TS 150 Class X^_^^A Rook -N ^b Copyright N^ COPYRIGHT DEPOSnV MENSURATION FOR SHEET METAL WORKERS AS APPLIED IN WORKING ORDINARY PROBLEMS IN SHOP PRACTICE By WILLIAM NEUBECKER Reprinted from The Metal Wokkfr, Plumber anu Steam Fitter NEW YORK DAVID WILLIAMS COMPANY 14-16 Park Place ^^^1 LIBRARY of OONeR^s| Two Copies ReceivdCi f JAN 28 ia08 (}«t>yag<>t tnto 0U8SA XXc, Nu. OOPY B. Copyright, 1907 David Williams Company \ f-y^-f^. CONTENTS Page. Length of material required for a round pipe 6 Amount of material for a square pipe 6 Length of perimeter of a hexagon 6 Square and circular sections of pipe of given dimen- sions 7 Length of side of a square to be inclosed by a given circle 8 Length of arc when angle and radius are known g Amount of material for an ellipse when length and width are given 9 Length of hypotenuse in a right angle triangle 9 Areas in square, rectangle and rhomboid lO Areas in a right and an oblique triangle 1 1 Area of triangle when three sides are given I2 Area of trapezoid 13 Area of trapezium 14 Area of circle 14 Area in circle and square 14 Area in a ring 15 2 Mensuration for Sheet Metal Workers. Page. Areas in sector and segment of a circle 15 Square whose area is equal to area of given circle 16 Circle whose area is equal to area of given square 17 Area of ellipse 17 Area of regular polygon 18 Area of sphere 18 Convex surface of cylinder 18 Convex surface of frustum of cylinder 19 Convex surface of elliptical cylinder and frustum of elliptical cylinder 19 Convex surface of right prism and of frustum of right prism 20 Convex surface of right cone 20 Convex surface of right pyramid 21 Convex surface of frustum of right cone 21 Convex surface of frustum of right pyramid 22 Capacities of tanks, etc 22 Contents of cube 23 Contents of hexagonal prism 24 Contents of c)'linder 24 Contents of sphere 25 Contents of cone 25 Contents of pyramid 25 Contents of hopper 2^ Contents of prism of pyramid 27 Contents. 3 Page. Contents of frustum of cone 29 Contents of prismoid , 29 Contents of wedge 30 Practical examples for the shop 30 Pipes equal to rectangular pipe 30 Boot 31 Chimney top , 31 Another form of boot 32 Pipe with branches 33 Triangular pipe 34 Elliptical pipe 34 Two pronged fork 35 Three pronged fork 35 Ascertaining the sizes of articles 36 A square tank 36 A sphere 37 An oil tank 39 Wash boiler 39 Flaring pail 40 Flaring measure 40 Flaring elliptical tub 41 Flaring pan 41 Short rules in computation 42 Diameter of circle 42 Height of inaccessible point 42 4 Mensuration for Sheet Metal Workers. Page. Use of steel square 44 Problems of steel square solution 44 Obtaining radius 45 Rectangle to square 47 Circle to square 47 Three squares . .47 Circle of four times the area 48 Circumference of an ellipse 48 1 Mensuration for Sheet Metal Workers, Very little has been written about mensuration for the sheet metal worker, and although various collections of tables have been published, the rules are not generally explained so as to be easily understood or to enable one to proceed intelligently with practical problems which come up. In the pages which follow examples in com- puting the circumferences, areas and capacities for va- rious shapes arising in practice are given in detail. The comprehension of these will enable the student to com- pute any ordinary problem in the shop. Besides methods for finding the lengths, areas and volumes of the simpler geometrical forms, examples are given in computing the areas of heating and ventilating pipes of all ordinary shapes, making their areas equal to those of pipes of other profiles. The use of the pris- moidal formula for obtaining the capacities of various shaped bodies is fully explained, as is a method of ob- taining the hight of any solid to hold a given quantity when the diameter is known, or vice versa. A short rule is illustrated for finding the diameters of branch pipes taken from a given main pipe, so that the areas of the branches will equal the area of the main, and many other problems are treated. One of the first problems arising in the shop is to find the true length of material required for round or other shaped pipes. The rule for obtaining the circumference 6 Mensuration for Sheet Metal Workers. of any circle is to multiply the diameter by 3.1416, or, as is sometimes used in practice, by 3 1-7. In Fig I, A represents a circle 4 inches in diameter. Then 4 inches X 3-i4i6 ^ 12.5664, or 12 9-16 inches CIRCUMFERENCE 12.5664 5 Fig. 1. — Finding Length of Matorial Required for a Round Pipe. approximately, the circumference, as shown rolled out from a to b. Referring to Fig. 2, let abed represent a 4-inch square. To obtain the perimeter or amount of material for this pipe, it is only necessary to multiply 4 inches by 4. which equals 16 inches. As shown from a to a', the distance represents the length of the sum of the sides of the square figure. In the same manner the length of the perimeter of the hexagon is obtained in Fig. 3. In this each side meas- o A^ PERIMETER J I d *" a 5' c' d' «' Fig. 2. — Finding Amount of Material for a Squai-e Pipe. ures- 3 inches. The figure has six sides, so we have 6 X 3 inches = 18 inches, the length shown from a to a'. Sometimes a round pi])e is to be formed to a square section at the opposite end, using the same amount of material as in the round pipe. This makes it desirable Mensuration for Sheet Metal Workers. 7 to know what the length of the sides at the square end will be. Knowing that the diameter of the circle a in Fig. 4 is 4 inches, and that the circumference is 12.5664, d 3" c PERIMETER 6' c d' e' f Fig. 3. — Finding Length of Perimeter of a Hexagon. it is only necessary to divide by four, which will give 3.1416 inches, the length of the sides of the square b. If the circumference is not known, multiply the diameter by 0.7854; thus, 4 inches X 0.7854 = 3.1416 inches, which 3.h41C ^'^vJBF^^' Figs. 4 and 5.— Finding Square and Cir- cular Sections of Pipe of Given Dimen- sions. T'ig. 6.— Finding Length of Side of a Square to Be Inclosed by a Given Circle. multiplied by four sides = 12.5664 inches, the perimeter for the square b, and is equal to the circumference of the circle a, proving the above rules. 8 Mensuration for Sheet Metal Workers. If, however, the conditions were reversed, and each side of a given square measured 5 inches, as shown at a in Fig. 5, making the perimeter of the square 20 inches, and it is desired to know what diameter a circle would have whose circumference would be equal to that peri- meter, to obtain this diameter multiply the length of one side, or 5 inches, by 1.2732, which equals 6.366 inches, the required diameter. ]\Iultiply this diameter,. 6.366 Fig. 7. — Finding Length of Arc Wlien Angle and Radius Are Known. Fig. 8. — Finding Amount of Material for an I'll ipse When Length and Width Are Given. Fi 9.— Find- ing Length of Hypotenuse in a Right Angle Triangle. inches by 3.1416, and the product will be 20 inches, the perimeter of the square thus proving the rule. When double ventilation pipes are constructed, as shown in Fig. 6, where the outer pipe is a true circle and the inner pipe a square, allowing an air space between the two pipes, and it is desired to know the length of the side of a square to pass inside of a given circle, it is only necessary to multiply the diameter of the given pipe by 0.7071. Suppose the round pipe a were 6 inches in diam- eter, then 6 inches X 0.7071 = 4,2426, or nearly 4j4 inches, for the side of the desired square. When it is desired to find the lengfth of an arc, when Mensuration for Sheet Metal JVorkers. 9 only the angle and radius are known, then multiply the number of degrees by the diameter and the product by 0.008727. In Fig. 7 it is desired to find the length of the 90-degree arc a b, whose radius is 4.5 inches, or diameter 9 inches. Following the above rule, we have 90 X 9 = 810. 810 X 0.008727 = 7.0688 inches, the length of a b. When the length and width of an ellipse are given, and it is desired to know how much material is required for its circumference, multiply half the sum of the two diameters by 3.1416. Thus, in Fig. 8, we have an ellipse be m Area 81 sq. inches lA Area / 130,5 sq, inches H ^9= sj i*-- ]Ai—- ~1 Fig. 10. — Areas in Square, Rectangle and Rhomboid. the major axis of which is 10 inches and the minor axis 10 + 6 6 inches. Following the above rule, = 8 X 3.1416 = 25.1328. It should be understood that the cir- cumference of an ellipse cannot be accurately determined, and the above rule is merely an approximation giving fairly close results. When a large smoke stack is to be carried up at an angle on a building and the vertical hight and horizontal projection are known, the length of the slant can be ob- tained by the rule illustrated in Fig. 9, which shows how the length of the hypotenuse is found in a right angle tri- angle. Add the square of the base to the square of the altitude, and the square root of the sum will be the hypo- tenuse. The base being 6 inches and the altitude 8 inches. 10 Mensuration for Sheet Metal JVorkers. we have V6- -|- 8" = V36 -|- 64 = Vioo = 10. The square root of 100 is 10, because 10 is the number which, when multiphed by itself, will equal 100. The area of any surface is the number of square inches or square feet within its outline. In connection with the illustrations, which immediately follow, the rules will be given for obtaining the areas of the various geo- metrical shapes, which will enable the student to proceed intelligently when computing areas and capacities of va- rious forms arising in practice. While many sheet metal workers understand how to compute the areas of the more common geometrical shapes, there are some who do Fig. 11. — Areas in a Right and an Oljlique Triangle. not know how this branch of mathematics can be applied in practical work. One of the most simple figures, the area of which the student usually masters first, is that of the square, shown hy a b c d in Fig. 10, each side of which measures 9 inches. To obtain the area simply multiply the length of the side by itself; thus: 9 X 9 = 81 square inches. To obtain the area of the rectangle e f h i, multiply the width by the length, thus: 9 X i4-5 = 130.5 square inches. Suppose a surface had to be covered with sheet metal, the shape being that of a rhomboid, shown by ni n h i, Mensuration for Sheet Metal Workers. 11 in which m n is parallel to and of the same length as h i; then knowing the perpendicular hight e i to be 9 inches, and the length to be 14^^ inches, we would have the same area as that shown in the rectangle e f h i, because the triangle / 11 h equals the triangle e m i. When the area of a triangle is to be found, whether right or oblique, the base and perpendicular hight be- ing known, the rule is to multiply the perpendicular hight by the base, and half the product is the area. In Fig. II, let a b c represent a right triangle, whose base is 8 inches and hight 12 inches; then 12 X 8 = 96. 96 -f- Flg. 12. — Area of Triangle When Three Si(le.s Are Given. 2 = 48 square inches, the area. If the product 96 were not divided by two it would represent the area of the rec- tangle a i c b, but by drawing the diagonal a c we divide the rectangle into two right triangles, each equal to one- half of 96, or 48, as shown. The diagram e d f represents an oblique triangle, whose base is 15 inches and perpendicular hight to the apex 12 inches. Now, following the above rule, we have 12 X 15 = 180. 180 -^ 2 = 90 square inches area of d e f. To prove this, construct the rectangle e in n f, and from d drop the vertical line d o. The distances ni d and d n are 9 and 6 inches, respectively. We then have two rectangles, one 9 x 12 and the other 6 x 12 inches. 12 Mciisiiratidii for SJicct Metal JVorkers. By drawing d e and d f we obtain two right triangles, doc and d o f. Following the explanation given in con- •^1 7 u 9 X 12 , 6 X 12 nection with a b c, we have = S4) and 2 ^^' 2 = 36 ; 36 -L- 54 = 90 square inches, the same as the area oi d e f. Sometimes an irregular shaped structure is to be cov- ered and none of the angles and only the dimensions of the three sides are known or can be obtained. The rule to be followed in this case is as follows : From half the 1^ 12'-0: ^ Fig. 115. — Area of Trapezoid. i'.O' Fig. 14. — Area of Trapezium. sum of the three sides subtract each side separately ; find the continued product of the half sum of the sides and the three remainders, and the square root of this product is equal to the area of the triangle. In Fig. 12 is shown a portion of a surface, each side measuring respectively 21, 17 and 10 feet. The sum of the three sides is 21 -{- 17 -f- 10 = 48. 48 ^ 2 = 24, the half sum. Subtract- ing each side separately from half the sum of the sides, we get the three remainders, 24 — 21 ^3 feet; 24 — 17 = 7 feet, and 24 — 10 = 14 feet. Now, 24 X 3 X 7 X 14 = 7056. V7056 = 84. The method of obtaining the square root is as fol- Mensuration for Sheet Metal Workers. 13 lows : Pointing off the number 7056 into periods of two figures each, we get 7o'^6, which shows that the complete part of the square root contains two figures. Then pro- ceed as below : Root. Trial divisor. Correct divisor. 70'.o6(84 ans. 160 164 64 656 656 It will be observed that the greatest number whose square is contained in 70 is 8 ; and, therefore, 8 is the first root of the figure; 8 X 8 = 64. Subtracting 64 from 70 and bringing down the next period, 56, we get the first partial dividend, 656. The double of 8, the partial root already found, is 16, and annexing a cipher to this we get 160 as the first trial divisor. This trial di- visor is contained in the partial dividend 656 four times, which suggests four as the second figure of the root. Adding 4 to 160 we obtain 164 as the correct divisor. When the product (4 X 164 = 656) is subtracted from the partial dividend 656 there is no remainder. Eighty- four is the required square root, and 84 X 84 =: 7056- The triangle in Fig. 12 contains 84 square feet area. In Fig. 13 is shown a trapezoid, and in Fig. 14 a trapezium. The difference between the two lies in the fact that the trapezoid has two sides parallel to each other, while the trapezium has no sides parallel. When computing the area of an irregular surface having two parallel sides use the rule for obtaining the area of a trapezoid, shown in Fig. 13, which is as follows: Mul- tiply half the sum of the parallel sides by the perpendic- ular hight. Half the sum of the parallel sides in Fig. 1 2 is 6 + 12 = 18 ; 18 -f- 2 = 9, the mean distance. Then li Mensuration for Sheet Metal JVorkers. 9X5 (the perpendicular higlit) = 45 square feet area in abed. In computing the area oi a d e c, in Fig. 14, we di- vide it into two triangles and a trapezoid, by drawing vertical lines from the angles a and d, at right angles to c c. The bases of the two triangles are 4 and 2 feet, re- FULL SIZE Fig. 15. — Area in Circle. Fig. 16. — Area in Circle and Square. spectively, and the altitudes 8 and 12 feet, respectively ; .1 ,• r , 8 + 12 _ 20 the mean distance of the trapezoid is — =10 Then 8X4 ^ 32 ^ J 6, area of triangle a b c; 24 — — =12, area of triangle d e f; 10 X 7 12X2 2 70, area of trapezoid b a d f. Then 16 -f 12 -f 70 = 98 square feet area m a d e c. To obtain the area of a circle, square the diameter and multiply by 0.7854. Fig. 15 shows a section of a pipe ioy2 inches in diameter, of which it is desired to find the area. Then, 10.5 X 10.5 = 110.25; 110.25 X 0-7854 = 86.59 square inches. Fig. 16 shows a square containing i square inch. The inscribed circle is i inch in diameter. Although it is i inch in diameter, it con- tains only 0.7854 square inch. Mensuration for Slicct Metal JJ^orkers. 15 When the area of a ring is to be determined, as shown in Fig. 17. in which the outside diameter is 12 inches and the inside diameter 5 inches, deduct the square of the small diameter from the square of the large diameter and m.ultiply the remainder by 0.7854. Following this rule, we have 12- — 5- = 12 X 12 — 5 X 5 = i44 — 25 = 119; 119 X 0.7854 =: 93.462 square inches in the ring. Sometimes a ventilating pipe is to be constructed, whose section is a sector of a circle, as shown hy a b c e in Fig. 18, and it is necessary to know its area. In the AREA OF SEGMENT 6. 136 8Q. INCHES Fig. 17. — Area in a Ring. Pig. 18.- -Areas in Sector and Segment of Circle. illustration, the radius ^ c is 3^ inches; the angle a e c, 105 degrees (which is determined by the use of the pro- tractor) ; the chord a c, 6 inches, and the rise or middle ordinate m b, i}^ inches. Before obtaining the area, it will be necessary to know the length of the arc a c, which is obtained by following the rule given in connection with Fig. 7, Part I, as follows: In Fig. 18 we have 105 X 7-5 X 0.008727 = 6.872 inches, the desired length of arc. Then to obtain the area oi a b c c, multiply the length of the arc by half the radius. The radius is 3.75 ; 3.75 -^ 2 = 1.875. Then 1.875 X 6.872 = 12.885, the area of the sector. 16 Mcnsurafioii for Sheet Metal Workers. If the area of the segment a b c a is desired, knowing the area of the sector, it is only necessary to deduct from this amount the area of the triangle a e e. Thus half the distance of the chord a c is 3 inches, and the hight from the chord in to the center e is 2}i inches. Then 3 X 2.25 = 6.75 square inches, the area of a c ^. Then 12.885 (the area of the sector) — 6.75 (the area of the triangle) = 6.135 square inches, the area of the segment. If the chord a c and the rise m b of the segment were K-- — 6 i 80. INCHES c AREA 78.64 8Q. INCHES B K — 8t862- Fig. 19.— Square Whose Fig. 20. — Circle Whose Area Is Equal to Area Area Is Equal to Area of Given Circle. of Given Square. Fig. 21. — Area of Ellipse. given, and it was desired to find the diameter of the circle of which the segment is a part, divide the sum of the squares of half the chord and the rise by the rise, and the quotient is the desired diameter. Half of the chord a c is 3 inches, and the rise m b is lYi inches. Fol- lowing the above rule, we have 3^ -f- 1.5- = 11.25; H-^S -^ 1.5 = 7.5 inches, the diameter. As twice the radius e c equals 73^ inches, the above rule is proven. It sometimes happens that a transition is to be made for a heating or ventilating pipe from round to square, Mensuration for Sheet Metal JVorkers. 17 and the square end is to have the same area as the given round end. Let A in Fig. 19 represent a section of a 10- inch heating pipe ; then, following the rule given in con- nection with Fig. 15, the area of A in Fig. 19 is 78.54 square inches. To find the dimensions of a square of equal area, multiply the given diameter by 0.8862. Thus 10 X 0.8862 = 8.862, or the side of the square B. If, however, the conditions were reversed and the square end of the transition piece were given, say 6 inches, as shown in C, Fig. 20, then multiply the given side by 1.1284, and the product will be the diameter of a i-.-IO-- Pig. 22. — Area in Regular Polygon. Fig. 2.3. — Area of Sphere. circle of equal area. Thus 6 X 1.1284 = 6.7704 inches, diameter of circle D. When the area of an ellipse is required, multiply the long diameter by the short diameter, and this product by 0.7854. In Fig. 21 is shown an ellipse, whose long diam- eter is 14 inches and short diameter 8 inches. Then 8 X 14 = 112; 112 X 0.7854 = 87.96 square inches area. The method for obtaining the dimensions of the opposite end of a transition piece, with one end a given ellipse and the opposite end to have similar area in either round, square or rectangular section, will be explained later un- der the head of Practical Examples for the Shop. When a surface is to be covered with sheet metal whose shape is that of any regular polygon, the rule to 18 Mensuration for Sheet Metal JJ'orkcrs. be followed in obtaining the area is to multiply the sum of the sides by half the perpendicular distance from cen- ter to sides. For example, there is shown in Fig. 22 a regular polygon having six sides, called a hexagon. The one side, c d, equals 10 inches, and the sum of the six sides, 6 X 10, or 60 inches. The whole perpendicular distance & a is 8.66 inches, which divided by two gives 4.33, or half the perpendicular distance. Then 60 X 4-33 = 259.8 square inches area in the hexagon. When a sphere is to be made of copper or sheet iron, CONVEX SURFACE ..I Fig.24. — Convex Surface of Cylinder. Fig. 25. — Convex Surface of Frustum of Cylinder. and it is required to know the amount of material it will take, the area is obtained by squaring the diameter of the sphere and multiplying by 3.1416. The sphere shown in Fig. 23 is 12 inches in diameter. To find its area, we have 12 X 12 = 144; 144 X 3.1416 = 452.39 square inches area. In Fig. 24, A is the plan of a cylinder 10 inches in diameter and B the elevation, 12 inches high. It is de- sired to find the convex surface of this cylinder. In this problem, as well as in others which will follow, the areas of the ends will not be considered, as this was explained in previous problems. The rule to be employed in obtain- Mensuration for Sheet Metal Workers. 19 ing the area of the convex surface of any cylinder is to multiply the circumference by the hight. As the cylinder is 10 inches in diameter and the hight 12 inches, then 10 X 3-1416 := 31.416, or circumference; 31.416 X 12 = 376.992 square inches in convex surface. In Fig. 25, A and B show the plan and elevation of a frustum of a cylinder. To obtain this convex surface, multiply one-half the sum of the greatest and least bights by the circumference. The greatest hight is 18, the least 28 10; 10 + 18 = 28; = 14; 14 X 3-1416 X 12 = 527.78 square inches area. CONVEX SURFACE 1006,31 SQ.IN, CONVEX > -r SURFACE ^ 703. ''.< B' SQ.IN ^\ r T'T J. J- 1 Fig. 26. — Convex Surface of Elliptical Cylinder and Frustum of Elliptical Cylinder. Another problem often arising in the shop is to find the convex surface of an elliptical tank. The same rule is employed as in a cylinder — that is, multiply the cir- cumference by the hight. In Fig. 26, A and B and A^ and B^ show, respectively, the plans and elevations of an elliptical cylinder and the frustum of an elliptical cylin- der, whose long diameter is 20 inches and short diameter 12 inches. Following the rule given in Fig. 8, Part I, for obtaining the circumference, we have 20 -h 12 32 = 16; 16 X 3-1416 = 50-2656; 50-2656 X 20 = 1005.31 square inches in the convex surface B. For the frustum 20 Mensuration for Sheet Metal Workers. B^ we have ^^ = 14; 14 X 50.2656 (the cir + 20 28 2 cumference) = 703.71 square inches. The same rules apphcable to the cyUnder are also ap- plicable to prisms whose bases are regular polygons, whether they be right prisms or frustums. In Fig. 2y, A and B represent, respectively, the plan and elevation of a prism 18 inches high, each side of the polygon being T' i •1 T c.s. 643 ig.iN. _l. _ _ MH Fig. 27. — Convex Surface of Right Prism and of Frustum of Right Prism. 1 c.s. \ • / 339.29 // SQ.fN. \ 1 1 Y L ,2— - \A^ J "'ig. 28. — Convex Surface of Right Cone. 6 inches. The perimeter of the hexagon is 6 X 6 = 36 inches ; 36 X 18 = 648 square inches in the convex sur- face B. For the convex surface of the frustum, we have 10 -f 18 _ 28 2 2 14; 14 X Z^ (the perimeter) = 504 square inches. When the convex surface of any right cone or pyra- mid is desired, then multiply the circumference, or pe- riphery, of base by half the slant bight. In Fig. 28 A and B show the plan and elevation of a right cone, which will serve as an example. The diameter of the cone at its base is 12 inches; its circumference is therefore 12 Mensuration for Sheet Metal JJ^orkers. 21 X 3-i4i6 = 37-6992; half of the slant hight is 9; then 9 X 37-6992 = 339-29 square inches in the convex sur- face of the cone B. Using the same rule for Fig. 29, whose base or plan A is a hexagon with 7-inch sides, we have 7 X 6 = 42 ; 9 X 42 = 378 square inches in the convex surface of the pyramid B. Suppose the elevations in Figs. 28 and 29 were cut off parallel to the base, forming the frustums of a cone Fig. 29. — Convex Surface of Right Pyramid. Fig. 30. — Convex Sur- Fig. 31. — Convex Sur- face of Frustum of face of Frustum of Righit Cone. Right Pyramid. and pyramid. The rule for obtaining the area of these convex surfaces is to multiply the circumferences or pe- ripheries of the two ends by half the slant hight. An example of this problem is given in Fig. 30, in which the plan of the base A is 10 inches and the diameter at the top 5 inches. Then 5 X 3-i4i6 == 15.708, or circum- ference at top, and 10 X 3-1416 = 31 -416, or circum- ference at base; 15.708 + 31.416 = 47.124; 47.124 X 5 (half the slant hight) = 235.62 square inches of convex surface. To prove this problem, we will assume that we have a right cone in Fig. 30, whose base is 10 inches and 22 Mensuration for Sheet Metal JVorkers. slant hight 20 inches. Then, following the rule given in connection with Fig. 28, we have 10 X 3-i4i6 X 10 = 314.16 square inches convex surface in whole cone. The area of the upper half of the cone shown by dotted lines, whose base is 5 inches, is 5 X 3-i4i6 X 5 = 78-54 square inches; 314.16 — 78.54 = 235.62 square inches, the area of the frustum of the cone, proving the problem. In Fig. 31, A is a regular polygon, with 6-inch sides, and the plan of the base of the frustum of a pyramid; Fig. 32. — Contents of Cube. J i \ CONTENTS = 15,588 CU.IN.. OR 67,48 GALLONS i >-■■ jL- \ w —10"^ Fig. 33.— Contents of Hexagonal Prism B the plan of the top, whose sides are 3 inches ; C, the elevation, has a slant hight of 8 inches. Then 6X6 = 36, or perimeter of base; 3 X 6 = 18, or perimeter of top; 36 -|- 18 = 54; one-half the slant hight is 4; 4 X 54 =: 216 square inches in C. Finding Capacities of Tanks, Etc. Some sheet metal workers would be at a loss to pro- ceed if a customer came in the shop and required a tank constructed of No. 20 galvanized sheet iron to hold 63^ gallons, the tank to fit in a space 51 inches high. Know- ing the number of gallons and the hight, it would be necessary to know the diameter before the tank could be Mensuration for Sheet Metal Workers. 23 laid out. While rules given in various publications are understood by those versed in mensuration, the less skilled do not know how to apply them practically. In computing the capacity of any vessel we deal with cubic and liquid measures, and therefore it may not be out of place to present the tables of cubic and liquid measures : Cubic or Solid Measure. 1728 cubic inches = 1 cubic foot, or 12 x 12 x 12. 27 cubic feet — 1 cubic yard, or 3 x 3 x 3. 231 cubic inclies = 1 United States gallon. 57.75 cubic inches = 1 United States quart. 28.875 cubic inches = 1 United States pint. 7.21875 cubic inches = 1 United States gill. Liquid Measure. 4 gills = 1 pint. 2 pints = 1 quart. 4 quarts = 1 gallon. 31% gallons = 1 barrel. 63 gallons, or 2 barrels = 1 hogshead. 1 gallon ■— 4 quarts — 8 pints = 32 gills. In the problems that follow practical exam.ples have been given as they are apt to arise in the shop, and the student should have no difficulty in learning to figure the capacity of any vessel which may arise in practice. One of the most simple forms to be computed is that of a cube or square tank, shown in Fig. 32. Here we have a tank 8x8 feet square and 8 feet high. The rule to be em- ployed in finding the solidity, whether the base is a square or rectangle, is to multiply the length of any one side by its adjoining side and multiply the product obtained by the hight. Then we have 8 X 8 r= 64; 64 X 8 = 512 cubic feet. As a cubic foot contains 1728 cubic inches, we have 1728 X 512, or 884,736 cubic inches. To find the number of gallons the tank will hold divide 884,736 by 231, the number of cubic inches in a gallon, and we get 884,736 -^ 231 = 3830 gallons, and 6 cubic inches over. 24 Mensuration for Sheet Metal JVorkers. Suppose a tank is to be constructed whose base is any regular polygon, as shown in Fig. 33, where a tank is shown 5 feet high whose base is a hexagon, each side measuring 10 inches. The rule to be used is to multiply the area of the base by the hight. Referring to Fig. 22, ;we find that the area of a hexagon whose side is 10 inches is 259.8 square inches, which multiplied by the hight in Fig. 33 will give the cubic contents. The hight shown is 5 feet. As the area of the base is in inches, then reduce the hight to inches; then 60 X 259.8 = 15,- -1_ Fig. 34.— Contents of Cylinder. CONTENTS = U,137.A CU..(N. OR 61.2 GALLONS Fig. 35. — Contents of Spliere. 588 cubic inches. For the number of gallons, 15,588 ~ 231 =: 67^ gallons, scant. In Fig. 34 is shown a cyclinder or round tank, the con- tents of which is obtained by the same rule as in the pre- ceding figure. The bottom of this round tank is 10 inches in diameter and its hight 303^ inches. Then 10- X 0.7854 equals the area of the base; or 10 X 10= 100; 100 X 0.7854 = 78.54 square inches ; 30.5 X 78-54 — 239547 cubic inches. For the number of gallons, divide the above product by 231 and the quotient will be 10.37 gallons. When a copper ball is made to use as a float in a large tank, it is sometimes desirable to know the number of Mensuration for Sheet Metal Workers. 25 cubic inches in same. Thus in Fig. 35 is shown a sphere 30 inches in diameter, whose capacity is formed by muhi- plying the cube of the diameter by 0.5236, or 30'"' = 2/,- 000; 27,000X0.5236=14,137.2 cubic inches; 14. 137.2 -^231 =61.2 gallons capacity. When the contents are required of a cone or pyramid, the rule to follow is to multiply the area of the base by one-third the perpendicular hight. In Fig. 36 is shown 1 CONTENTS = 5,026.66 CO. IN. 21.76 GALLONS Fig. 36. — Contents of Cone. CONTENTS = 2,598 CU.IN. Il-Z't GALLONS Fig. 37. — Contents of Pyramid. cone, whose base is 20 inches and whose vertical hight is 48 inches. Following the above rule, we have 20 X 20 = 400; 400 X 0-7854 = 314.16 square inches, or area, mul- tiplying which by one-third the vertical hight, 48, or 16, we have 314.16 X 16 = 5026.56 cubic inches. Dividing this by 231, the number of cubic inches in a gallon, we get 21.76 gallons. The method of computing the solidity of a pyramid is shown in Fig. ^7, which shows a pyramid whose base is a hexagon, each side being 10 inches, and whose vertical hight is 30 inches. As the area of a hexagon each side of which is 10 inches is 259.8 square inches, then multiply 26 Mensuration for Sheet Metal Workers. this amount by one-third the vertical hight (30), or 259.S X 10 ^ 2598 cubic inches, which divide by 231 and we get 11.24 gallons. The prismoidal formula can be used in calculating the volume or capacity of a prism, cylinder, cone, pyramid, frustum of a cone or pyramid, wedge, as well as many irregular shaped bodies. A prismoid is by definition a solid whose bases are polygons and lie in parallel plans and whose faces are quadrilaterals or tri-angles. To find Fig. 38. — Contents of Hopper. Fig. 30. — Contents of Frustum of Pyramid. the contents of a prismoid, or any of the above mentioned solids, add together the areas of the two parallel planes and four times the area of a section taken midw^ay be- tween and parallel to them, and multiply the sum by one- sixth of the perpendicular distance between the parallel planes. Applying this rule for obtaining the volume of a wedge, pyramid or cone, the area of the upper base is o, because it runs to an apex. For prisms or cylinders the areas of the upper, lower and middle planes are equal. The prismoidal formula when applied to the frustum of a pyramid saves the labor of extracting the square root, as required under the old rule. Mensuration for Sheet Metal Workers. 27 As an example let us apply the above formula in ob- taining the solidity of the cone shown in Fig. 36. Follow- ing the above rule, we have as the area of the lower base 314.16 and the area of the upper base o. Four times the area of the middle section, which is 10 inches in diameter, is 10 X 10 X 0.7854 = 78.54 X 4 = 314-16. The sum is 314.16 + 314.16 = 628.32. One-sixth the perpendicular bight is 48 = 8; 8 X 628.32 = 5026.56 cubic inches. In Figs. 38 and 39 is shown the method of computing the volumes of the frustums of a square and hexagonal pyramid. Fig. 38 shows an inverted hopper, such as is usually made of heavy galvanized iron. Whether the shape is regular or irregular, the same method is used in determining the capacity. The top opening is 28 x 28 inches, the bottom 10 x 10 inches, and the sides, a section taken midway between the top and bottom, would equal ■ — ^ or 19 ; then 19 x 19 inches would be the size for the middle section, abed. The area of the upper plane equals 28 x 28 inches, or 784 square inches ; of the lower plane 10 x 10 inches, or 100 square inches, and of the middle plane 19 x 19 inches, or 361 square inches. Four times the middle plane is 1444. Then, following the rule, we have upper plane, 784, -{- lower plane, 100, -f four times middle plane, 1444, = 2328. This is multiplied by 39 one sixth the vertical bight, 39. Then 2328 X-^= I5'i32 cubic inches. This divided by 231, the number of cubic inches in a gallon, gives 65 >4 gallons and i>^ cubic inches over. The same rule is applied to Figs. 39 to 42, inclusive. Fig. 39 shows a hopper hexagonal in shape, each side of 28 Mensuration for Sheet Metal Workers. the lower base of which is 20 inches, the upper base 10 inches, and a section taken midway between the two — or 15 inches, as a b. To obtain the area of these planes multiply the square of one of the sides of the regular poly- 1,1,064,022.2208 ou. in. 4,006 gallons and 5 gills Fig. 40. — Contents of Frustum of Cone. gon by the multiplier giyen in the following table for the proper polygon : Name. Sides to polygon. Multiplier. Triangle 3 0.433 Square 4 1.000 I'entagon 5 1.720 Hexagon 6 2..598 Heptagon 7 3.634 Octagon 8 4.828 Nonagon 9 6.182 Decagon 10 7.694 As the shape in question is a hexagon, we find the multiplier in the table to be 2.598. Then, following the above rule, the area of the lower plane is equal to 20 X 20 ==400; 400X2.598^1039.2. The area of the upper plane, 10 X 10 X 2.598 = 259.8, and four times the area of the middle plane, 15 X 15 X 2.598 X 4 = 2338.2. 24 Theii 1039.2 + 259.8 -f 2338.2 = 3637.2. 3637.2 X ^ = 14,548.8 cubic inches contents. This divided by 231 = 63 gallons, or 2 barrels, less 4.2 cubic inches. Mensuration for Sheet Metal Workers. 29 Fig. 40 shows the frustum of a cone. A tank of this form is usually made from ^^^-i^ch metal, riveted and re- inforced with angles and tees. The top diameter is 6 feet, bottom 10 feet and the middle diameter- 10 , or 8 feet. The area of the top section is 6 X 6 X 0.7854 = 28.2744 square feet ; the bottom area, 10 X 10 X 0.7854 = 78.54 square feet ; the middle area, 8 X 8 X 0.7854 = 50.2656. Then 28.2744 + 78.54 + (50.2656 X 4) = 307-8768 X 12 v-= 615.7536 cubic feet capacity. Multiply this amount by 1728, the number of cubic inches in a foot, and divide ^ Contents 5640 cu. In. Fig. 41. — Contents of Prismoid. k-14^ Contents 0,76S cu. In. 29 gallons .-* \ — ^X +69 ou. In Fig. 42. — Contents of Wedge. by 231, the number of cubic inches in a gallon, and we obtain the capacity in gallons. Thus, 615.7536 X 1728 = 1,064,022.2208 cubic inches, which divided by 231 = 4606 gallons and 5 gills. Fig. 41 shows an odd shaped vessel, whose parallel bases are right angle triangles. The bottom base is 20 x 32 and contains 320 square inches ; the top 8 x 20, and contains 80 square inches. The size of the middle section is obtained by adding 20 and 32 and dividing by 2, which gives the length b c, or 26 inches. In similar manner --^ == 14, length of a b. Then the area oi a b c is. 30 Mensuration for Sheet Metal Workers. ^^ ^^ = t82. Then 80 + ( 182 X 4) + 320 = 1 128. 30 ii28X-^ = 5640 cubic inches. Divide this amount by 231 and we get 24 gallons and 96 cubic inches over. In Fig. 42 is given another example in obtaining the capacity of odd shape solids, this example being in the form of a wedge. The rectangular base is 12 x 40 inches and contains in arear 480 square inches. The top runs to an apex 14 inches wide, whose area is o. A section taken midway between the top and bottom equals 6 X 27 inches and has an area of 162 square inches. Then o -|- (162 X 06 4) + 480 =1128. 1 1 28 X -^= 6768 cubic inches, 6 which, divided by 231=29 gallons, i quart and 11.25 cubic inches over. Practical Examples for the Shop. The section of an 8 x 32 inch rectangular pipe is rep- resented in Fig. 43 of the diagrams. If a transition is made to a perfectly square pipe, what must each side measure? In solving this problem proceed as follows: Extract the square root of the area of the rectangular pipe, 256 square inches, which is 16 inches, the size of the square pipe. Supposing this 8 x 32 inch pipe was to form a transi- tion to another rectangular pipe, the width of which was 12 inches, what must be its length to have the same area? Simply divide 256 by 12, and the quotient will be 21 1-3, making the size of the pipe 12 x 21 1-3 inches. If this 8 X 32 inch pipe were to form a transition to an oblong pipe with semicircular ends, 8 inches in diam- Mensuration for Sheet Metal Workers. 31 eter, as shown at A, what must the length of the distance be, shown from a to h? As two semicircles make a full circle, then deduct the area of the 8-inch circle from 256, and divide the remainder by eight, as follows : Area of 8- inch circle = 50.26; 256 — 50.26 = 205.74; 205.74-^ 8 = 25.72, or 25% inches scant, the length from a to b. In Fig. 44 is shown a fitting in hot air piping known as a boot. With it a 14-inch round horizontal pipe and a vertical rectangular pipe whose width is 7 inches are y' 41!" — AREA 256 SQ. IN. \ 16 X 16 SQUARE V =12"x 21j' RECTANGULAR ' s'x+lT ROUND ENDS, SEE A T Fig. 43. — Pipes Equal to Rectangular Pipe. Pig. 44.— Boot. joined. What must the length of this rectangular pipe be so that it will have the same area as the 14-inch round? The area of a 14-inch round pipe is 153.93, or approxi- mately 154 square inches. Divide 154 by 7, and the quo- tient will be 22 inches, the desired length. Suppose the size of the rectangular pipe is given, say 7 X 22, and it is desired to know what size round pipe will have similar area? Divide the area of the given pipe, or 154, by 0.7854, and extract the square root of the product. Thus 154 -^ 0.7854 =: 196. Vi96rri4. Then 14 inches is the diameter of the round pipe. In Fig. 45 is shown a chimney cap, which measures 8 X 6^ inches at the bottom. The cap is to form a transi- tion from square to round. What must be the diameter at the top so that the area will be similar to the base? 32 Mcnsnraiion for Sited Metal Workers. The same rule could be used as given in connection with Fig. 44, but by using a table of areas and circumferences of circles, much labor in computing can be saved. These tables are found in some text books, in engin- eers' pocket books and in the " Tinsmith's Helper and Pat- tern Book," published by the David Williams Company. In the following examples the use of these tables will be explained when applied to sheet metal work. Refer- ring to Fig. 46, the base measures 8 x 6^ inches and equals 50 square inches area. Instead of computing, to k- 8--J 120.27 SQ. IN Fig. 45. — Chimney Top. Fig. 4G. — Boot. obtain the diameter of the top of the cap, simply refer to the table of areas and circumferences of circles, following the column under areas until 50.2655 is reached (the near- est to the required number 50), when we find it is the area of an 8-inch circle, which will be the top diameter of the cap. This method could have been used in Fig. 44, in which the area of the rectangular opening is 154 square inches. Follow the area column in the table until 153.9380 is reached, which suggests a 14-inch circle, as shown. In Fig. 46 is shown another form of boot. In this case the inlet A is 12^ inches in diameter and equals Mensuration for Sheet Metal Workers. 33 120.27 square inches. If the outlet B is to be oblong in shape with semicircular ends 6 inches in diameter, what must the length be of the straight side a h? Find the area of a 6-inch circle, which is 28.27 square inches. De- duct this from 120.27, which leaves 92. Divide this amount by six, the given width of the outlet, and the quo- tient will be 15 2-T) inches, the desired length of a b. Both sections then equal 120.27 square inches. In Figs. 47 and 48 is shown how to compute the amount that a duct must be increased in size in propor- tion to the number and size of inlets connected. Fig. 47 AREA 48 SQ. IN. i' I DEEP Fig. 47. — ripe with Branches. AREA 50.27 SQ. IN. Fig. 48.— Pipe with Branches. shows a rectangular duct, whose depth is 6 inches throughout. Ducts of this kind are usually employed in ventilating work, and are increased in size every time a register is connected. A shows the first inlet, 6 x 15 inches, having 90 square inches area. The second inlet, B, is 6 X 8 inches, and has 48 square inches area. The combined area of A and B is 138 square inches. Now what size must the duct be beyond B so as to have the area of the two inlets A and B? Divide 138 by 6 and the cjuotient will be 23, the width of the duct. This same rule is followed no matter how many inlets are taken up. In Fig. 48 is shown the method employed when the pipe is round. The first inlet is 14 inches and the second 8. They contain, respectively. 153.93 and 50.27 square inches area. The combined area of the two inlets is 204.20 34 Mensuration for Sheet Metal Workers. square inches. The circle that has this area is found to be 16% inches in diameter. This is then the diameter of the large pipe. A triangular ventilating pipe is shown in Fig. 49. The size of the pipe is 16 x 32, containing one-half of the prod- uct of its dimensions, or 256 square inches area. What must the size of the various pipes be if a transition is desired to square, to rectangular 8 inches wide, and round ? To obtain the size of the square pipe, simply ex- -16- — J OF EQUAL AREA 16"x 16" SQ. 8"X 32" RECTANGLE ISnr ROUND Fig. 49. — Triangular Pipe. 10 ROUND 8. 862* SQUARE e'x 13.09° RECT. * I Fig. 50. — Elliptical Pipe. tract the square root of 256. Then 16 x 16 is the size of the square pipe. The length of the section of the rectang- ular pipe, whose width is desired to be 8 inches, is ob- tained by dividing 256 by 8. The quotient will be 32, or the required length. For the size of the round pipe, which should have an area equal to the triangular, follow the column of areas in the table of circle areas until 256.2398 is reached, which will be found to equal the area of a cir- cle 18 1-16 inches in diameter and is the desired round pipe. In Fig. 50 is shown the section of an elliptical pipe measuring 8 x I2>^ inches. The area of this ellipse is 12.5 X 8 X 0.7854 = 78.54 square inches. Suppose a transi- tion is to be made to round, square or rectangular pipe, 6 Mensuration for Sheet Metal Workers. 35 inches wide, whose areas must be similar to the ellipse, what must their sizes be? Following the column of areas in the table, we find the area 78.54 is for a lo-inch round pipe. For the size of the side of a square pipe we have V78.54 = 8.862 -)- inches, or about 8% inches. Thus the square pipe would be 8% x 8% inches. For the length of the rectangular pipe, whose given width is 6 inches, divide 78.54 b}^ 6, and the quotient will be 13.09. Then 6 X 13.09 inches will be the size of the rectangular pipe. In putting up ventilating, blower and blast pipes, it is AREA 194 SQ. IN. Fig. 51. — Two-Pronged Fork. A-B-C EACH 23r DIAM. COMBINED AREA 1273.6 1*. — 4or — >j AREA 1272.4 I Fig. 52. — Three-Pronged Fork. often the case that a number of branches are connected to one main, and the main pipe must have the combined area of the branches. An example of this is shown in Fig. 51, where two round branches are connected to a rectangular main pipe, whose width must be 121^ inches. What must the length of the section of this main pipe be to have a combined area of the two branches? The area of the lO-inch pipe is 78.54 square inches. The area of the i2i/^-inch pipe is 115.46 square inches. The sum of these two areas is 194 square inches ; 194 ^ 12.125, the width of the main pipe, = 16. Therefore the size of the main pipe is I2>^ x 16 Fig. 52 shows three 36 Mensuration for SJicct Metal JVorkers. branches of round pipe, connecting in fork shape to a round main. Each of the branches, A, B and C, is 23^4 inches diameter, the combined area of which equals 1273.6 square inches. What must the size of the main be? Fol- lowing the table of areas, we find the nearest number to be 1272.4, which is the area of a pipe 4034 inches diame- ter. By using the tables of areas and circumference much time and labor are saved in computing. If these tables Fig. .53.— A Square Tank. Fig. 54.— A Sphere. Fig. .5.5.— An Oil Tank are not available, the calculations must be made by meth- ods described in the earlier installments of this series. Ascertaining Sizes of Articles. What follows is devoted to explaining how to obtain the unknown size of an article when the bight and capacity are given, or z'ice versa. The .first problem, shown in Fig. 53, represents a water tank. Assume that a customer has ordered an 8-gallon tank, whose base is to measure 11 x 14 inches. How high must it be to have the desired capacity ? The rule to follow in any square or rectangular tank is : Reduce the gallons to cubic inches; divide this amount by the area of the base, the quotient being the desired bight. As there are 231 cubic inches in the United States gallon, Mensuration for Sheet Metal Workers. 37 then in 8 gallons there will be 8 X 231 or 1848 cubic inches. The base is 11 x 14 and contains 154 square inch- es area. Then 1848-^-154=12 inches, the required hight of the tank. Suppose the hight and length of one of the sides and the capacity are given. What will the size of the remain- ing side be? Assuming the capacity to be 8 gallons, the hight 12 inches and the given side 11 inches, then divid- ing the number of cubic inches in 8 gallons by 12 and the quotient by 1 1 the result will be the required side. Thus, .:i_ Pig. 56. — Wash Boiler. Fig. 57. — Flaring Pail. Fig. 5S. — Flaring Measure. 1848 -^ 12 = 154; 154 ^ II = 14, the required length of side. A copper ball Fig. 54, to be used as a float must con- tain 22,449 35-100 cubic inches. What must the diameter be to contain the above number of cubic inches ? The rule to follow is to divide the number of cubic inches by 0.5236, and from the quotient thus obtained extract the cube root. As the sphere is to contain 22,449.35 cubic inches, then 22,449.35 ^ 0.5236 = 42,875. The next step is to extract the cube root of this quotient. When a number is multiplied by itself, as 5X5, the product, 25, is called the square of that number. When 38 Mensuration for Sheet Metal Workers. a number is multiplied by itself twice, as 5 X 5 X 5, the product, 125, is called the cube of that number. There- fore the extraction of the cube root is nothing more than the finding of that number which, when multiplied by itself twice will result in the given number. To extract the cube root of 42,875, start at the decimal point and, counting to the left, separate the number into periods of three figures, as shown by 42'875. Cube Number, root. 42'875.(35 27 Trial divisor. 2700 15875 42875 Find the greatest number whose cube is contained in the first or left hand period, 42 ; 4 X 4 X 4 = 64, and is too great. Then take 3 ; 3 X 3 X 3 = 27. Therefore 3 is the first figure of the root. Subtracting 2y from 42 we obtain 15. Bring down the next period, 875, obtaining the first partial dividend, 15,875. Take three times the square of the root already found, which is 3- X 3 ^= 3 X 3X3 =^2y. Annex two ciphers to it, and we have 2700 for the trial divisor. Divide the trial divisor into 15,875, which suggests 5 as the second figure of the root. Prove this by multiplying 35 X 35 X 35, which equals 42,875, as shown, and leaves no remainder. Then 35 is the cube root of 42,875. Therefore the sphere in question must be 35 inches in diameter. The trial divisor is sometimes contained in the partial dividend, a higher number than required. Whether it is too high or not can be ascertained by cubing the root found, and if its product is higher than the partial divi- dend a lower number must be taken, w'hose cube will be equal to or smaller than the partial dividend. If there Mensuration for Sheet Metal JJ'orkers. 39 had been a remainder in the problem just shown, and it was desired to continue the root, periods of three ciphers each would have to be added to the whole number, 42,875, and continued as above described, so as to obtain the deci- mal part of the root. Tn Fig. 55 is shown an oil tank. Assume a tank whose diameter is 16 inches that must hold one barrel : Then what must be the hight of the tank ? Reduce the barrel to cubic inches, into which divide the area of the 16-inch circle. The quotient will be the required hight. One barrel equals 31^ gallons, or 7276.5 cubic inches. The area of a 16-inch circle is 201.062 square inches; 7276.5 -— 201.062 =: 36.19 -[-. Therefore 36 1-5 inches is the de- sired hight. Suppose the tank is to hold the same quantity and the hight is to be 403/2 inches. What must the diameter be? In this case multiply the number of cubic inches by the hight and divide the quotient by 0.7854. From the quo- tient thus obtained extract the square root, which will be the desired diameter. 7276.5 cubic inches, the capacity of the tank, -f- 40.5 inches, the hight, = 179.6667; 179.6667 jo.7854 = 228.758 -f ; V^8.758= 15.124, or 15^3 inches, the desired diameter, as shown at B. In Fig. 56 is shown a lo-gallon wash boiler, whose base is 10 inches wide, with semicircular ends, the length of the straight part of the side from A to B being 954 inches. What must be its hight? Reduce the capacity to cubic inches and divide it by the area of the bottom,. the quotient giving the required hight. Ten gallons equal 2310 cubic inches. The area of the bottom equals the area of a lo-inch circle, which is 78.54, plus 92.5, the area of the rectangle 9^ X 10, = 171.04 square inches. 2310 -r- 171.04 =. 13.5 inches, the hight of the boiler. 40 Mensuration for Sheet Metal Workers. If for any reason the hight of the boiler and the diam- eter of the semicircular ends are given, and it is desired to know the width of the straight side A B, then find the capacity in cubic inches and divide by the hight ; from the quotient obtained subtract the area of the two semicir- cles and divide the remainder by the given width of the base, and the quotient will be the desired width. Thus: 2310 ^ 13.5 = 171.11; 171-11 — 78-54 = 92.57; 92.57 4- 10 = 9.26, the desired width of A B. In Figs. 57 and 58 are shown a flaring pail and a meas- sure whose capacities and top and bottom diameters are given, and it is required to find their hight. The rule ap- plicable to any form of flaring ware whose section is round, no matter what its capacity may be, is to find the number of cubic inches in the given capacity, which divide by the sum of the areas of the top and bottom diameters and four times the area of the middle section and multiply the quotient by six. As one quart contains 57.75 cubic inches, a lo-quart pail, shown in Fig. 57, will contain 577.5 cubic inches. The area of the top diameter equals 78.54, the bottom diameter 50.26, and the middle section, whose diameter is 9 inches, 63.61 square inches; 63.61 X 4 = 254.44 square inches. Then 254.44 + 50.26 + 78.54 = 383.24 ; 577.50 -:- 383.24 = 1.506 X 6 = 9.036 inches, the required hight. The measure shown in Fig. 58 is to hold one quart. Its top and bottom diameters are 2}^ and 4^4 inches, re- spectively. What must the hight be ? Following the same rule as above we have : Area of top equals 4.90. Area of bottom equals 14.18. The middle diameter equals £- 5 + 4.25 _ 2^2,7S' Its area equals 8.94. 8.94 X 4 = 35.76. Mensuration for Sheet Metal Workers. 41 Capacity of one quart equals 57.75 cubic inches ; di- vided by combined areas, or 54.84, leaves a quotient 1.053, which multiplied by 6 equals 6.318 inches, or hight. In Fig. 59 is shown the method of finding the hight in elliptical flaring ware when the top and bottom dimen- sions and capacity are given. The tub in this case is to hold 2i^ pints and 21^ cubic inches, the top dimensions to be II X 15)^ and the bottom 8 x I2>^ inches. What must the hight be? The rule to follow is the same as in the preceding problem. It should be remembered that Vig. 59. — Flaring Elliptical Tub. Fig. 60.— Flaring Pan. the area of an ellipse is found by multiplying the short and long diameters together and their product by 0.7854. Working this out, it will be found that 9 inches is the desired hight of the tub. Another problem where the same rule is employed is shown in Fig. 60, in which a drip pan to hold 29 quarts and 1314 cubic inches is illustrated. The top is to be 16 X 22 and the bottom 12 x 18 inches. The middle sec- ,• .• r , r „ 16 +12 18 4- 22 tion'is found as follows: ! r=: 14- ^ =20 2 ^2 or 14 X 20 inches. Following the rule as before we have 6, the number of inches in the hight of the pan. 42 Mensuration for Sheet Metal Workers. Short Rules in Computation. We now come to a point where a few short rules in computation may be of value to the sheet metal worker. In Fig. 6i let A B represent a wall with a rounded corner from a to b, on which a molding, gutter or cornice is to be placed, and it is desired to find the radius. To do this measure the distance from a to b, which is 5 feet. Bisect a b, obtaining d, and from d measure the distance, at Pig. <>1. — Finding Diametei' of Circle. right angles to a b, to the outside of the curve at e, which in this case measures i foot 3 inches. Divide the sum of the squares of one-half the chord and the rise by the rise, and one-half the cjuotient will be the desired radius. Thus, reducing to inches, we have, 30-4- IS" Qoo-(-225 1 125 . . , -^^^ — ^—^ =^ ' ^= =375^ mches 15X2 30 30 -^^^ or 3 feet i^ inches radius. In Fig. 62 is shov/n how the area of a given object can be obtained, even though it is so far away that meas- urements cannot be taken. This is obtained by propor- Mensuration for Sheet Metal Workers. 43 tion of triangles. Let us assume in this case that the spire shown at A is to be covered with metal or slate. If we obtain the contract, we will erect scaffolding to do the w^ork ; but it will not pay to erect scaffolding to take the measurements and measurements must be obtained to es- timate on the job. As we can get to the ridge of the roof, B C, the number of square feet in the four sides of the tower is obtained as follows : Measure a given dis- Fig. 62. — Finding Hight of Inaccessible Point. tance from the bottom of the roof of the spire, b — in this case 60 feet— to a. At a place a rod so high as to make the imaginary line b a horizontal. Take another rod and place it vertically at a distance of about 5 feet from a, as shown at C. Then, with the eye at a, cast the line of sight from a to the top of the spire, at d, and mark the second rod where the line of sight crosses it at C, and measure the distance from C on a c? to