« ΠροηγούμενηΣυνέχεια »
On the diameter BC describe a semi-circle, and divide the quadrant into any number of equal parts, ef, fg, gh, &c., and from the points e, f, g, &c., draw lines, parallel to the axis Fk, to meet the seat ab of the groined line, or line of intersection of the two surfaces. From the points k, l, m, &c. of intersection, draw the lines KQ, IR, MS, &c., parallel to the axis of the other vault, to meet the line VQ, perpendicular to that other axis in the points Q, R, S, &c. Then, upon any line, DE, transfer the points Q, R, S, &c. to q, r, s, &c., and draw qu, rw, sx, &c. perpendicular to DE, and transfer the ordinates Fe, Gf, Hg, &c., of the semi-circle, to qu, rw, sx, &c., and through the points v, w, x, &c. draw a curve; then quЕ will be half of the section required.
To find the covering of the semi-cylinder. Upon any straight line, YZ, No. 2, set off the distances lm, mn, no, &c., each equal to the chord ef or fg, &c., in No. 1; and draw (K, mL, nM, &c., in No. 2, perpendicular to YZ. Make IK, ML, NM, &c., No. 2, equal to Lk, MI, Nm, &c., of No. 1, and through the points K, L, M, &c., No. 2, draw a curve. Then will the figure KIZ be half of the covering of the cylinder.
To construct the covering, No. 3, for the great opening.
In the straight line vq, No. 3, make vu, ut, ts, &c., equal to the parts, Ex, xy, yx, &c., of the elliptic curve, No. 1. In No. 3, draw vB, 10, tN, sM, &c., and make vB, u0, tN, sM, &c., No. 3, equal Vb, Vo, Tn, Sm, &c., No. 1; and in No. 3, draw a curve through the points B, O, N, M, &c.; then quBKq will be the covering required.
GROINS AND ARCHES.—PROBLEM 2. (fig. 2, pl. VIII.) To find the groin of a cylindro-cylindric arch.
Let A, A, A, A, be the plans of four piers, which form the openings of different widths.
On the lesser opening PM, as a diameter, describe a semicircle. Divide the quadrant next to P into any number of equal parts, and through the points of section draw the lines 16, 2H, 31, &c., perpendicular to PM, cutting PM in B, C, D, &c., and through the same points 1, 2, 3, &c., draw the lines la, 2b, 3c, &c., parallel to PM, cutting a line qe perpendicular to PM in the points a,b,c; produce the line which contains the points a, b, c, through the greater opening; and upon the part of the line thus produced, which is intercepted between the piers A, A, describe a semi-circle. Produce the line MP to k, and from q describe arcs af, bg, ch, &c., cutting Bk in the points f,g,h, &c. Draw fk,gl, hm, &c., parallel to the base of the greater semicircle, to cut the arc of the same in the points, k, l, m, &c. From the points k, l, m, &c., draw the lines kG, TH, MI, &c., parallel to PM; then, through the points G, H, I, K, L, draw a curve GHIKL, which will be the seat of the groin.
The covering to coincide with the groin is shown at No. 1. Draw pm, No. 1, and make pb, bc, cd, &c., each equal to P1; 1,2; 2, 3, &c., in the semicircular arc.
In No. 1, draw pq, bg, ch, &c., respectively equal to BG, CH, DI, &c., and through the points q, g, h, i, &c., draw a curve; then will ponm be the covering required.
GROINS AND ARCHES.--PROBLEM 3. (fig. 3, pl. VIII.) To find the diagonal or groin-rib of a Vault, of which the lesser openings are semi-circles, and the groins, in vertical planes, passing through the diagonals of the piers.
On ah, (fig. 3, pl. VIII,) the perpendicular distance between two adjacent piers of the lesser opening, describe a semi-circle, abh; and, in the arc, take 1, 2, 3, &c., any number of points, and draw the lines 11, 2m, 3n, &c., cutting the diagonal ik, in l, m, n, &c. Draw, as before, lq, mr, ns, &c., perpendicular to ik, and through the points i,q, r, s, &c. draw a curve; then iuk will be the edge of the rib to be placed in the groin.
The edge of the rib, for the other opening, will be found thus: From the points l, m, n, &c., draw the lines, II, mK, nL, &c., parallel to the axis of the opening of the larger body, cutting HB at the points C, D, E, &c. Make CI DK, EL, &c., each equal to cl, d2, e3, &c.; then, through the points B, I, K, L, &c., draw a curve ; and the line thus drawn will be in the surface of the greater opening, so that BNH will be one of the ribs of the body-range.
The method of placing the ribs is exhibited at the lower end of the diagram, fig. 3, the ribs of each opening being placed perpendicular to the axis of each groin.
GROINS AND ARCHES.—-PROBLEM 4. (fig. 4, pl. VIII.) To find the groined and side ribs of a LUNETTE, where the groined ribs are in vertical planes upon the straight lines ag, gl, (fig. 4, pl. VIII, the principal arch being a semi-circle.
Let AC be the base of one of the principal arches, perpendicular to one of the sides of the main vault, the points A and C being in the same range with those sides. Let
mg be the opening of one of the lunette windows. From the point g, the meeting of the two seats of each groin, draw gr perpendicular to mq, cutting my at n; draw g3 parallel to mq, cutting the semi-circular arc ABC at 3. Between A and 3 take any number of intermediate points, 1, 2, &c., and, through the points 1, 2, &c., thus assumed, draw le, 2f, &c., cutting the seat ag, of the first groin, in the points e, f, &c., and AC in b, c, d, &c. Perpendicular to ag draw eh, fi, &c., and make eh, fi, gk, each equal to 61, c2, d3, &c.; then, through the points g, h, i, k, draw a curve, which will form the groin belonging to the seat ag. From the points e, f, &c., draw lines et, fs, &c., cutting qm in the points p, o, &c.; and, through the points q, t, s,r, draw the curve qtsr, which will be one of the ribs of the lunette.
GROINS AND ARCHES.—PROBLEM 5. (fig. 5, pl. VIII.) Given one of the ribs of a LUNETTE, and a rib of the main arch, to determine the seat of the groin, or the seat of the intersection of the two surfaces.
This is, in fact, a cylindro-cylindric arch; we shall therefore refer the reader to Problem 2, for the geometrical construction of the same.
LUNETTES are used in large rooms or halls, and are made either in waggonheaded ceilings, or through large coves, surrounding a plane ceiling: they have a very elegant effect when they are numerous, and disposed at equal distances. Though it is not necessary to have the axes of the lunettes and the axes of the quadrantal cylindric surfaces in the same plane, they have the best effect when executed so; as the groin, formed by the meeting of the two surfaces, has, in this case, less projection : 'and, though the groins are eurves of double curvature, their seats are perfect hyperbolas, and may be described independent of the rules of projection, the summit or vertex of the curve being once ascertained: by these means we shall have the abscissa and double ordinate ; the transverse axis being the distance between the opposite curves.
GROINS AND ARCHES.-PROBLEM 6. (fig. 1, pl. IX.) To find the groin of a CYLINDRO-CYLINDRIC ARCH, and the moulds for the boarding
A Cylindro-cylindric Arch is the intersection of one semi-cylinder, of a less diameter, with another of a greater diameter. The principal objects to be found are, the seat of the curve on the plan, and the moulds for terminating the ends of the boards.
For this purpose, on any straight line, which has A at one of its ends, as a diameter, describe a semi-circle, as at No. 1, in the figure, terminating in A, for the section of the greater vault, or semi-cylindric arch. As the axis of the one cylinder is supposed to cut the axis of the other at right angles, the sides of the cross-vaults will also be at right angles to each other : therefore draw the diameter AC, of the lesser vault, perpendicular to the diameter of the greater vault; and on AC, as a diameter, describe the semi-circle ABC: divide the quadrantal arc AB into any number of equal parts, as here into five. Draw Ae perpendicular to AC, and produce CA to'k. Through the points of division, in the quadrantal arc AB, draw la, 2b, 3c, 4d, Be, cutting Ae, in a,b,c,d,e. Again, through the same points 1, 2, 3, 4, B, in the quadrantal arc AB, draw straight lines 19, 2r, 35, 4t, BD, perpendicular to AC. From the point A, as a centre, with the several distances Aa, Ab, Ac, Ad, Ae, describe the arcs ek, di, ch, bg, af, cutting Ak in f, g, h, i, k.
Parallel to the diameter of the greater semi-circle, or parallel to Ae, (fig. 1, No.1,) draw fl,gm, hn, io, kp, cutting the greater semi-circular arc in the points 1, m, n, o, p. Through the points l, m, n, o, p, draw lq, mr, ns, ot, pD, parallel to AC, cutting the perpendiculars 19, 2r, 38, 4t, BD, in the points q, r, s, t, D. Through the points A, q, r, s, t, D, trace a curve by hand, or put in nails at the points A, q, r, s, t, D, and bend a thin slip of wood so as to come in
contact with all the nails; then, by the edge of this slip, which touches the nails, draw a line with a pencil, or find points; and the curve thus drawn will be half the seat of the rib. The other half, being exactly the reverse, may be found by placing the distances of the ordinates at the same distance from the centre, upon the diameter AC, and setting up the perpendiculars by making them respectively equal to the others.
It will perhaps be eligible to make the whole curve ADC at once.
The mould for cutting the ends of the boards, which are to cover the centres of the lesser openings, will be found as follows:
On any straight line, C5, as on the diameter AC produced, set off the equal parts A1; 2,3; 3,4; 4B; of the quadrant AB,on the straight line C5, from Cto 1, from 1 to 2, from 2 to 3, from 3 to 4, from 4 to 5, and draw the straight lines lu, 20, 3w, 4x, 5y, perpendicular to C5. Make lu, 2v, 3w, 4x, 5y, each respectively equal to each of the ordinates comprehended between the base AC, and the seat AD; then, through all the points C, u, v, w, x, y, draw a curve Cuvwxy, as before; then the shadowed part, of which the curve line Cuvwxy is the edge, is the mould for one side, which may also be made use of for the other.
To apply this mould, all the boards should be laid together, edge to edge, on a flat or plane surface, to the breadth C5. Draw a straight line C5, perpendicular to the edge of the first board, at the distance of 5y from the end. At the distance C5 draw a perpendicular 5y, and set off the distance 5y, Then apply the proper edge of the mould from C to y, as exhibited in the plate, and draw a curve across the boards, and cut their ends off by the line thus drawn; then the ends, thus formed of the remaining parts, will fit upon the boarding of the greater vault, after being properly bevelled, so as to fit upon the surface of the said boarding.
No. 4, of fig. 1, exhibits the curve, in order to draw or discover the line on the boarding of the greater vault, in order to place the boarding of the lesser vault.
Nos. 2 and 3, fig. 1, show the method of forming the inner edges of the ribs, so as to range with the small opening. The under edge of the rib must be