Page:Sheet Metal Drafting.djvu/142

whose height is 5′–0″. Allowing ¾″ for locks what is the area of its pattern?

Problem 30B.—The funnel of the pump is a frustum of a cone whose lower base diameter is 8″, upper base diameter 4″, and altitude 4″. What is its area?

Sample Problem.—Substituting for the above dimensions 6″ upper base, 3″ lower base, and 3″ altitude, we would have

Original Formula:

${\frac {{\text{Circumference of upper base}}+{\text{Circumference of lower base}}}{2}}\times {\text{slant height.}}$

Substituting,

${\frac {(6\times \pi )+(3\times \pi )}{2}}\times {\sqrt {3^{2}+1.5^{2}}}$

The slant height of any frustum of a right cone is equal to the hypotenuse of a right triangle whose base is the difference between the radius of the upper base and the radius of the lower base, and whose altitude is the altitude of the frustum. The slant height of this frustum is, therefore, equal to the hypotenuse of a right triangle whose base is 3″–1½″=1½″, and whose altitude is 3″. But the hypotenuse is equal to ${\sqrt {{\text{base}}^{2}+{\text{altitude}}^{2}}}$; therefore, the slant height for this frustum is ${\sqrt {3^{2}+1.5^{2}}}$

$14.13\times 3.35{=}47.33{\text{sq. in.}}$

Problem 30C.—The spout, Fig. 171, is also a frustum of a cone whose lower base has a diameter of 3⅛″, and upper base a diameter of 2½″. The slant height may be called 7½″. What is its area?

Problem 30D.—What is the combined area of Problems 30A, 30B, and 30C?

Problem 30E.—Allowing 10 per cent for waste, what will be the cost of the material for one pump at 8 cents per square foot?