Page:Great Neapolitan Earthquake of 1857.djvu/394

commencement of its motion, may be assumed to have moved horizontally and free of restraint from the jamb stones. The height of its centre of gravity from the floor, was 4.33 feet; but inasmuch as the small bit of pyramidal brickwork above it, was separated and began to move along with it, we may consider the centre of gravity of the lintel raised by it to 5 feet. The block was thrown from this height to a horizontal distance of 7.2 feet, including that of its having once turned over, towards the south upon its front arris on reaching the floor.

We may further conclude, that as it so came to rest, the velocity impressed, at the first moment of its motion in an horizontal direction, was not greater than sufficient to make it overset upon that edge.

The difference of the side and diagonal of the lintel (in transverse section) = 0.62 feet; 4/3 of this is = $H$, the height due to the horizontal velocity of overthrow.

Then

$b=5{\text{ feet. }}a=7.2{\text{ feet}},$

and

$-b=a\tan {e}-{\frac {a^{2}}{4\mathrm {H} }}$

from which putting in the values we obtain

$\mathrm {Tan\,} {e}=1.472$

and $e=55^{\circ }49^{\prime }$, the angle of emergence.

But $V^{2}=2g\mathrm {H}$, the horizontal velocity; the total velocity, or that in the direction of the wave-path is therefore

$\mathrm {V} ^{2}=2g\mathrm {H} .\sec ^{2}{e}$

whence

$\mathrm {V} =12.863{\text{ feet per second.}}$