Pattern of Body.—An arc of stretchout, Fig. 234, should be drawn whose radius is equal to the distance from Apex No. 1 to the top of the body. Six spaces each of which are equal to the radius of the top are set off upon this arc. The first and last points should be connected to the center from which the arc of stretchout was drawn. Another arc is now drawn by using the same center and a radius equal to the distance from Apex No. 1 to the bottom of the body in Fig. 231. A ¼-inch lap is added to one side of the pattern, and a ¼-inch wire edge to the top edge.
Figure 233 shows the pattern of the perforated tin strainer. The diameter of this blank is equal to the diameter of the body, Fig. 231, with a -inch lap added all around.
Figure 235 is the pattern of the rim. Since the rim is a cylinder, its pattern will be a rectangle whose length is equal to 2⅜ in×π, and whose height is equal to ½ in. plus a ⅛-inch hem. A quarter-inch lap should be added to one side of the pattern.
73. Related Mathematics on Cup Strainer.—Problem 38A.—What is the area in square inches of the body pattern?
Problem 38B.—What is the area of the pattern of the handle?
Problem 38C.—Show by means of a sketch a method of cutting the blanks required for the manufacture of twelve cup strainers, that will leave a minimum amount of waste.
Note.—The formula for the frustum of a cone is given in Chapter V.
