The Theory and Practice of Perspective Part 24

OF ARCHES, ARCADES, BRIDGES, &C.

[Ill.u.s.tration: Fig. 232.]

For an arcade or cloister (Fig. 232) first set up the outer frame _ABCD_ according to the proportions required. For round arches the height may be twice that of the base, varying to one and a half. In Gothic arches the height may be about three times the width, all of which proportions are chosen to suit the different purposes and effects required. Divide the base _AB_ into the desired number of parts, 8, 10, 12, &c., each part representing 1 foot. (In this case the base is 10 feet and the horizon 5 feet.) Set out floor by means of distance. Divide it into squares of 1 foot, so that there will be 8 feet between each column or pilaster, supposing we make them to stand on a square foot. Draw the first archway _EKF_ facing us, and its inner semicircle _gh_, with also its thickness or depth of 1 foot. Draw the span of the archway _EF_, then central line _PO_ to point of sight. Proceed to raise as many other arches as required at the given distances. The intersections of the central line with the chords _mn_, &c., will give the centres from which to describe the semicircles.

CXXVII

OUTLINE OF AN ARCADE WITH SEMICIRCULAR ARCHES

This is to show the method of drawing a long pa.s.sage, corridor, or cloister with arches and columns at equal distances, and is worked in the same way as the previous figure, using distance and base.

The floor consists of five squares; the semicircles of the arches are described from the numbered points on the central line _OS_, where it intersects the chords of the arches.

[Ill.u.s.tration: Fig. 233.]

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SEMICIRCULAR ARCHES ON A RETREATING PLANE

First draw perspective square _abcd_. Let _ae_ be the height of the figure. Draw _aefb_ and proceed with the rest of the outline. To draw the arches begin with the one facing us, _EoF_ enclosed in the quadrangle _EefF_. With centre _O_ describe the semicircle and across it draw the diagonals _eF_, _Ef_, and through _nn_, where these lines intersect the semicircle, draw horizontal _KK_ and also _KS_ to point of sight. It will be seen that the half-squares at the side are the same size in perspective as the one facing us, and we carry out in them much the same operation; that is, we draw the diagonals, find the point _O_, and the points _nn_, &c., through which to draw our arches. See perspective of the circle (Fig. 165).

[Ill.u.s.tration: Fig. 234.]

If more points are required an additional diagonal from _O_ to _K_ may be used, as shown in the figure, which perhaps explains itself. The method is very old and very simple, and of course can be applied to any kind of arch, pointed or stunted, as in this drawing of a pointed arch (Fig. 235).

[Ill.u.s.tration: Fig. 235.]

CXXIX

AN ARCADE IN ANGULAR PERSPECTIVE

First draw the perspective square _ABCD_ at the angle required, by new method. Produce sides _AD_ and _BC_ to _V_. Draw diagonal _BD_ and produce to point _G_, from whence we draw the other diagonals to _cfh_.

Make s.p.a.ces 1, 2, 3, &c., on base line equal to _B 1_ to obtain sides of squares. Raise vertical _BM_ the height required. Produce _DA_ to _O_ on base line, and from _O_ raise vertical _OP_ equal to _BM_. This line enables us to dispense with the long vanishing point to the left; its working has been explained at Fig. 131. From _P_ draw _PRV_ to vanishing point _V_, which will intersect vertical _AR_ at _R_. Join _MR_, and this line, if produced, would meet the horizon at the other vanishing point. In like manner make O2 equal to B2. From 2 draw line to _V_, and at 2, its intersection with _AR_, draw line 2 2, which will also meet the horizon at the other vanishing point. By means of the quarter-circle _A_ we can obtain the points through which to draw the semicircular arches in the same way as in the previous figure.

[Ill.u.s.tration: Fig. 236.]

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A VAULTED CEILING

From the square ceiling _ABCD_ we have, as it were, suspended two arches from the two diagonals _DB_, _AC_, which spring from the four corners of the square _EFGH_, just underneath it. The curves of these arches, which are not semicircular but elongated, are obtained by means of the vanishing scales _mS_, _nS_. Take any two convenient points _P_, _R_, on each side of the semicircle, and raise verticals _Pm_, _Rn_ to _AB_, and on these verticals form the scales. Where _mS_ and _nS_ cut the diagonal _AC_ drop perpendiculars to meet the lower line of the scale at points 1, 2. On the other side, using the other scales, we have dropped perpendiculars in the same way from the diagonal to 3, 4. These points, together with _EOG_, enable us to trace the curve _E 1 2 O 3 4 G_. We draw the arch under the other diagonal in precisely the same way.

[Ill.u.s.tration: Fig. 237.]

The reason for thus proceeding is that the cross arches, although elongated, hang from their diagonals just as the semicircular arch _EKF_ hangs from _AB_, and the lines _mn_, touching the circle at _PR_, are represented by 1, 2, hanging from the diagonal _AC_.

[Ill.u.s.tration: Fig. 238.]

Figure 238, which is practically the same as the preceding only differently shaded, is drawn in the following manner. Draw arch _EGF_ facing us, and proceed with the rest of the corridor, but first finding the flat ceiling above the square on the ground _ABcd_. Draw diagonals _ac_, _bd_, and the curves pending from them. But we no longer see the clear arch as in the other drawing, for the s.p.a.ces between the curves are filled in and arched across.

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A CLOISTER, FROM A PHOTOGRAPH

This drawing of a cloister from a photograph shows the correctness of our perspective, and the manner of applying it to practical work.

[Ill.u.s.tration: Fig. 239.]

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THE LOW OR ELLIPTICAL ARCH

Let _AB_ be the span of the arch and _Oh_ its height. From centre _O_, with _OA_, or half the span, for radius, describe outer semicircle. From same centre and _oh_ for radius describe the inner semicircle. Divide outer circle into a convenient number of parts, 1, 2, 3, &c., to which draw radii from centre _O_. From each division drop perpendiculars.

Where the radii intersect the inner circle, as at _gkmo_, draw horizontals _op_, _mn_, _kj_, &c., and through their intersections with the perpendiculars _f_, _j_, _n_, _p_, draw the curve of the flattened arch. Transfer this to the lower figure, and proceed to draw the tunnel.

Note how the vanishing scale is formed on either side by horizontals _ba_, _fe_, &c., which enable us to make the distant arches similar to the near ones.

[Ill.u.s.tration: Fig. 240.]

[Ill.u.s.tration: Fig. 241.]

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