Here is a view of a 5 piece tapering elbow, having a round base and a elliptical top. This form is generally known as a ship ventilator. The principles shown in this problem are applicable to any form or shape no matter what the prospective profiles may be at the base or the top.
The first step is to draw the correct side view of the elbow as shown in Fig. 1. The outline A B C D E F can be drawn at pleasure, but for practice, dimensions are given.
First draw the vertical line A F equal to 4 1/2 inches. On the same line extend measure down 1 1/4 inches to f and draw the horizontal line H B. From f set off a distance of 1 1/4 inches at G, and using G as a center and G F as a radius describe the arc F E intersecting H B at E, from which draw the vertical line E D equal to 1 inch. Draw D C equal to 1 3/4 inch, then draw C B. From B layoff 5 3/4 inches and using this point (H) as a center and H B as a radius describe the arc B A. The portion shown B E D C is a straight piece of pipe whose section is shown by I J K L. Now divide the two arcs B A and E F into the same number of parts that the elbow is to have pieces (in this case four) and draw the lines of joint or miter lines as shown by U V, etc.
Bisect each one of the joint lines and obtain the points a b c d and e. Then A B C D E F will be the side view.
The patterns will be developed by triangulation, but before this can be done, true sections must be obtained on all of the lines in the side elevation. The true sections on the lines B E and C D are shown by I J K L. The length of the sections are shown by the joint lines, but the width must be obtained from the front outline of the elbow, which is constructed as follows: In its proper relation to the side elevation, draw the center line M R upon which draw the ellipse M N O P (by the methods given here) which represents the section on A F in side. Take half the diameter I K in section and place it on either side of center line M R as R T or R S. Then draw the outline O S and T N in a convenient location. While this line is drawn at will it should be understood that when once drawn it will become a fixed line. Now from various intersections a b c d and e in the side elevation, draw lines through and intersecting the front outline as shown on one side of b´ c´ d´ and e´. Then these distances will represent the widths of the sections shown by similar letters in side. For example, the method will be shown for obtaining the true section on U V, and and the pattern for piece 1 in side elevation. To avoid confusion of lines take a tracing of A F V U and place it as shown by 1, 13, 12, 0 in Fig. 2. On 1-13 place the half profile M N P of Fig.1. Bisect 0-12 in Fig.2 and obtain the point 6; at a right angle to 0-12 from 6 draw the line 6 6´ equal to b´ b? in front outline in Fig 1. Then through the three points 0, 6´ and 12 in Fig.2 draw the semi ellipse, which will represent the half section on U V. The other sections on the joint lines in side elevation are obtained in the same manner.
If the sections were required for piece 2 in side it would be necessary to use only 0 6´ 12 in Fig.2. and place it on U V in Fig.1. and on a perpendicular line erected from c, place the width c´ c? shown in front and through the three points obtained again draw the semi-elliptical profile or section. Now divide the two half sections (Fig.2) into equal parts as shown by the small figures, from which the right angles to 1-13 and 0-12 draw lines intersecting these base lines from 1-13. Connect opposite points as 1 to 2 to 3 to 4 to 5, etc, to 12. Then these lines will represent the base of the sections whose altitudes are equal to the heights in the half section. For these heights proceed as follows:
Take the various lengths from 1 to 2, 2 to 3, 3 to 4, 4 to 5, ect., to 11 to 12 and place them on a horizontal line in Fig.3 as shown by similar figures; from these points erect verticle lines equal in height to similar figures, in the half section in Fig. 2 as shown by similar figures in Fig.3. For example: take the distance from 7 to 8 in Fig.2 and place it as shown from 7 to 8 in Fig.3 and erect verticle lines 7-7´ and 8-8´ equal to 7-7´ and 8-8´ in Fig.2. Draw a line from 7´ to 8´ in Fig.3 which is the true length on 7-8 in Fig.2 For the pattern take the distance of 1-0 and place it as shown by 1-0 in Fig.4. Now using 0 as a center and 0 2´ in Fig.2 as a radius, describe the arc 2 in Fig.4 and intersect by an arc struck from 1 as a center with 1-2´ in Fig.3 as a radius. Now with 1-3´ in Fig.2 as a radius and 1 in Fig.4 as a center, describe the arc 3, and and intersect it by an arc struck from 2 as center and 2´ - 3´ in Fig.3 as a radius. Proceed thus, using alternately as radii, first the divisions in 0-6´-12 in Fig.2, then the proper line in Fig.3, the divisions 1-7´-13 in Fig.2 and again the proper line in Fig.3, until the line 12-13 in Fig.4 is obtained which equals 12-13 in Fig.2. In this manner all the sections are obtained, to whaich laps must be allowed for wiring and seaming.
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