an oblique section of a cone. The difference
between the two figures consists in their centres not being in the same
place, that of the perspective circle being at _c_, higher up than _e_
the centre of the ellipse. The latter being a geometrical figure, its
long diameter is exactly in the centre of the figure, whereas the centre
_c_ and the diameter of the perspective are at the intersection of the
diagonals of the perspective square in which it is inscribed.

[Illustration: Fig. 167.]

XC

FURTHER ILLUSTRATION OF THE ELLIPSE

In order to show that the ellipse drawn by a loop as in the previous
figure is also a circle in perspective we must reconstruct around it the
square and its eight points by means of which it was drawn in the first
instance. We start with nothing but the ellipse itself. We have to find
the points of sight and distance, the base, &c. Let us start with base
_AB_, a horizontal tangent to the curve extending beyond it on either
side. From _A_ and _B_ draw two other tangents so that they shall touch
the curve at points such as _TT'_ a little above the transverse diameter
and on a level with each other. Produce these tangents till they meet at
point _S_, which will be the point of sight. Through this point draw
horizontal line _H_. Now draw tangent _CD_ parallel to _AB_. Draw
diagonal _AD_ till it cuts the horizon at the point of distance, this
will cut through diameter of circle at its centre, and so proceed to
find the eight points through which the perspective circle passes, when
it will be found that they all lie on the ellipse we have drawn with the
loop, showing that the two curves are identical although their centres
are distinct.

[Illustration: Fig. 168.]

XCI

HOW TO DRAW A CIRCLE IN PERSPECTIVE WITHOUT A GEOMETRICAL _PLAN_

Divide base _AB_ into four equal parts. At _B_ drop perpendicular _Bn_,
making _Bn_ equal to _Bm_, or one-fourth of base. Join _mn_ and transfer
this measurement to each side of _d_ on base line; that is, make _df_
and _df'_ equal to _mn_. Draw _fS_ and _f'S_, and the intersections of
these lines with the diagonals of square will give us the four points _o
o o o_.

[Illustration: Fig. 169.]

The reason of this is that _ff'_ is the measurement on the base _AB_ of
another square _o o o o_ which is exactly half of the outer square. For
if we inscribe a circle in a square and then inscribe a second square in
that circle, this second square wil