ROOFING
55. Kinds of Roof Covering.—The roof coverings most generally used are shingles, slate, tin, tile, and tarred paper and gravel (known as gravel roofing). While there are slight variations in the methods of measuring the different kinds, they are all based on the square of 100 square feet.
56. Shingles.—In measuring shingle roofing, it is necessary to know the exposed length of a shingle. This is found by deducting 3 inches (the usual cover over the head of the lowest shingle in the four overlapping courses) from the length and dividing the remainder by 3. Thus, in [Fig. 5], the distance b that one shingle is overlapped by the third above it is usually made equal to 3 inches, and the remaining length of the lowest shingle may be divided into three equal portions, each equal to a. The lowest of these three portions is the part exposed to the weather. Multiplying the length exposed to the weather by the average width of a shingle will give the exposed area. Dividing 14,400, the number of square inches in a square, by the exposed area of 1 shingle, in square inches, will give the number of shingles required to cover 100 square feet of roof. For example, it is required to compute the number of shingles 18 in. × 4 in. needed to cover 100 square feet of roof. With a shingle of this length, the exposure will be
| 18 - 3 | = 5 inches; |
| 3 |
then, the exposed area of 1 shingle is 4 in. × 5 in., or 20 square inches, and 1 square requires
14,400 ÷ 20 = 720 shingles.
Fig. 5
An allowance should always be made for waste in estimating the number of shingles required.
[Table X] is arranged for shingles from 15 to 27 inches in length, 4 and 6 inches in width, and for various lengths of exposure.
57. Shingles are classed as shaved, or breasted, and sawed shingles.
Shaved shingles have fallen almost into disuse, owing to the difficulty of manufacturing them. These shingles vary from 18 to 30 inches in length, and are about ½ inch thick at the butt and ¹/₁₆ inch at the top.
Sawed shingles are usually from 14 to 18 inches long and of various thicknesses. In the case of 18-inch shingles, five shingles, at their butts, will make 2¼ inches; that is, the thickness of one shingle at the butt is 2¼ ÷ 5 = .45, or about ⁷/₁₆ inch. At the top, each shingle is ¹/₁₆ inch thick. With 16-inch shingles, however, five of them make only 2 inches. Therefore, the thickness of a 16-inch shingle at the butt is 2 ÷ 5 = .4, or about ⅜, inch.
TABLE X
DATA FOR ESTIMATING SHINGLES
| Exposure to Weather Inches | Number of Square Feet of Roof Covered by 1,000 Shingles | Number of Shingles Required for 100 Square Feet of Roof | ||
|---|---|---|---|---|
| 4 Inches Wide | 6 Inches Wide | 4 Inches Wide | 6 Inches Wide | |
| 4 | 111 | 167 | 900 | 600 |
| 5 | 139 | 208 | 720 | 480 |
| 6 | 167 | 250 | 600 | 400 |
| 7 | 194 | 291 | 514 | 343 |
| 8 | 222 | 333 | 450 | 300 |
White-pine and white-cedar shingles are graded alike. The shingles made of No. 1, or clear, stock are designated XXXX. Those made of No. 2 stock, with 6-inch clear butt, are given the brand XX, while those made of mill cull, with sound butt, are called X.
Red-cedar shingles are graded differently. Their grade depends on their length. Thus, 18-inch, No. 1 shingles are termed “Perfection,” while 18-inch, thin butt are termed “Eureka.” Red-cedar shingles 16 inches long, if made of No. 1, or clear, stock, are designated “Extra * A *.” If they are 16-inch, thin butt, they are termed simply “* A *.”
Sawed shingles are made up into bundles of 250, and are sold on a basis of 4 inches width for each shingle. Shingles cost from $4 to $6.75 per thousand, according to material and grade. Dimension shingles—those cut to a uniform width—if of prime cedar, shaved, ½ inch thick at the butt and ¹/₁₆ inch at the top, will cost about $7.75 per thousand, but since such shingles are usually 6 inches wide, less will be required per square.
A fairly good workman will lay about 1,000 shingles per day of 8 hours, on straight, plain work; while in working around hips and valleys, the average will be about 700 per day.
58. Slating.—In measuring slating, the method of determining the number of slates required per square is similar to that given for shingling; but in slating, each course overlaps only two of the courses below, instead of three, as in shingling. The usual lap, or cover, of the lowest course of slate by the uppermost of the two overlapping courses, is 3 inches; hence, to find the exposed length, deduct the lap from the length of the slate, and divide the remainder by 2. The exposed area is the width of the slate multiplied by this exposed length, and the number of slates required per square is found by dividing 14,400 by the exposed area of 1 slate in square inches. Thus, if 14" × 20" slates are to be used, the exposed length will be
| 20 - 3 | = 8½ inches; |
| 2 |
the exposed area will be 14 × 8½ = 119 square inches; and the number per square will be 14,400 ÷ 119 = 121 slates.
The following points should be observed in measuring slating: Eaves, hips, valleys, and cuttings against walls are measured extra, 1 foot wide by their whole length, the extra charge being made for waste of material and the increased labor required in cutting and fitting. Openings less than 3 square feet are not deducted, and all cuttings around them are measured extra. Extra charges are also made for borders, figures, and any change in color of the work; and for steeples, towers, and perpendicular surfaces.
[Table XI], which is based on a lap of 3 inches, gives the sizes of the American slates and the number of pieces required per square. The cost of slating varies from 9 to 15 cents per square foot, depending on the class of work.
The thickness of stock slate varies from five to 1 inch to ⅜ inch and special thicknesses up to 1 inch are made to order. For ordinary dwellings, the usual thickness used is five to 1 inch, which gives a thickness of a little more than ³/₁₆ inch, and the size used for this class of work is generally 8 in. × 12 in. or 9 in. × 18 in., the price being the same.
TABLE XI
NUMBER OF SLATES PER SQUARE
| Size Inches | Number of Pieces | Size Inches | Number of Pieces | Size Inches | Number of Pieces |
|---|---|---|---|---|---|
| 6 × 12 | 533 | 9 × 16 | 246 | 14 × 20 | 121 |
| 7 × 12 | 457 | 10 × 16 | 221 | 11 × 22 | 138 |
| 8 × 12 | 400 | 9 × 18 | 213 | 12 × 22 | 126 |
| 9 × 12 | 355 | 10 × 18 | 192 | 13 × 22 | 116 |
| 7 × 14 | 374 | 11 × 18 | 174 | 14 × 22 | 108 |
| 8 × 14 | 327 | 12 × 18 | 160 | 12 × 24 | 114 |
| 9 × 14 | 291 | 10 × 20 | 169 | 13 × 24 | 105 |
| 10 × 14 | 261 | 11 × 20 | 154 | 14 × 24 | 98 |
| 8 × 16 | 277 | 12 × 20 | 141 | 16 × 24 | 86 |
In [Table XII] is given a list of the different colors of slate used in the Eastern and Middle States, the quarries from which they are obtained, and the cost of slate, labor, etc. per square, pertaining to each variety. The prices in the table are based on the 8 in. × 12 in. or 9 in. × 18 in. sizes, thickness five to 1 inch, and for quantities of not less than 50 squares. The cost of labor, etc. being based on current prices in the aforementioned territory.
Slate ¼ inch thick cost about 20 per cent. more than the five to 1 inch for the material, and about 5 per cent. more for laying and freight.
Slate ⅜ inch thick cost about 45 per cent. more than the five to 1 inch for the material, and about 15 per cent. more for laying and freight.
When copper nails are specified obtain current prices.
TABLE XII
APPROXIMATE COST OF SLATING,
PER SQUARE
| Classification | Cost of Slate F. O. B. Quarries | Cost of Laying, Including Roofing Felt, and Freight | Total Cost, Exclusive of Builder’s Profit |
|---|---|---|---|
| Black Slate | |||
| Brownville, Maine | $8.00 | $5.00 | $13.00 |
| Monson, Maine | 7.00 | 5.00 | 12.00 |
| Peach Bottom, Pennsylvania | 5.50 | 4.00 | 9.50 |
| Chapman (Hard Vein), Pennsylvania | 4.50 | 3.50 | 8.00 |
| Bangor, Pennsylvania | 4.50 | 3.50 | 8.00 |
| Lehigh, Pennsylvania | 4.00 | 3.50 | 7.50 |
| Buckingham, Virginia | 3.75 | 4.00 | 7.75 |
| Red Slate | |||
| Vermont | 12.00 | 4.25 | 16.25 |
| Green Slate | |||
| Vermont | 5.50 | 4.25 | 9.75 |
| Purple Slate | |||
| Vermont | 5.00 | 4.25 | 9.25 |
| Mottled Slate | |||
| Vermont (Purple and Green) | 4.00 | 4.25 | 8.25 |
59. Sheet-Metal Roofs.—In estimating sheet-metal roofs, the hips and valleys are measured extra their entire length by 1 foot in width, to compensate for increased labor and waste of material in cutting and laying. Gutters and conductor pipes, or leaders, are measured by the linear foot, 1 foot extra being added for each angle. All flashings and crestings are measured by the linear foot. No deductions are made for openings (chimneys, skylights, ventilators, or dormer-windows) if they are less than 50 square feet in area; if between 50 and 100 square feet, one-half the area is deducted; if over 100 square feet, the whole opening is deducted. An extra charge is made for labor and waste of material to flash around openings.
60. There are two regular sizes of roofing plates, namely, 20 in. × 28 in. and 14 in. × 20 in. The larger size is generally used on common work, owing to the fact that it requires fewer seams on the roof and consequently cheapens the cost of laying. A third size, namely, 10 in. × 20 in., is also supplied, and is used generally for gutters and leader pipes. Sheets 10 in. X 14 in. are sometimes used for laying roofs, as they can be cleated better than the larger sizes. Such small sheets, however, cost more to lay.
Two thicknesses of roofing plates are commonly recognized. One is the IC, or No. 29 gauge, and weighs 8 ounces to the square foot; the other is the IX, or No. 27 gauge, and weighs 10 ounces to the square foot. Sometimes, a still heavier plate is called for, and it is therefore kept in stock by the best manufacturers. This plate is known as IXX, or No. 26 gauge, and is used for especially heavy work.
Formerly, the standard net weight per box of IC, 14" × 20" roofing tin was 112 pounds, or 1 pound per sheet, making 112 sheets to the box; but now this weight is reduced to 108 pounds. The old standard for IX plates was 140 pounds, but very few brands now weigh more than 135 pounds per box. The most reliable manufacturers guarantee the weights for the different boxes of tin, and if the material does not come up to the guaranteed weight, it can be returned. The best sheets in the market today are stamped with the mark of the brand and the designation IC or IX of the thickness.
61. Using standing joints, a 14" × 20" sheet of roofing tin will cover about 235 square inches of surface, or one box of such tin will cover about 182 square feet. With a flat, lock seam, a sheet will cover 255 square inches, allowing ⅜ inch all around for joints; or a box will lay 198 square feet. These figures make no allowance for waste.
Two good workmen can put on from 250 to 300 square feet of tin roofing per day of 8 hours; this also includes painting the outside of the tin. Tin roofing will cost from 8 to 10 cents per square foot, depending on the quality of material and workmanship.
62. Tile Roofs.—Since tile roofs are constructed of so many styles of tile, no general rules of measurement can be given. Every piece of work must be estimated according to the particular kind of tile used and the number of sizes and patterns. Information on all these points is to be found in the catalogs of tile manufacturers.
TABLE XIII
APPROXIMATE COST OF ROOF TILING,
PER SQUARE
| Classification | Cost of Tile Delivered | Cost of Laying, Including Ashphalt Felt | Total Cost, Exclusive of Builder’s Profit |
|---|---|---|---|
| Shingle tile (rectangular), 6" × 12" | $13.00 | $ 7.50 | $20.50 |
| Shingle tile (rectangular), 8" × 12" | 14.00 | 6.50 | 20.50 |
| Shingle tile (geometric shapes) | 12.00 | 7.00 | 19.00 |
| Conosera (interlocking), 8" × 12" | 14.00 | 5.50 | 19.50 |
| Conosera (interlocking), 10" × 15" | 12.00 | 5.00 | 17.00 |
| Conosera, combination 8" × 12" and 2" × 12" | 16.50 | 8.00 | 24.50 |
| French A (interlocking), size 10" × 15" | 12.00 | 5.00 | 17.00 |
| Spanish, 8" × 12" | 13.00 | 6.50 | 19.50 |
| Old Spanish, semicircular, channels | 22.50 | 10.00 | 32.50 |
| laid alternately, concave and convex | |||
| Roman, pan and semicircular roll, | 17.00 | 8.00 | 25.00 |
| laid 7½ in. center to center of rolls | |||
| Greek, pan and semihexagonal cap, | 17.00 | 8.00 | 25.00 |
| laid 7½ in. center to center of caps | |||
| Promenade for flat roofs, laid on 5 layers | 7.00 | 13.00 | 20.00 |
| of asphalt felt in asphalt pitch |
In [Table XIII] is given a list of the prevailing styles of roof tiling, the cost of tiling, labor, etc. per square, pertaining to each variety. The prices in the table are based on the natural red color of the clay when burnt; extra prices are asked for glazed-surface finish which can be obtained in different colors. The prices in the table are based on quantities of not less than 30 squares, as less than a minimum carload means increased freight rates. The prices given cover railroad delivery to points in the Eastern and Middle States. Labor, etc. being based on current prices.
The above prices are figured on the tile being laid on wooden sheathing; if laid on book tile or cement add 20 per cent.
If copper nails are used, care must be taken in figuring the number of nails, as well as their length and gauge, for the special forms of tile specified. Fluctuating values of copper make this an item of much importance.
Ridges, hip rolls, barge tile, and finials are charged as extras and due allowance must be made for cutting at valleys and hips.
63. Gravel Roofs.—In gravel roofing, the cost per square depends on the number of thicknesses of tarred felt and the quantity of pitch used per square. A value of 4 cents per square foot for four thicknesses may be considered an average.
ROOF MENSURATION
64. While a knowledge of how to apply the ordinary principles of mensuration is all that is necessary to calculate any roof area, yet the modern house, with its numerous gables and irregular surfaces, introduces complications that render some further explanation of roof measurement desirable. The most common error made in figuring roofs—and one that should be carefully guarded against—is that of using the apparent length of slopes, as shown by the plan or side elevations, instead of the true length, as obtained from the end elevations.
Fig. 6
65. The area of a plain gable roof, as shown in end and side elevations in [Fig. 6], is found by multiplying the length g j by the slope length b d, and further multiplying by 2, for both sides. The area of each gable is found by multiplying the width of the gable a d by the altitude c b, and dividing by 2.
Fig. 7
Fig. 8
Fig. 9
66. In [Fig. 7] is shown the plan and elevation of a hip roof, having a deck z. The pitch of the roof being the same on each side, the line c d shows the true length of the common rafter l m.
In [Fig. 8] is shown the method of developing the true lengths of the hips and the true size of one side of the roof. Let a b c d represent the same lines as the corresponding ones in [Fig. 7]. From the line a d, [Fig. 8], through b and c, draw perpendiculars, as g h and e f; lay off from g and e on these lines, the length of the common rafter c d, [Fig. 7], and draw the lines a h and d f, [Fig. 8]; then the figure a h f d will represent the true shape and size of the side of the roof shown in the elevation in [Fig. 7]. The area of the triangle d e f is equal to the area of the triangle a g h or a similar triangle a i h. Hence, the portion of the roof a h f d is equal in area to the rectangle a i f e, the length of which is half the sum of the eave and deck lengths, while its breadth is the length of a common rafter.
67. A method of obtaining the lengths of valley rafters, applicable also to hip rafters, is shown in [Fig. 9], which is the plan of a hip-and-gable roof. To ascertain the length of the valley rafter a b, draw the line a c perpendicular to a b and equal in length to the altitude of the gable; then draw the line c b, which will represent the true length of the valley rafter a b.
68. As an example of roof mensuration, the number of square feet of surface on the roof shown in [Fig. 10] will be calculated.
Fig. 10
The area of the triangular portion a c b is equal to the slope length of d c (found by laying off c′ c equal to the height of the ridge above the eaves and drawing c′ d) multiplied by the length of the eaves line a b and divided by 2. Multiplying the dimensions 13.5 feet and 23 feet, respectively, and dividing by 2, the area is found to be 155.3 square feet.
The area of the trapezoid g f i h is half the sum of f i and g h (shown in their true length on the plan) multiplied by the true length of h i. The latter is found by marking the height of the gable i i′ on the ridge line, and drawing the line i′ h, which measures 10.6 feet. Performing these operations, there results
| 5 + 14 | × 10.6 = 100.7 square feet |
| 2 |
for each side, or 201.4 square feet for both. As each of the side gables is the same size, the area of the two roofs is 201.4 × 2 = 402.8 square feet.
The area of the polygon q p n k is equal to the triangle q p w minus the triangle k n w, the area covered by the intersecting gable roof. The former is equal to the triangle a c b, the area of which is 155.3 square feet. The area of k n w is equal to half of n w, or 6.5 feet, multiplied by the true length of k s or the altitude of the triangle; the latter is obtained by laying off k k′ equal to the height of the gable, 5.5 feet, at right angles to k s, and drawing s k′, which is the required altitude and which measures almost 7.4 feet. Then k n w = 6.5 × 7.4 = 48.1 square feet; whence q p n k equals 155.3 - 48.1 = 107.2 square feet.
The area of a p q c is
| a p + q c |
| 2 |
multiplied by the true slope length of t v, or t v′, which measures 15.2 feet. Substituting dimensions, the area is found to be
| 6 + 24 | × 15.2 = 228 square feet. |
| 2 |
From this deduct the area of y z u, which is the portion covered by the intersecting gable roof. The true length of t u along the slope is t u′, measuring 12 feet; hence, the area of y z u is
| 14 × 12 | = 84 square feet. |
| 2 |
The net area of a p q c is therefore 228 - 84 = 144 square feet; b c q w being equal to a p q c, its area is the same, making the area of both sides 288 square feet.
The area of k n m l is
| m n + l k | × m l′, |
| 2 |
the slope length of m l. Substituting dimensions, the area is
| 11 + 16 | × 8.5 = 114.8 square feet. |
| 2 |
As k l x w is equal to k n m l, the area of both is 229.6 square feet.
Adding the partial areas thus obtained, the sum is 155.3 + 402.8 + 107.2 + 288 + 229.6 = 1,182.9 square feet, or approximately 11.9 squares.