Chapter X: Preface (10)

To take a more practical case, suppose we are called upon to estimate exactly, without any allowance for waste, the amount of lumber in a packing case built of one-inch stock, whose outside dimensions are 4 feet 8 inches × 3 feet 2 inches × 2 feet 8 inches. Referring to the drawing (Fig. 235, _d_), we draw up the following bill of material:

2 pieces (top and bottom) 4 ft. 8 in. × 3 ft. 2 in.
2 " (sides) 4 ft. 8 in. × 2 ft. 6 in.
2 " (ends) 3 ft. 0 in. × 2 ft. 6 in.

The top and bottom, extending full length and width, are the full dimensions of the box, while the sides, although full length, are not the full height, on account of the thickness of the top and bottom pieces--hence the dimensions, 2 feet 6 inches. From the ends must be deducted two inches from each dimension, for the same reason. In multiplying, simplify as much as possible. There are four pieces 2 feet 6 inches wide; as their combined length is 15 feet 4 inches, we have 15-1/3 feet × 2-1/2 feet = 38-1/3 square feet. The combined length of top and bottom is 9 feet 4 inches = 9-1/3 × 3-1/6 = 29-5/9, and 38-1/3 + 29-5/9 = 67-8/9 or 68 board feet, ignoring such a small amount as 1/9 of a foot. This is close figuring, too close for practical work, but it is better to figure the exact amount, and then make allowances for waste, than to depend on loose methods of figuring, such as dropping fractions, to take care of the waste.

As a good example of estimation, take the hexagonal tabourette shown in Fig. 178; the five pieces, aside from the hexagon under and supporting the top, which may be made from scrap lumber, are shown laid out in Fig. 236. The board must be at least twelve inches wide in order to get out of it the large hexagon. The legs may be laid out as shown with space left between for sawing, yet even by this method considerable waste will result, and it should be kept constantly in mind that as far as possible waste is to be reduced to a minimum. "Wood butcher" is the common shop name for the workman who spoils more material than he uses.

The great advantage of making out a bill of material before starting is that it not only makes you study your drawing, but causes you to consider the best method of laying out the blank pieces.

It is often necessary to find the areas of figures other than the square or parallelogram. Assume that we are to floor a room in an octagonal tower or summer house. If the distance across the flat sides of the octagon is sixteen feet, leaving out the item of waste, how many square feet will be required?

The octagon may be drawn in a square and its area will be that of the square, less the four triangles in the corners. (Fig. 237). So the problem resolves itself into finding the area of one of these triangles. If we knew the length of one of the sides of the octagon, the solution would be simple, but we only know that the eight sides are equal. The following method may be worked out: Find the diagonal of the sixteen foot square. It is 22.6+. Deduct the distance across the flats, 16, leaving 6.6 feet equally divided between _a_ and _b_; _a_ = 3.3 and it may be proved that _c_ = _a_ = _d_. So in each corner we have a triangle whose base is 6.6 × 3.3. The area of a triangle equals half its base by the altitude. Therefore the area of each triangle is 3.3 × 3.3 and 3.3 × 3.3 × 4 equals 43.56 square feet, the combined area of the four corners. This deducted from the area of the square leaves the area of the octagon, or 256-43.56 = 212.44 square feet.

Assume that our problem is to find the narrowest board we can use to cut out a hexagon whose diameter is fourteen inches. As shown in