wall so as to be incapable of tilting, and another edge resting either on the ground itself, or on the edge of the preceding stair stone or platform, as the case may be. The stones which form a platform are generally of the same thickness as those formning the steps.

ON THE SCIENTIFIC OPERATIONS OF STONE CUTTING.

1930. The operations by which the forms of stones are determined, so as to combine them properly in the various parts of an edifice, are founded on strictly geometrical principles, and require the greatest care and exactness in execution. It is only by a thorough know. ledge of the nature of these operations that the master mason is able to cut and carve the parts which, when joined together, compose the graceful arch, the light tracery of the Gothic vault, or the graceful and magnificent dome. The method of simple walling, and its general principles, have been given in this book, chap. i. sect. x. In what follows we propose to confine ourselves, 1st, to the leading operations necessary to set out the simple arch or vault, and the groins formed by it; 2d, to the forms produced by vaults with plain and curved surfaces intersecting; 3d, and lastly, to dome vaulting; giving such examples as will so initiate the student that he may, we trust, have little, if any, difficulty in resolving any case that may occur, and reminding him that if he well understand the section already submitted to him on Descriptive Geometry, his labour will be much abridged, not only in what immediately follows, but in that section which treats hereafter on Carpentry.

1931. I. OF THE CONSTRUCTION OF ARCHES AND SIMPLE VAULTS, AND THE GROINS FORMED BY THEIR INTERSECTION. In arches and simple vaults we have to ascertain the exact form of the arch in all its parts, and the direction of its joints; both which points are dependent on the geometrical properties of the curve used for the arch.

1932. To find the joints of a flat arch without using the centre of the circle of which the arch is a part. Divide the arch AB (fig. 619.)

into as many equal parts as there are intended to be arch stones, at the points 1, 2, 3, &c.

From

def

*

3

45

Fig. 619.

XD

A, with any convenient radius, describe an arc at a, and from 2, with the same radius, describe another arc, crossing the first at a, and join al; then I is the first joint from A. To find the joint passing through 2; with the same radius as before, from the joints 1 and 3 as centres, describe arcs cutting each other at b, and draw 2b; then 2b is the second joint. In the same manner all the other joints between A and B will be found. To find the skew backs, or abutting joints AC and DB; with a radius equal to la, from the centre A describe an are at C; from the centre 1, with the radius Aa, describe an arc cutting the former at C, and draw the line AC, which will be the springing bed of the arch. In the same manner the joint BD may be found.

1933. The joints of any arch may be drawn with considerable accuracy by setting off at equal distances a point in the curve on each side of the place for the joint, and from these points, as centres, with any radius, arcs to intersect, through whose intersections lines being drawn, will give the directions of the joints.

1934. To draw an elliptical arch to any two dimensions by circular arcs. Draw the straight line AB (fig. 620.). Bisect AB in C by the perpendicular Dg, make CA and CB each

equal to half the span of the arch, and make CD equal to the height, and Aj parallel and equal to CD. In Cg make Ck equal to CD. Divide Aj and AC each into two equal parts. Through 1 in AC draw kn, and through 1 in Aj draw 1D, cutting kn at n. Bisect nD by the perpendicular lg, and from g with the radius gn or gD describe the arc n Dih. Draw gh parallel to AB, and join h B, and produce hB to meet the arc n Dh in i. Join gi cutting AB in f and make Ce equal to Cf. Join ge, and produce it to meet the arc n Dh in n. From ƒ with the radius fi describe the arc iB, and from e with the radius eA describe the arc Amn. Then Am DiB is the arch required.

1935. An elliptical arch ADB (fig_621.) being given, to draw the joints for a given number

of arch stones. Find the centres e, f, g in the same manner as if the arch were to be drawn; join ge and produce it to meet the arch; also join g, f and produce it to meet the are in i. Divide the elliptical curve ADB into as many equal parts as the number of arch stones. From the centre e draw lines through the points of division in the curve between A and where ge meets the curve and from the

centre g draw lines through all the intermediate points between ge and gf, and lastly draw lines from ƒ through all the intermediate points between i and B, and the parts of the lines thus drawn on the outside of the curve will be the joints of the arch stones.

1936. In very large arches it will be desirable to find five centres, as in fig. 622., and these will be obtained by finding two intermediate points in each half of the curve instead of one; then bisecting each pair of adjacent points by a perpendicular, we shall have the centres e, h, g, i, f, to be used for drawing the joints in the same manner as in the preceding figure.

Fig. 622.

D

1937. The above methods are sufficient for ordinary purposes; but where strict accuracy is required, the following method is mathematically true. Suppose any joint, as gk, is required to be drawn (fig. 623.), and that the point D is the middle of the arch and the point C the middle of the springing line; then with the distance CA or CB, from the point D describe an arc at e and another at f to cut AB ate and f. Draw eg and fy; produce eg to i and fg to h, bisect hgi by the straight line gk, which will be the joint required. In the same manner, by drawing

Fig. 623.

lines from e and f to each point of division, and bisecting the angle, lines for the other joints may be drawn. 1938. To draw a Gothic arch to any given dimensions (fig. 624.). Draw the straight line

AB equal in length to the span of the arch. Bisect AB in C by the perpendicular DI. and draw AG and BH parallel to DI. Make CD equal to the height of the arch, and the angles CDG and CDH each equal to half the vertical angle; make CF equal to the dif ference between CD and AG and join FA and FB. Divide AG and AF each into the same number of equal parts, counting each from the point A. Through the points 1, 2, 3, 4 in AF draw Ia, Ib, Ic, Id, and through the points 1, 2, 3. 4 in AG draw 1D, 2D, 3D, 4D cutting Ia, Ib, Ic, Id at the points a, b, c, d, then through the points Aabed D draw a curve; which will be half of the Gothic arch required. (Other methods. par. 1943a., et seq.) 1939. To draw the joints of the arch stones of a Gothic arch (fig. 625.). Having formed the angles CDG and CDH as before, make A equal to AG and draw Di perpendiculat to DG. In Dl make Dk equal to Ai and join k. Bisect ik by a perpendicular meeting Dl in l. Produce li to p. Divide the curve into as many equal parts as the arch stones are to be in number. Then i will be the centre of the joints which pass through all the

points between A and p, and I will be the centre for drawing the joints of the arch stones which pass through all the points between p and D.

1940. The reason for the foregoing rule is obvious; for the joints are merely made to radiate to the centres of the arcs of circles whereof the arches themselves are formed; as in subsections 1934, 1935, they were drawn to the centres of the approximating circles wherefrom the elliptical curves were struck.

5

D

3

e

1941. To describe a parabolic curve for a pointed or Gothic arch by means of a series of lines touching the curve, the dimensions of the arch and the angles it forms at the crown being given. Draw the straight line AB (fig. 626.) and draw CD perpendicular to AB. Make CD equal to the height of the arch, CA and CB each equal to half the span. Make the angles CDe and CDf each equal to half the vertical angle. Divide Ae and eD each into the same number of equal parts, and through the corresponding points of division draw lines

Fig. 626.

which will form one half of the arch: the other half DB may be found in the same manner. 1942. To draw the joints of the arch stones Draw the chords

to the above sort of arch.

AD, DB for each half of the arch (fig. 627.); divide the arch into as many equal parts as there are to be arch stones. Let it now be required to draw a joint to any point h: bisect AD in k, and join ek cutting the curve in l. Draw hg parallel to Ak, cutting ek in g, and in el make li equal to lg. Join hi and draw hm perpendicular to hi. Then hm is the joint required. In the same manner all the remaining joints will be found.

E

1943. To describe a rampant pointed arch, whose span, perpendicular height, and the height of the ramp are given. Draw the straight line AB (fig. 627.), and make AB equal to the span of the arch. Draw BC perpendicular to AB, and make BC equal to the height of the ramp. Bisect AC in D, and draw DE perpendicular to AB. Make DE equal to the height of the arch; draw Af and Cg parallel to DE, and make Af and Cg equal to about two thirds of DE. Join ƒE and Eg. Divide Af and fE each into the same number of equal parts, and through each two corresponding points of division draw a straight line. All the lines thus drawn will give one half of the curve. The other half may be

Fig. 628.

J

R

drawn in the same manner. To find the joints of the arch-stones to this sort of arch, proceed as for a plain arch in the last example, as shown by fig. 629.

Fig. 6296

1943a. Besides the rule given in par. 1938, To draw a Gothic arch to any given dimensions, the following plan has been put forward for finding the curves of arches and ribs The fundamental rule that the curves should spring from the line of the impost has been abandoned by many; one centre was to be taken a little above this line, another a little below it, and so on. The following rule furnishes a principle which gives the centres of all these curves with perfect certainty and perfect harmony, at the same time furnishing

what is a further requisite, an independent projection for each rib. The author insists that these curves were always elliptical. If the arch to be drawn be less in height than the half width, let AB, (fig. 629a.) be the half width; BC the height; join AC; draw lines from B and A perpendicular to AC, and the points E and D are those required. Then EC will be the smaller radius, and EC added to AD will be the longer radius. For arches whose height is greater than their half width (fig. 629b.), draw CF and BE perpendicular to AC, then EC will be the smaller radius; and EC added to CF will be the longer radius. The author of this theory is Thos. L'Aker, as read at the Liverpool Arch. Society, 16th October, 1850, and printed in the Civil Engineer, vol. xiii. p. 365. See par. 2002d.

19436. Although the term are en tiers point is still used in France for an arch enclosing an equilateral triangle, as it was in the time of De Lorme, that architect, in his work entitled Nouvelles inventions pour bien bastir, published in 1578, showed that the arc en tiers point was obtained by division of the space into three equal portions, of which two gave the radius. The arc en quatre points was obtained by division into four, three of thent giving the radius. This mediaval mode of determining some of the shapes of pointed arches, was noticed by Professor Willis in his elucidation of Wilars de Honecort's Sketch-book, 1859, p. 138-40. He is disposed to call the equilateral arch, the arch of two points; mentions arches of six points; and instances cases with a radius of five-sevenths and a radius of five-eighths, besides the occurrence of a centre placed to the extent of half the span outside the springing point. The same authority observes that the true method was forgotten soon after the disuse of medieval art, as Viola Zanini, in his book Della Architettura, published 1629, defines the terzo acuto as the arch on an equilateral triangle, the quarto acuto as the arch on a square with the diameter for radius, and the quinto acuto on a pentagon: these last are respectively rather higher and lower than the true arch of four points. The term point is here used as meaning a division, and not a puncture.

8

12

1943c. The Professor has also explained, p. 141, that to know the extra length of a voussoir at the top of an arch of 2, 3, 4, or 5 points, the radius may be prolonged through the point P (fig. 629c.) of the arch to any extent S; then PS being divided into twenty-four parts, a line from S may be drawn parallel with the springing line to T, and respectively 12, 6, 8, or 9 of those parts in length; which will give a point V, so that PV will be the line of the central joint.

1943d. The construction of ogee arches is very simple; but as will presently be shown, the rule is open to judicious variation. The general principle is to draw the line of the Fig. 629c. nose Z, (fig. 629d.) of the hoodmould; to take a point upon that line; to draw from the springing a line through that point to the centre line, to accept the place where the centre line is cut as the height of the ogee, and to find in the usual manner the centre for the upper part of the ogee. The following directions are chiefly taken from Viollet le Duc, Dict.

1943e. To draw an ogee arch of one point (fig. 629d.). Bisect the span in D, draw the

H

K

centre line CD, describe the arc AG, bisect AG in E, and through E from A draw a line cutting the centre line in H; through H draw FK parallel to the springing line, and through E from D draw a line cutting FK at M, which will be the centre for the upper part of the ogee arch. In some cases, as in the figures 629e, f, and h, the three points KMN form an equilateral triangle.

1643f. To draw an ogee arch of two points (fig. 629e.). Bisect the span AB, draw the centre line, and describe G the arcs AGB; then divide G B into five parts G1, &c., and proceed as before.

1943g. To draw an ogee arch of three points (fig. 629f.). Repeat the above operations, observing to divide the span AB into three parts, AE, EF, FB, and to divide GB into four parts, G1, &c. It will be observed that in fig. 629g. (from Pugin), AB is divided into three parts, and the centres E with

F serve to describe the arcs on their own sides of the centre line; that the distances A., II, and 1B are equal, and that EH is equal to EI.

centre line, fix

the four points, and describe the arcs AG, GB; then divide GB into four parts, and

proceed as above indicated. But a difference is taught by an illustration adduced by Viollet le Duc, to show another feature of mediaval art. In fig. 629i. it will be observed that the arch GA is divided into five portions, and that

the line AH is drawn through the second division. The line F2 produced, cuts the horizontal line JH in M; or 2H may be bisected, and a perpendicular obtained meets in the point M, for the ogee line 2H. A centre N has been assumed for the line RR; and also another centre, O, for the line PP, both lines being drawn each way from I; from which arrangement it results that the lines A2H, RIR, and PIP, are not parallel for their whole lengths. In some cases the line of work must be the centre of a fillet or of a boltel. It should be noticed that some very good decorated work of the middle of the 14th century, uses five-eighths of the space for the radius, and finds the centre of the ogee curve upon a line drawn from that central point of radius at an angle of 45° with the hori

D21A
Fig. 629k.

1944. II. OF THE CONSTRUCTION OF INTERSECTING VAULTS OR GROINS. The forms of vaults may be so adapted to one another that the lines of intersection shall be in planes, and these planes the diagonals of the plan of the intersecting part of the vaults; if, however, they be not so adapted, the lines of intersection will be curved on the plan, and these curves it is necessary to ascertain in making both the moulds and the centerings for executing the work. 1945. To determine the form of a vault to intersect with a given one in the plane of the diagonal, and also to find the diagonal rib for the centering. Let the given vault be EIF (fig. 630.) and AC and BD the diagonals, crossing in f. Draw f I perpendicular to EF, eutting EF in c. In the arc IF take any number of points ab, and draw ag, bh parallel to If, cutting EF in d, e, and the diagonal AC in g, h. Draw fp, gq, hr parallel to EF, cutting the base GH at m, n, o. Make mp. nq, or, each respectively equal to ci, du, eb. Draw fl', gk, hl, perpendicular to AC, and make fl, gk, hl respectively equal to cl, da, eb. Make