The adversary says that this arch must be more than half a circle,
and that then it will not need a tie, because then the ends will not
thrust outwards but inwards, as is seen in the excess at _a c_, _b
d_. To this it must be answered that this would be a very poor
device, for three reasons. The first refers to the strength of the
arch, since it is proved that the circular parallel being composed
of two semicircles will only break where these semicircles cross
each other, as is seen in the figure _n m;_ besides this it follows
that there is a wider space between the extremes of the semicircle
than between the plane of the walls; the third reason is that the
weight placed to counterbalance the strength of the arch diminishes
in proportion as the piers of the arch are wider than the space
between the piers. Fourthly in proportion as the parts at _c a b d_
turn outwards, the piers are weaker to support the arch above them.
The 5th is that all the material and weight of the arch which are in
excess of the semicircle are useless and indeed mischievous; and
here it is to be noted that the weight placed above the arch will be
more likely to break the arch at _a b_, where the curve of the
excess begins that is added to the semicircle, than if the pier were
straight up to its junction with the semicircle [spring of the
arch].

AN ARCH LOADED OVER THE CROWN WILL GIVE WAY AT THE LEFT HAND AND
RIGHT HAND QUARTERS.

This is proved by the 7th of this which says: The opposite ends of
the support are equally pressed upon by the weight suspended to
them; hence the weight shown at _f_ is felt at _b c_, that is half
at each extremity; and by the third which says: in a support of