lines, and fixed upon the position
of the point of sight, we next mark the position of the spectator by
dropping a perpendicular, _S ST_, from that point of sight, making it
the same length as the distance we suppose the spectator to be from the
picture, and thus we make _ST_ the station-point.

To understand this figure we must first look upon it as a ground-plan or
bird's-eye view, the line V2V1 or horizon line representing the picture
seen edgeways, because of course the station-point cannot be in the
picture itself, but a certain distance in front of it. The angle at
_ST_, that is the angle which decides the positions of the two vanishing
points V1, V2, is always a right angle, and the two remaining angles
on that side of the line, called the directing line, are together equal
to a right angle or 90 deg. So that in fixing upon the angle at which
the square or other figure is to be placed, we say 'let it be 60 deg and
30 deg, or 70 deg and 20 deg', &c. Having decided upon the station-point
and the angle at which the square is to be placed, draw TV1 and TV2,
till they cut the horizon at V1 and V2. These are the two vanishing
points to which the sides of the figure are respectively drawn. But
we still want the measuring points for these two vanishing lines. We
therefore take first, V1 as centre and V1T as radius, and describe arc
of circle till it cuts the horizon in M1, which is the measuring point
for all lines drawn to V1. Then with radius V2T describe arc from centre
V2 till it cuts the horizon in M2, which is the measuring point for all
vanishing lines drawn to V2. We have now set out our points. Let us
proceed to draw the square _Abcd_. From _A_, the nearest angle (in this
instance touching the base line), measure on each side of it the equal
lengths _AB_ and _AE_, which represent the width or side of the square.
Draw EM2 and BM1 from the two measuring points, which give us, by their
intersections with the vanishing lines AV1 and AV2, the perspective
lengths of the sides of the square _Abcd_. Join _b_ and V1 and dV2,
which intersect each other at _C_, then _Adcb_ is the square required.

This method, which is easy when you know it, has certain drawbacks, the
chief one being that if we require a long-distance point, and a small
angle, such as 10 deg on one side, and 80 deg on the other, then the size
of the diagram becomes so large that it has to be carried out on the
floor of the studio with long strings, &c., w