[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_. II. 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. I ask, given a weight at _a_ what counteracts it in the direction _n_ _f_ and by what weight must the weight at _f_ be counteracted. 778. ON THE SHRINKING OF DAMP BODIES OF DIFFERENT THICKNESS AND WIDTH. The window _a_ is the cause of the crack at _b_; and this crack is increased by the pressure of _n_ and _m_ which sink or penetrate into the soil in which foundations are built more than the lighter portion at _b_. Besides, the old foundation under _b_ has already settled, and this the piers _n_ and _m_ have not yet done. Hence the part _b_ does not settle down perpendicularly; on the contrary, it is thrown outwards obliquely, and it cannot on the contrary be thrown inwards, because a portion like this, separated from the main wall, is larger outside than inside and the main wall, where it is broken, is of the same shape and is also larger outside than inside; therefore, if this separate portion were to fall inwards the larger would have to pass through the smaller--which is impossible. Hence it is evident that the portion of the semicircular wall when disunited from the main wall will be thrust outwards, and not inwards as the adversary says.