25 sq. in.; if the side is 5 ft., the area is 25 sq. ft. If we simply have the number 5, its square is 25, no matter what kind of things the 5 may refer to. As mentioned before, 1 sq. ft. is the area of a square 1 ft. on each side and, if divided into square inches, will be found to contain 122 or 144 sq. in. Likewise, a square yard is 3 ft. on each side and, therefore, contains 329 sq. ft. The following table gives the relation between the units ordinarily used in measuring areas: FIG. 18. MEASURES OF AREA (SQUARE MEASURE) 144 square inches (sq. in.) = 1 square foot (sq. ft.) 640 acres 1 square mile (sq. mi.) = 68. Area of a Circle.-If a circle is drawn in a square as shown in Fig. 19, it is easily seen that it has a smaller area than the square because the corners are cut off. The area of the circle is always a definite part of the area of the square drawn on its diameter, the area of the circle being always .7854 times the area of the square. This number .7854 happens to be just onefourth of the number 3.1416 given in Chapter VII. Just why this is so will be shown later on. If the diameter of the circle = 10 in., as in Fig. 19, the area of the square is 100 sq. in. and the area of the circle is .7854 × 100=78.54 sq. in. You can prove this to your own satisfaction in the following manner. Cut a square of cardboard of any size, and from the center describe a circle as shown just touching on all four sides. Weigh the square, and then cut out the circle and weigh it. The circle will weigh .7854 X weight of the square. A pair of balances such as are found in a drug store are the best for this experiment. AREA 78.54 10" FIG. 19. Rule for Area of Circle.-The area of any circle is obtained by squaring the diameter and then multiplying this result by .7854. If written as a formula this rule would read If you think a little you will see that, if the diameter is doubled, the area is increased four times. This can also be seen from Fig. 20. The diameter of the large circle is twice that of one of the small circles, but its area is four times that of one of the small circles. This is a very important and useful law and may be stated as follows: "The areas of similar figures are to each other as the squares of their like dimensions." A 2 in. circle contains 22X.7854 3.1416 sq. in., while a 6 in. circle contains 62X .7854 28.2744 sq. in., or nine times as much. This we can find = 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" 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 2X3-6 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" 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 10×10 × 10=1000 cubic inches, and since this is obtained by "cubing" 10 (103-10X10X10=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 5X5×5=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.) (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 3×4=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. =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 = .7854XD2 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 × L V.7854 X 22 X 144 V=.7854×4×144=452.39 cu. in., Answer. |