(201 × 165)/144 = 130 - 45/144 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. [sqrt]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
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.
[Illustration: Fig. 235. A packing case]
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.
[Illustration: Fig. 236. Laying out the pieces for a tabourette]
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.
[Illustration: Fig. 237. Finding the area of an octagon]
[Illustration: Fig. 238. Problem of the hexagon]
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
Chapter IV, the hexagon is drawn in a circle. One of the sides is equal
to the radius or half the diameter. This gives us the arrangement shown
in Fig. 238, in which our problem is confined to the right-angled
triangle whose base is seven and hypothenuse fourteen. From our
knowledge of triangles, we deduct the square of seven (49) from the
square of 14 (196), leaving 196-49 = 147, which is the square of the
altitude. Then [sqrt]147 = 12.12, which is the narrowest board from
which we can obtain a hexagon 14 inches in diameter.
These examples are given to show the close connection between woodwork
and arithmetic.