OF THE COVERINGS OF SOLIDS. PROBLEM 1. To find the covering of the frustum of a right cone. Let ABCD (fig. 1, pl. VII,) be the generating section of the frustum. On BC describe the semi-circle BEC, and produce the sides BA and CD, of the generating section ABCD, to meet each other in F. From the centre F, with the radius FA, describe the arc AH; and, from the same centre, F, with the radius FB, describe the arc BG; divide the arc, BEC, of the semi-circle, into any number of equal parts; the more, the greater truth will result from the operation; repeat the chord of one of these equal arcs upon the arc BG, as often as the arc BEC contains equal parts ; then, through G, the extremity of the last part, draw GF, cutting the arc AH at H; then will ABGH be the covering required. PROBLEM 2. To find the covering of the frustum of a right cone, when cut by two concentric cylindric surfaces, perpendicular to the generating section. Let ABCD (fig. 2, pl. VII,) be the given section, and AD, BC, the line on which the cylindric surface stands. Find the arc BG, as before, in problem 1, and mark the points, 1, 2, 3, &c. of division, both in the arc BG, and in the semi-circumference; from the points 1, 2, 3, &c. draw lines to F; also from the points 1, 2, 3, &c. in the semi-circumference, draw lines perpendicular to BC; so that each line thus drawn may meet or cut it. From the points of division in BC, draw more lines to F, cutting the arc BC in a, b, c, &c. From the points a, b, c, &c. draw lines parallel to BC, to cut the side BA: from the centre F, through each point of section in BA, describe an arc, cutting the lines drawn from each of the points 1, 2, 3, &c., in BG, at a, b, c, &c.; then will BeG be the curve, which will cover the line BC on the plan, or BC will be the seat of the line BeG. In the same manner AH, the original of the line AD, will be found ; and, consequently, BeGHA will form the covering over the given seat, ABCD, as PROBLEM 3. To find the covering of a right cylinder. Let ABCD (fig. 3, pl. VII,) be the seat or generating section. Produce the sides DA and CB to H and G, and on BC describe a semi-circle, and make the straight line BG equal to the semi-circumference : draw GH parallel to AB, and AH parallel to BG; then will ABGH be the covering required. PROBLEM 4. To find the covering of a right cylinder contained between two parallel planes, perpendicular to the generating section (fig. 4, pl. VII). Through the point B draw IK, perpendicular to AB, and produce DC to K; on BK describe a semi-circle, and make BI equal to the length of the arc of the semi-circle, by dividing it into equal parts, and extending them on the line BI. Through the points of section, 1, 2, 3, &c., in the line BI, draw lines, la, 2b, 3c, &c., parallel to BA, and through the points 1, 2, 3, &c., in the arc of the semi-circle, draw the other lines la, 2b, 3c, &c., parallel to BA, cutting AD in a, b, c, &c. Draw aa, bb, cc, &c., parallel to BK ; then, through the points, a, b, c, &c., draw the curve AH, and AH will be the edge of the covering over AD. In the same manner the other opposite edge BG will be found, and the whole covering will therefore be ABGH. PROBLEM 5. ABCD (fig. 5, pl. VII,) being the seat of the covering of a semi-cylindric surface, contained between the surfaces of two other concentric cylinders, of which the axis is perpendicular to the given seat ; it is required to find the covering Through B draw IK, perpendicular to AB; and produce DC to K. On BK describe a semi-circle, and divide its circumference into equal parts, at the points 1, 2, 3, &c.; the more of these the truer will be the operation; and repeat the chord on the straight line BI, as often as the arc contains equal parts, and mark the points 1, 2, 3, &c., of division. Through the points 1, 2, 3, &c., in the arc of the semi-circle, draw the lines la, 26, 3c, &c., parallel to BA; and, through the points 1, 2, 3, &c., in BI, draw lines la, 26, 3c, &c., parallel to BA. Draw aa, bb, cc, &c., parallel to. KI, and through all the points a, b, c, &c., draw the curve line AH, which is one of the edges of the covering In the same manner the other edge BG will be found ; and, consequently, the whole covering ABGH. PROBLEM 6. To find the covering of that portion of a semi-cylinder contained between two concentric surfaces of two other cylinders, the axis of these cylinders being perpendicular to ABCD (fig. 6, pl. VII). Join BC, and, in this case, BC will be perpendicular to AB. Produce CB to G; and, on BC, describe a semi-circle. Divide the arc of the semi-circle into any number of equal parts, and extend the chords upon the straight line BG, marking the points of section both in the semi-circle and in the straight line BG. Through the points, 1, 2, 3, &c., in the arc of the semi-circle, draw lines la, 2b, 3c, &c., parallel to AB; and through the points 1, 2, 3, &c., in BG, draw the lines la', 20', 3c', &c., parallel to AB; also draw aa', bb', cc', &c., parallel to BG, and, through the points a,b,c, &c, draw a curve, which will form one of the edges of the soffet; the opposite edge is formed in the same manner. GROINS AND ARCHES. Groins are the intersections of the surfaces of two arches crossing each other. CONSTRUCTION OF GROINED ARCHES. GROINED Arches may be either formed of wood, and lathed over for plaster, or be constructed of brick or stone. When constructed of brick or stone, they require to be supported upon wooden frames, boarded over, so as to form the convex surface, which each vault is required to have, in order to sustain the cross arches during the time of turning them. This construction is called a CENTRE, and is removed when the work is finished. The framing consists of equidistant ribs, fixed in parallel planes, perpendicular to the axis of each body; so that, when the under sides of the boards are laid on the upper edges of the ribs, and fixed, the upper sides of the boards will form the surface required to build upon. In the construction of the centering for groins, one portion of the centre must be completely formed to the surface of its corresponding vault, without any regard to the cross-arches, so that the upper sides of the boards will form a complete cylindric or cylindroidic surface. The ribs of the cross-vaults are then set at the same equal distances as that now described; and parts of ribs are fixed on the top of the boarding at the same distances, and boarded in, so as to intersect the other, and form the entire surface of the groin required. Groins constructed of wood, in place of brick or stone, and lathed under the ribs, and the lath covered with plaster, are called plaster-groins. PLASTER-GROINS are always constructed with diagonal ribs intersecting each other, then other ribs are fixed perpendicular to each axis, in vertical planes, at equal distances, with short portions of ribs upon the diagonal ribs ; so that, when lathed over, the lath may be equally stiff to sustain the plaster. When the axis and the surface of a semi-cylinder cuts those of another of greater diameter, the hollow surface of the lesser cylinder, as terminated by the greater cylinder, is called a cylindro-cylindric arch, and, vulgarly, a Welsh groin. : CYLINDRO-CYLINDRIC ARCHES, or Welsh groins, are constructed either of brick, stone, or wood. If constructed of brick or stone, they require to have centres, which are formed in the same manner as those for groins; and, if constructed of wood, lath, and plaster, the ribs must be formed to the surfaces. In the construction of groins, and of cylindro-cylindric arches, the ribs that are shorter than the whole width are termed jack-ribs. Cellars are frequently groined with brick or stone, and sometimes all the rooms of the basement stories of buildings, in order to render their superstructures proof against fire. The surfaces of brick or stone, on which the first arch stones, or course of bricks, are placed, are called the springing of the arches. It is evident that the more weight that is put on the side-walls, which sustain arches, the more will they be able to sustain the pressure of the arches; therefore the higher a wall is, the greater the weight will be on each of the side-walls ; and for this reason groins are often constructed of wood in upper stories, instead of brick or stone, as not being liable to thrust out the walls, or bulge them, by the lateral pressure of the arches. The upper stories of buildings are never groined with stone or brick, unless when the walls are sufficiently thick to sustain the lateral pressure of the arches. The ceilings of Gothic buildings were frequently constructed with groined arches of stone, which were obliged to be supported with buttresses, at the springing points of the arches. GROINS AND ARCHES.—PROBLEM 1. (fig. 1, pl. VIII.) Given the plan of a rectangular groined arch or vault, of which the openings are of different widths, but of the same height, and a section of one of the arches, as also the seats of the groins, to find the covering of both arches, so as to meet their intersection. In fig. 1, pl. VIII, let A, A, A, &c., be the plan of the piers, and ab, cd, the seats of the groins. Let the section of the arch, standing upon the lesser opening, BC, be a semi-circle: it is required to find the section upon the greater opening and the ends of the boards, so as to meet the groin, or line of intersection, of the two surfaces. * The difference between the plan of any body and the seat of a point or line is distinguished thus : The plan is a figure upon which a solid is carried up, so that all sections, parallel to the plan, are equal and similar to that plan, and the surfaces are perpendicular; but the seat of a line is not in contact with the line itself; but a perpendicular erected from any point in the seat will pass through its correspond |