Page:Great Neapolitan Earthquake of 1857.djvu/451

which is the horizontal velocity necessary to overturn the shaft only.

But it was also projected, so that a point taken at the lower arris of the overthrown shaft had moved horizontally a distance of 9 inches from the corresponding point of the base above $r$ (Fig. 6, Diagram No. 221), while the same point had descended vertically 6 inches from the arris of the base, as already described. Calling the horizontal ordinate A, and the vertical one B, we obtain the horizontal velocity of projection only from the equation

$V^{2}={\frac {g}{2}}\times {\frac {a^{2}}{b-a\tan {e}}}.$

Here $a={\text{0.75 feet, }}b={\text{0.50 feet, }}e=25^{\circ }\,30^{\prime }$

$V^{2}={\frac {32.2}{2}}\times {\frac {0.563}{0.500-0.750\times 0.4769}}$

$=16.1\times {\frac {0.563}{0.142}}$

$V^{2}=63.756{\text{ and }}V={\text{7.98 feet per second;}}$

adding the two horizontal velocities thus found, we obtain the total horizontal velocity impressed upon the shaft; or

$V=0.38+7.98={\text{11.36 feet per second.}}$

But this must be resolved to the direction of the wave-path, whose angle of emergence here we found to be

$e=25^{\circ }\,30^{\prime }.$

The velocity in this direction therefore is

$V\sec {e}=11.36\times \sec {e},$

or finally

$V=11.36\times 1.108={\text{12.586 feet per second,}}$