WX. Transfer the distances 1A, 26', 3c', 4d', 6B upon the perpendiculars to WX. From Wn, from a to e, from b to f, from c to g, from d to h, and from X to v, and join ne, ef, fg, gh, hv. In the same manner draw the line yiklmz, which will complete the sofites. 1979. To find the moulds of the joints. Transfer the distances vß, wy, rd, ye, on the line XW from a to a, from b to B, from e to y, and from d to 8, and through the points a, B, y, d draw the lines ny, os, Pt, du perpendicular to WX. Find the points n,o, p, q, as also r, s, t, u, as in the preceding examples; then the moulds of the joints will be eirn, fkso, htig, houg. It must be observed that the boundaries, or extrados and intrados, DRI, ACB of the ring of the arch, do not stand in a plane perpendicular to the plan, but are supposed to be the lines which are drawn on the wall itself; and this is the reason why arcs are described between the perpendiculars DF and the line of slope DE. It must also be observed, that the voussoirs of this arch must be cut by the moulds of the heads of the straight arch, and the moulds of the sofite must be applied on the voussoirs before the sofite is hollowed. Thus, let the first voussoir on the right hand be cut by the head mould on that face of the stone intended for the sofite; apply the first sofite mould, and its upper bed the first joint mould, and on its under bed the plan of the impost. Then cut the two heads according to these moulds, and hollow the sofite square to its arrisses, using for this purpose the curved bevel. 1980. To find the moulds for executing a semicircular-headed arch in a mass of masonry, of which one of its faces is a battering plane upon an oblique plan, and the other opposite face a portion of a cylindric surface. Describe the intrados and extrados of the eleva tion; From 1981. To find the plan of any joint, as that for the line or joint ch in the elevation. Bisect ch in v, draw cm', vw', and hq' perpendicular to AB, intersecting the line VD in the points quc. the points cvh, in the joint ch, draw cc, vv, hh, meeting the line AS in the points cvh, and intersecting the line AR in the points 1, 3, 2 by three intervals, 1c, 3v, 2h. Find the places hvc of the three h S R R S'I points hve on the elevation. In the same manner find the places q'w'm' of the three corresponding points; then will c'v'h'q'w'm' be the plan of the joint required. The plans of the other joints will be found in the same manner. W 1982. To find the joint mould itself. Draw the line HI (fig. 654.) equal in length to the developement of the intrados, and let He be the developement of the arc ac; draw em" perpendicular to HI. Draw any line WX in the plan parallel to VD, intersecting the lines c'm', v'w', h'q', in the points 1, 2, 3. Draw W'X' in the developement or sofite parallel to HI, and at the same distance from HI that WX is from VD in the plan, and let WX intersect the line c'm" in 1. Make the distances 1-2, 2-3 respectively equal to cv, vh, in the joint ch in the elevation, and through the points 1, 2, 3, just found, draw VW, h"q", parallel to C'm". From the plan transfer the distances 2v', 2w', 3h', 3q" to the sofite from 2 to V, and from 2 to W; also from 3h" and from 3 to g the points cup" will be in a straight line, because they correspond to the straight face of the wall, and the points m", w, q" will be in a curve, because they correspond to the cylindric surface. Draw, therefore, the straight line c"h", and draw the curve line m"wq", which will be a portion of an ellipsis, differing in its curvature but in a very small degree from that of a circle drawn through the same three points. However, if more exactness be required, we may find as Fig. 654. many points in the joints of the surface of the wall and in the cylindric surface as we please; then c"m"p"h" is the joint required, which serves for the upper and under beds of the two stones that unite together in that joint. 1983. Find all the other joint moulds b'l'p'g', d'n"v"t", e"o"s"k", in the same manner, and find the points a"f" in the developement. Through the points a"b"c"d"e"f" draw a curve line by hand, or by a ruler bent to the points, and this will be the front curve of the sofite. Find the points k'p" in the developement corresponding to the points a' and ƒ on the plan, and through the points corresponding to the points a' and f on the plan, and the points klm"n"o"p", draw another curve, which will be the developement of the other side of the sofite. The developements of each of the parts of the sofite and of the two adjacent joint moulds give the three moulds for working one stone and the adjacent joints of the stone on each side of it. The angle which each of these joints makes with the sofite is found by making a bevel with one of its edges, circular for the intrados of the arc of the elevation, and the other to coincide with the joint line adjacent. 1984. To find the moulds for executing a gateway in the quoin of a sloping wall. Let ABCD (fig.655.) be the plan of the angle in which the arch is to be constructed, whereof AB is the span. Draw the centre line EL, to which draw the perpendicular FG. Prolong the line CA to F, and DB to G; then from the point L, as a centre, describe the sofite FHG and its extrados. Divide these arcs into equal parts for the arch stones, and from the divisions let fall perpendiculars, and also from the middle of the sofites to EC, ED. From the summits of the perpendiculars draw lines parallel to FG terminated by the lines of slope. Set off the slope at the different heights al, a2, a3, a4 respectively at right angles to the lines on the plan, on d1, b2, d3, b4, K5; also on the opposite side.lay a2, a4 on d2, 64; then on one side draw the curve AbbK, and on the other, to abridge the work, join Bb, bb, bK. Again, for the outer curve, or extrados, set off c1, c2, cG on di, d2, N3. On both sides draw the curve MddO on the one side, and to abridge the labour, draw the straight lines Od, dd, dN. 1985. To find the moulds of the sofites. Draw the line PQ (fig. 656.), on which lay the arc of the sofite FHG in the usual manner, making the points 1, 2, 3, which correspond to the points dividing such are into equal parts; then on the lines of the sofite lay the distances 1986. To find the back curve of the sofite. Lay the distances eo, fp, gq, kr, LE on PT, 10, 2t, 3u, 4v, QU, and trace the curve Totuv U. Q43 21 F n m R Fig. 656. 1987. To find the moulds of the beds or joints. The sofite lines to which the beds belong are 2t and 4v. Draw the straight lines eb, fd parallel to QU, respectively distant from 2t, 4v by the breadth GI of the joint, and let the lines be, fd meet PQ in e and f; make ea equal to gd, and ab equal to dw, and join al, bt; make fe equal to hd, and cd equal to dr, and join nc, vd. To trace the stones by moulds, prepare the voussoirs with the head of the moulds of the straight arch FHG. The sofite should be hollowed in each voussoir by its particular mould: the rest is done as usual; but it must be observed, that if the sofite moulds are made with straight lines in front and near the sofite, it must not be hollowed till the last. The voussoirs may be worked by bevels, preparing the stones by the plans ACVM, BDWO, as for common imposts. Although the arch in each front be not absolutely necessary here, we shall give the method of constructing it. Let the line mn be drawn apart, on which lay the distances L5, L4, L2, LA on the lines ns, nq, no, nm square to mm. Draw the perpendiculars op, qr, st, on which lay the heights of the joints of the straight arch taken on the line of slope; that is, lay 12, on op, 14 on qr, 15 on st, and draw the line nt, which is the slope. Then draw the curve mprt, and from the point n draw the joint lines pr and rX. The centre of this gate is represented (in the upper part of the diagram) with voussoirs, and the keystone placed behind to show the mitre of the centre. The sofite moulds serve for curving the ends of the stone where the intrados meets the surface of the two walls. It must, however, be observed, that, previous to the application of the sofite mould, the concave surface of the intrados must be formed by a mould with a convex edge, and then the sofite mould or moulds of developement must be bent into the hollow, so that the two parallel edges may coincide with the corresponding edges of the stone. The angles which the intrados makes with the joints are taken from the elevation of the face of the arch. This elevation is no more than a section of the arch perpendicular to the axis of the cylinder which forms the intrados. I P30 w 1988. To construct a semicircular-headed arch in a round tower or circular wall. Let ABDC (fig. 657.) be the plan of the tower. Bisect the are AB, and through the point of bisection draw EF parallel to the jamb line AC or BD. Through any point a in EF draw GH perpendicular to EF. Produce the lines CA and DB to meet GH in the points G, H, and GH will be bisected in a. From a, as a centre, and with the radius aG or aH, describe the semicircular arc GFH. Also describe the are of the extrados and divide the arcs each into five equal parts, and let fall the perpendiculars of the joint lines, and those of the middles of the sofite curves to the inside circular line CED of the tower. Having extended the arcs of the intrados curve on the line IK, and having drawn the lines of the sofites and those in the middle of each sheet as before directed, lay off the distances between the right line GH and the circular outside line AbB, viz. GA on IX and on KZ, ed on ef, Vg on hi, Sk on lm, Mn on op, ab on qr; then trace the front curve on the sofite XrZ. To find the rear curve, lay GC on IY, cC on eS, &c., by which the rear curve will be obtained. 1989. We do not consider it necessary to pursue the construction of the moulds, the operations being very similar to those already given in the previous examples. K 22 E y w น Fig. 657. The 1990. To find the moulds for an oblique semicircular arch in a circular tower. construction of this differs from the preceding only in the bevel or obliquity of the tower; hence it requires no particular description; only observing, that the bevel causes the mould to be longer on one side than on the other (see fig. 658.), as is evident from the plan; therefore the distances taken between the right line AB and the circular line of the tower CDE, being unequal, must be transposed each on its particular line of the mould and joint to which it corresponds in the sofite, that is, the distance AC must be laid on FG, BE on HI, and so of the rest. To work the stones, dress the beds, then apply the proper moulds and cut the head and tail circular as before. Trace the breadth of the sofite on the upper bed, then hollow the sofite, and cut the joints by the bevel. cases. Н 1991. To construct an oblique arch in a round sloping tower intersecting a semicircular arch within it. This is nearly the same as the two preceding On one side draw the line of slope (fig. 659.) AB, and on the other the arc CD. Draw parallels from the divisions of the sofites and their middles, as in the figure, in order to cut the line of slope and arc. To work for the slope, set off all the retreats comprised between the perpendiculars AH and the line of slope AB on the perpendiculars of the sofite, square to the front line of the tower F 19 G, as follows: Transfer the retreat 9-10 on 19-20 by placing the compasses so that the line 19-20 would pass through the centre of the tower, and the point 20 fall on the centre of the gate 0—75, and 7-8 on 17-18, and on 21-22 in the same manner (only terminated by the lines Fig. 658, from the sofite instead of the centre line of the arch), set also 5-6 on 15-16, and on 23-24, 3-4 on 13-14 and on 25-26, and lastly 1-2 on 11-12 and on 27-28, and through these points trace the sofite 28-20 -11. The extrados is found in like manner, and the middles of the joints 47, 49, 53; which done, draw the plan of the joints 14-47-35, 18-19-37, 22-51-39, and 26-53-41. 1992. To find the curve of the plan which terminates the tails of the moulds. Set the projections of the buttress of the semicircular arc at right angles to the inside line of the tower; viz. 64-65 on 74-75; 62-63 on 72-73 and on 76-77; 60-61 on 70-71, and on 78-79, 58-59 on 68-69. and on 80-81; 56-57 on 66-67 and on 82-83; then trace by hand the curve 83-75-66. The curves of the extrados and joints are found in the same manner. 1993. To find the moulds of the sofite. Draw the line of direction 94-84 (fig. 660.) as before, below which set off the distances I-11 or 84-85, K-12 on 86-87, L-14 on 88-89, M-16 on 90-91, N-18 on 92-93, O-20 on 94-95, and then trace the front of the sofite moulds 85-95-99. To find the rear, set I-66 on 84-33, K-67 on 86-36, L-69 on 88-100, M-71 on 90-98, N73 on 92-97, 0-75 on 94-96, and trace the rear curve of the mould 101-96-33. E Fig. 660. F 1994. To find the moulds of the joints. Transfer P-19 on 31-54, Q-37 on 32-48, I-47 on 42-52, R-35 on 43-40, and through these points trace the front joint or bed moulds 93-54-48, 89-92-40. To find the rear, make 31-50 equal to PV, 32-38 equal to QX, 42-46 equal to IT, and 43-34 equal to RS; which done, trace the curve lines 97-50-38 and 100-46-34. The two other joints are found by the same method. We do not consider it necessary further to multiply examples of the kind here given the latter sort, especially, rarely occur in practice; and if they should, all that will be necessary to master the operations will be the application of a little thought and study. 1995. III. OF DOME VAULTING. whatever direction a hemispherical dome is cut, the section A is always the same. B represents one half (see fig. 661.) of the same in the plane of projection. The construction is sometimes such that the plan is only a semicircle, as B, as in the termination of the choir of a church: in which case the French call it a cul-de-four; with us it is called a semi-dome. In 1996. Through the extremities of the joints, and through the middle of each sofite of the section A, let fall on the line ab, perpendiculars, whereof all the distances de from the centre c will be the radii of the arcs, which will serve for the developement of the sofites, of the joints, and for the construction of the arch stones. The method which follows, though it will not perhaps give the sofites and joints strictly accurate, will do so sufficiently for all practical purposes. Upon the developement C make SC equal to the arc MDGC, then set out to the right of the points of division the parts ST equal to st on the plan B; then raise through the points T upon the line SC perpendiculars equal Fig. 661. to the correspondents e, t, d of the plan B, and draw the curve ESD through the points so found. 1997. The sofites are terminated by four curves, whereas the joints have two right sides, as DI, EI, and DO, EO, and two curved sides, as II, DE, and OO, DE; the widths DI, DO of the joints are equal to DI, GE of the section; in one direction they are curved only one way, but as respects their sofites they are so in every way. The heights of the voussoirs are given by the section A, their bases on the plan B. Thus G, I, in the voussoir next the keystone, being the most opposite points, the base of it on the plan will be comprised between the two arcs dte, which answer to the perpendiculars let fall from G and I. The base of the first voussoir, according to the first method, will be equal to the surface comprised between the arc aof and the arc dse, which answers to the perpendicular let fall from the point D. 1998. EF and GH are the diameters of the upper and lower bases of a truncated cone, whose lower surface is hollowed out spherically. After working the voussoirs, so as to make their bases such as we have just indicated, they must be worked to sofite moulds for giving them the hemispherical form of the section; after which the angles of the moulds are joined by arcs parallel to the arrisses of each stone, or by applying a general mould of the form of the section, that is, circular, of the radius of the dome. A Section C Elevation с A 1999. For the pendentives formed in an hemispherical dome. The piers D and E are supposed those of half the dome pierced by the pendentives. If we suppose the face or elevation B (fig. 662.) to make one quarter of a revolution about the point A, we obtain the elevations B and C. Through the points of division on the elevation C draw to the are AD right lines perpendicular to CA. On the extremities of these lines upon CA, and from C, as centre, describe arcs in the plan F, by which the plan of the projection on F is obtained, whose intersections with the right lines drawn from B will give the joints and faces for the level beds. The lines HF, FE, ED are right lines. The spaces GAEF, FHIK are pieces of cylindrical vaulting, so that the only difficulty is in joining to each of their voussoirs their correspondent parts in ELMHFE. 2000. The elevation B gives the height of the voussoirs ; Plan B G NK Fig. 662. their bases, as seen in the preceding example, will be OPQRNO, GSTUVKFG. length of the keystone will be XY, and a-A will be half its width. The 2001. The part FQR is the plan of the springing stones of the pendentive in the elevation A. The remaining parts of the construction are sufficiently shown by the lines of the diagram, which will be understood by the student if he has previously made himself acquainted with the previous portions of this section. 2002. We should willingly have prolonged this part of our labours, if space had permitted us to do so without sacrificing other and important objects. If the subject be one in which more than the ordinary practice of the architect is called upon to put into execution, we refer him to Simonin, Coupe des Pierres, Paris, 1792, and Rondelet's Art de Bâtir, which we have used with much freedom, and in which many more interesting details will be found than we have thought it absolutely necessary here to introduce, though we believe we have left no important point in masonry untouched. We cannot close this section without paying our tribute of respect to the masons of this country, who are among the most intelligent of the operative builders employed in it. A very great portion of them are from the north of the island, and possess an astuteness and intelligence which far exceeds that of the other classes of artisans. We must not, however, altogether do this at the expense of those employed in carpentry, which will form the subject of our next section, among whom there will be found much skill and intelligence, when the architect takes the proper means of drawing it out; and we here advise him never to be ashamed of such |