2. How many cubic inches in a blank for a boiler head 60 in. in diameter

72. Weights of Metals.-The chief uses in the shop for calculations of volume are in finding the amount of material needed to make some object; in finding the weight of some object that cannot be conveniently weighed; or in finding the capacity of some bin or other receptacle. Having obtained the volume of an object, it is only necessary to multiply the volume by the known weight of a unit volume of the material to get the weight of the object. In the case of the shaft of which we just got the volume, 1 cu. in. will weigh about .283 lb., so the total weight of the shaft will be

452.39X.283=128.0+ pounds.

The weight of the boiler head will be

1237.283=350+ pounds.

The following table gives the weights per cubic inch and per cubic foot for the most common metals and also for water:

73. Short Rule for Plates.-A flat wrought iron platein. thick and 1 ft. square will weigh 5 lb., since 12×12×}=18 cu. in., and 18.278-5 lb. The rule obtained from this is very easy to remember and is very useful for plates that have their dimensions in exact feet.

Rule.-Weight of flat iron plates = area in square feet X number of eighths of an inch in thickness X5. This rule can also be used for steel plates by adding 2 per cent. to the result calculated from the above rule.

Example:

Find the weight of a steel plate 30 in. X96 in. X

30 in. =2ft.,
-2 ft., 96 in. -8 ft.,in. =3 eighths.

in.

28×3×5=300 (weight if it were of wrought iron).
2% of 300-6 lb.

300+6=306 lb.; weight of steel plate, Answer.

If this weight is calculated by first getting the cubic inches of steel, we get:

1080.283=305.64 lb. weight of steel plate, Answer.

We see that the results check as closely as could be expected and, in fact, different plates of supposedly the same size would differ as much as this because of differences in rolling.

74. Weight of Casting from Pattern.-In foundry work, it is often desired to get the approximate weight of a casting in order to calculate the amount of metal needed to make it. The prob

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 21 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 a in. by 3 in. wagon tire for a 40 in. wheel? (Length of stock-circumference of a 39 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

10021'00'00

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.