outline of one-half of the pattern. This is copied on the other side of line 1-A in order to produce the whole pattern.
86. Related Mathematics on Oval to Round Transition.—Equivalent Areas.…When two dissimilar profiles contain the same number of square inches of surface area, they are said to have equivalent areas. When any change of profile occurs in a system of piping, the areas must be equivalent.
Area of an Oval.—An oval is a rectangle having semicircular ends; therefore, its area is equal to the area of some rectangle, plus the area of some circle. In any oval the diameter of this circle is equal to the width of the oval. The rectangle has for its dimensions the width of the oval, and the difference between the width and the length of the oval.
Example.—What is the area in square inches of an oval profile 4" wide and 14" long?
| .7854 | |
| 16 | |
| 47124 | |
| 7854 | |
| 12.5664 | sq. in |
Area of rectangle = 4×10 = 40 sq. in.
Combined areas = 40+12.57 = 52.57 sq. in. Ans.
Problem 47A.—What are the areas of the following sizes of oval profiles?
- (a) 3½"×15"
- (b) 4½"×14½
- (c) 3"×11"
- (d) 3¾"×15½
- (e) 6"×13¾"
Problem 47B.−An 8" round pipe is to be "ovaled down" to a width of 3⅝". What must be the length of the oval?
(Hint: Subtract the area of a 3⅝" circle from the area of the 8" circle and divide the remainder by the width of the oval.)
