by saying the 6 in. circle has three times the diameter of the 2 in. circle and, therefore, the area is 32, or nine times as great. A piece of steel plate 6 in. in diameter weighs nine times as much as a piece 2 in. in diameter of the same. thickness. Likewise a 10 in. square has four times the area of a 5 in. square.

If we

FIG. 20.

let A represent the area of the larger circle, a the area of the smaller circle, D the diameter of the larger circle, and d the diameter of the smaller circle, then we have the direct proportion:

A:a=D2:d2

69. The Rectangle. When a four-sided figure has square corners it is a Rectangle. Each side of a brick is a rectangle.

3"
FIG. 21.

A Square is a special kind of rectangle having all the sides equal. The area of a rectangle is obtained by multiplying the length by the breadth. In Fig. 21 the area is 2×3=6 sq. in., as can be seen by counting the 1-in. squares, which each contain 1 sq. in.

70. The Cube.-Just as the square of a number is represented by the area of a square, one side of which represents the number,

10"

10

so the cube of a number is represented by the volume of a cubical block, each edge of which represents the number. The volume of a cube which is 10 in. on each edge is 10 X 10 X 10 = 1000 cubic inches, and since this is obtained by "cubing" 10 (103=10×10×10=1000), we can see that the cube of a number can be represented by the volume of a cube, the edge of which represents the number. If the edge of the cube is one-half as long, that is 5 in., the volume is 5X5X5=125 cubic inches, or only the volume of the 10 in. cube.

FIG. 22.

FIG. 23.

MEASURES OF VOLUME (CUBICAL MEASURE)

1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.)
27 cubic feet=1 cubic yard (cu. yd.)

(Larger units than cubic yards are seldom, if ever, used.)

71. Volumes of Straight Bars.-A piece 1 in. long cut from a bar will naturally contain just as many cu. in. as there are sq. in. on the end of the bar. In the billet shown in Fig. 24, there are 3X4 12 sq. in. on the end of the bar, and a piece 1 in. long contains 12 cu. in. The entire billet contains 10 slices just like this one, so there are 12×10=120 cu. in. in the entire billet.

=

Therefore, we see that to find the number of cu. in. in any straight bar we proceed as follows:

Calculate the area of one end of the bar in square inches; then multiply this result by the length of the bar in inches; the result will be the number of cubic inches in the bar. For bars of square or rectangular section, the volume is the product of the three dimensions, length, breadth, and thickness. If L, B, and T represent the length, breadth, and thickness, and V stands for the volume, then

V=LXBXT

10".

FIG. 24.

Example:

How many cubic inches of steel in a bar 2 in. square and 4 ft. long?
4 ft. 48 in.
V=LXBXT

=48×2×2=192 cu. in., Answer.

For round bars, the area on the end is .7854 times the square of the diameter, and this, multiplied by the length, gives the volume. Then, if D represents the diameter of the bar and L its length, the volume V will be

V = .7854X D2 XL

This will apply equally well to thin circular plates or to long bars or shafting. With thin plates, we would naturally speak of thickness (T) instead of length (L). Fig. 25 shows that the two objects have the same shape except that their proportions are different.

Examples:

1. How many cubic inches of steel in a shaft 2 in. in diameter and 12 ft. long?

12 ft. =12×12=144 in., length of bar.

V = .7854 X D2 XL

V = .7854 X 22 X 144

V = .7854X4X144-452.39 cu. in., Answer.

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

1237X.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 plate in. thick and 1 ft. square will weigh 5 lb., since 12X12X18 cu. in., and 18X.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 Xnumber 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.

X8×3×5=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:

1080X.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