Occupations > Agriculturist's calculator

TABLE XLIV. SOLID CONTEXT OF CYLINDRICAL TREES, ETC. 559
If there are odd inches in the length, add a proportional part
of one foot, after multiplying by the feet.
Thus, if the length had been 15 feet 3 inches,
content of 15 feet in length, as above, . 103.980
3 inches equal J of C.932 = • 1.733
105.713
12
8.556
4
2.224
Content, 105 feet inches.
If a given girth exceeds the Table, multiply the number oppo- |;
site its half by 4, the product is the content of 1 foot in length i
by the given girth, which multiply by the given length.
Example.
What is the solid content of a tree 12 feet in length, and girth
16 feet 6 inches.
Opposite 8 feet 3 inches, the half of 16 feet 6
inches, stands 5.416
Which multiply by .... . 4
21.664
To be multiplied by the length, 12
259.968
and .968 = 11J inches, gives the content 259 feet 11J inches.
The Table is calculated as follows:—The area of a circle,
whose circumference is 1 foot = .0795775, is divided by 144, the
number of square inches in a square foot, = .000552622, = the
area of a circle whose circumference is 1 inch, in the decimal of a
foot. Then, as the areas of circles are to one another as the
squares of their circumferences, the square of any given circum¬
ference or girth, multiplied by this fraction and by the length,
gives the solid content.
The content of the first example is thus found:—The girth is
112 inches, then 1122 = 125 44 X .000552622 = 6.932090368
the tabular content of 1 foot, which is to be multiplied by 15, the
length in feet, and the product is 103.98135552, the content in
cubic feet.
| To save trouble in the multiplication, only three places of
j decimals are retained in the Table, and the third figure is in¬
creased by 1, when the figure in the fourth place is above 5.