# Sheet metal drafting/Chapter 5

CHAPTER V CONES OF REVOLUTION
Problem 19 CONICAL FLOWER HOLDER
37. The Conical Flower Holder.—The sketch, Fig. 115, shows a flower holder, such as is often carried in stock by florists. The body of this holder is a right cone.
Solid of Revolution.—Any plane surface rotating about a fixed point, or a line, generates a solid. For instance, a rectangle rotating about one of its sides generates a cylinder. A right-angled triangle rotating about its altitude generates a cone. Such a cone is known as a Right Cone or a cone of revolution.
Axis of the Cone.—The line about which the generating surface revolves in forming a solid of revolution is called the axis. It is the shortest distance between the apex (point) and the base, and forms a right angle to the plane of the base.
Elevation of the Cone.—The elevation of the cone, Fig. 116, is drawn in the following manner: Draw a horizontal line four inches long; from the center of this line drop a perpendicular seven inches long. Connect the ends of the four-inch line to the end of the seven-inch line by straight lines. The four-inch line is the Base of the Elevation. The seven-inch line is the Altitude of the Cone. The straight lines connecting the ends of the base and the altitude are the Slant Height lines of the elevation. Complete the elevation by drawing a wire nail, Fig. 116, which is to be "soldered" in after the cone is formed.
Profile of the Cone.—The profile of any cone of revolution is a true circle. This circle may be divided into two equal parts; therefore, it is necessary to draw but one-half of the profile as shown in Fig. 116. This profile is divided into equal parts and each division numbered. Extension lines are carried downward until they meet the base line of the elevation.
Elements of a Surface.—Dotted lines are shown in Fig. 116 running from each intersection of the base, to the apex. These represent imaginary lines drawn upon the surface of the cone. If lines were drawn upon the surface of the cone, until the surface was completely covered, each one of these lines would become a part, or an element, of the surface. Any surface may be regarded as being made up of an infinite number of lines placed side by side, each line being an element of the surface.
​The slant height lines are also elements of the surface of the cone, and are the only elements shown in the elevation that represent the true distance from the base to the apex, along the surface of the cone.
The Arc of Stretchout.—With one point of the compass on the
apex, and a radius equal to the slant height, an arc, Fig. 117, is drawn. Upon this arc, as many spaces are laid off as there would be in the whole profile, Fig. 116. Since this arc answers the same purpose as did the line of stretchout in parallel line drawing, it ​can be called the arc of stretchout. The intersections on the arc of stretchout are numbered to correspond to the profile. Points 1 are connected with the apex and ¼-inch edges are added for a tin lock. Notice that the locks are parallel to lines number 1 and do not connect with the apex until they are "notched." The notch at the apex of any cone is made very long in order to bring the cone to a sharp point. The elements of the surface should be drawn, as shown in Fig. 117, and the length of these compared with the length of the corresponding foreshortened elements of Fig. 116.
38. Related Mathematics on Conical Flower Holder.—Area of a Sector.—A sector is a part of a circle set off by two radii and an arc. Is Fig. 117 a sector? What is the length of its radius? What is the length of its arc? If the length of the arc is multiplied by one-half of the radius, the result will be the area of a sector. Thus the area of a sector whose arc measures 14" and whose radius is 7" would equal 14×3½=49 sq. in.
The length of the arc, Fig. 117, is equal to the circumference of the base of the cone (the profile). In addition, the radius of pattern, Fig. 117, is equal to the slant height of the cone, Fig. 116. Because of these facts, the formulæ for the area of a sector and for the lateral area of a right cone are very much alike.
Lateral Area of a Cone.—The lateral area of a right cone is equal to the circumference of the base times one-half the slant height.
| Then using the formulae for area | ${\begin{aligned}\scriptstyle {A}&\scriptstyle {=C\times {\frac {\text{Slant Height}}{2}}}\\\scriptstyle {A}&\scriptstyle {=15.708\times {\frac {12}{2}}}\\\scriptstyle {A}&\scriptstyle {=94.258{\text{ sq. in.}}}\end{aligned}}$ |
Problem 19A.—How many square inches of surface area (lateral) has a right cone whose base is 7" in diameter and whose slant height is 11¼"?
Problem 19B.—The base of a cone has a circumference of 96" and a slant height of 102½". What is the area of its lateral surface in square inches?
Problem 19C.—How much would 1000 flower holders, Fig. 115, weigh if made from No. 28 galvanized steel (.7812 lb. per square foot)? Allow 5 per cent of total weight for waste.
Problem 20 PITCH TOP COVER
39. The Pitch Top Cover.—The cover of any receptacle should be made with a pitched top, such as considered in this problem, in order to obtain the necessary rigidity.
The Elevation.—The elevation of the pitch top cover appears as shown in Fig. 119. This cover consists of a cone top, joined to a cylindrical rim by a "clinched" seam. The rim has a No. 12 wire rolled into the bottom edge. A semicircular wired handle is drawn with a 2-inch radius by using the apex of the cone as a center. The distances C to B, and J to K, are straight lines connecting the semicircle and the slant height lines. The distance from A to B is 1 inch. The handle is joined to the cover by 1½ lb. rivets.
The Profile.—A half-profile should be drawn, using extension lines to locate the view properly. The half-profile is divided into equal parts and each division numbered. The profile is equal in diameter to that of the rim inside of the wire. Extension lines are carried upwards from each division of the half-profile, to the base of the cone, and thence to the apex.
Drawing the Pattern.—With a radius equal to the slant height of the cone, and any point as a center, the arc of stretchout is drawn. The spacing of the half-profile is transferred to the arc of stretchout, doubling the number of spaces in order to obtain the whole pattern. The divisions are numbered as shown in Fig. 120. A ⅜-inch edge parallel to the arc of stretchout is added to allow for joining the rim. The locks are drawn parallel to lines 1 and 1, Fig. 120. These locks are notched as indicated. The rivet holes may be located on any two elements that are opposite each other, viz., on 4 and 4, 3 and 5, 2 and 6, etc., but in shop practice they are generally placed 90° from the lock seam. This would bring them on lines 4 and 4 as in Fig. 120. The distance from the center of the pattern to the center of the holes is found by measuring downwards on the slant height. Fig. 120, from the apex of the cone to the center of the rivet as shown in the elevation.
$\scriptstyle {\frac {3}{16}}$
Pattern for the Handle.—Any straight line, Fig, 122, may be used as a line of stretchout. The profile of the handle, Fig, 122, is divided into equal parts and the divisions lettered A, B, C, etc. These divisions are transferred to the line of stretchout. Perpendiculars to the line of stretchout are erected at points A, B, K, and M. Distances of ½ in. and ¼ in. are set off on lines B and K, on each side of the line of stretchout. These intersections are connected by straight lines to form the body of the pattern. Lines A and M set off that part of the handle that laps and rivets to the cover, and should be notched as shown in Fig, 122. Rivet holes are located in the exact center of these laps.
40. Related Mathematics on Pitch Top Cover.—Problem 20A.—How much would 50 cone tops, shown in Fig. 118 (no rims or handles), weigh if made from No. 26 galvanized steel (.9062 lb. per square foot)? Add 25 per cent for waste.
Problem 20B.—What would be the weight of the cone top of a cover to fit over a 14" garbage pail? Allow ½" clearance between pail and cover on all sides, making the diameter of the base of cone 15", and the slant height 9". Cover to be made of No. 26 galvanized steel.
Problem 21 VEGETABLE PARER
41. The Vegetable Parer.—Figure 124 shows an elevation of a vegetable parer which is in the form of a right cone cut by a curved plane.
The Elevation.—A "vertical" line which is to be used for the center line, or altitude of the cone, is drawn first. At right angles to the lower end of this line, the base of the cone is drawn. This base is to be ¾ in. long and is to have ⅜ in. on each side of the center line of the cone. A distance of 17 in., which will locate the apex of the cone, is set off upon the center line. The apex and the ends of the base line are connected by the slant height lines. At an altitude of 4 in. a curve that cuts the cone, as shown in Fig. 124, is drawn in. This curve may be drawn to suit the ideas of the designer, and is in reality the miter line.
The Profile.—A whole profile, using extension lines to locate the view, is drawn. The profile is divided into twelve equal parts and each division numbered. Extension lines are carried from each division of the profile upwards to the base of the cone, and thence to the apex. Each one of these extension lines intersects the miter line at some point. Horizontal lines from each of the intersections of the miter line, Fig. 124, are drawn over to the slant height.
$\scriptstyle {\frac {1}{16}}$
$\scriptstyle {\frac {1}{32}}$
Problem 22 CONICAL ROOF FLANGE
42. The Conical Roof Flange.—Whenever a smoke pipe is to pass through a roof, it is necessary that a hole much larger than the pipe be cut in the roof in order to lessen the fire risk. In order to render this construction water-tight, a conical roof flange, as shown in Fig. 126, must be used.
The Elevation.—First, the roof line is drawn at the angle demanded by the job specifications (in this drawing 30°). The roof fine immediately becomes the miter line. Next, a "vertical" center line, line 4 in Fig. 127, is drawn in. Upon each side of this center line a distance equal to one-half the diameter of the smoke pipe is set off. A short horizontal line is put in to represent the joint between the pipe and the flange. One-half the diameter of the hole in the roof is set off on each side of the vertical center line. This will locate the low point (point 7) of the miter line. From this point the base of the cone should be drawn at right angles to the center line. The slant height lines may now be drawn by connecting the ends of the base and the ends of the short horizontal line of the junction between the flange and pipe. These lines must be prolonged until they meet at the apex.
The Profile.—A half-profile. Fig. 127, is drawn and divided into equal spaces. The divisions are numbered and an extension carried upwards from each division as far as the base of the cone. From each intersection thus obtained, extension lines are drawn to the apex of the elevation. Where these extension fines cross the miter line, numbers that correspond to the numbering of the profile are placed. Horizontal extension lines from each intersection of the miter line are drawn over to the right-hand slant height line.
The Pattern.—With the apex of the elevation as a center, and a radius equal to the slant height of the full cone, the arc of stretchout is drawn. The spacing of the profile is transferred to this line, and the number of spaces doubled to provide for a whole pattern. The divisions are numbered to correspond. The measuring lines of the stretchout are drawn from each division of the arc of stretchout to the center point (apex).
43. Related Mathematics on Conical Roof Flange.—Area of Frustum.—If a right cone is cut by a plane parallel to that of the base, the top section will still be a right cone although of small dimensions, and the lower part will be a frustum of a cone. The profiles of both bases, or ends, will be circles. The smaller circle is generally called the upper base, and the larger circle the lower base of the frustum. The lateral area of a frustum of a cone is found by adding together the circumferences of the upper and lower bases, dividing the sum by 2, and then multiplying by the slant height. This is often expressed as a formula for area of a frustum:
| | $\scriptstyle {A={\frac {B+b}{2}}\times H_{s}}$ |
| in which | |
| | ${\begin{aligned}\scriptstyle {B}&\scriptstyle {={\text{Circumference of lower base}}}\\\scriptstyle {b}&\scriptstyle {={\text{Circumference of upper base}}}\\\scriptstyle {H_{s}}&\scriptstyle {={\text{Slant height of frustum}}}\end{aligned}}$ |
${\frac {40+22}{2}}\times 16=496{\text{ sq. in.}}$
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?
Problem 23 APRON FOR A CONICAL ROOF FLANGE
44. The Apron for a Conical Roof Flange.—When any solid is cut by a plane that is inclined to the plane of the base, the shape or section thus formed is not the same as the profile of the base.
The Elevation.—The elevation of the roof flange used in Problem 22 can be reproduced.
The Pattern.—Any straight line, Fig. 130, may be drawn and the exact spacing of the miter line set off upon it. Perpendicular measuring lines, Fig. 130, are erected through each point and are numbered to correspond to the miter line.
Returning to the elevation, Fig. 129, an extension line is dropped from point 1, down to the horizontal center line of the half-profile. From point 2 of the miter line, the horizontal extension line is followed over to the slant height. From this point, a perpendicular to the horizontal center line of the profile is dropped, and with one point of the compass on the center of the profile, this line is extended by an arc until it strikes a radial line from point 2 of the profile at the point B.
The perpendicular distance from point B to the horizontal center line should be measured and placed on each side of the line of stretchout. Fig. 130, on measuring line number 2. In like manner, points C, D, E, and F are located and their distances placed on measuring lines 3, 4, 5, and 6 respectively. A curve traced through the points thus obtained will give the shape of the hole in the apron as well as that of the hole to be cut in the roof.
A rectangle representing the shape of the apron should be drawn, allowing a space of at least 6 in. "up the roof," and at least 3 in. on the other sides. A hem should be added to three sides to turn or direct the flow of any roof water that might leak in. A $\scriptstyle {\frac {3}{16}}$-inch single edge should be allowed around the inside of the hole, in order to double seam the body to the apron.