Page:Great Neapolitan Earthquake of 1857.djvu/176

128 therefore

substituting in (Eq. II.) we find

or,

Let $$\delta$$, as before, denote the difference between the diagonal and altitude of the parallelopiped,

$$\mbox{and since } \frac{\alpha^2 + \beta^2}{\alpha^2} = \sec^2 \phi$$

we have the following resulting theorem—

"The height due to the horizontal velocity of wave that will overturn a rectangular parallelopiped is two thirds of the difference of the diagonal and altitude, multiplied by the square, of the ratio of the diagonal to the altitude, or of the secant of the angle $$\phi$$."

3rd. In the case of a solid right cylinder overturned.

The height of the cylinder, or altitude, being $$\alpha$$, and the diameter, or base, $$\beta$$, as before, we have

and

therefore