series of 12th parts, and decimals by a series of 10th parts, in multiplying each of the parts of the former, the trouble of dividing by 12 will then be unavoidable, and more burdensome to the mind than if the operation had been done by the latter, where there is no such division to be made, but merely to multiply, as in common multiplication, and point off the decimal places in the product. This last method I shall always prefer, being the most natural, as well as the most easy of the two. OF BRICKLAYERS' WORK. PROBLEM I. To find the number of rods contained in a piece of brick work. METHOD I. If the wall is at the standard thickness, that is, one brick and a half thick, divide the area, or superficial content, by 272, and the quotient, if any, will be the answer in rods; and the remainder, if any, in feet: but if the wall is more or less than one brick and a half in thickness, multiply the area of the wall by the number of half bricks contained in its thickness; divide the product by 3, and the wall will be reduced to the standard thickness of a brick and a half in thickness; then divide the quotient by 272, and the last quotient will be the number of rods of brick work required. METHOD II. Multiply the length of the wall by its breadth, and the product by its thickness; that is, to find the solidity of the wall: then multiply the solidity by 8, and divide the product by 9, and the quotient will be reduced to the standard thickness of one brick and a half in thickness; then divide by 272, as before, and the answer will be the number of rods contained in the brick work. EXAMPLE I. How many rods are there in a wall 621ft. long, 14ft. Sin. high, and 2 bricks thick? In a piece of brick work, of the aforesaid dimensions, there are 5 rods, 167 feet, 9 inches, and 4 seconds; but in finding the number of rods in a piece of brick work, neither inches nor seconds need be taken any account of, as such an addition is so very trifling. EXAMPLE II. A triangular gable end is raised to the height of 15ft. above the end wall of a house, whose width is 45ft. and the thickness of the wall is 2 bricks, required the content at the standard measure. This figure being a triangle, its contents must be found according to the rules of measuring that kind of figure; then the operation will be as follows: 45 width of the house, or base of the triangle. 225 45 2)675 337.5 area of the gable, or of the triangle. To find the quantity of materials for building any wall in brick work. Take the dimensions of a building, by measuring half round the outside, and half round the inside, for the whole length of the wall; that being multiplied by the height, will give the area, and proceed as before to find the number of rods or feet reduced to the standard thickness; out of this must be deducted the number of rods or feet, reduced to the same thickness as before, that would be contained in all the vacuities, such as doors, windows, window backs, chimnies, chimnies, &c. or any other open space in the wall, and the remainder will give the true quantity of rods. Now the quantity of materials that are generally allowed to a rod of brick work, is 4500 bricks, one hundred and a quarter of lime, and two loads and a half of sand; therefore, if the number of rods or feet are multiplied by 4500, the product will be the number of bricks to a rod of brick work; or if the number of rods, feet, &c. are multiplied by 1 hundred weight of lime, will give the quantity of lime; and also if the number of rods are multiplied by 2 loads of sand, will give the quantity of sand. EXAMPLE I. Plate 55, Fig. 1 and 2, let ABCD be the plan or horizontal section of the carcase of any story in a building, the length of the front on the outside is 41ft. 6in.; the length of the end on the inside is 25ft.; the height of the wall or story is 15ft. and 2 bricks thick; in this story are 7 windows, whose vacuities on the outside are 8ft. high, 3ft. 6in. wide, and 1 brick thick; the vacuity in the inside is 4ft. 3in. wide, by 11ft. high, and 1 brick thick; there is also a door 4ft. 6in. wide, by 11ft. high, and 2 chimnies; the breast of each project from the face of the wall 1ft. 6in.; the breadth of the breast is 8ft., the height is 15ft. or the height of the story; the width of the chimney is 3ft. 6in. by 4ft. high; and its depth, from the front to the back, is 3 bricks; required the number of bricks, and the quantity of sand and lime to build the said wall. 3 6 8 0 28 0 area of the vacuity of 1 window on the outside, 7 at 1 brick thick. 196 0 arca of the vacuities for 7 windows on the outside, at 1 brick thick. Continued. 4 S 11 0 46 9 area of the vacuities of 1 window in the inside, at 1 327 3 area of the vacuity in the inside for 7 windows, at 1 brick thick. 4 6 11 49 6 area of the vacuity of the door, at 2 bricks thick. 2 99 0 area of the vacuity of the door, at 1 brick thick. O area of the vacuity of 1 chimney, within the face of the wall, at 1 brick thick. 28 area of the vacuities of the 2 chimnies within the 99 face of the wall, at 1 brick thick. 327 3 196 0 650 3 area of all the vacuities, at 1 brick thick, included in the thickness of the wall. 3 6 4 0 14 O area of one of the vacuities of the chimney, included within the thickness of the breast, at 1 brick thick. 15 8 120 area of one breast, at 2 bricks thick, as if it were solid, 14 deduct the area of the vacuity, at the same thickness. 106 true area of one breast, at 2 bricks thick. 2 12 true area of the two breasts, at 2 bricks thick. Con |