4 of 5 32 5 " Relation of, to 89 specific heat... 5 34 90 " weights, Table of 5 28 49 Atomicity '... 5 20 17 " Table of - 5 21 18 Atoms and molecules 5 18 37 " " " Nomencla- ture of.. 5 19 71 " Motion of 4 3 3 Attraction, gravitation, and co- 3 hesion, Relation 4 between 5 8 1 " of matter 4 2 6 Avoirdupois weight 2 9 ix INDEX. B. Sec. Page. Balance, The 5 13 " Torsion 4 105 Barometer, Aneroid 4 23 Barometers 4 22 Base (Percentage) 2 2 Bases and alkalies 5 41 Basic, normal, and acid salts 5 45 Basicity of acids 5 44 Battery, Electrostatic 4 115 Voltaic 4 118 Berthollet's laws 5 58 Binary compounds, Formation of 5 38 " Nomencla- ture of 5 36 Binomial, Definition of 3 6 Bodies, Solid, liquid, and gaseous 4 3 Brace 1 49 Brackets 1 49 British thermal unit 4 58 Brittleness, Definition of 4 6 Bunsen burner 5 99 Burettes 5 93 " Practical suggestions concerning 5 108 C. Sec. Page. Calculation of atomic group 5 67 " percentage com- position 5 64 Calculations from equations 5 68 Calorie, The 4. 58 Cancelation 1 19 " inequations 3 53 Rule for 1 21 Capacity, Measures of 2 10 Cause and effect, Principle of... 2 54 Cell, Voltaic or galvanic 4 118 Chain, Engineer's. . 2 9 " Gunter's.... 2 8 Charges, Positive and negative 4 109 Chemical action, Modes of 5 59 " " Relation to force 5 7 " and physical changes 5 4 " proper- ties. ... 5 5 " operations.... 5 81 Chemistry, Object of 5 4 Cipher 1 2 Circular polarization 4 98 Clearing equations of fractions 3 54 Coefficient, Definition of 3 4 " of expansion 4 49 Cohesion, attraction, and gravi- tation. Relation between 5 8 Combination by volume 5 50 ** . Chemical, and mechanical mix- ture 5 5 Sec. Page. Combination in definite propor- tions 5 25 " in multiple propor- tions 5 25 Combining power, Quantity of. . 5 21 Combustion 5 86 Supporters of 5 87 Common denominator 1 26 Compass, The 4 97 Complement of an angle 3 62 Complex fractions 3 45 Composition, percentage, Calcu- lation of 5 64 Compound denominate number 2 8 proportion 2 55 Rule for.. 2 56 " radicals, Names of... 5 40 Compounds, binary, Nomencla- ture of 5 36 " Table of solubility of 5 60 " Ternary 5 40 Compressibility, Definition of . . . 4 5 Concrete numbers 1 1 Condenser, Electrical 4 114 Liebig's 5 84 Conductivity, Electrical 4 108 Conductors, Electrical 4 108 Consequent of a ratio 2 43 Cosines and sines, natural, Table of.- 3 77 Cosine of an angle 3 63 Cotangent of an angle 3 63 Cotangents and tangents, nat- ural, Table of 3 85 Couplet of a proportion 2 47 " (Ratio) 2 43 Crystallization 5 84 Crystallography 5 74 Crystals, Classification of 5 79 . " Systems of. 5 79 Cube of a number 2 23 ! root 2 25 " 2 32 " Proofof 2 38 " Rule for 2 38 Cubic measure 2 9 Currents, electric, Production of 4 116 D. Sec. Page. Dates, To find interval of time between 2 18 Decantation 5 82 Decimal point 1 36 " To express approxi- mately as a fraction having a given denom- inator 1 48 INDEX. XI Decimal, To reduce a fraction to Sec. Page. To reduce, to a fraction Decimals Addition of Division of how read Multiplication of.... 1 Subtraction of 1 39 Reduction of 1 46 Denominate number, Compound 2 8 numbers 2 7 " Addition of 2 15 44 Division of 2 .19 " 44 Multiplica- tion of.... 2 19 44 " Reduction of 2 12 44 Simple 2 7 Subtrac- tion of.... 2 17 44 " To reduce, to higher denomi- nations... 2 13 44 " To reduce, to lower denomina- tions 2 12 Denominator 1 22 Density and molecular weight.. 5 47 44 Definition of 5 20 41 Magnetic 4 101 Destructive distillation 5 87 Determination of sodium car- bonate 5 95 44 of sulphuric acid 5 94 Difference 1 9 (Percentage) 2 2 Diffusion of gases 5 49 Digits 1 2 Direct proportion..^ 2 47 ratio 2 43 Directions for laboratory 5 96 Dispersion of light 4 83 Distillation 5 84 " Destructive 5 87 Dividend 1 16 Divisibility, Definition of 4 4 Division 1 16 * 4 in algebra. 3 25 of decimals. 1 42 ** " denominate numbers 2 20 * "fractions 1 32 3 42 * "monomials 3 25 " "polynomials. 3 Rulefor 1 1 Sec. Page. Division, Sign of ................. i 16 Divisor ............................ i je Dollars and cents, how written 49 decimally 44 Sign for Double refraction Dry measure Ductility, Definition of Dynamical theory of heat E. Sec. Page. Elasticity, Definition of 4 5 Electric currents, Production of 4 116 44 series 4 104 Electrical conductivity 108 44 conductors 108 44 non-conductors 108 44 potential 115 44 resistance 108 Electricity 101 Electrochemical character of elements, Table of 31 Electrodes or poles 118 Electrodj-namics 115 Electromagnet, The 127 Electromagnetism 122 Electromotive series 119 Electrophorus, The 110 Electropositive and electronega- tive elements 30 Electroscope, The 104 Electrostatic battery 115 44 machines 4 112 Electrostatics 4 102 Elemental molecules, Number of 5 18 Elements, Electropositive and electronegative 5 30 Meta- 5 30 44 Nascent condition of 5 87 44 Table of 5 28 " Table of electro- chemical character of 5 31 Engineers' chain 2 9 Equality, Sign of 1 4 Equations, Calculations from;.. 5 68 44 Cancelation in 3 53 Changing signs in ... 3 53 Chemical 5 56 Clearing of fractions 3 54 Definition of 3 3 Literal 3 56 Membersof 3 51 Simple 3 55 Transformations in.. 3 52 " with one unknown quantity 3 56 Evaporation 5 82 Xll INDEX. Sec. Page. Evolution 2 Expansibility, Definition of 4 Expansion, Coefficient of 4 of bodies by heat 4 44 gases 4 Explosion 5 Exponent 2 44 3 44 Fractional 3 44 Negative 3 Exponents, Literal 3 14 Theory of 3 Expressions, algebraic, Reading of 3 Extension, Definition of 4 44 Measures of 2 Extremes (Proportion) 2 F. Sec. Factor, Prime . 1 Factoring 3 Factors ... 1 44 Equal 3 Fathom 2 Figures 1 44 Local or relative values of 1 44 Simple value of 1 Filtration 5 Force 5 44 Magnetic 4 44 of gravity 4 44 Relation of chemical action to 5 Formulas and symbols 5 44 for gravity problems 4 Foucault's prism 4 Fraction 1 44 Improper 1 44 Lowest terms of 1 44 Proper 1 44 Signs of 3 44 Terms of 1 44 To invert a 1 44 To reduce a decimal to 1 44 To reduce, to a decimal 1 44 To reduce, to a higher term 1 44 To reduce, to an equal fraction with a given denominator 1 " To reduce, to lower terms 1 44 Value of 1 Fractional exponent 3 Fractions, Addition of 1 Page. 20 28 19 4 29 11 2 2 2 98 34 Fractions, Addition and subtrac- tion of 44 Clearing of, in equa- tions 44 Complex 14 Division of. . . Sec. Page. in algebra Multiplication of, 44 Reduction of 1 44 " 44 3 44 Roots of 2 44 Subtraction of 1 To reduce, to common denominator 1 Function, To find angle from 3 Functions, To find, from angle. . 3 Trigonometric 3 Fusion and vaporization, Table of temperature of 4 Latent heat of 4 G. Sec. Gain or loss per cent 2 Gallon, C ubic inches in 2 44 Weight of 2 Gallons in cubic foot 2 Galvanic or voltaic cell, The 4 Gaseous bodies 5 Gases, Diffusion of 5 44 Expansion of 4 44 Mixtures of. . . 4 Permanent and Pressure, volume, temperature of 5 44 Properties of 4 Specific heat of 4 44 Table of specific gravity of 44 Tension of Gay-Lussac's laws Glass 44 Directions for bending 44 " cutting 44 u drawingout Gravitation, cohesion, and at- traction, Rela- tion between 5 14 Law of 4 Gravity, Force of 4 44 problems, Formulas for 4 44 Specific 4 44 of gases, De- termination of... 4 65 Page. 6 11 11 11 118 2 49 25 35 72 18 61 11 18 30 50 99 101 100 102 Gravity, specific, of liquids, De- termination of 44 44 44 solids, De- termination of 44 44 Tables of Group, atomic, Calculation of... Gunter's chain 2 H. Hardness, Definition of 4 Heat " conduction by metals, Table of 41 Dynamical theory of 4 41 Expansion of bodies by 44 Latent '. 41 44 of fusion 4 44 44 vaporization 44 Measurement of 4 44 Nature of 44 Propagation of " Quantity of 44 Sources of 44 Specific 44 of gases, Table of.. 44 4l 44 liquids, Table of 44 l4 44 solids, Table of 44 44 Relation of, to atomic weight... Hydrogen, Absolute weight of.. 44 unit, Relation to air unit Hydrometers 4 I. Ignition 5 Impenetrability, Definition of... Improper fractions, To reduce, to mixed numbers 1 Inches, To reduce, to decimal parts of a foot 1 Indestructibility, Definition of. . 44 of matter Index of a root Inertia, Definition of Inflammable bodies Insulators, List of 4 Integer 1 Integral expression 3 Inverse proportion 2 2 44 ratio 2 Involution 2 K. Known quantity . INDEX. Kill Sec. Page. L,. Sec. Page. Laboratory directions 5 96 Latent heat 4 64 4 12 44 of fusion 4 65 1 44 vaporization 4 66 Law, Mariotte's 4 26 4 13 44 44 5 72 4 10 44 of gravitation 4 7 5 67 44 44 refraction 4 76 2 8 Laws, Berthollet's 5 58 44 Gay-Lussac's 4 30 Sec. Page. 44 5 50 4 5 44 of electrostatics 4 104 4 41 44 44 reflection 4 74 44 44 weight 4 7 4 56 League 2 11 4 56 Least common denominator, To 4 47 find 1 26 4 64 Lenses 4 81 4 65 Letters, Use of, in algebra 3 1 4 66 Light 4 72 4 57 ' dispersion 4 83 4 41 44 Pencil of 4 73 55 44 Propagation of 4 73 57 44 Ray of 4 73 69 44 Theory of 4 72 58 4 ' Velocity of 4 73 61 Like numbers 1 1 61 44 quantities, Addition of 3 12 60 Subtraction of 3 14 44 terms 3 5 5 34 Linear measure 2 8 5 48 44 Surveyors' 2 8 Liquid bodies 5 2 5 71 44 measures 2 10 4 16 Liquids, Specific heat of 4 61 Sec. Page. 44 Table of specific grav- ity of 4 10 5 86 Literal equations 3 56 4 4 44 exponents 3 47 Litmus 5 43 1 25 " 5 91 Local and relative values of fig- 1 47 ures 1 2 4 5 Long-ton table 2 10 5 6 Loss or gain per cent 2 6 2 25 4 4 M. Sec. Page. 5 87 Machines, Electrostatic 4 112 4 108 Magnetic density 4 101 1 2 44 field 4 98 3 44 44 force 4 98 2 47 44 metals 4 98 2 50 44 poles 4 97 2 43 Magnetism 4 96 2 22 Magnets 4 96 Malleability, Definition of 4 6 s*. Page. Manometer, The 4 27 3 52 Mariotte's law 4 26 XIV INDEX. Sec. Page. Mariotte's law 5 72 Mass, Definition of 4 2 44 5 2 44 Motions of 4 2 44 of a body 4 6 Matter, Attraction of 4 2 44 Definition of 4 1 44 5 1 Divisions of 4 3 44 5 2 44 Indestructibility of 5 C Motions of 4 2 44 Physical state of 5 1 Properties of 4 4 Means (Proportion) 2 47 Measure, Cubic 2 9 Definition of 2 8 Dry 2 10 44 Linear 2 8 44 of angles or arcs 2 11 44 money 2 11 44 44 time 2 11 44 Square 2 9 44 Surveyors' linear 2 8 square 2 9 Measures and weights 5 8 Classification of 2 8 44 Liquid 2 10 44 Miscellaneous 2 11 44 of capacity 2 10 44 u extension 2 8 44 4l weight 2 9 44 Standard units of 2 8 Members of an equation 3 51 Mercurial barometer 4 22 Meta-elements 5 30 Metals and non-metals 5 26 44 Magnetic 4 98 44 Table of conduction by.. 4 56 44 44 specific gravity of 4 10 Meter 2 11 Methyl-orange 5 91 Metric system of weights and measures 5 8 Minuend 1 9 Minus 1 9 Mixed number 1 23 44 To reduce, to im- proper fraction 1 25 44 quantities 3 44 Mixture, Mechanical, and chemi- cal combination 5 5 Mixtures of gases 4 35 44 Temperature of 4 62 Mobility, Definition of 4 4 Molecular stabil ity 5 55 44 volume . . , 5 47 Molecular weight and density. . . Molecule, Definition of . . . Molecules and atoms 44 Chemical definition of 44 Classification of compound, Classifica- tion of 44 Saturated and unsat- urated ... Money, Measure of U. S Monomial, Definition of 14 To extract root of Monomials, Addition of Division of Multiplication of Subtraction of Motion, Atomic 44 Molecular 44 of matter Multiplicand Multiplication in algebra of decimals 44 d enom in ate Sec. 5 4 5 5 5 5 Page. 47 2 2 18 18 18 35 39 11 11 6 50 12 25 21 13 3 2 2 11 11 19 40 numbers.. . . . 2 19 44 fractions 1 31 u n u 3 41 44 monomials 3 21 44 polynomials... 3 21 Rule for 1 14 Sign of 1 11 " table 1 12 Multiplier 1 11 X. Sec. Page. Nascent condition of elements.. 5 87 Naught 1 2 Negative and positive charges. . . 4 109 44 44 quantities 3 8 44 exponent 3 48 Nicholson's hydrometer 4 17 Nicol's prism 4 92 Nomenclature of acids and salts 5 43 44 binary com- pounds 5 36 44 elemental mole- cules and atoms 5 19 Non-conductors, Electrical 4 108 Non-metals and metals 5 26 Normal, acid, and basic salts 5 45 Notation 1 1 44 Arabic 1 2 44 in algebra 3 3 Number 1 1 Abstract 1 1 INDEX. XV Sec. Page. v Sec. Page. Number, Concrete 1 1 Pressure, Absolute 4 2? " Denominate 2 7 volume, and tempera- Mixed 1 23 ture of gases 5 72 44 Prime 1 20 Prime factor i 20 Unit of 1 1 " number 1 20 Numbers, Like 1 1 Prism, Foucault's 4 93 Reading 1 3 44 Nicol's 4 92 Unlike : 1 1 Prisms 4 80 Numeration 1 1 Product 1 11 H 1 22 Propagation of heat 4 55 14 light 4 72 0. Sec. Page. Proper fractions 1 23 Oblique-angled triangles 3 72 Properties, Physical and chemi- Opaque bodies 4 72 cal 5 5 Oxidation 5 89 Proportion 2 46 Oxidizing agents 5 89 44 Compound 2 55 44 Direct 2 47 P. Sec. Page. how read 2 46 Parenthesis 1 49 44 written 2 46 Removal of 3 17 44 Inverse 2 47 Pencil of light 4 73 " " 2 50 Per cent., Sign of 2 1 44 Operations in 2 48 Percentage 2 1 44 Powers and roots in. . 2 52 " 2 2 44 Rules for 2 47 composition, Calcu- 44 Simple 2 55 lation of 5 64 Pumps, Air 4 37 Meaning of 2 1 Permanent gas 4 3 Q. Sec. Page. " " 5 2 Quantities, Known and unknown 3 52 Phenol-phthalein 5 91 Mixed 3 44 Physical and chemical changes . . 5 4 44 Positive and nega- 44 properties 5 5 tive 3 8 44 science, Definition of... 4 1 Unlike 8 14 Physics, Province of 4 3 Quantity, Definition of 3 3 Pite, Voltaic 4 120 41 Factors of 3 4 Pipettes 5 103 Quotient 1 16 Plane trigonometry Plus 3 1 62 4 R. Sec. Page. Polarization 4 90 Radical sign 2 25 Angle of 4 90 " " 8 5 44 Circular 4 93 Radicals, compoimd, Names of 5 40 in . opposite direc- Rate (Per cent) 2 2 tions 4 91 Ratio 2 42 Poles, Magnetic 4 97 44 Direct 2 43 44 or electrodes 4 118 44 Inverse 2 43 Polynomials, Addition and sub- 44 Operations upon 2 45 traction of 3 15 44 Reciprocal 2 43 44 Definition of 3 6 44 Terms of 2 43 44 Division of 3 26 44 To invert 2 44 " Multiplication of.. 3 21 44 Value of 2 43 Porosity, Definition of 4 4 Ray of light 4 73 Positive and negative charges. . . 4 109 Reaction 5 90 '' " " quantities 3 8 44 chemical, Facility of 5 57 Potential 4 115 Reactions 5 56 Power, Definition of 3 5 44 Classification of 5 57 " of a number 2 22 Reagent 5 90 Precipitation Prefixes, Use of 5 5 58 37 Reagents Reciprocal of a quantity 5 3 56 34 XVI INDEX. Sec, Reciprocal ratio 2 Reducing agents 5 Reduction of decimals 1 " denominate num- bers 2 " " fractions... 1 Reflection and refraction ........ 4 " Laws of ............... 4 Refraction and reflection ........ 4 Double ............... 4 Irregular ............. 4 Law of ................ 4 " Results of ............ 4 Table of indices of... 4 Total . . . . ! ............ 4 Relative and local value of figures ........................... 1 Remainder ........... , .......... . . 1 Resistance, Electrical ............ 4 Right-angled triangles ........... 3 Root, Cube ........................ 2 " Definition of ................ 3 u Index of .................... 2 " of a number ................ 2 " Square ...................... 2 Roots of fractions ................ 2 " other than square and cube ....................... 2 Rule for addition ................. 1 " " cancelation ............. 1 compound proportion.. 2 " cube root " division '' multiplication " square root " subtraction of three Rules for percentage ............. 2 Page. 43 90 46 12 23 35 74 74 108 68 25 5 25 23 25 40 41 8 21 56 38 18 14 30 10 47 2 S. Sec. Page. Saccharimeter, The 4 94 Salt, Definition of v 5 42 Salts, Acid, normal, and basic. . . 5 45 " and acids, Nomenclature of 5 43 Saturated and unsaturated molecules 5 39 Science, natural and physical, Definition of 4 1 Score 2 11 Separation, Chemical 5 81 Series, Electric 4 104 " Electromotive 4 119 Sign for dollars 1 48 " of addition 1 4 '* "division 1 16 " " equality 1 4 Sec. Page. Sign of multiplication 1 11 " "percent 2 1 " "subtraction 1 9 . " Radical 2 25 3 5 Signs, Changing, in equations... 3 53 " used in algebra 3 4 Simple denominate number 2 7 " equations 3 55 " proportion 2 55 Sine of an angle 3 63 Sines and cosines, natural, Table of 3 77 Sodium carbonate, Determina- tion of 5 95 Solubility of compounds, Table of 5 60 Solution 5 82 " of right-angled triangle 3 68 " "triangles 3 68 Solutions, Standard 5 92 Specific gravity, Definition of . . . 4 8 " of gases, Deter- mination of... 4 16 " of liquids, De- termination of 4 12 of solids, Deter- mination of . . . 4 13 " " Tables of 4 10 44 heat 4 58 44 " Relation of, to atomic weight 5 34 44 " Tables of 4 60 Spectra, Dispersion 4 83 Spectroscope, The 4 85 Spectrum, The 4 83 Sprengel's air pump 4 39 Square measure 2 9 " Surveyors' 2 9 " root 2 25 " Proof of 2 30 44 " Rule for 2 30 44 " Short method for... 2 31 Stability, Molecular 5 55 Standard solutions 5 92 44 units of various measures 2 8 Stoichiometry 5 63 Sublimation 5 86 Subtraction 1 9 of decimals 1 39 44 denominate num- bers 2 17 44 fractions 1 29 3 39 44 like quantities.. .. 3 14 44 " monomials... 3 13 INDEX. xvii Sec. Page. Subtraction of polynomials 3 15 44 unlike quantities 3 14 44 Rule for 1 10 Sign of 1 9 Subtrahend 1 9 Sulphuric acid, Determination of 5 94 Supplement of an angle 3 62 Supporters of combustion 5 87 Surveyors' linear measure 2 8 square measure 2 9 Symbols and formulas 5 24 44 of aggregation 1 49 44 44 3 17 44 " Order of prece- dence in 1 49 44 elements, Table of.. 5 28 Synthesis and analysis 5 7 T. Sec. Page. Table for conversion of English measures into metric 5 10 "~ 44 conversion of metric measures into Eng- lish 5 11 44 of atomic weights 5 28 44 44 atomicity 5 21 44 44 coefficients of expan- sion 4 50 44 44 conduction of heat by metals 4 56 44 44 electrochemical char- acter of elements 44 44 elements 44 44 indices of refraction. .. 44 44 metric measures 44 44 metric weights 44 44 natural sines and co- sines ' 4 44 natural tangents and cotangents 3 44 44 solubility of com- pounds 5 60 44 44 temperatures of fusion and vaporization 4 67 44 44 valence 5 23 Tables of specific gravity of sub- stances 4 10 44 trigonometric, Use of 3 64 Tangent of an angle 3 63 Tangents and cotangents, nat- ural, Table of 3 86 Temperature 4 42 Absolute 4 31 44 Measures of 4 43 44 of mixtures 4 62 pressure, and vol- ume of gases 5 72 Sec. Page. Temperatures of fusion and vaporization, Table of .......... 4 67 Tenacity, Definition of ........... 4 5 Tension of gases .................. 4 18 Terms, Algebraic ................ 3 5 " of a fraction .............. 1 23 44 ' 4 an equation, Changing signs of .............. 3 53 4< 44 a ratio ................. 2 43 Ternary compounds .............. 5 40 Theory of exponents ......... .. .. 3 47 Thermal units .................... 4 58 Thermometers .................... 4 43 Rules for com- paring ---- ...... 4 46 Time, Measure of ................. 2 11 Torsion balance .................. 4 105 Transformations in equations. . . 3 52 Translucent bodies .............. 4 72 Transparent bodies. . '. ........... 4 72 Transpositions in equations ..... 3 53 Triangles, Right-angled ......... 3 68 44 Solution of ............ 3 68 Trigonometric functions ......... 3 62 44 tables, Use of.... 3 64 Trigonometry, Plane ............ 3 62 Trinomial, Definition of ......... 3 6 To factor ............. 3 30 Troy weight ...................... 2 10 U. Sec. Page. 5 5 31 28 44 Hydrogen, relation to air unit , 5 71 4 78 44 of a number 1 1 6 9 Unknown quantity 3 52 6 10 Unlike numbers 1 1 44 quantities g 14 8 77 3 5 8 86 Unsaturated and saturated molecules.. . 5 39 U. S. money 2 V. Sec. Vacuum, Definition of 4 Valence 5 44 Graphic description of. . 5 Table of 5 Value of a fraction 44 44 44 ratio Vapor Vaporization and fusion, Table of temperatures of 14 Latent heat of Vary, Meaning of the term Vinculum 11 Page. 21 21 87 xvm INDEX. Sec. Page. Voltaic battery, The 4 118 " elements, List of 4 119 " or galvanic cell 4 118 pile 4 120 Volume, Combination by 5 50 Molecular 5 47 " temperature, and pres- sure of gases 5 72 AV. Sec. Page. Weighing, Method of 5 16 Weight, Absolute, of hydrogen.. 5 48 atomic, Definition of.... 5 21 " Determination of ... 5 32 Sec. Page. Weight, atomic, Relation of, to specific heat 5 34 " Avoirdupois 2 9 " Definition of 4 4 Laws of 4 7 " Measures of 2 9 " Molecular, and density 5 47 Weights and measures, English 5 12 " " " Metric.. 5 8 " atomic, Table of 5 28 Wood, Table of specific gravity of... 4 11 Zero. Sec. Page. 1 2 O:> :> S2:>; THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. "1 J f y r. 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