Sheet metal drafting/Chapter 13
CHAPTER XIII
TRIANGULATION OF SCALENE CONES
Problem 45
SCALENE CONE
83. The Scalene Cone.—Figure 264 shows the top view of a scalene cone whose base is represented by a circle and whose apex falls in a line perpendicular to the plane of the base at the point 2.
Figure 265 is an elevation of the cone, showing the apex, point 3, on the perpendicular line drawn from point 2. The line 2-3 is the altitude of the cone.
The top view is drawn and the circumference of the circle divided into eight equal parts. Straight lines are then drawn connecting point 2 with points A, B, C, and D. These are known as base lines, since they are equal in length to similarly drawn lines on the model; that is, they are true lengths.
Four horizontal lines, Fig. 266, are now drawn equal in length to lines A–2, B–2, C–2, and D–2 of Fig. 264. Corresponding letters and numbers are placed at the extremities of these lines. Perpendicular lines are erected at point 2 of each line, equal in length to line 2–3 of Fig. 265. Points A–3, B–3, C–3, and D–3 may now be connected by straight lines, thereby forming four right triangles. The hypotenuses of these triangles are elements of the surface of the cone; that is, they are equal in length to lines similarly drawn on the surface of the model. That branch of pattern drafting known as triangulation takes its name from the fact that the surfaces are developed from a series of triangles whose hypotenuses are equal to certain elements—straight lines drawn on the surface of the cone.
A vertical line, Fig. 267, is now drawn equal in length to the altitude, line 2–3, of Fig. 265. With point 3 as a center and radii equal to hypotenuse 3–D,3–C, 3–B, and 3–A, arcs are drawn to the left of line 2–3. Point D is located by an arc drawn from point 2 whose radius is equal to distance 2–D of Fig. 266. In like manner points C, B, and A are located by arcs drawn from points D, C, and B respectively. All of these intersecting arcs have the same radii, since the base of the cone was equally divided. A straight line 3–A and a curved line A, B, C, D, 2 completes the half pattern, which may now be copied on the other side of line 2–3 to obtain the full pattern.
It is advisable to make a model by cutting out the triangles of Fig. 266, attaching them to the base lines of Fig. 264, and slipping the envelope, Fig. 267, over this framework.
Problem 46
SQUARE TO ROUND TRANSITION
84. Square to Round Transition.—The sheet metal worker is often called upon to make square to round transitions. In heating and ventilation, square and rectangular pipes are changed to round pipes, and ventilators with round shafts are mounted on rectangular bases. Wherever the cross-section of a pipe is changed to another shape the transformation should be gradual in order to avoid excessive friction.
Figure 268 is a pictorial view of a square to round transition. The transition may be considered as being made up of a rectangular prism, having a portion of a scalene cone at each corner, the spaces between these being filled by triangular-faced pieces.
Figure 269 shows one-quarter of the transition removed and the triangles that are to be used in the development of the pattern drawn in their respective positions.
The Plan.—The plan, Fig. 271, is the first view to be drawn. The plan may be divided into four equal parts. It is necessary to treat but one part. The center points of two sides of the square are first determined as shown by points 1 and 3. That part of the circumference between the horizontal and vertical diameters is now divided into four equal parts as shown by points A, B, C, and D.
The base lines are now drawn in, but before drawing them the draftsman must determine the order in which he intends to develop the pattern. It will simplify the study of triangulation if a standard method of development is adopted. Every line should be considered as running in but one direction; for instance, the line AB should be considered as running from point A to point B and not from point B to point A. Furthermore, this line should always be read as A to B, and not simply AB. By pursuing this method the draftsman is enabled to leave his drawing at any time and pick up the "thread" where he left off, upon his return. The letters should be confined to one base and the figures to the other. Thus in Fig. 271, the order would be 1–A, 2–A, 2–B, 2–C, 2–D, and D–3.
The elevation, Fig. 270, may now be drawn, but since the only added information it contains is the altitude of the triangles the experienced draftsman rarely draws this view.
The Diagram of Triangles.—A series of short horizontal lines, Fig. 272, are drawn equal in length and numbered to correspond to
the base lines of Fig. 271. Perpendiculars are erected at points 2 and 3, and the several hypotenuses are drawn in as shown.
The Pattern.—A horizontal straight line is drawn equal in
length to line 1–2 of Fig. 271. With point 1 as a center and a radius equal to the hypotenuse of triangle 1–A, an arc is drawn below line 1–2. This is intersected by an arc drawn from point 2 with a radius equal to the hypotenuse of triangle 2–A. The intersection locates the point A on the pattern. With point 2 as a center and a radius equal to the hypotenuse of triangle 2–B, an arc is drawn bearing away from point A. This arc is intersected by another drawn from point A, whose radius is equal to line AB of Fig. 271. This intersection is lettered B. In like manner, points C and D are located. Then with point D as a center and a radius equal to the hypotenuse of triangle D–3, an arc is drawn bearing away from point 2. This arc is intersected by another drawn from point 2, whose radius is equal to line 2–3 of Fig. 271. Straight and curved lines connecting these points give the outline of the quarter pattern. This is now duplicated on the other side of line 1–A to obtain the half pattern.
Half-inch locks are added to each side of the pattern, but the workman, in forming the locks, should turn but in. It is advisable to construct a model from the plan and diagram of triangles, in order to aid in the visualization of the project.
Problem 47
OVAL TO ROUND TRANSITION
85. Oval to Round Transition.—The oval to round transition is extensively used in hot air furnace heating. In Fig. 275 the oval and the circle have the same center, but often the job demands that the center of the circle be placed to one side of the oval. However, the method of developing the pattern is the same in all cases, as long as the planes of the top and the bottom are parallel.
The Plan (Fig. 275).—The profiles of the upper and lower bases should be drawn in their proper positions with a horizontal center line for each. Since these profiles have the same center, the line AJ divides the figure into two equal parts and, therefore, but one-half need be treated.
Both half-profiles are now divided into equal spaces and each division numbered or lettered as shown. The order of development, as explained in Problem 46, should now be determined.
The Diagram of Triangles (Fig. 276).—Having determined the order in which the base lines are to be taken from Fig. 275, short horizontal lines equal to each base line are drawn. These are shown in Fig. 276, and the order should be carefully studied. Perpendicular lines equal in length to the altitude of the fitting, as shown in Fig. 276, are erected at one end of each of these lines. The hypotenuses of the several triangles are then drawn in.
The Pattern (Fig. 277).—A distance equal to the hypotenuse of triangle AD is set off upon any vertical line. These points are lettered A and 1. With point 1 as a center and a radius equal to the hypotenuse of triangle 1–B, an arc is drawn bearing away from point A. This is intersected by an arc drawn from point A, whose radius is equal to line AB of Fig. 275, thereby locating point B.
With B as a center and a radius equal to the hypotenuse of triangle B–2, an arc is drawn bearing away from point 1. This is intersected by an arc drawn from point 1, whose radius is equal to the line 1–2 of Fig. 275, thereby locating point 2.
In like manner all points of the pattern may be located. Attention is called to the space between letters E and F of Fig. 275. Since this is a straight line it is not divided and, therefore, the space EF is greater than any of the others.
Curved lines passing through the points thus obtained give the 
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.) Problem 47C.—A furnace man finds the following sizes of oval risers have been installed in the partitions: One 3½"×13⅝"; one 6"×9¼"; one 3½"×15⅛"; and one 3½"×14¼". What are the equivalent round pipe areas for cellar mains to supply each of these risers? What is the nearest diameters of the cellar mains if the diameters increase by half inches?