2 becomes 
TURPENTINE. 25 
Since the values of V, and d are known in this case, V, being equal 
to V», the volume of the bilged end of the tank (as calculated on p. 20), 
k, or the depth of dishing for a cone having the same capacity as the 
spherical end of the tank, can be determined. Transposing, formula 
12 CV, 3319 
iO d? 
In this example, V,=15,116 cubic inches and d=78 inches. Sub- 
stituting these values in formula 9, 
BRIO 15116) >) : 
k= 6084 =9.5 inches (approximately). 
This value for & is the length of the line ob. 
Having found the value of k, the next step is to ascertain the maxi- 
mum distance for any particular outage from the end of the straight 
part or body of the tank to the surface of the imaginary conical end, 
i. e., line pg in figure 4. aob and apq are similar triangles, and the 
ratio of ap to pq is the same as that of ao to 0b. ‘That is, 
ap - GO = = oo Xap 
DY oh. CN XO Beane 
(formula 9) 
k= 
(formula 10) 
The values of ap, ob and ao are known, being, respectively, 6 inches 
(the measured outage), 9.5 inches (the value for /:), and 39 inches 
(the radius of the tank). Substituting these values in formula 10, 
se 
pe = 1247 inches: 
The second assumption to : made in this solution is that the out- 
age in the end (represented by the shaded portion in figure 4) bears 
the same ratio to the volume cf one-half the entire dished end as 
the area of the triangle apq bears to the triangle aob. That is, 
V, 
Outage sought: _" : area apg : area aob 
V, 
2 X area apg (formula 11) 
area aob 
V, = volume of the bilged end of the tank head. 
The area of the triangle a py— ed OO a square inches, 
aoXob 39X95 
De 2 
inches. Substituting these values in formula 11, this becomes 
7581 X 4.4 
185.2 
Since this is the outage at one end only, the total outage at the two 
ends of the tank is 360 cubic inches, which is equivalent to approxi- 
mately 1.5 gallons. 
474°—20———_4 
Or, outage = 
and the area of the triangle aob= =185.2 square 
Tank end outage = = 180 cubic inches (approximately). 
