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.
| Fig. 237. | Fig. 238. |
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.