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.
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_.
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 spaces between the curves are filled in and arched across.
This drawing of a cloister from a photograph shows the correctness of our perspective, and the manner of applying it to practical work.
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.