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XIII. Theoretical Writings on Architecture Index

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That part of the arch which is nearer to the horizontal offers least resistance to the weight placed on it.

When the triangle *a z n*, by settling, drives backwards the 2/3 of
each 1/2 circle that is *a s* and in the same way *z m*, the reason
is that *a* is perpendicularly over *b* and so likewise *z* is above
*f*.

Either half of an arch, if overweighted, will break at 2/3 of its
height, the point which corresponds to the perpendicular line above
the middle of its bases, as is seen at *a b*; and this happens
because the weight tends to fall past the point *r*.--And if,
against its nature it should tend to fall towards the point *s* the
arch *n s* would break precisely in its middle. If the arch *n s*
were of a single piece of timber, if the weight placed at *n* should
tend to fall in the line *n m*, the arch would break in the middle
of the arch *e m*, otherwise it will break at one third from the top
at the point a because from *a* to *n* the arch is nearer to the
horizontal than from *a* to *o* and from *o* to *s*, in proportion
as *p t* is greater than *t n*, *a o* will be stronger than *a n*
and likewise in proportion as *s o* is stronger than *o a*, *r p*
will be greater than *p t*.

The arch which is doubled to four times of its thickness will bear four times the weight that the single arch could carry, and more in proportion as the diameter of its thickness goes a smaller number of times into its length. That is to say that if the thickness of the single arch goes ten times into its length, the thickness of the doubled arch will go five times into its length. Hence as the thickness of the double arch goes only half as many times into its length as that of the single arch does, it is reasonable that it should carry half as much more weight as it would have to carry if it were in direct proportion to the single arch. Hence as this double arch has 4 times the thickness of the single arch, it would seem that it ought to bear 4 times the weight; but by the above rule it is shown that it will bear exactly 8 times as much.

The column *c b*, being charged with an equal weight, [on each side]
will be most durable, and the other two outward columns require on
the part outside of their centre as much pressure as there is inside
of their centre, that is, from the centre of the column, towards the
middle of the arch.

Arches which depend on chains for their support will not be very durable.

The arch itself tends to fall. If the arch be 30 braccia and the interval between the walls which carry it be 20, we know that 30 cannot pass through the 20 unless 20 becomes likewise 30. Hence the arch being crushed by the excess of weight, and the walls offering insufficient resistance, part, and afford room between them, for the fall of the arch.

But if you do not wish to strengthen the arch with an iron tie you
must give it such abutments as can resist the thrust; and you can do
this thus: fill up the spandrels *m n* with stones, and direct the
lines of the joints between them to the centre of the circle of the
arch, and the reason why this makes the arch durable is this. We
know very well that if the arch is loaded with an excess of weight
above its quarter as *a b*, the wall *f g* will be thrust outwards
because the arch would yield in that direction; if the other quarter
*b c* were loaded, the wall *f g* would be thrust inwards, if it
were not for the line of stones *x y* which resists this.