The contents of any piece of timber reduced to cubic inches can be found in board feet by dividing by 144, or from cubic feet by multiplying by 12. As simple examples: How many board feet in a piece of lumber containing 2,880 cubic inches? 2880⁄144 = 20 board feet. How much wood in a joist 16 feet long, 12 inches wide and 6 inches thick? 16 × 1 × 1⁄2 = 8 cubic feet: 8 × 12 = 96 board feet. A simpler method may be used in most cases. How much wood in a beam 9 inches × 6 inches, 14 feet long? Imagine this timber built up of 1-inch boards. As there are nine of them, and each 14 ft. × 1⁄2 foot × 1 inch and contains 7 board feet ([Fig. 235]), 7 × 9 = 63 board feet. Again, how much wood in a timber 8 inches × 4 inches, 18 feet long? This is equivalent to 4 boards 1 inch thick and 8 inches or 2⁄3 foot wide. Each board is 18 × 2⁄3 × 1 = 12 board feet, and 12 × 4 = 48, answer. (See b, [Fig. 235]).
To take a theoretical case: How much wood in a solid circular log of uniform diameter, 16 inches in diameter, 13 feet and 9 inches long? Find the area of a 16-inch circle in square inches, multiply by length in inches and divide by 144.
| 16 × 16 × .7854 = 201 | 13 ft. 9 in. = 165 inches |
| (201 × 165)⁄144 = 13045⁄144 board feet | |
It is not likely that a boy would often need to figure such an example, but if the approximate weight of such a timber were desired, this method could be used, reducing the answer to cubic feet and multiplying by the weight per cubic foot.
A knowledge of square root is often of great value to the woodworker for estimating diagonals or squaring foundations. The latter is usually based on the known relation of an hypothenuse to its base and altitude. It is the carpenters' 3-4-5 rule. The square of the base added to the square of the altitude = square of the hypothenuse. 3² = 9, 4² = 16; 9 + 16 = 25. The square root of 25 is 5. (See [Fig. 235]). To square the corner of his foundation the carpenter measures 6 feet one way and 8 the other. If his 10-foot pole just touches the two marks, the corner is square. 6² = 36, 8² = 64; 36 + 64 = 100. √100 = 10. This method was used in laying out the tennis court, the figures being 36, 48, 60—3, 4, and 5 multiplied by 12.
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 151⁄3 feet × 21⁄2 feet = 381⁄3 square feet. The combined length of top and bottom is 9 feet 4 inches = 91⁄3 × 31⁄6 = 295⁄5 , and 381⁄3 + 295⁄5 = 678⁄5 or 68 board feet, ignoring such a small amount as 1⁄5 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.
