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 k ገዛ Fig. 702. points join; then either aC or as represents the length of the hip rafter, and joining the several lines aqb, bpoc, end, and dmla, they will be the skirts of the roof. 2054. To find the back of the hip. touching the hip as, and cutting at in u. hip rafter required. 2055. Fig. 703. represents, in abed, the 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 7. By the same kind of operation oc will be Join ge, and from r as a centre describe an arc plan of a building whose sides are bevel to each found, as also the other parts of the skirts of the roof. P Fig. 703. m 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 Eqro F. Produce AB to p, and FE to m, cutting Ap in y. Through the centre x of the semi Fig. 705. DF 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 are 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 Bedef 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 BedefA draw the straight lines cq, dr, es cutting the diagonal HD at q, r, s. Draw qh, ri, sh parallel to al cutting the chord BG at the points 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 tc, 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 construction 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 curvature (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 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 Divide hB into any d, B, p describe the semicircle ABC for the head of the window. 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, So 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. 2066. The construction of a niche, which is a portion of a spherical surface, and stands on a plan formed by the segment of a circle, is simple enough; for the ribs of a niche are all of the same curvature as the plan, and fixed (fig. 710.) in planes passing through an axis corresponding to the centre of the sphere and perpendicular to the plane of the wall. If the plan of the niche be a Fig. 710. 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. 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. Let us suppose 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 ÿjkl. Produce the line e A to 8, 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 a 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 n 19 A h B m 19 Pg. 14. (fig. 714.), the plan being ABCDEFGH. Let i be the centre of the circle in which the hexagon may be inscribed. Draw the half diagonal iA, iB, iC perpendicular to any side AB of the plan. Draw the straight line ih, cutting AB in h. Let himZ be the outline of one of the ribs of the dome, which is here supposed to be the quadrant of a circle. Divide the arc hZ into any number of equal parts from h at the points Imn, and through these points draw lx, my, nz, cutting Bi at the points ryz, and ih at the points 1, 2, 3. the arcs hl, Im, mn, on the line hn, from h to o, from c to p, from p to q, and through the points opq draw the straight lines ru, sv, tw perpendicular to hn. Make ou, pv, qw, as also or, ps, qt, respectively equal to 1x, 2y, 32; then through the points Arst draw a curve, and through the points uvw draw another curve, meeting the former one in the point n. Thus will be formed the gore or cover-ing of one side of the hexagonal dome. A D Fig. 715. Extend 2071. When the plan of the base is a rectangle, as fig. 715., draw the plan ABCD and the diagonals AC and BD, cutting each other in E. Through E draw EI perpendicular to AB cutting AB in F, and through E draw EJ perpendicular to BC, cutting BC in G. Let the height of the dome be equal to half its breadth, and the section over the straight line EF a quadrant of a circle; then from the centre E describe the arc FH, its base being EF, and with the straight line EG as half the major axis of an ellipsis, and EF the minor axis, describe the quadrant GF of an ellipsis. Produce EF to I, and EG to J. Divide the arc of a quadrant FH from F into any number of equal parts, and extend the parts on the line FI to klm, through which draw the lines kq, Ir, ms, &c. perpendicular to FI. Through the points 1, 2, 3, &c. draw wt, xu, yo, &c., cutting AE at w,x,y, and FE at t,u,v. Make k'n', l'o', m'p', also kq, lr, ms, respectively equal to tw, ux, vy, and through the points n'o'p' draw a curve, also through the points qrs draw another curve meeting the former in I; then these two curves with the line AB will form the gore or boundary of the building of two sides of the dome. Also in the elliptical arc GF, take any number of points 1, 2, 3, and draw the lines lw', 2x', 3y', parallel to BC, cutting EC in the points w'x'y', and GE in the points t', u', . Extend the arcs G1, 12, 23 from Gk', k'l', I'm', upon the straight line GJ, and through the points k'l'm' draw the lines n'q', o'r', p's'. Make k'n', l'o', Fig. 716. Eg h i j F 772 A B m'p', also k'q', l'r', m's' respectively equal to t'w', u'x', v'y, and through the points Bn'o'p' draw the curve BJ, and through the points Cq'r's' draw the curve CJ; then BJC will be the gore required, to which the boards for the other two sides of the dome must be formed. Let 2072. A general method of describing the board or half gore of any polygonal or circular dome is shown in fig. 716. DE be half either of the breadth of a board or of one of the sides of a polygon, EF the perpendicular drawn from the centre. Draw the straight line AB parallel to EF, and draw EA and FB perpendicular to EF; then upon the base AB describe the rib AC of the vertical section of the dome. Divide the curve AC into the equidistant arcs A1, 12, 23, and through the points of division draw the lines Ig, 2h, 3i perpendicular to AB cutting EF at ghi and DF at klm. Produce FE to V and extend the arcs A1, 12, 23 upon the straight line EV from E successively to the points opq. Through the points opq draw the lines or, ps, qt parallel to ED. Make or, ps, qt respectively equal to gk, hl, im; then through the points rst draw a curve, and DEV will |