Page:Sheet Metal Drafting.djvu/106

whose upper base has a circumference of 22" will have a surface area of

${\frac {40+22}{2}}\times 16=496{\text{ sq. in.}}$

if the slant height of the frustum is 16".

Also, a frustum having an upper base diameter of 8", a lower base diameter of 12", and a slant height of 10" will have a surface area of:

Circumference of upper base=8″×3.1416=25.133″
Circumference of lower base=12″×3.1416=37.669″
Formula, $A={\frac {B+b}{2}}\times H_{s}$
Substituting, $A={\frac {37.699''+25.133''}{2}}\times 10$
A=314.16 sq. in.

Problem 22A.—The roof flange, Fig. 126, would be treated by any estimator as a frustum of a right cone, although in reality its surface area is less. The estimator would take the diameters of the upper and lower bases from the elevation. Fig. 127, calling the upper base 2″ and the lower base 6″ in diameter. How much would the conical part of this roof flange weigh if it were made from No. 26 galvanized iron?