TABLE XUV. SOLID CONTENT OF CYLINDRICAL TREES, ETC. 550 



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 G.932 = . 1.733 



105.713 

 12 



8.55G 

 4 



2.224 

 Content, 105 feet 8J 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 

 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 C 



inches, stands ,*.*.. 5.416 



Which multiply by ..... 4 



21.664 

 To be multiplied by the length, ... 12 



259.968 

 and .968 = 11 J inches, gives the content 259 feet 11 J 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 112 2 = 12544 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 

 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. 



