Page:Great Neapolitan Earthquake of 1857.djvu/111

and

$2y={\frac {L\sin \rho }{\sin \theta }}$

but as ${\text{L}}+{\text{M}}:{\text{L}}-{\text{M}}::\tan {\frac {1}{2}}\rho :\tan {\frac {1}{2}}(\sigma -\tau )$

$\tan {\frac {1}{2}}(\sigma -\tau )={\frac {({\text{L}}-{\text{M}})\tan {\frac {1}{2}}\rho }{{\text{L}}+{\text{M}}}}$,

and $\angle$$\tau ={\frac {1}{2}}\rho$, + arc corresponding to, $\tan {\frac {1}{2}}(\sigma -\tau )$
$\angle$$\sigma ={\frac {1}{2}}\rho$ – the same arc.

Again, $\sin \tau :{\text{L}}::\sin \phi :2x$

$2x={\frac {{\text{L}}\sin \phi }{\sin \tau }}$ = the distance from one fracture to the other diagonally opposite

$2y$ = the resultant of $f$ and $f'$

${\text{N}}:y::1:\tan \theta ''$

$\tan \theta ={\frac {y}{\text{N}}}$

which gives the angle of emergence, or that made by the polar $a,~b,$ of the subabnormal wave, with the horizon.

${\text{Cos}}\theta '':y::1:z$

$z={\frac {y}{\cos \theta ''}}$

$2\times {\frac {y}{\cos \theta ''}}$ = R, the common resultant in the polar $ab,$ and, $2y\tan \theta ''=f''$ the vertical component.

An extremely easy method may be practised of finding the path of a subabnormal wave by an observer in the field.

Referring to Fig. 44. Let a line be stretched across the top of the walls (or anywhere below that, but horizontally), from the exterior or interior angle of fracture,