Fig. 641. 1954. Where the piers supporting groins (fig. 642.) are made octangular, the angles of the groins should be cut off or arched as ribs, by which they are rendered much stronger than when they are square. In stone groins, where the arch is cut off, there is no advantage in point of strength, and rather a defect in point of appearance, to the groined angles. 1955. Arches intersecting a coved ceiling are similar to groins. Such arches are called lunettes, and are generally practised for semicircularheaded windows piercing the coves in the ceiling: fig. 643. exhibits a plan and section of such arches. 1956. A dome is a solid, which may be conceived to be generated by the figure of the base diminishing as it rises, till it becomes a point at the summit; and when a dome has a polygonal base, the arches are plain arches, and the construction is similar to that of a groin. A domed ceiling of this kind upon a rectangular plan is shown in plan B (fig. 644.); the sections AA being elliptical in the top, and with lunette windows. C shows the geometrical construction. Fig. 642 Fig. 644. 1957. When arches intersect an inclined vault, and the projections of the arrisses cross each other at right angles, and the angle of elevation of one of the semicircular vertical ribs of the ascending avenue or opening is given to obtain the geometrical construction; so that the cross arches may be cylindrical surfaces. Draw the straight line AB (fig. 645.) to represent the axis of the inclined vault, and draw CD perpendicular to AB. Produce D to e and h; make AC and AD each e, E 3 2 1 h Fig. €45. L, a, b, c draw the curve Laber, and this will be the pliable mould for forming the angle or groin over the plan, and for working the arch stones. Draw Dk parallel to Az. Let E divide the circumference CED into the two equal parts EC, ED; divide the arcs DE, EC into the same number of equal parts as uv at the points 1, 2, 3, and draw 1w, 2x, 3y, Ez, parallel to AB; also through the points 1, 2, 3 in the quadrant qu draw gk, lw, 2x, 3y, uz, perpendicular to yN; then through the points k, w, x, y, z, draw a curve, which will be the plan of the groin whereof the stretch-out is Labcv. In the same manner the other half of the plan will be found, as also the whole of the other parts. right angles, and the plan Let AB, BC (fig. 646.) E C b A D K V 1958. The form of an arch crossing an inclined groined vault at of the diagonal ribs being given; to find the arch of the level vault. be the plan of the axis of the vaults. Through any point A in AB draw DF perpendicular to AB, and make AD and AF each equal to the horizontal breadth of the vault. Draw DG and FH parallel to AB; draw also any line LK parallel to AB, cutting BC in C, and make the angle KIL equal to the inclination of the axis represented by its plane AB. Make CM and CK equal to the breadth of the level vaults; draw KG and MN parallel to BC, and let MN cut DG in N, and FH in P. Draw the diagonals PG and NH. Produce GK to cut IL in L, and NM to cut IL in Q. In the curve DEF take any number of points a, b, c, and draw ad, be, cf parallel to AB, cutting DF in the points p, q, r, and the diagonal GP in d, e, f, and the diagonal HN in the points d', e', f'. Produce BA to E, draw dl, em, fn, Bo parallel to BC, cutting QL in the points g, h, i, k; make gl, hm, in, ko equal respectively to pu, qb, rc, AE; then through the points l, m, n, o, draw the curve QoL. Draw HR perpendicular to NH, and make HR equal to KL, and join NR; then will HR be the line of ramp for the diagonal rib over its plan HN. Perpendicular to HN, draw d'v, e'w, f'y, BG cutting the line of ramp RN in the points s, t, u, v. Make sv, tw, uy, vG respectively equal to pa, qb, rc, AE. Then through all the points v, w, y draw a curve, which will be the angle rib standing over HN, and which will also serve for the angle rib standing over GP. All the groined vaults continued in the same range may be constructed by the same moulds. Fig. 646. 1959. To make the working drawings for a semicircular arch with a straight face, und to describe the moulds for the voussoirs. This simple case will serve as a rule for those following; hence the explanation should be perfectly understood, as all the other cases differ from it only according to the different kinds of arches to be constructed; such as the bevelled arch, that in a battering or sloping wall, and that on a circular wall. 8 G 7 C 5 H E 1960. Draw two lines (fig. 647.) perpendicular to and crossing each other, as BA, GD From the point E, as a centre, describe the sofite curve ACB, and the extrados or upper curve FGH. Divide each of these arcs into two equal parts, as the dotted arc abc. Draw LM parallel to AB, and make the distance A, L equal to the thickness of the wall wherein the arch is to be constructed. Draw the outer and inner lines of the plan F'K, A'L, B'M, HN parallel to CD. Divide the arc ACB into the proper number of equal parts for the arch stones or voussoirs, suppose five, by the joint lines 1, 2, 3, 4; from the point E draw the joints 1-5, 2-6, 3-7, 4-8; then from every point where the joints cut the arcs ACB, FGH, &c. draw the lines 8c4, 7h3, &c. On KN let fall the perpendiculars 8d, cM, 4f, hi, 3k, 21, mn, 60, 1p, aL, and 5s. Divide the sofite of each voussoir A1, 1—2, 2—3, &c. into two equal parts in t, u, v, w, from which all also let fall the perpendiculars Y, uX, vV, wT. H N E' K MTef Vik 1961. To draw the moulds of the 80fite below NK. Draw the line OP parallel to the line KN; prolong ED to Z and make the distance QZ equal to ED. Through Z draw RS parallel to OP, and on each side of QZ lay off the distances C3, 3v, v4, 4w, and wB respectively on Qx, xy, ya, ab, and bP. On the other side lay off C2, 2u, ul, It and tA on Qc, cd, de, ef, and fO. Through the points O, e, c, x, a let fall on RS the perpendiculars OR, ea', cd', xc', ad, PS, and through the points f, d, y, b let fall the perpendiculars from the middle sheetings fe', df', yg', bh'; the distances of the dark lines give the breadth of the sofite of each stone in the sofite curve. S p ht d m gn z d' I a' rf Fig. 674. 1962. To draw the moulds of the joints: lay off the distance 1-5 on eg, ch, xi, ak, and through the points ghin draw the lines gq, hl, im, kp, parallel to QZ. To find the middle of the joint divide the distances eg, ch, xi, an, each into two equal parts, as in k', m', g', s, through which draw the lines k'l', m'n', q'r', s't parallel to QZ. 1963. The elevation is a section of a hollow cylinder, of which the concave or interior surface forms the intrados of the arch, and the convex or exterior surface the extrados, and of which the cutting plane of the section is perpendicular to the common axis of the cylinder. 1964. The angles of the stone are found from the angle which the arc of this section makes with any joint, and the curving of the sofite of the stone is found by a ruler or mould, the edge of which is made to the curve. The ends of the sofite are found by its developement. 1965. When the stones are shaped according to the moulds, and joined together in consecutive order, the whole mass, thus united, will form the solid arch as required. 1966. These separate operations being properly attended to, every difficulty will be removed, and no confusion will arise during the process, which can, in any degree, tend to per- AC plex the delineator. 1967. To find the bevels and moulds for the joints and sofites of an elliptical arch cutting obliquely through a straight wall, the joints radiating to the centre of the opening. Draw the axis EN of the arch (fig. 648.), and therein take any point E, through which draw AB perpendicular to EN; make EA and EB each equal to half the space of the extrados or centre line of the arch; also make EC and ED each equal to half the span of the inner arch. Produce the diameter NE to G; G make EF equal to the height of the inner arch and EG equal to the height of the outer arch. On the major axis AB, and semi-minor axis EG, describe the semi-ellipsis AGB, which is the extrados of the arch. Also, on CD as the major axis, and EF the semi-minor axis, describe the semi-ellipsis CFD. 1968. Make the angle ABH equal to the angle which the wall makes with the right section of the arch, and let BH cut the axis in K. Draw ML at such a distance from BH that they may comprehend between them the thickness of the wall, and let ML cut the axis in N. The intrados of the arch on the one side of the wall is OPR, and the extrados is LQM; they are both ellipses respectively of the same height as the intrados and extrados of the right arch, but with the axes OR and LM. 1969. To find the bevel of the angle of the arch stones corresponding to the joint ab tending to the centre E. Describe the arc be from E with the radius Eb cutting AB in Draw bg parallel to EN cutting BH in g, and draw cd parallel and gd perpendicular to EN, and join KD; then EKD is the angle or bevel required. c g A B 1970. The sofite of the arch is drawn according to the general principles of developement. 1971. To make the working drawings for an arch in a sloping wall, as, for instance, an arch in a terrace wall. To draw the elevation; from any convenient point o in the line AB (fig. 649.), describe the arc of the intrados $ aPf and the arc of the extrados AQB: divide each of these arcs into odd numbers of equal parts (for the arch stones in this example five), and draw the joints bg, ch, di, ek. For the plan of the arc of the intrados draw AR perpendicular to AB, and draw the line of slope or batter AS. In the arc of the intrados take any number of points bed, &c. and draw the lines bb, cc, intersecting AR in the points 1, 2, &c. and meeting the line of batter AS in the points bc. Draw CD parallel to AB, and at any convenient distance from it draw aubvcw perpendicular to CD, intersecting it in the points e, l, m, n, &c. Find the points b', c', d' in the straight lines bv, mw, nz, such that the distance of those points from the line ED may be respectively equal to the intervals 1b, 1c, &c. between the perpendicular AR and the line of batter AS, and draw the curve a' b' c' d' e' f', which will be the plan of the arc of the intrados. In the same manner the curve Eg'h'ikD may be described; which being done, the plan of the are of the extrados will be obtained. E L Pe 19m is Di a b w m" M all 2011 Fig. 649. k N 1972. To find the moulds of the sofites and beds. Draw any straight line HI in a separate place, and extend the arc of the intrados abcdef upon the line HI from H to I; divide it into the same number of parts that the arc aPf of the intrados is divided into (in this instance five), and mark the points of division l', m', n', c'. Transfer the distances ea', lb', mc' between the line CD and the plan of the are of the intrados, to the perpendiculars n"a", "b", m"c", n"d", c"e", and through the points a"b"e"d"e"f" draw a curve, which will be the developement of the arc of the intrados. Pro duce the lines l"b", m'c", n"d", to v", w", x", and transfer the distances b'v, c'w, d'r from the plan to the sofite on the lines b'v", c"w", d′′x". Draw ga", hb', ic", kd" perpendicular to HI; transfer the distances g'a, h'b, i'c from the plan to the sofite upon ga", hb", ic", and join a"v", b"w", x"c", which will complete the moulds of the joints. 1973. To make the drawings for an oblique arch by an abridged method. The following method is said to be abridged, because, by one very short operation the moulds of the sofites and joints are found within the plan of the arch ABDC (fig. 650.). Divide AB in E into two equal parts, and draw EF parallel to AC. From the point A draw AG perpendicular to AC; prolong DB to G; divide AC into two equal parts in the point H. From H, as a centre, describe the arc AFG, which divide into voussoirs, and draw the joints from the centre H. Draw lines from each sofite parallel to EF, and below the line CD; the moulds for the sofites are comprised between the parallels of the key, and those of the joints are traced on the sides of the plan, as follows: 1974. To find the moulds of the sofites. Through the point Q draw QN parallel to GH. To find on RS the point N, through the point K draw KL also parrallel to GH. To find on QT the point M, and on RS the point L, draw the front line of the second sofite MN, and the front line of the first IL. The back of this sheeting sofite is found by the same operation below the plan. The mould of the key is formed by two lines RS, QT, and the front and back lines of the plan AB, CD; the two moulds of the sofites NMTS and LIXV serve to trace the two stones on each side, observing only that the lower arrisses of the sofite on the side AC become those of the top on the side BD; or that the under arriss of one side may be that on the other side by reversing the mould, which will have the same effect. 2 3 G 1975. To find the moulds of the beds or joints. Prolong NQ to meet DG, to find the point P, and through it and the point E draw the front of the second joint P2; prolong LM to GD to find O, through which and the point E draw the front of the joint 03. Proceed in the same manner to find the backs of the other joints, which are sufficient also to trace the stones by reversing them. It is not absolutely necessary to cut out the moulds of the sofites and joints, but the angles may be taken by bevels and applied to stones. The heads are prepared, as usual, with the moulds of the heads of the straight arc. must be observed, that in this arch the face or front differs from a straight arch, being formed by different sections of a cylinder. It DA D R g 78 0 a Fig. 650. 1976. To find the moulds for an oblique arch, whereof the front slopes and the rear are per. pendicular to the axis. Let A'B'GH (fig. 651.) be the plan of the imposts. From the point a', as a centre, describe the arcs ACB, DRI, which divide into five or more equal parts for the arch stones. Draw the joint lines from the centre, and the perpendiculars from the joints below the line AB. From the summits of the perpendiculars, draw lines parallel to AB, to terminate in the perpendiculars DF. From the point D, as a centre, describe arcs from the points which terminate in DF, to meet the line of slope DE in the points m, l, k, E. Draw the lines mr, ls, kt, EF parallel to AB, meeting the perpendicular DF in the points rst F; transfer the distances rm, tk, uP from n to b', from o to c', from a' to s', and through the points A'b'c'd'e' B' draw the curve. Find the extrados or outer line Dfghi in a manner similar to that in which the inner curve has been found. Draw the joints b'f, c'g, d'h, e'i. Prolong AH and BG to K and L, and draw the lines b'b, ce, d'd parallel to AK. g 2 3 Fig 651. 1977. To make the straight arches. Draw KL perpendicular to AK, and produce KL to f and g. Transfer the distances between the points m, l, k, D, and the line QD to the ordinates of the lower are from b to v, from c to w, from d to x, and from Q to Y, and draw the curve Kvwxy L. Also find the outer curve in the same manner, and draw VT at right angles to AH. Draw 1978. To find the moulds of the sofites. the line WX (fig. 652.) in any convenient surface, and lay the breadths of sofite, not from the arc ACB as before, but from those of the right arc KvwxyL, that is, transfer the distances Kv, vw, wx, xy, yL to the line WX upon Wa, aa, Bb, gyod, and 8X. Through the points Wabgd X, draw the lines ny, ei, fk, gl, hm, dz, perpendicular to Fig. 652. |