# Sheet metal drafting/Chapter 8

CHAPTER VIII SECTIONS FORMED BY CUTTING PLANES
Problem 29
SPRINKLING CAN
55. The Sprinkling Can.—Figure 164 represents a section of a sprinkling can with the names of the various parts. Special attention should be given to the methods of assembly.
An elevation, Fig. 165, should be drawn according to the dimensions given in Fig. 164,
The Spout.—The sides of the spout, Fig. 165, should be carried upwards until they meet at the apex. The top side of the spout, Fig. 165, is extended into the elevation so that the distance from the apex to point 1 will be equal to the distance from the apex to point 5. The straight line 1–5 is drawn to serve as the base of the cone. A half-profile of the cone (spout) is drawn and divided into four equal spaces. Extension lines are carried from each division, meeting the base line at right angles. From the intersections of the base lines, extension lines are carried to the apex of the cone. The profile of the body of the sprinkling can, Fig. 166, is drawn next. The horizontal center line is extended to the right of the profile indefinitely. An extension line is dropped from the apex of Fig. 165 until it intersects the horizontal center line of Fig. 166. This will locate the apex of Fig. 166. Extension lines are dropped from the base of the cone (spout) in Fig. 165 until they intersect the horizontal center line of Fig. 166. These intersections should be numbered 1, 2, 3, 4, and 5 to correspond to the half-profile. The distances a, b, and c taken from corresponding lines in the half-profile of Fig. 165 are set off upon these extension lines. Lines are drawn from each of these points to the apex of Fig. 166. These lines intersect the profile of the body at three different points D, E, and F. Extension lines should be carried upward until D intersects the line from 2, F intersects the line from 3, and E intersects the line from 4. A curved line passing through these points will give the developed miter line between the body and the spout.
With a radius equal to the distance from the apex to point 5, Fig. 165, and with the apex as a center, the arc of stretchout. Fig. 167, is drawn. Twice as many spaces as there are in the half-profile of the spout are set off upon this arc. Lines from each of these points should be drawn to the apex. Straight lines are drawn ​parallel to the base line 1-5 across the elevation of the cone, Fig. 165, from each intersection of the developed miter line. These lines intersect the side or slant height of the cone at five different points. With the apex as a center, arcs are carried from each of these points over into the stretchout. Starting with point 1 of
the half-profile, the extension line is traced to the base of the cone, thence to the developed miter line, and from this point to the correspondingly numbered lines of the stretchout. In like manner the remaining intersections may be traced. A curved line passing through these points will give the miter cut of the pattern. An ​arc drawn from the apex as a center, with a radius equal to the distance from the apex to the top of the spout, will complete the pattern. A $\scriptstyle {\frac {3}{16}}$-inch flange is added outside of the miter cut, as shown in Fig. 167, and a ¼-inch lap parallel to one side of the spout is also added.
Pattern of Body.—The pattern for the body is a rectangular piece of metal the length of which is equal to Diameter×π+½in., and the width of which is equal to the height of the can plus ⅜in. for the wire edge, plus ⅛in. for the single edge at the bottom of the can. This feature of the problem is so familiar to the student that it is omitted from the drawing. Manufacturers generally punch a round hole at the point where the center line of the spout intersects the side of the can. The hole may, however, be developed according to the principles laid down in Chapter V.
The Breast.—The breast is a portion of a cylindrical surface cut by two inclined planes. The top line of the breast in Fig. 165 is extended indefinitely to the left. At right angles to this line another line is drawn to serve as a center line for the half-profile. The line H-K is drawn at right angles to the center line. This line, if prolonged, should pass through the intersection of the breast and the top of the can. A distance equal to one-half the diameter of the can is set off upon H-K. The connecting points P and K of the half-profile should be drawn. The center from which this arc is drawn should fall on the center line of the half-profile. The arc is divided into four equal spaces. Extension lines are carried with the elevation of the breast as shown. A line of stretchout is drawn and the spacing of the half-profile with letters to correspond is transferred to it. From each intersection of the miter lines, extension fines are carried over into the stretchout. Starting from the half-profile, each extension line can be traced first to the miter lines and thence to a correspondingly lettered line in the stretchout. Curved lines passing through these points will give the half pattern of the breast. Fig. 168. A ¼-inch lap is added to the side that adjoins the body, and a ⅜-inch wire edge to the other side of the breast.
The Handle.—The handle or bail of the can is a straight piece of metal 1¼ in. wide that is formed to the profile shown in Fig. 169. The handle extends down below the top of the can and is riveted and soldered to the body. A double hem or a wire may be used to stiffen each edge of the handle. A boss is soldered into the upper ​part of the handle to aid the hand in gripping it. The handle does not require a pattern since it is a straight strip of metal whose length is equal to the perimeter of the profile, Fig. 169, and whose width is equal to 1¼ in. plus the allowances for stiffening each side. The pattern of the boss is obtained in exactly the same manner as the pattern of the breast, and needs no further description. Figure 170 shows the profile of the S handle together with its boss. The pattern for the S handle is a tapering strip of metal whose length is equal to the distance around the profile, Fig. 170, and whose large end is 1¼ in. wide, while the small end is ⅝ in. wide. A ¼-inch wire edge is added to each side for the stiffening wire.
56. Related Mathematics on Sprinkling Can.—Volume of Sprinkling Can.—In Chapter II it was learned that the volume of a cylinder is equal to the area of the base multiplied by the height, or V=A×H. It was also found that the area of a circle was equal to the diameter squared, multiplied by .7854, or D$^2$×.7854.
Sprinkling cans are generally listed according to their holding capacity in gallons, quarts, and pints.
| Formula: | V=D$^2$ | ×.7854 | ×H |
| Substituting, | V=8$^2$ | ×.7854 | ×10.5. |
| | V=64 | ×.7854 | ×10.5. |
| .7854 |
| 64 |
| 31416 |
| 47124 |
| 50.2656 |
| 10.5 |
| 2513280 |
| 5026560 |
| 527.78880 | cu. in. |
There are 231 cu. in. in one gallon; therefore,
| 527.789 | gallons |
| 462 |
| 657 |
| 462 |
| 1958 |
| 1848 |
| 1109 | Ans. 2.28 gal. or 9 qt. |
But, 2.28 gallons is approximately 9 quarts, the extra contents being allowed for carrying nine full quarts without danger of spilling.
| | Diameter of Bottom. | Diameter of Bottom. | Height of Body. | Height of Body. |
| (a) | 4 | ″ | 4 | ⅝″ |
| (b) | 5 | ⅝″ | 7 | $\scriptstyle {\frac {3}{16}}$′ |
| (c) | 7 | ⅛″ | 9 | ″ |
| (d) | 8 | $\scriptstyle {\frac {13}{16}}$″ | 11 | $\scriptstyle {\frac {5}{16}}$″ |
Problem 29B.—Regarding the body blank for the can as a rectangle whose height is equal to 10½″+⅜″+$\scriptstyle {\frac {3}{16}}$″ and whose length is (8×3.1416)+½″ what is its area in square inches?
Problem 29C.—Regarding the bottom of the can as a circle whose diameter is equal to 8″+⅜″+⅜″, what is its area in square inches?
Problem 29D.—Regarding the whole pattern of the breast, Fig. 168, as the combined area of two right triangles, having bases 5½″ long and altitudes 8″ high, what is the total area of the breast?
Problem 29E.—Regarding the spout as a trapezoid whose upper base is 1¾″, whose lower base is 9″, and whose altitude is 13½″, what is its area in square inches?
Problem 29F.—Regarding the semicircular handle as a rectangular piece of metal whose length is $\scriptstyle {{\frac {8\times \pi }{2}}+4''}$ and whose width is 1¼″+¼″+¼″, what is its total area?
Problem 29G.—Regarding the "boss" in Fig. 169 as a rectangle whose length is 4⅞″, and whose width is 2″, what is its total area?
Problem 29H.—Regarding the S handle as a trapezoid whose upper base is ⅝″, whose lower base is 1¼″, and whose altitude is 5¼″, what is its total area?
Problem 29I.—Regarding the "boss" in Fig. 170 as a rectangle whose width is $\scriptstyle {1{\frac {1}{16}}}$″ and whose length is 2⅛″, what is its area?
Problem 29J.—Adding together the answers to problems, 29B, C, D, E, F, G, H, and I, what is the combined area of all the parts?
Problem 29K.—A sheet of IXXXX charcoal tin measuring 20″×28″, costs 50 cents. Adding 5 per cent to the answer of Problem 29J, how much will the stock required for one sprinkling can cost?
Problem 30 BOAT PUMP
57. The Boat Pump.—A boat pump consists of a straight piece of pipe called the barrel, to which is attached a frustum of a cone (the funnel). A tapering spout is riveted and soldered over an opening in the pump barrel. Into the lower end of the pump barrel a "lower box" is soldered. The "lower and upper boxes" may be obtained from almost any supply house. The upper box is threaded to receive the pump rod. The pump rod has an oval handle formed upon its upper end. Measurements are usually given from the under side of the spout to the lower end of the pump, for the diameter of the barrel, and for the length of the spout. The other details of construction are left to the discretion of the designer.
The Barrel.—The barrel of the pump is made from one piece of metal, if possible, in order to avoid any possibility of the upper box catching as it works up and down in the cylinder. The pattern is a rectangular piece of metal the length of which is shown in the elevation, Fig. 171, and the width of which is equal to (Diameter×π)+½in. for locks.
The Spout.—An elevation should be drawn according to the dimensions given in Fig. 171. The center line of the spout is next drawn and prolonged to the right indefinitely. The sides of the spout are extended until they meet the center line at the apex. They are also extended to the left until they meet the center line of the barrel. This will give a right cone whose base is the line 1–5. A half-profile for this cone should be drawn. It should be divided into four equal parts, and each division numbered. A profile of the barrel, Fig. 172, should next be drawn. Using the same center, a half-profile of the spout, Fig. 172, should be put in. A horizontal center line that is long enough to receive an extension line dropped from the apex of Fig. 171 should also be drawn. This half-profile is divided into four equal parts and these divisions numbered to correspond to those of the half-profile in Fig. 171. These numbers change their positions as can be seen. From each division in both half-profiles, lines are drawn to the bases of the cones forming angles of 90°. From each intersection thus found, lines are drawn to the apex. In Fig. 172, the line 1 intersects the ​profile of the barrel at point a, line 2 at point b, and line 3 at point c. Extension lines are carried up into Fig. 171 from points a, b, and c of Fig. 172. At the points where these lines intersect
corresponding lines in Fig. 171, will be the location of the new points a, b, and c of Fig. 171. A curved line traced through these points will give the developed miter line.
​Pattern of Spout.—The arc of stretchout, Fig. 173, is drawn with a radius equal to the slant height of the cone, and with the apex as a center. The spacing of the half-profile is transferred to this arc with numbers to correspond. Where extension lines from points a, b, and c cut the slant height of the elevation, extension arcs are drawn over into the stretchout. All intersections should be traced out by starting from the profile, following each extension line to the miter line, and thence to a correspondingly numbered line in the stretchout. The miter cut of the pattern is obtained by tracing a curved line through these intersections. An arc whose radius is equal to the distance from the apex to the end of the spout completes the pattern. A lock on each side of the spout and a hem on the small end of the spout are added. A flange (not shown) should be added to the miter cut.
The Opening.—Upon any straight line, the distances a to b and b to c of Fig. 172 are set off. Since this is but half of the opening, this operation must be repeated as shown in Fig. 174. Measuring lines are drawn through each of these points. Upon the lines a the distance from point 2 to the center line of the half-profile of Fig. 172 is set off. Upon the lines a, the distance from point 1 to the center line of the half-profile of Fig. 172 is set off. Points c fall upon the center line of Fig. 174. A curve traced through the points thus obtained will give the shape of the opening.
The Funnel.—One side of the funnel should be extended inward until it meets the center line of the barrel. This will locate the apex of the whole cone, of which the frustum is a part. With any convenient point as a center, and a radius equal to the distance from this apex to the large end of the funnel, an arc of stretchout. Fig. 175, is drawn. A quarter-profile is placed above the elevation of the funnel in Fig. 171. This is divided into three equal parts. Since this is but a quarter-profile, twelve spaces must be transferred to the arc of stretchout in Fig. 175. The first and last points are connected to the apex by means of straight lines. The pattern is completed by an arc, drawn from the center of Fig. 175, whose radius is equal to the distance from the apex of the funnel to the point F in Fig. 171. The necessary locks and wire edge should be added.
58. Related Mathematics on Boat Pump.—Problem 30A.—The barrel of the boat pump is a cylinder whose diameter is 4″ and ​whose height is 5′–0″. Allowing ¾″ for locks what is the area of its pattern?
Problem 30B.—The funnel of the pump is a frustum of a cone whose lower base diameter is 8″, upper base diameter 4″, and altitude 4″. What is its area?
Sample Problem.—Substituting for the above dimensions 6″ upper base, 3″ lower base, and 3″ altitude, we would have
Original Formula:
${\frac {{\text{Circumference of upper base}}+{\text{Circumference of lower base}}}{2}}\times {\text{slant height.}}$
Substituting,
${\frac {(6\times \pi )+(3\times \pi )}{2}}\times {\sqrt {3^{2}+1.5^{2}}}$
The slant height of any frustum of a right cone is equal to the hypotenuse of a right triangle whose base is the difference between the radius of the upper base and the radius of the lower base, and whose altitude is the altitude of the frustum. The slant height of this frustum is, therefore, equal to the hypotenuse of a right triangle whose base is 3″–1½″=1½″, and whose altitude is 3″. But the hypotenuse is equal to ${\sqrt {{\text{base}}^{2}+{\text{altitude}}^{2}}}$; therefore, the slant height for this frustum is ${\sqrt {3^{2}+1.5^{2}}}$
$\scriptscriptstyle {{\frac {18.85+9.42}{2}}\times {\sqrt {(9+2.25}}{=}}$
$14.13\times 3.35{=}47.33{\text{sq. in.}}$
Problem 30C.—The spout, Fig. 171, is also a frustum of a cone whose lower base has a diameter of 3⅛″, and upper base a diameter of 2½″. The slant height may be called 7½″. What is its area?
Problem 30D.—What is the combined area of Problems 30A, 30B, and 30C?
Problem 30E.—Allowing 10 per cent for waste, what will be the cost of the material for one pump at 8 cents per square foot?
Problem 31 ROOF FLANGE
59. The Roof Flange.—Case I. The Roof Having One Inclination.—The measurements usually given for a roof flange are the diameter of the pipe, and the pitch of the roof. The roof pitch is generally given as "so many inches to the foot." Figure 176 shows the rise and run of the roof line. If the job called for a roof pitch of 4 in. to 1 ft., the line marked run, Fig. 176, would measure 12 in. and the line marked rise would measure 4 in.
Pattern of Pipe.—An elevation is first drawn according to the dimensions given in Fig. 176. A profile is drawn above this view and is divided into equal spaces. Each division is numbered. Extension lines from each division of the profile are carried down to the roof line. A line of stretchout is drawn, and the spacing from the profile transferred to this line. The spaces are numbered to correspond. The measuring lines of the stretchout are drawn in. From each intersection of the miter (roof) line of Fig. 176, extension lines are carried over into the stretchout. Starting from point 1 of the profile, the extension lines should be traced downward to the miter line, and thence to a correspondingly numbered line in the stretchout, Fig. 177. In like manner all points of intersection can be located in the stretchout. A curved line passing through these points will give the miter cut of the pattern. An extension line drawn from the top of the roof flange elevation completes the pattern. A ½-inch lock is added to each side of the pattern and a ⅜-inch double edge to the miter cut in order to join the pipe to the apron by double seaming.
Opening in Apron.—Any straight line is drawn to serve as a line of stretchout, Fig. 178. The exact spacing between the intersections of the miter line is transferred to this line, and these spacings numbered to correspond. A measuring line is drawn through each point as shown in Fig. 178. Upon line 2 of Fig. 178, a distance equal to line a of Fig. 176 is set off. Similarly, line 3 would receive the length of line b from Fig. 176, line 4 would receive c, line 5 would receive d, and line 6 would receive e. These distances may now be transferred to the opposite side of the line of stretchout, since both parts of the opening are exactly equal. A curved line drawn through the points thus located will give the ​shape of the opening. The apron of the flange may now be drawn around the opening, allowing at least 3 in. on the sides and
and 6 in. on the top. A hem should be added to the long sides of the apron to direct the flow of water.
​Case II. The Flange Fitting over the Ridge of a Roof.—Figure 179 shows an elevation and profile of a roof flange fitting over the ridge of a 90° or "square-pitch" roof. Figure 180 shows the pattern for the pipe. It will be noticed that this type of flange cannot be easily double seamed; therefore, a ½-inch edge is added to the miter cut. This edge is turned off, and riveted and soldered to the apron. Figure 181 shows the pattern of the apron. A bend of 90° must be made on line 4 in order to fit the apron over the ridge of the roof. The description for Case I will apply to this problem, the only difference being the shape of the miter cut in Fig. 176.
Case III. The Flange Fitting over the Ridge and Hips of a Roof.—An elevation is first drawn according to the dimensions given in Fig. 182. Above this elevation a half-profile is drawn and divided into equal spaces and numbered as shown. Extension lines are carried from each division down to the roof line. A plan, Fig. 183, is drawn in the following manner: Extend the center line of the elevation downward indefinitely. Construct a rectangle, Fig. 183, using this line as a center line. This rectangle will represent the top view or plan of the apron of the finished flange. Draw two lines at an angle of 45° to represent the hips of the roof. The center line becomes the ridge. With the point where the ridge and hips meet, as a center, and a radius equal to that of the half-profile. Fig. 182, draw a circle. Divide this circle into twice as many equal parts as there are in the half-profile. Number these divisions to correspond. The hips cross this circle halfway between points 2 and 3 and 5 and 6. Number these points 2½ and 5½ respectively. Carry extension lines upward from each division of the circle, to the roof line.
That part of the miter line not already shown in elevation can be developed. Points 2½ and 5½ in Fig. 182 occur where the lines from these points in Fig. 183 intersect the roof line in Fig. 182. Since this is the highest point of the miter line on the hips, point 3 must be opposite point 2, point 4 opposite point 1, and point 5 opposite point 6. This is indicated by horizontal dotted lines in Fig. 182. A curved line drawn through these points will be the developed miter line. A line of stretchout, Fig. 184, is now drawn. The exact spacing is transferred from the circle in Fig. 183 to this line, and numbered to correspond.
The measuring lines of the stretchout are drawn. Starting at ​point 1 of the plan, the extension line is traced to the miter line, and thence to a correspondingly numbered line in the stretchout. In like manner, all points of intersection can be located in the stretchout. A curved line passing through these points will give the miter cut. The pattern is completed by an extension line drawn from the
top of the pipe. A lock is added to each side of the pattern. A flange (not shown) should be added to the miter cut.
The Apron.—The exact spacing of points 10, 11, 12, 1, 2, and 2½, as shown on the center line of Fig. 183 by lines a, b, c, d, and e, are set off upon any straight line, Fig. 185. These points also ​bear the numbers 9, 8, 7, 6, and 5½ as shown in Fig. 185. Measuring lines are drawn through each of these points. The distances a, b, c, etc., are taken from lines, a, b, c, d, and e of Fig. 183 and are set off on the measuring line. A curved line passing through these points will give the shape of the opening for the main part of the flange that is to fit over the ridge. The distance 10–M, Fig. 185, is made equal to 10–N of Fig. 183. A perpendicular is drawn at point M. The distance MN is made equal to 10–N of Fig. 182. Another perpendicular line is erected at point N. Distance NF is made equal to NF of Fig. 183. The bending line FG is drawn in. With F as a, center, arcs are drawn from points H, J, and K to the left an indefinite distance. With G as a center, the distance GH is set off on the other side of point G. Similarly, the distances GJ and GK are set off on the other side of point G. The straight lines KF and KJ will complete the pattern of one side of the flange, with the exception of the curve JHG. The other half of the pattern is exactly equal to the one already drawn and is produced by the same method.