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).

[Illustration: 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 spaces 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.

[Illustration: Fig. 236.]

CXXX

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.

[Illustration: 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