2050. The largest roof that was, perhaps, ever executed, was over a riding-house at Moscow, built in 1790, by Paul I. Emperor of Russia, the representation whereof may be seen in Kraaft, Recueil de Charpente. The span is 235 feet, and the slope with the horizon about 19 degrees. The external dimensions of the building were 1920 feet long by 310 feet wide. It was lighted by a lantern at top, and had an interior gallery round the building for spectators. The contrivance is exceedingly ingenious; but, from the great extent of the span, considerable settlement took place, and alterations, or rather strengthening ribs, became necessary. 2051. We shall close this part of the section with a diagram (fig. 700.) of the roof of -78 ft. 6 ins.- Fig. 700. the basilica of S. Paolo fuorì le murà, executed in the fifteenth century. The trusses are double, each consisting of two similar frames, nearly 15 inches apart, at intervals from each other of about 10 feet 6 inches. The principal rafters abut on a short-king post k. Between the trusses a piece of timber S is placed and sustained by a strong key of wood passing through it and the short king-posts. This piece sustains the beams by means of another strong key at a. The tie beams are in two lengths, and scarfed together, the scarf being held together by three iron straps. The scantlings of the timbers are as follow beams t, 22 in. full by nearly 15 in.; principal rafters p, 21 in. by nearly 15 in.; auxiliary rafters b, full 133 in. by full 13 in.; straining beam C, near 15 in. by full 12 in.; purlines d, 8 in. square and 5 ft. 7 in. apart; common rafters, full 54 in. by 44 in., and 8 in. apart. The roof, which is constructed of fir, is nearly 78 ft. 6 in. span, and is covered with the Roman tile, the exact dimensions and form whereof will be found, under the head TILE, in the Glossary appended to this work. The roof is ingeniously and well contrived, and, with a different covering, would suit other climates. It was consumed by fire about two or three and twenty years ago. (275.) Philibert Delorme, in his work entitled "Nouvelles Inventions pour bien bâtir a petits Frais," Paris, 1561, gives a mode of constructing domes without horizontal cross ties, when the springing of each rib is well secured at the foot. It is a very simple method, and of great use in domes, even of large diameter, the principle being that of making the several ribs in two or more thicknesses, which are cut to the curve in lengths not so great as to weaken the timber, and securing these well together by bolts or keys, and observing especially to break the joints of the several thicknesses. This method was adopted in the large Halle aux bleds at Paris, which was many years since destroyed by fire, and has been replaced by an iron-ribbed dome. The fig. 701. will explain the construction; and, if necessary, an iron Fig. 701. hoop passed round at different heights will add much to the strength. 36 feet diameter, 60 feet diameter, 108 feet diameter, 10 in. deep, and 1 in. thick. 13 in. deep, and 2 in. thick. The work of the author from which we have given this short and summary account deserves the study of every one that seeks to be an architect, though in these unfortunate days for the art the reward of study and reading is very doubtful; patronage being of much more importance to the professor than a profound knowledge of construction and design. 2053. The following instructions relative to the lines necessary to be found in the framing of roofs are from Francis Price's British Carpenter; and though published long since, now nearly 100 years, we have not found that any subsequent work on this particular point gives us more information than is to be there found. Let abcd (fig. 702.) be a plan to be inclosed with a hipped roof, whose height or slope is Cb. Divide the plan lengthwise into two equal parts by the line ef, which produce indefinitely at both ends. Make ag equal ea, and dk equal to df; and through k and g, parallel to ab or cd, draw lines indefinitely mo, lp. With the distance de or Cc, either of which is equal to the length of the common rafters, set off qe, as also from h to p, from i to o, and from fton; from k to m, and from g to l. Make ts equal to Cb, and ab equal to ta, which points join; then either aC or as represents the length of the hip rafter, and joining the Join ge, and from r as a centre describe an arc 2055. Fig. 703. represents, in abcd, the plan of a building whose sides are bevel to each other. Having drawn the central line ef indefinitely, bisect the angle rag by the line ae, meeting ef in e. From e make eg equal to re, and rg perpendicular to ea; then, if e a be made equal to ea, ra or aq, it will be the length of the hip rafter from the angle a. Through e and f, perpendicular to the sides db, ca, draw the lines np, mq indefinitely; and from a, as a centre with the radius aq, describe an arc of a circle, cutting mq in q, and er (perpendicular to ba) produced in l. By the same kind of operation oc will be Fig. 703. m found, as also the other parts of the skirts of the roof. The lines nt, tfv, and vp are introduced merely to show the trouble that occurs when the beams are laid bevel. The angle of the back of the hip rafter, rwg, is found as before, by means of u as a centre, and an arc of a circle touching aq. The backs of the other hips may be found in the same manner. 2056. Fig. 704., from Price's Carpentry, is the plan of a house with the method of placing the timbers for the roof with the upper part of the elevation above, which, after a perusal of the preceding pages, cannot fail of being understood. The plan F is to be prepared for a roof, either with hips and vallies, or with hips only. The open spaces at G and H are over the staircases: in case they cannot be lighted from the sides, they may be left to be finished at discretion. The chimney flues are shown at IKLMNO. Then, having laid down the places of the openings, place the timbers so as to lie on the piers, and as far as possible from the flues; and let them be so connected together as to embrace every part of the plan, and not liable to be separated by the weight and thrust of the roof. P is a trussed timber partition, to discharge the weight of the roof over a salon below. 2057. Q is the upper part of the front, and R a pediment, over the small break, whose height gives that of the blank pedestal or parapet S. Suppose T to represent one half of the roof coming to a point or ridge, so as to span the whole at once, “which," as Price truly observes, "was the good old way, as we are shown by Serlio, Palladio," &c., or suppose the roof to be as the other side U shows it, so as to have a flat or sky-light over the lobby F, its balustrade being W; or we may suppose X to represent the roof as spanning the whole at three times. If X be used, the valley and hip should be framed as at Y; if as T, the principal rafters must be framed as at Z, in order to bring part of the weight of the roof and covering on the partition walls. The remainder needs not further explanation. 2058. We shall now proceed to the method of forming the ribs for groined arches, niches, &c. The method of finding the shape of these is the same, whether for sustaining plastering or supporting the boarding of centres for brick or stone work, except that, for plaster, the inner edge of the rib is cut to the form, and, in centering, the outer edge. Groins, as we have already seen, may be of equal or unequal height, and in either case the angle rib may be straight or curved; and these conditions produce the varieties we are about to consider. 2059. To describe the parts of a groin where the arches are circular and of unequal height, commonly called WELSH GROINS. We here suppose the groin to be right-angled. Let AB (fig. 705.) be the width of the greater arch. Draw BD at right angles to AB, and in the straight line BD make CD equal to the width of the lesser arch. Draw DF and CE perpendicular to BD and EF parallel to BD. On AB describe the semicircle Bghi A, and on EF describe the semicircle EqroF. Produce AB to p, and FE to m, cutting Ap in y. Through the centre a of the semi Fig. 705. P circle Eqrs F draw ts perpendicular to BD, cutting the circumference of the semicircle in s. Draw sp parallel to BD. From the centre y, with the distance yp, describe the quadrant pm. Draw mi parallel to AB, cutting the semicircle described upon AB in the point i. In the arc Bi take any number of intermediate points g, h, and through the points ghi draw it, hu, gv, parallel to BC. Also through the points ghi draw gk, hl, im parallel to AB, cutting FE produced in k and l. From the centre y describe the arcs kn, lo, cutting AB produced in mo. Draw nq, or, parallel to BD, cutting the lesser semicircular arc in the points q, r. Through the points q, r, s draw qu, ru, st parallel to AB; then through the points tuv draw the curve tuvc, which will be the plan of the intersection of the two cylinders. The other end of the figure exhibits the construction of the framing of carpentry, and the method in which the ribs are disposed. 2060. To describe the sides of a groin when the arches are of equal height and designed to meet in the plane of the diagonals. Let af and al (fig. 706.) be the axes of the two vaults, meeting each other in a, perpendicular to af. Draw AB cutting af in w, and perpendicular to al, draw BG cutting al in b. Make wA and wB each equal to half the width of the greatest vault, and make bB and bG each equal to half the width of the lesser vault. Draw AH and BE parallel to af, and draw BH and DF parallel to al, forming the parallelogram DEHF. Draw the diagonals HD, FE. On the base AB describe the curve Bcdef A, according to the given height wf of the required form, which must serve to regulate the form of the other ribs. Through any points cde in the arc BcdefA draw the straight lines cq, dr, es cutting the diagonal HD at q, r, s. Draw gh, ri, sh parallel to al cutting the chord BG at the points H x, y, z, b. Make xh, yi, zk, bi each respectively equal to tc, ud, ve, wf, and through the points Ghikl to B, draw the curve Ghikl B. Draw qm, rn, so, ap perpendicular to HD. Make F qm, rn, so, ap respectively equal to te, ud, ve, wf, and through the points D, m, n, o, p, H draw a curve, which will be the angle rib of the groin to stand over HD; and if the groined vault be rightangled, all the diagonals will be equal, and consequently all the diagonal ribs may be made by a single mould. Fig. 706. 2061. The upper part of the above figure shows the method of placing the ribs in the con. struction of a groined ceiling for plaster. Every pair of opposite piers is spanned by a principal rib to fix the joists of the ceiling to. 2062. The preceding method is not always adopted, and another is sometimes employed in which the diagonal ribs are filled in with short ribs of the same curva. ture (see fig. 707.) as those of the arches over the piers. 2063. The manner of finding the section of an aperture of a given height cutting a given arch at right angles of a greater height than the aperture is represented in fig. 708. Fig. 707. 2064. When the angle ribs for a square dome are to be found, the process is the same as for a groin formed by equal arches crossing each other at right angles, the joints for the laths being inserted as in fig. 707.; but the general construction for the angle ribs of a polygonal dome of any number of sides is the same as to determine the angle rib for a cove, which will afterwards be given. 2065. When a circular-headed window is above the level of a plane gallery ceiling, in a church for example, the cylindrical form of the window is continued till it intersects the plane of the ceiling. To find the form A of the curb or pieces of wood employed for completing the arris, let dp (fig. 709.) be the breadth of the window in the plane of the ceiling. Bisect dp in h, and draw h4 perpendicular to dp. Make h4 equal to the distance the curb extends from the wall. Produce 4h to B. Make hB equal to the height of the window above the ceiling, and through the three points d, B, p describe the semicircle ABC for the head of the window. Divide hB into any number of equal parts, as 4 at the points k, l, v; and h4 into the same number of equal parts at the points 1, 2, 3. Through the points klu draw the lines et, fu, gw parallel to dp, and through the points 1, 2, 3 draw the lines mg, nr, os. Make 1m, 2n, 30 respectively equal to ke, lf, vg; as also 1q, 2r, 3s equal to kt, lu, vw; that is, equal to ke, lf, vg. Then through the points dmno4, and also through pqrs4, draw a curve which will form the curb required. In the section X of the figure, AC shows the ceiling line, whereof the length is equal to h4, and AB is the perpendicular height of the window; hence BC is the slope. semicircle (fig. 711.) the ribs may be disposed in vertical planes. Fig. 711. 2067. In the construction of a niche where the ribs are disposed in planes perpendicular to the horizon or plan, and perpendicular to the face of the wall, if the niches be spherical all their ribs are sections of the sphere, and are portions of the circumferences of different circles. If we complete the whole circle of the plan (fig.712.), and produce the plan of any rib to the opposite side of the circumference, we shall have the diameter of the circle for that rib, and, consequently, the radius to describe it. 2068. Of forming the boards to cover domes, groins, &c. The principles of determining the developement of the surface of any regular solid have already been given in considerable detail. In this place we have to apply them practically to carpentry. The boards may be applied either in the form of gores or in portions of conic surfaces; the latter is generally the more economical method. Let us suppose 2069. To describe a gore that shall be the form of a board for a dome circular on the plan. Draw the plan of the dome ABD (fig. 713.), and its diameter BD and Ae a radius perpendicular thereto. If the sections of the dome about to be described be semicircular, then the curve of the vertical section will coincide with that of the plan. the quadrant AB to be half of the vertical section, which may be conceived to be raised on the line Ae as its base, so as to be in a vertical plane, then the arc AB will come into the surface of the dome. Make Ai equal to half the width of a board and join ei. Divide the arc AB into any number of equal parts, and through the points of division draw the lines li, 2j, 3k, 41, cutting Ae in the points efgh and ei in the points ijkl. Produce the line e A to s, and apply the arcs A1, 12, 23, 34 to Am, mo, oq in the straight line As. Through the points mnoq draw the straight lines tn, up, vr, and make mn, op, qr, as also mt, ou, qv, respectively equal to ei, fj, gk; then through the points inpr to s, and also through the points xtuv to s, draw two curves from the points r and i so as to meet each other in s; and the curves thus drawn will include one of the gores of the dome, which will be a mould for drawing the boards for covering the surface. 2070. In polygonal domes the curves of the gore will bound the ends of the boards; as, for example, in the hexagonal dome H 19 u h B m E Fg. 714. |