Page:Great Neapolitan Earthquake of 1857.djvu/103

Rh parallel to its plane, $$c h$$, and both through the centre of gravity or of inertia.

The component $$c h$$ produces the fissure $$n$$. The resistance of the base of the wall, $$e h$$, in the directions $$d e$$ and $$g h$$, uniformly along its length, may be resolved into a parallel force at $$f c$$ in the centre of the length and at the level of the base, which is below the the [sic] level of the component force, $$c f$$, through the centre of gravity, there is therefore a dynamic couple, tending to turn the wall to the left round $$e h$$, and this produces the fissure $$w$$.

The fissure $$n$$ may take place anywhere along the wall $$e h$$, but usually occurs towards $$e$$, at a distance from $$h$$, that is to the distance of $$w$$ from $$h$$, approximately as the component $$f c$$, is to that $$c h$$, so that a line drawn across from $$n$$ to $$w$$, either at the inside or outside faces of the walls, will cross the path of the wave at right angles more or less exactly. The position of $$n$$, however, is subject to many disturbing circumstances, and it never occurs (as the proportion of the figures might seem to infer) so far from $$h$$ towards $$e$$, that the masonry of the wall $$e h$$, cannot yield sufficiently to the shearing strain in the direction of its length, to admit of the opening of the fissure.

Now from the observation of the widths of these respective fissures, we are in a position to find the angular direction in which the path of the wave has traversed the building. For referring to Fig. 36, let $$w$$ be the wider and $$n$$ the narrower fissure, whose widths of ope are proportionate to the components $$a$$ and $$b$$; the path of the wave, being $$a$$ to $$b$$, $$\theta$$ and $$\theta'$$ the angles (together = 90°) made by these components with it, and which we require to find, then—