Page:Sheet Metal Drafting.djvu/58

Problem 8C.—What is the diameter of the can mentioned in Problem 8B?

Illustrative Examples

Example of Problem 8B.

What is the area of the bottom of a can 12″ high holding 20 qts.?

20 qt.=5 gal. (volume)
5×231=1155 cu. in. (volume)
Formula (b) would apply here, Area=Volume÷Height
Substituting known values, Area=1155 cu. in.÷12″
121155.0096.25 sq. in.

Ans. 96.25 sq. in.

Example of Problem 8C.

What is the diameter of the bottom of the can mentioned in the preceding example?

Formula (d) would apply	$\scriptstyle {{\text{Diameter}}={\sqrt {{\text{Area}}\div .7854}}}$
Substituting,	$\scriptstyle {{\text{Diameter}}={\sqrt {96.25\div .7854}}}$

.7854	96.250000	122.54	sq. in.
	78 54
	17710
	15708
	020020
	015708
	0043120
	0039270
	00038500
	00031416

Extracting square root.,	$\scriptstyle {\sqrt[{2}]{}}$	122.5411.0
		1
21		022
		021
220		00154

Ans. 11″, diameter.