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fourth sort, the most noted examples are those of Westminster Hall, 68 feet span (fig. 196.); Hampton Court, 40 feet span; Eltham Palace, 36 feet 3 inches; Beddington Hall;

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South Wraxhall, 19 feet 9 inches; Croydon, 37 feet 9 inches, &c. It will be well to notice, what is not usually known, or shown, in the sections of the Westminster roof, that the main purlines over the strut, are upheld with the collar-beam by an intermediate rafter of great strength. Mr. S. Smirke has observed (Archæologia, xxvi. page 417-18), that "this roof is the common collar-beam roof, and of extremely simple construction; the whole pressure is carried by the straight lines of the principal rafter, and (curved) brace, above alluded to, directly into the solid wall, where it ought to be." The examples of lesser importance, as regards span, are not all of the same elegance as that of Westminster, which, at the same time that it is the largest and best, is also the earliest (1397) of the series. Some examples present double hammerbeams, forming a sort of corbelling over up to the ridge or to the collar-beam. Fig. 701t., from Knapton Church, Norfolk, is 32 feet span, and is a fair specimen of such roofs. The wall is 2 feet 10 inches in thickness. For all these examples, we are indebted to the excellent publication by Brandon, Mediaval Roofs, to which work we must refer the reader for details of decoration and painting, as the above figures are only here introduced to show the principles of construction displayed in such roofs.

15.3

Fig. 701t. KNAPTON, NORFOLK.

2052u. In fig. 701u, we give the modern roof, of 31 feet 2 inches span, over the nave of Bickerstaffe Church, Yorkshire, designed by Sydney Smirke, R. A., as a good specimen of the adaptation of modern science to mediæval structures. The collar-beam is double, each 9 in. by 3 in., through which the king-post is tenoned and strapped. The purlines are 7 in. by 4 in.; the brace, 9 in. by 7 in.; and the corbels are 11 in. wide, being also tailed in 11 in.

Fig. 701u.

-15.7

BICKERSTAFFE, YORKSHIRE,

2052v. V. Aisle, or Lean-to, roofs, may be described as usually consisting of strong timbers, answering the purpose of principal rafters, laid at each end on plates, the lower plate resting on the external wall, the upper one either supported on corbels projecting from the nave wall, or inserted therein. Wall-pieces are tenoned into the upper and lower extremities of the principals, and curved braces springing from the feet of these meet in the centre of the principal, forming a perfect arch, having the spandrils generally filled in with tracery. A purline is usually framed into the principal, and on this and the plates the common rafters are supported (see also fig. 7010.). In aisle roofs the whole of the timbers, even to the common rafters, were frequently found more richly moulded than those of the nave, possibly from being nearer the eye of the spectator.

2053. The following instructions relative to the lines necessary to be found in the framing of roofs are from Price's British Carpenter; and although published nearly 100 years, few equent works on this subject give more complete information. 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, c
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 1. Make ts equal to Cb,
and ab equal to ta, which

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will be the skirts of the roof.

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 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 abcd, 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 . By the same kind of operation oc will be

Join ge, and from r as a centre describe an arc
Then join gu and ue, and gue is the back of the

plan of a building whose sides are bevel to each

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found, as also the other parts of the skirts of the roof. The lines nt, tfv, and up 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 are 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, wnich, 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.

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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 r of the semi

Fig. 705

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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 ghe 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 turc, 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 B 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 uf of the required form, which must serve to regulate the form of the other ribs. Through any points cde in the arc Bcdef A draw the straight lines cq, dr, es cutting the diagonal HD at q, r, s. Draw gh, ri, sk parallel to al cutting the chord BG at the points I x, y, z, b. Make xh, yi, zk, bl each respectively equal to te, ud, ve, wf, and through the points Ghikl to B, draw the curve Ghikl B. Draw qm, rn, 80, 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.

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Fig. 706.

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B

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.

TT

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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 klo draw the lines et, fu, gw parallel to dp, and through the points 1, 2, 3 draw the lines mg, nr. os. Make Im, 2n, So respectively equal to ke, lf, rg; as also 1q, 2r, 3s equal to kt,

lu, vw; that is, equal to ke, If, 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 perpendicula 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.

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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, 4l, 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 octagonal dome

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