able weight of the casting can be obtained closely enough by weighing the pattern and multiplying this weight by the proper number from the following table. In case the pattern contains core prints, the weight of these prints should be calculated and subtracted from the pattern weight before multiplying; or else the total pattern weight can be multiplied first and then the weight of metal which would occupy the same volume as the core print be subtracted from it.

131. Find the weight of a piece of steel shafting 2 in. in diameter and 20 ft. long.

132. What is the weight of a billet of wrought iron 4 in. square and 2 ft. 8 in. long?

133. What would a steel boiler plate 36 in. by 108 in. by in. weigh? 134. A cast steel cylinder is 42 in. inside diameter, 4 ft. 6 in. long and 1 in. thick. Find its weight.

135. A steam engine cylinder 4 in. inside diameter has the cylinder head held on by four studs. When the pressure in the cylinder is 125 lb. per square inch, what is the total pressure on the cylinder head and what is the pull in each stud?

136. 50 studs 2 in. long and 1 in. in diameter are to be cut from cold rolled steel. Find the length and weight of bar necessary, allowing in. per stud for cutting off.

137. What would be the weight of ain. by 3 in. wagon tire for a 40 in. wheel? (Length of stock=circumference of a 391⁄2 in. circle.)

138. A copper billet 2 in. by 8 in. by 24 in. is rolled out into a plate of No. 10 B. & S. gage. The thickness of this gage is .1019 in. What would be the probable area of this plate in square feet?

139. The steel link shown in Fig. 26 is made of in. round steel (round steel in. in diameter). Find the length of bar necessary to make it and then find the weight of the link.

140. A steel piece is to be finished as shown in the sketch below (Fig. 27). The only stock available from which to make it is 4 in. in diameter. Compute the length of the 4 in. stock which must be upset to make the piece and have extra stock all over for finishing.

75. The Meaning of Square Root.-The previous chapter showed the usefulness of squares in finding areas and of cubes in finding volumes. Problems often arise in which it is necessary to find one edge of a square or cube of which only the area or volume is given. For instance, what must be the side of a square so that its area will be 9 sq. in.? The length of the side must be such that when multiplied by itself it will give 9 sq. in. A moment's thought shows that 3X3=9, or 32=9. Therefore, 3 is the necessary side of the square. Finding such a value is called Extracting the Square Root, and is represented by the sign

called the square root sign or radical sign. Thus √9=3; 16-4. To make clear the idea of extracting square roots, the student should consider it as the reverse or "the undoing"

of squaring, just as division is the reverse of multiplication or as subtraction is the reverse of addition.

52=25, and its reverse is: √25=5.

The square roots of some numbers, like 4, 9, 16, 25, 36, 49, 64, 81, etc., are easily seen, but we must have some method that will apply to any number. There are several methods of finding square root, of which two are open to the student of shop arithmetic: (1) by actual calculation; (2) by the use of a table of squares or square roots. A third method which uses logarithms will be explained in the chapters on logarithms. In many handbooks will be found tables giving the square roots of numbers, but we must learn some method that can be used when a table is not available and the method that will now be explained should be used throughout the work in this chapter.

76. Extracting the Square Root.-The first step in finding the square root of any number is to find how many figures there are in the root. This is done by pointing off the number into periods or groups of two figures each, beginning at the decimal point and working each way.

12=1

102=1'00

1002=1'00'00

From these it is evident that the number of periods indicates the number of figures in the root. Thus the square root of 103684 contains 3 figures because this number (10'36'84) contains three periods. Also the square root of 6'50'25 contains three figures since there are three periods. (The extreme left hand period may have 1 or 2 figures in it.) We must not forget that, for any number not containing a decimal, a decimal point may be placed at the extreme right of the number. Thus the decimal point for 62025 would be placed at the right of the number (as 62025.)

The method of finding the square root of a number can best be explained by working some examples and explaining the work as we go along. The student should take a pencil and a piece of paper and go through the work, one step at a time, as he reads the explanation.

Example:

Find the square root of 186624.

Point off into periods of two figures each (18'66'24) and it will be seen that there are 3 figures in the root. The work is arranged very similarly to division.

2X40=80

18'66'24(432

16
2 66

3

83

2 49

2 862

2 X 430=860

17 24

17 24

Explanation: First find the largest number whose square is equal to or less than 18, the first period. This is 4, since 52 is more than 18. Write the 4 to the right for the first figure of the root just as the quotient is put down in long division. The first figure of the root is 4. Square the 4 and write its square (16) under the first period (18) and subtract, leaving 2.

Bring down the next period (66) and annex it to the remainder, giving 266 for what is called the dividend. Annex a cipher to the part of the root already found (4) giving 40; then multiply this by 2, making 80, which is called the trial divisor. Set this off to the left. Divide the dividend (266) by the trial divisor (80). We obtain 3, which is probably the next figure of the root. Write this 3 in the root as the second figure and also add it to the trial divisor, giving 83, which is the final divisor. Multiply this by the figure of the root just found (3) giving 249. Subtract this from the dividend (266) leaving 17.

Bring down the next period (24) and annex to the 17, giving a new dividend 1724. Repeat the preceding process as follows: Annex a cipher to the part of the root already found (43) giving 430; and multiply by 2, giving 860, the trial divisor. Divide the dividend by this divisor and obtain 2 as the next figure of the root. Put this down as the third figure of the root and also add it to the trial divisor, giving 862 as the final divisor. Multiply this by the 2 and obtain 1724, which leaves no remainder when subtracted from the dividend. As there are no more periods in the original number, the root is complete.

77. Square Roots of Mixed Numbers.-If it is required to find the square root of a number composed of a whole number and a decimal, begin at the decimal point and point off periods to right and left. Then find the root as before.

Example:

Find the square root of 257.8623 2'57.86'23'00(16.058+, Answer.

20

1
1 57

6

26

3200

5

3205

32100

8

32108

1 56

1 86 23 (a)

1 60 25

25 98 00

25 68 64

29 36

78. Square Roots of

Explanation: 1 is the largest number whose square is equal to or less than 2, the first period. Proceeding as before, we get 6 for the second figure. After subtracting the second time (at a) we find that the trial divisor 320 is larger than the dividend 186. In this case, we place a cipher in the root, annex another cipher to 320 making 3200, annex the next period, 23, to the dividend and then proceed as before. If the root proves, as in this case, to be an interminable decimal (one that does not end) continue for two or three decimal places and put a + sign after the root as in division. In this example the decimal point comes after 16, because there must be two figures in the whole number part of the root since there are two periods in the whole number part of our original number.

Decimals. Sometimes, in the case of a decimal, one or more periods are composed entirely of ciphers. The root will then contain one cipher following the decimal point for each full period of ciphers in the number.

Example:

Take .0007856 as an example.

Beginning at the decimal point and pointing off into periods of two figures each, we have .00'07'85'60. Hence, the first figure of the root must be a cipher. To obtain the rest of the root we proceed as before.

=

It will be noticed that the square root of a decimal will always be a decimal. If we square a fraction, we will get a smaller fraction for its square, (); or as a decimal, .25 .0625. Therefore, the opposite is true; that, if we take the square root of a number entirely a decimal, will get a decimal, but it will be larger than the one of which it is the square root. Notice the example just given: .0007856 is less than its.square root .028. 79. Rules for Square Root.-From the preceding examples the following rules may be deduced:

1. Beginning at the decimal point separate the number into periods of two figures each. If there is no decimal point begin with the figure farthest to the right.

2. Find the greatest whole number whose square is contained in the first or left-hand period. Write this number as the first figure in the root; subtract the square of this number from the first period, and annex the second period to the remainder.

3. Annex a cipher to the part of the root already found and multiply by 2; this gives the trial divisor. Divide the dividend by the trial divisor for the second figure of the root and add this figure to the trial divisor for the complete divisor. Multiply the complete divisor by the second figure in the root and subtract this result from the dividend. (If this result is larger than the dividend, a smaller number must be tried for the second figure of the root.)

Bring down the third period and annex it to the last remainder for the new dividend.

4. Repeat rule 3 until the last period is used, after which, if any additional decimal places are required, annex cipher periods and continue as before. If the last period in the decimal should contain but one figure, annex a cipher to make a full period.