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while c stands perpendicularly over d.]
7
76.
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.
7
77.
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
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