.7854 4712400 23562 2827.4400=area of the base. 20=3 of the height (30). 56548.8000=solidity required. 2. In an hexagonal spherical dome, one side of the base is 20 feet: what is the solidity ? Ang. 12000 feet. PROBLEM IV. RULE.* Multiply the area of the base by 2, and the product will be the superficial content required. EXAMPLES. : 1. What will the painting of an hexagonal spherical dome come to, at 1s. per yard; each side of the base being 20 feet? 2.598076=area of a hexagon whose side is 1. 400=square of 20. 1039.230400=area of the base. 2 9)2078.460800=superficial content required. 2.0)230.940088 11.5470044=111. 108. 11d. the expense of painting. * The practical rule for elliptical domes is as follows : RULE. Add the height to half the diameter of the base, and this sum multiplied by 1.5708 will give the superficial content nearly. PROBLEM V. To find the solid content of a saloon. RULE.* 1. Multiply the height of the arc, its projection, one fourth of the perimeter of the ceiling, and 3.1416 continually together, and call the product A. 2. From a side or diameter of the room take a like side or diameter of the ceiling, and multiply the square of the remainder by the proper factor, (page 63,) and this product again by two-thirds of the height, and call the last product B. 3. Multiply the area of the flat ceiling by the height of the arch, and this product added to the sum of A and B will give the content required. EXAMPLES 1. What is the solid content of a saloon with a circular quadrantal arch of 2 feet radius, springing over a rectangular room of 20 feet long, and 16 feet wide ? * To find the superficial content of a saloon. 2. Find the convex surface of a cylinder, or cylindroid, whose length is equal to one-fourth the perimeter of the ceiling, and its diameters to twice the height and twice the projection of the arch. 3. Find the superficial content of a dome of the same figure as the arch, and whose base is either a square, or a figure similar to that of the ceiling; the side being equal to the difference of a side of the room and a side of the ceiling. 4. Add these three articles together, and the sum will give the superficial content required. Note.--In a rectangular, circular, or polygonal room, the base or the dome will be a square, a circle, or a like polygon. Here the flat part of the ceiling is 16 feet by 12; and 4)56 14=1 of the perimeter. 28 56 56 188496 157080 175.9296=A. 4. 4 16 16.000 1=$ of the height. 16.000 21.333=B. 16 192=area of the flat ceiling. 2=height of the arch. 384 581.2629=solid content required. 2. A circular building of 40 feet diameter, and 25 feet high to the ceiling, is covered with a saloon, whose circular quadrantal arch is 5 feet radius; required the capacity of the room in cubic feet. Ans. 30779.45948 feet. PROBLEM VI. To find the solid content of the vacuity formed by a groin arch, either circular or elliptical. RULE. Multiply the area of the base by the height, and the product again by .904, and it will give the solidity required. EXAMPLES. 1. What is the solid content of the vacuity formed by a circular groin, one side of its square base being 12 feet? Here 122 x 6x.904–781.056-solidity required. 2. What is the solid content of the vacuity formed by an elliptical groin, one side of its square base being 20 feet, and the height 6 feet? Ans. 2169.6. PROBLEM VII. To find the concave superficies of a circular groin. RULE.* Multiply the area of the base by 1.1416, and the product will be the superficies required. * This rule may also be observed in elliptical groins, the error being too small to be regarded in practice. In measuring works where there are many groins in a range, the cylindrical pieces between the groins, and on their sides, must be computed separately. And to find the solidity of the brick or stone work, which forms the groin arches, observe the following. RULE. Multiply the area of the base by the height, including the work over the top of the groin, and this product lessened by the solid content, found as before, will give the solidity required. The general rule for measuring all arches, is this: From the content of the whole, considered as solid, from the springing of the arch to the outside of it, deduct the vacuity contained between the said springing and the under side of it, and the remainder will be the content of the solid part. And because the upper sides of all arches, whether vaults or groins, are built up solid, above the haunches, to the same height with the crown, it is evident that the area of the base will be the whole content above mentioned, taking for its thickness the height from the springing to the top. And for the content of the vacuity to be deducted, take the area of its base, accounting its thickness to be two-thirds of the greatest inside height. But it may be noted that the area used in the vacuity, is not exactly the same with that used in the solid; for the diameter of the former is twice the thickness of the arch less than that of the latter. And although I have mentioned the deduction of the vacuity as common to both the vault and the groin, it is reasonable to make it only in the former, on account of the waste of materials and trouble to the workman, in cutting and fitting them for the angles and intersections. Whoever wishes to see this subject more fully handled, may consult La Théorie et la Pratique de la Geométrie, par M. l'Abbé Deidier ; a work in which several parts of Menstruation and Practical Geometry are skilfully handled, the examples being mostly wrought out in an easy, familiar manner, and illustrated with observations, and figures very neatly executed. |