Page:Sheet Metal Drafting.djvu/186

Problem 40B.—A customer wishes a 2½ gallon flaring measure. The bottom of this measure is to be 9″ in diameter, and the top 7″ in diameter. How high must the measure be made to fulfill these requirements?

Example of Problem 40B.—Suppose the dimensions were 6″ diameter of bottom, 4″ diameter of top, and the measure was to hold 3 quarts.

Original Formula	$\scriptstyle {V=(B+b+{\sqrt {(B\times b)}}H\div 3}$
Transposing,	$\scriptstyle {H\div 3=V\div (B+b+{\sqrt {B\times b)}}}$
In this formula,	H=Altitude, which is the unknown
	V=Volume=3 quarts or 173.25 cu. in.
B=area of lower base=62×.7854=28.274 sq. in.
b=area of upper base=42×.7854=12.566 sq. in.

Substituting known values in above formula,

and

${\begin{aligned}\scriptstyle {H\div 3}&\scriptstyle {=173.25\div (28.274+12.566+{\sqrt {28.274\times 12.566}}}\\\scriptstyle {H\div 3}&\scriptstyle {=173.25\div 59.69}\\\scriptstyle {H\div 3}&\scriptstyle {=2.9}\\\scriptstyle {H}&\scriptstyle {=2.9\times 3=8.7{\text{ in.}}}\end{aligned}}$

Ans. 8.7 in. high.