Sheet metal drafting/Chapter 2

by Ellsworth M. Longfield

 

CHAPTER II
WIRED CYLINDERS

Objectives of Problems on Wired Cylinders.

Problem 5
GALVANIZED CHIMNEY TUBE

17. The Chimney Tube.—In the problem of the chimney tube, Fig. 39, the student will get a clear understanding of the idea of unrolling the envelope of an object to get the pattern.

A chimney tube is a short piece of pipe intended to be built into a chimney. If the tube is held on a level with the eye, the sides, the top, and the bottom will appear as four straight lines. The elevation of Fig. 40 shows this view. It is impossible to tell whether this elevation represents a flat or a curved surface unless another view is drawn. For this reason a profile should be drawn. This profile will show that the elevation is that of a cylinder. Extension lines are used to locate this profile properly. Fig. 41. Three dimensions are given in the elevation. The elevation shows a ⅜-inch flange on the top end of the tube. It is unnecessary to draw this flange in the profile.

Figure 43 represents a cylinder placed on its side. The profile appears on the left-hand end. Straight lines are drawn on the body from each division of the profile, parallel to the sides of the cylinder. If each line left a mark as the cylinder was rolled along a flat surface, we would obtain the stretchout as shown. The lines running from the top to the bottom are the measuring lines of the stretchout, since upon these lines any point on the surface of the cylinder can be measured (located). This illustration also makes plain the reason for drawing the line of stretchout at right angles to the view from which the pattern is to be taken.

The profile, Fig. 41, is divided and the divisions numbered as shown. The line of stretchout should be drawn and the spacing of the profile transferred to this line. The numbers should correspond. It should be remembered that it is necessary to start with the number of the profile at which the seam is to occur in the finished article. Perpendicular lines should be erected at points 1 and 1 of the stretchout. Extension lines drawn from the elevation complete the stretchout. A ⅜ inch edge is added to allow for the flange called for in the elevation. On the right and left edges of the stretchout, ½-inch edges for locks are added. These locks must be turned in the stove-pipe folder. The top of each lock is notched to reduce the thickness of the seam on the flanged end.

Figs. 39–43.—Galvanized Chimney Tube.

18. Related Mathematics on Chimney Tube.—Circumference of a Circle.—A circle is a portion of a flat surface bounded by a curved line, every point in which is the same distance from a point within, called the center. The circumference is simply the curved line that is drawn with the compass. The diameter of a circle is any straight line that passes through the center of the circle and has its ends in the circumference. It is possible to draw any number of diameters in the same circle, but they will all have the same length.

Value of $\scriptstyle {\pi }$.—The girth or distance around a cylinder can be found by wrapping a narrow strip of newspaper around it. The point where the strip overlaps the end should be marked before the paper is removed from the cylinder. The distance from the end of the paper to the mark will give the distance around the cylinder, or the girth.

If the diameter of the cylinder be accurately measured and the circumference or girth divided by the diameter, the answer will be about $\scriptscriptstyle {3{\frac {1}{7}}}$. Regardless of the size of the cylinder this experiment will always produce the same result. Mathematicians have proved that the exact relation of circumference to diameter cannot be found. The value 3.1416 is near enough for most purposes. Some sheet metal workers use $\scriptscriptstyle {3{\frac {1}{7}}}$ or $\scriptscriptstyle {\frac {22}{7}}$ in their computations. This relation between circumference and diameter is indicated by the Greek letter $\scriptstyle {\pi }$ (pronounced pi).

Suppose it is desired to find the circumference of a 7" circle: $\scriptstyle {7''\times {\frac {22}{7}}={\frac {154}{7}}=22''}$, or 7″ × 3.1416 = 21.9912″. If the circumference is given and the diameter is wanted, the process is reversed; i.e., with a circumference of 26″, $\scriptstyle {26''\div {\frac {22}{7}}=26''\times {\frac {7}{22}}=8{\frac {3}{11}}''}$; or $\scriptstyle {26''\div 3.1416=8.2442''}$.

Lateral Area of a Cylinder.—Lateral means pertaining to the side. Lateral area is the area of the side wall of a cylinder. The pattern of the side wall of a cylinder is a rectangle whose length is equal to the circumference of the profile, and whose height is the height of the cylinder. The area of the pattern is equal to the length times the height; therefore, the lateral area of any cylinder must be equal to the circumference of the base times the height.

Problem 5A.—Compute the circumference of the chimney tube, Fig. 42, and compare the answer to the length of the line of stretchout between points 1 and 1. These should agree or a mistake has been made.

Problem 5B.—Compute the lateral area of the cylinder shown in Fig. 42, without the locks and flange.

Problem 5C.—Compute the area of the pattern with locks and flange added, Fig. 42.

Note.—The lateral area of a cylinder does not include any locks, laps, or flanges, and in order to arrive at the cost of material these must be added to the lateral area.

Problem 6
HALF-PINT CUP

19. The Half-pint Cup.—This problem is intended to bring out the method of notching employed when a wire is rolled into a cylinder, to describe the standard "tin lock," and to show how a bottom is snapped on.

In drawing the elevation of the half-pint cup, special attention should be given to the following items: The lines representing the wire must be $\scriptstyle {\frac {3}{16}}$ in. apart. The lines at the bottom must be ⅛ in. apart. The handle must be drawn according to the dimensions given in Fig. 49. The profile is located by dropping extension lines from the elevation. At a distance of 1¾ in. from the lower line of the elevation, the horizontal center line of the profile should be drawn. The extension lines dropped from the elevation should intersect the center line, thereby setting off the horizontal diameter of the profile. The profile is drawn with the compass after the center of this diameter is located. The handle of the cup is shown attached to the profile, but it is not essential that this be drawn, since the pattern of the handle is a regular taper from a width of ½ in. at the top to ¼ in. at the lower end.

The profile is divided into equal spaces and each division numbered. After the line of stretchout is drawn, the spacing of the profile is transferred to this line and the divisions numbered to correspond. At the points 1 and 1 of the line of stretchout perpendicular lines are erected. The stretchout is finished by extension lines carried over from the elevation. The wire edge which must be computed is added to the top edge of the stretch-out. A ¼-inch edge is added to each side for a standard "tin lock." Since a lock has three thicknesses the full allowance is never turned. For a tin lock $\scriptstyle {\frac {5}{32}}$ in. must be turned in a bar folder. The notching of the wire edge in Figs. 50 and 51 never goes in as far as the circumference line, and always goes down below the top line of the stretchout a distance equal to the diameter of the wire. This notch removes the thick seam on the body at the point where the wire crosses. The bottom of the lock is notched as shown in Fig. 51.

The pattern of the bottom of the cup is drawn by first reproducing the profile and adding a ⅛-inch edge all around. This edge

Figs. 44–51.—Half-pint Cup.

is turned up in the "thin edge" and is "snapped on" over the lower edge of the body. The profile of the handle is shown in elevation. This profile is divided into equal spaces. This spacing is transferred to any straight line and perpendiculars erected at the first and last points. Using the line of stretchout as a center line, ¼ in. is set off on each side for the width of the top and ⅛ in. on each side for the width of the bottom. The pattern of the handle is completed by connecting these points with straight lines. The handle is intended to be made from No. 20 gage iron, tinned. Should the handle be made from lighter material, it would be necessary to add a hem to the long sides of the pattern in order to gain the necessary rigidity.

20. Related Mathematics on Half-pint Cup.—Problem 6A.—How many sheets of tin plate measuring 20″×28″ would be required to make fifty half-pint cups? Treat the bottom of the cup as a square piece of metal.

Problem 6B.—What would be the percentage of waste for the entire job?

Problem 6C.—20″×28″ IX "Charcoal Tin, Bright" is packed by the manufacturers in boxes containing 112 sheets. If this grade of tin plate is selling for $26 per box, how much will the tin required for fifty half-pint cups (Problem 6A) cost?

Area of a Circle.—The method of calculating the area of a circle will be thoroughly understood by the student if he will go through the following exercise:

Draw a 5″ square. Draw straight lines connecting opposite corners of this square. These lines are called the diagonals of the square. The diagonals of a square, or rectangle, always divide each other into two equal parts. Using the point where the diagonals cross each other (intersect) as a center, draw a circle that will just touch the center of each side of the square. What is the diameter of this circle? How does this diameter compare with the length of the sides of the square? You have drawn what is known as an inscribed circle; that is, a circle whose circumference touches all sides of the containing figure but does not pass beyond the sides. What is the area of this square? Would you get the same answer if you simply multiplied the diameter by itself? This operation is known as "squaring the diameter" and is always written D². Look up a table of areas of circles and you will find the area of a 5″ circle given as 19.635″. Now, divide the area of the 5″ circle by the area of the 5″ square. Is your answer .7854? If you should try this experiment with a circle of any diameter you would get the same result. Therefore, by squaring the diameter of any circle and multiplying by .7854, you can find its area. You will often see this rule written $\scriptstyle {A=D^{2}\times .7854}$. Does the method of arriving at this result resemble the one employed in establishing the rule for finding the circumference of a circle? In each case did we divide one quantity by another? Dividing one quantity by another establishes a comparison of the size of one to the size or the other. This comparison is called a ratio. For instance, the ratio of the foot to the inch is 12, and is found by dividing the foot by the number of inches in a foot. What is the ratio of the yard to the foot?

Problem 6D.—What is the area of the pattern for the bottom of the half-pint cup, Fig. 48. Compute the area of a 7″ circle. Compute the area of an 8″ circle. Compute the area of a 9⅞″ circle.

Problem 7
PAINTER'S PAIL

21. The Painter's Pail.—The Painter's Pail, Fig. 52, is generally made of No. 28 Black Iron. The bottom of the pail is double seamed but it is not soldered. The wire bail is formed with a hook on each end. These hooks are inserted in holes punched through the sides of the pail.

A full size elevation, using the dimensions given in Fig. 53, should first be drawn and dimensioned. The lines representing the wire at the top of the pail should be slightly more than ⅛ in. apart. Two lines at the bottom represent the double seam and should be $\scriptstyle {\frac {5}{32}}$ in. apart. The upper left-hand corner of the elevation should be "broken" in order to determine accurately the profile of the "hook" on the end of the bail. Extension lines drawn downward from the elevation locate the profile, Fig. 56. The horizontal center line of the profile should be drawn at a distance of three inches from the elevation. By means of the "T-square" and triangle a vertical center line of the profile is put in. The profile is then completed. The center lines will indicate four points on the circumference. These points are to be numbered 1, 5, 9, and 13 as shown. In order to divide the circumference into sixteen equal parts, as indicated, the student should proceed as follows:

With points 1 and 5 as centers, draw two arcs that cross each other as at A. You may use any radius in drawing these arcs. Carefully connect point A with the center of the profile by a straight line. This line will divide that part of the profile between points 1 and 5 into two equal parts. Number this center point 3. With points 1 and 3 as centers repeat this operation, thereby obtaining point 2. The space between points 1 and 2 may be used to divide the profile into sixteen equal parts.

The straight line from point A to the center of the profile also divides the angle formed by the horizontal and vertical center lines into two equal parts. The angles shown in Fig. 54 are to be bisected. Since these angles have no arc shown, it will be necessary to draw one. The corner (vertex) of the angle should be used as a center. The radius should be as large as possible and yet have the arc cut the sides of the angle. This arc will give points corresponding to points 1 and 5 of the profile. These points are to be used as centers from which to draw the intersecting arcs.

The line of stretchout, Fig. 55, can now be drawn and the entire

Figs.52–58—Painter's Pail.

stretchout developed. A standard tin lock is added to each side of the stretchout. Why does the stretchout start with point 1? Why do the holes for the bail occur on lines 5 and 13? A ⅛-inch single edge is added to the bottom of the stretchout. The pattern for the bottom, Fig. 57, may be drawn by reproducing the profile and adding a ¼-inch double edge all around. The double seam on this pail is of the same construction as the one shown in Fig. 36. The wire edge is added to the top of the stretchout. Using the bail shown in the elevation as a profile, the stretchout for the wire blank, Fig. 58, can be determined in the usual manner.

22. Related Mathematics on Painter's Pail.—Volume of a Cylinder.—The volume of a cube is equal to the length of the base, times the width of the base, times the height of the cube. This is written $\scriptstyle {V=L\times W\times H}$. It has also been found that length times width gives the area. Because of this it can be said that volume equals area times height, and that the volume of a cylinder is equal to the area of the base times the height. The base of a cylinder is a circle, the area of which equals D²×.7854. Therefore, for a cylinder, Volume equals Diameter squared times .7854 times the height, or $\scriptstyle {V=D^{2}\times .7854\times H}$.

Sample Problem.—Find the volume of a cylinder 4″ in diameter and 6″ high.

${\begin{array}{l}\scriptstyle {V=D^{2}\times .7854\times H}\\\scriptstyle {V=4^{2}\times .7854\times 6}\\\scriptstyle {V=16\times .7854\times 6}\\\scriptstyle {V=75{\text{ cu. in.}}}\end{array}}$

Ans. 75 cu. in.

Problem 7A.—Compute the volume of the Painter's Pail, Figs. 52 and 53.

Cubic Inches in One Gallon.—It is established by law that one gallon of liquid measure shall contain 231 cubic inches.

Problem 7B.—How many cubic inches are there in one quart? In one pint?

Problem 7C.—What is the exact capacity of the pail. Fig. 53, in quarts and fractional parts of a quart?

Problem 7D.—If a job called for a pail 8″ in diameter and 7¼″ high, what would be its exact capacity in quarts?

Problem 8
GARBAGE CAN

23. The Garbage Can.—In developing the patterns for large objects such as a garbage can, Fig. 59, it is necessary to make the drawings to scale. Drawing to scale means making each line on the drawing one-half, one-third, one-quarter, or other fractional part of its true length. Scale rules are provided to assist the draftsman in this work. In order to understand the principles of the scale rule, the student should construct a model of one according to the following directions:

Suppose we wish to make a drawing to a scale of three inches to the foot (3″=1′0″). This would call for the use of a rule having a three-inch scale. We will proceed to construct such a scale. Procure a strip of thin cardboard twelve inches long and one inch wide. Lay off along one edge spaces three inches apart. Each of these spaces will represent twelve inches on the finished article. How many feet of the finished product are represented by the entire rule? Mark these divisions 0, 1, 2, and 3. Divide the first space into twelve equal parts. Each new space represents one inch. Mark the third, sixth, and ninth spaces as shown. Each space representing one inch can now be divided into four, eight, or sixteen equal parts, in order to represent ¼ in., ⅛ in., or $\scriptscriptstyle {\frac {1}{16}}$ in. respectively. The rule shown in Fig. 60 is measuring a distance of three feet, four inches (3′-4″). Notice that the feet are read to the right of the zero mark and the inches to the left of the zero mark.

The elevation of the garbage can. Fig. 61, should be drawn according to the dimensions shown. This elevation shows two O-G beads on the body of the pail. The size of these beads is not given, because the equipment of beading rolls varies with different shops. The necessary rigidity will be obtained regardless of the size of bead used. Bail ears are used on this job. The student's attention is called to the location of these ears. This can is designed for a cover that has a rim fitting inside of the body. If the cover rim were fitted over the outside, the ears would have to be placed below the bead and the bail lengthened to correspond. The development of the patterns is similar to the preceding problem and needs no additional description. All dimensions on the elevation and the patterns should be full size. These are known as Witness Dimensions, and the workman always follows these while manufacturing from scale drawings. The use of witness dimensions relieves the workman from responsibility for errors, transferring this responsibility to the draftsman or designer.

Figs 59–65.—Garbage Can.

24. Related Mathematics on Garbage Can.—Problem 8A.—Compute the exact capacity of the garbage can, Fig. 61, in gallons.

Transposing a Formula.—Very often a customer requires a certain capacity in a vessel but does not care about the dimensions. In the example of Article 22 the diameter and the height were given and the formula $\scriptstyle {V=A\times H}$ was used. If the volume and the height were given, to find the area the volume would be divided by the height. If the volume and the area were given, to find the height the volume would be divided by the area. If the volume and the diameter were given, the preceding formula would be used first, finding the area corresponding to the given diameter.

(a)	Volume	=Area	×Height
	$\scriptstyle {V=A\times H}$
(b)	Area	=Volume	÷Height
	$\scriptstyle {A=V\div H}$
(c)	Height	=Volume	÷Area
	$\scriptstyle {H=V\div A}$

Formula (a) is the original, or basic, formula while (b) and (c) are obtained by changing the position of certain quantities.

What happens to the multiplication sign when area is carried over to the left-hand side in place of volume? Does the same thing happen when height and volume are interchanged? This process of changing the location of the terms in a formula is called transposing the formula. When any terms are transposed, the sign must also be changed to the opposite; that is, multiplication becomes division, addition becomes subtraction, and so on.

Problem 8B.—What is the area of the bottom of a garbage can 16″ high, the volume of which is 42 qts.?

Finding the Diameter of a Circle from the Area.—The formula for the area of the base of a cylinder is $\scriptstyle {A=D^{2}\times .7854}$. Problem 8B gives the area of the bottom of the can. Before the bottom can be made, its diameter must be found. There are two ways of doing this: by using printed tables giving this information; or by transposing the formula for area and finding the diameter by square root. A sheet metal worker must know how to find the square root; consequently, the student is advised to become familiar with this process.

	Original formula	$\scriptstyle {\text{Area}}$	$\scriptstyle {=D^{2}\times .7854}$
	Transposing,	$\scriptstyle {D^{2}}$	$\scriptstyle {={\text{Area}}\div .7854}$
(d)	Extracting square root,	$\scriptstyle {D}$	$\scriptstyle {={\sqrt {{\text{Area}}\div .7854}}}$

Problem 8C.—What is the diameter of the can mentioned in Problem 8B?

Illustrative Examples

Example of Problem 8B.

What is the area of the bottom of a can 12″ high holding 20 qts.?

20 qt.=5 gal. (volume)
5×231=1155 cu. in. (volume)
Formula (b) would apply here, Area=Volume÷Height
Substituting known values, Area=1155 cu. in.÷12″
121155.0096.25 sq. in.

Ans. 96.25 sq. in.

Example of Problem 8C.

What is the diameter of the bottom of the can mentioned in the preceding example?

Formula (d) would apply	$\scriptstyle {{\text{Diameter}}={\sqrt {{\text{Area}}\div .7854}}}$
Substituting,	$\scriptstyle {{\text{Diameter}}={\sqrt {96.25\div .7854}}}$

.7854	96.250000	122.54	sq. in.
	78 54
	17710
	15708
	020020
	015708
	0043120
	0039270
	00038500
	00031416

Extracting square root.,	$\scriptstyle {\sqrt[{2}]{}}$	122.5411.0
		1
21		022
		021
220		00154

Ans. 11″, diameter.