Sheet metal drafting/Chapter 11

by Ellsworth M. Longfield

CHAPTER XI
FRUSTUMS OF CONES

Objectives of Problems on Frustums of Cones.

Problem 38
CUP STRAINER

72. The Cup Strainer.—This problem introduces the principles that apply when two cones intersect each other. Figure 231 is an elevation of the cup strainer. The body is a frustum of a cone whose apex is noted upon the drawing as Apex No. 1. The handle of the strainer is also a cone and miters upon the conical body as shown in Fig. 231. When two cones miter upon each other in this manner, the miter line must be developed.

Developing the Miter Line.—The elevation of the cup strainer, Fig. 231, should first be drawn according to the dimensions given. The body and handle should have their sides extended to determine the apex of each. The half-profile of the handle is then drawn and divided into equal spaces. After numbering each space, perpendicular lines are drawn to the base of the cone. From these points, extension lines are drawn to Apex No. 2. Directly above the elevation, a half plan of the handle is drawn using extension lines to locate the view properly. A quarter-profile, Fig. 230, is drawn and the spacing of the half-profile of Fig. 231 is transferred in such a way that point 1 falls on the horizontal center line. An extension line is drawn from point 2 perpendicular to the base line and thence to Apex No. 3. Perpendicular lines are now drawn from the points of intersection of extension lines 2 and 3 of Fig. 231 and the slant height of the body, until they meet the horizontal center line of Fig. 230. With one point of the compass on the center of the profile of the body, Fig. 230, arcs A and B are drawn. Extension lines should now be carried back from the points of intersection of arcs A and B with line 2 until they intersect lines 2 and 3 in Fig. 231. The curved miter line is now drawn in as shown.

Developing the Pattern for Handle.—From each intersection of the miter line in Fig. 231, lines intersecting the slant height should be drawn parallel to the base of the cone. These intersections are shown by letters, c, d, e, and f, in Fig. 231. The arc of stretchout is now drawn and the spacing of the profile transferred with numbers to correspond.

From each point on the arc of stretchout, Fig. 232, lines are drawn to Apex No. 2. Arcs should now be drawn from points c, d, e, and f of Fig. 231 over into the stretchout. Points of intersection can be determined by starting from the half-profile, tracing the extension line to the miter line, and thence to a correspond-

 

Figs. 230-235—Cup Strainer.

 

ingly numbered line in the stretchout. A curved line drawn through these intersections will give the miter cut of the pattern. A ¼-inch lap should be added to one side of the pattern.

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 $\scriptstyle {\frac {3}{32}}$-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.

Problem 39
SHORT HANDLED DIPPER

74. Short Handled Dipper.—The dipper presents a problem in which three right cones are mitered. The elevation, Fig. 236, is drawn and dimensioned. The handle of the dipper is raised about 10° above the horizontal, although there is no set rule governing this feature. The boss is also drawn to suit the ideas of the designer.

Pattern of Boss.—Figure 237 is a reproduction of the elevation of the boss and that part of the dipper that is adjacent to it. The sides of the boss are extended to form a right cone. This right cone is cut by two planes, the surface of the body being one cutting plane, and the line of junction between the boss and the handle, the other. A half-profile is drawn directly upon the base of the cone, and divided into equal parts. These parts are numbered and extension lines drawn from each division, perpendicular to the base of the cone. From each intersection of the base, lines are drawn to the apex.

A half plan of the boss is drawn directly above the elevation. A quarter-profile is attached to the half plan, with divisions and numbers that correspond to the profile in the elevation.

From each point of intersection on the slant height of the body dotted extension lines are carried up to the horizontal center line of the plan. Using the center of the top as a center point, arcs are drawn from each intersection of the dotted lines and the horizontal center line. These arcs intersect extension lines drawn from the base to the apex in the half plan. Perpendicular lines are now carried back to correspondingly numbered extension lines in the elevation. A curved line passing through the points thus obtained will be the developed miter line.

An arc of stretchout is drawn using a radius equal to the distance from the apex to point 4. Upon this arc are placed twice as many spaces as there are in the half-profile, with numbers to correspond. Measuring lines are drawn from each of these divisions, to the apex.

From each intersection of both miter lines, extension lines are drawn parallel to the base of the cone until they intersect the slant height. From each of these points, extension arcs are drawn until they meet correspondingly numbered measuring lines in the stretchout. Curved lines drawn through these intersections will give the miter cuts of the pattern, A small lap is added to one side of the pattern.

The pattern of the handle, Fig. 239, is obtained in exactly the

 

Figs. 236-239.—Short Handled Dipper.

 

same way as was the handle for the Cup Strainer, and needs no further explanation.

The pattern for the body is that of a frustum of a right cone. This development has been explained in previous problems and will not be shown on this drawing. It may, however, be mentioned that the bottom of a dipper is always double seamed to the body, and, therefore, proper allowances must be added to the pattern for this purpose. 75. Related Mathematics on Short Handled Dipper.—Volume of a Frustum of a Cone of Revolution.—The frustum of a cone has a circular top and a circular base. These are known as the upper and lower bases of the frustum. The altitude of the frustum is the shortest distance between the upper and lower bases, and is always measured perpendicularly. The volume of a frustum is found by adding together the area of the upper base, the area of the lower base, and the square root of the product of the upper base area times the lower base area; the sum of these quantities is then multiplied by one-third of the altitude. Expressed as a formula

In applying this formula to Fig. 236, the areas of the upper and lower bases must first be computed.

Area of Circle	=	D2×.7854 (Chapter II)
6¼2×.7854	=	30.68 area of upper base
4½2×.7854	=	15.904 area of lower base

Known values can now be substituted in this formula

$\scriptstyle {V=(30.68+15.904+{\sqrt {30.68\times 15.904}})\times 3{\frac {3}{4}}\div 3}$

Performing the arithmetic:

$\scriptstyle {3{\frac {3}{4}}\div 3={\stackrel {\scriptstyle {~5}}{\frac {{\cancel {1}}{\cancel {5}}}{4}}}\times {\frac {1}{\cancel {3}}}=1{\frac {1}{4}}}$

15.904     30.68    127232    95424 477120 487.93472

4!√487.934722.08+ 4|√4 42|087 42|084 4408|3 9347 4408!3 5264 4408 !34083

The formula will now stand,

V=(30.68+15.904+22.08)×1¼
or 85.83 cu. in.

Solution

	15.904	68.664
	30.68	1.25
	22.08	343320
	68.664	137328
		68664
		85.83000	Ans. 85.83 cu. in.

There are 231 cubic inches in one gallon and the capacity of this dipper would be $\scriptstyle {\frac {85.83}{231}}$ or .37 gallon.

Problem 39A.—What would be the capacity in cubic inches of a dipper whose dimensions are as follows: Diameter of top, 7¼″; diameter of bottom, 5¼″; altitude or depth, 4½″?

Problem 39B.—What would be the capacity in quarts of the dipper described in Problem 39A?

Problem 40
LIQUID MEASURES

76. Liquid Measures.—The body of the measure should first be drawn and its side extended to locate the apex. Figure 240 shows such a view of the body with a half-profile attached to its top edge. This half-profile is divided into equal parts, and extension lines carried from each division to the top edge of the body.

Pattern of Body.—With a radius equal to the distance from the apex to point 7 of Fig. 240, an arc of stretchout, Fig. 241, is drawn. The spacing of the half-profile is transferred to the arc of stretchout with numbers to correspond. Straight lines are now drawn from the apex through points 1 and 7, continuing downward indefinitely. An arc drawn from the apex, with a radius equal to the distance from the apex to the point J, completes the half pattern of the body. A ¼-inch wire edge is added to the top of the pattern. A ¼-inch lock and a ⅛-inch single edge for double seaming are added as shown in Fig. 241. The half pattern is revolved about line 7 of Fig. 241 in order to obtain the full pattern.

Elevation of Lip.—The lip should now be added to the elevation of the body. In constructing the elevation of the lip, a point H is selected 1⅛ in. below the top edge of the body on the center line. Straight lines are drawn from this point through and beyond points 1 and 7. Point A is located ½ in. from point 1, and point G is located 1½ in. from point 7. The line AG completes the elevation. Extension lines must now be drawn from the apex (point H) cutting the top of the lip at points A,B,C,D,E,F, and G. From these points, horizontal fines are now shown intersecting the line HG, which is the slant height of the cone.

Pattern of Lip.—In any convenient space, an arc of stretchout, Fig. 242, is drawn with a radius equal to line H–1 of Fig. 241. Twice as many equal spaces are set off on this arc as there are spaces in the half-profile, with numbers to correspond. Measuring lines are now drawn from the point H through and beyond each division of the arc of stretchout. The distance 1–A, on fine HA of Fig. 242, is exactly equal to distance 1–A of Fig. 240. Distances HB, HC, HD, HE, and HF of Fig. 242 are obtained by measuring the distances from point H to the intersections of the horizontal lines that were previously drawn from points B,C, D, E, and F, and the slant height (line HG) of Fig.

 

Figs. 240-244.—Measure for Liquids.

240. Distance HG is taken directly from line HG of Fig. 240 because it is the slant height, and, therefore, a true length. A curved line traced through the points thus obtained completes the pattern for the lip. A ¼-inch wire edge and a ¼-inch lap should be added as shown.

Pattern of Boss.—The pattern of the boss, Fig. 244, is obtained by drawing a profile in its proper location as shown. This profile is divided into four equal spaces and extension lines carried from each division, intersecting the elevation of the handle. A line of stretchout is next drawn with spacing and numbers to correspond to the profile. Measuring lines should now be drawn and intersected by extension lines drawn from the elevation. A curved line drawn through the points thus obtained will give the pattern of the boss.

Pattern of Handle (Fig. 243).—Upon any straight line, the distance around the handle profile, shown in Fig. 240, is set off. Perpendiculars are erected at the first and last points. The upper end of the handle is obtained by setting off $\scriptstyle {\frac {7}{16}}$ in. on each side of the horizontal line, and the lower end by setting off $\scriptstyle {\frac {5}{16}}$ in. on each side of the horizontal line. The pattern is completed by adding $\scriptstyle {\frac {3}{16}}$-inch wire edges to each side of the handle. The pattern for the bottom (not shown) is a circle whose diameter is equal to the finished diameter (4⅛ in.) plus a ¼-inch edge for double seaming to the body. The actual diameter of the pattern would be 4⅛ in.+¼ in.+¼ in.=$\scriptstyle {4{\frac {5}{8}}}$ in.

77. Related Mathematics on Liquid Measures.—Problem 40A.—Compute the capacities of the measures given in the following table:

Standard Sizes For Flaring Liquid Measures

Diameter of Bottom		Diameter of Top.		Height. (Altitude).		Capacity.
2¼	in.	$\scriptstyle {2{\frac {1}{16}}}$	in.	2	in.
2½	''	2¼	''	3¼	''
3	''	$\scriptstyle {2{\frac {11}{16}}}$	''	$\scriptstyle {4{\frac {9}{16}}}$	''
4	''	3¼	''	4⅝	''
$\scriptstyle {5{\frac {3}{16}}}$	''	3¾	''	$\scriptstyle {7{\frac {5}{16}}}$	''
6½	''	5	''	$\scriptstyle {8{\frac {7}{16}}}$	''
8¾	''	6¾	''	9¾	''
10½	''	8	''	10¼	''
11	''	8½	''	$\scriptstyle {12{\frac {5}{16}}}$	''
12½	''	9½	''	$\scriptstyle {12{\frac {1}{16}}}$	''

Problem 40B.—A customer wishes a 2½ gallon flaring measure. The bottom of this measure is to be 9″ in diameter, and the top 7″ in diameter. How high must the measure be made to fulfill these requirements?

Example of Problem 40B.—Suppose the dimensions were 6″ diameter of bottom, 4″ diameter of top, and the measure was to hold 3 quarts.

Original Formula	$\scriptstyle {V=(B+b+{\sqrt {(B\times b)}}H\div 3}$
Transposing,	$\scriptstyle {H\div 3=V\div (B+b+{\sqrt {B\times b)}}}$
In this formula,	H=Altitude, which is the unknown
	V=Volume=3 quarts or 173.25 cu. in.
B=area of lower base=62×.7854=28.274 sq. in.
b=area of upper base=42×.7854=12.566 sq. in.

Substituting known values in above formula,

and

${\begin{aligned}\scriptstyle {H\div 3}&\scriptstyle {=173.25\div (28.274+12.566+{\sqrt {28.274\times 12.566}}}\\\scriptstyle {H\div 3}&\scriptstyle {=173.25\div 59.69}\\\scriptstyle {H\div 3}&\scriptstyle {=2.9}\\\scriptstyle {H}&\scriptstyle {=2.9\times 3=8.7{\text{ in.}}}\end{aligned}}$

Ans. 8.7 in. high.