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*a b* the new wall, *c* the old wall, which
has already settled; and the part *a b* settles
afterwards, although *a*, being founded on *c*,
the old wall, cannot possibly break, having a stable
foundation on the old wall. But only the remainder
*b* of the new wall will break away, because
it is built from top to bottom of the building; and
the remainder of the new wall will overhang the gap
above the wall that has sunk.

774.

A new tower founded partly on old masonry.

775.

OF STONES WHICH DISJOIN THEMSELVES FROM THEIR MORTAR.

Stones laid in regular courses from bottom to top and built up with an equal quantity of mortar settle equally throughout, when the moisture that made the mortar soft evaporates.

By what is said above it is proved that the small
extent of the new wall between *A* and *n*
will settle but little, in proportion to the extent
of the same wall between *c* and *d*.
The proportion will in fact be that of the thinness
of the mortar in relation to the number of courses
or to the quantity of mortar laid between the stones
above the different levels of the old wall.

[Footnote: See Pl. CV, No. 1. The top
of the tower is wanting in this reproduction, and
with it the letter *n* which, in the original,
stands above the letter *A* over the top of the
tower, while *c* stands perpendicularly over
*d*.]

776.

This wall will break under the arch *e f*, because
the seven whole square bricks are not sufficient to
sustain the spring of the arch placed on them.
And these seven bricks will give way in their middle
exactly as appears in *a b*. The reason is,
that the brick *a* has above it only the weight
*a k*, whilst the last brick under the arch has
above it the weight *c d x a*.

*c d* seems to press on the arch towards the
abutment at the point *p* but the weight *p
o* opposes resistence to it, whence the whole pressure
is transmitted to the root of the arch. Therefore
the foot of the arch acts like 7 6, which is more
than double of *x z*.

ON FISSURES IN NICHES.

777.

ON FISSURES IN NICHES.

An arch constructed on a semicircle and bearing weights
on the two opposite thirds of its curve will give
way at five points of the curve. To prove this
let the weights be at *n m* which will break
the arch *a*, *b*, *f*. I say that,
by the foregoing, as the extremities *c* and
*a* are equally pressed upon by the thrust *n*,
it follows, by the 5th, that the arch will give way
at the point which is furthest from the two forces
acting on them and that is the middle *e*.
The same is to be understood of the opposite curve,
*d g b*; hence the weights *n m* must sink,
but they cannot sink by the 7th, without coming closer
together, and they cannot come together unless the
extremities of the arch between them come closer, and
if these draw together the crown of the arch must
break; and thus the arch will give way in two places
as was at first said &c.