# Great Neapolitan Earthquake of 1857/Part I. Ch. XVIII

CHAPTER XVIII.
FOMURLÆ REFERRING TO CAP. D.—BODIES OR STRUCTURES PROJECTED.
Let a body, $\mathrm {A}$ (Fig. 105), such as the coping-stone of a wall, a church bell, a ball or finial, upon a tower summit, &c.,
be thrown from its place by the earthquake wave in its first semiphase (direct shock), in the direction of wave-path $\mathrm {OC}$, and be found projected to the ground at $\mathrm {B}$, in a direction contrary to the wave transit. It is required, if the angle of emergence $e$, of the wave-path at the place, be known, to determine the velocity of projection, and vice versâ. The ​body is projected downwards throughout its trajectory; motion is imparted to it, in virtue of the grasp that its base had of it, by adhesion or otherwise; and the velocity of projection impressed, or that which, of the total velocity of the wave at its moment of maximum, is effective in projection, is the difference, between the maximum velocity of the wave, and that which is destroyed by adhesion, or other equivalent resistances. The larger the mass the greater is the proportion of the total velocity effective.
Were there no adhesion or equivalent resistance (as in the case of a ball balanced on the top of a staff), the body would drop plumb or nearly so, and might be struck by the base (the wall in Fig. 2) in the second semiphase of the wave; or if the velocity of the wave were infinite or extremely great in relation to $g$, the body might, whether adherent or not, be displaced and replaced, without projection. These, however, do not occur. The relation in nature between $\mathrm {V}$ and $g$ is such, that bodies are projected from buildings, &c., in both semiphases of the wave, and the adhesion of the base is most generally of such a nature as to impart a movement of rotation to the body thrown, which is sufficient to turn it over, more or less (usually from the forms found either through 90° or 180° during its descent, notwithstanding its high vertical velocity downwards.
Let the axis of $y$ (Fig. 105), be measured downwards vertically, and that of $x$ horizontally from the origin, at the centre of gravity of the body projected, the trajectory described is
| $y=x\tan {e}+{\frac {x^{2}}{4\mathrm {H} \cos ^{2}{e}}}$ | (XXXVII.) |
$\mathrm {H}$
​If $b$ denote the height through which the centre of gravity has descended, to reach the ground, or the horizontal plane pessing through the centre of gravity, when so deposited, and $a$ the horizontal distance, traversed by the same centre, on striking the ground, then
* $b=a\tan {e}+{\frac {a^{2}}{4\mathrm {H} \cos ^{2}{e}}}$
| $\mathrm {Tan\,} {e}={\frac {-2\mathrm {H} \pm {\sqrt {4\mathrm {H} (\mathrm {H} +b)-a^{2}}}}{a}}$ | (XXXVIII.) |
| $\mathrm {V} ^{2}={\frac {a^{2}g}{2\cos ^{2}{e}(b-a\tan {e})}}$ | (XXXIX.) |
$e=0$
In the case of projection by subnormal wave, observing that the axis of $y$ is measured vertically upwards, and that of $x$ horizontally, from the origin, in the centre of gravity of the body as before, we have for the trajectory (Fig. 106)
| $y=x\tan {e}-{\frac {x^{2}}{4\mathrm {H} \cos ^{2}{e}}}$ | (XL.) |
* $y=-b$
* $x=+a$
​we find
* $-b=a\tan {e}-{\frac {a^{2}}{4\mathrm {H} \cos ^{2}{e}}}$
| $\mathrm {Tan\,} {e}={\frac {2\mathrm {H} \pm {\sqrt {4\mathrm {H} (\mathrm {H} +b)-a^{2}}}}{a}}$ | (XLI.) |
| $\mathrm {V} ^{2}={\frac {a^{2}g}{2\cos ^{2}{e}(b+a\tan {e})}}$ | (XLII.) |
$\mathrm {H}$
$\mathrm {V}$
$e$
Let $\mathrm {A}$ (Fig. 107) be the top of any tower or other elevation from which a body has been projected. From $\mathrm {A}$ draw $\mathrm {AB}$ vertical and $=4\mathrm {H}$ ($\mathrm {H}$ being the height due to the velocity, supposed given). Through $\mathrm {A}$ draw $\mathrm {AX}$ horizontal. Bisect $\mathrm {CB}$ in $\mathrm {Y}$, and on $\mathrm {BC}$ describe the semicircle $\mathrm {BXC}$. Bisect $\mathrm {BA}$ in $\mathrm {O}$, and with $\mathrm {O}$ as centre and $\mathrm {OX}$ as radius ​describe the circle $\mathrm {XT}$. This is the locus circle (i.e. that in which the line of aim—in our case the wave-path, shall cut the vertical drawn through the point $\mathrm {D}$, at which the projectile falls—whether the angle $e$, be above or below, the horizontal line through the point of projection).
Let $\mathrm {D}$ be the point at which the projectile falls to the ground; draw $\mathrm {DFE}$ vertical through its centre of gravity. The directions $\mathrm {AE}$ and $\mathrm {AF}$, formed by its intersections with the vertical, give the superior, and inferior angles of elevation, for the given horizontal range and elevation, and coincide in result with Eq. XXXVIII. and XLI.
The wave-path must always, be either horizontal or emergent. Hence in the first semiphase of the wave, although the motion of the projectile is contrary to that of the wave transit, the angle $e$, given by the above construction, will be the superior one, and also in the second semiphase of the wave, in which the motion of the projectile is in the same direction with the wave transit, the angle $e$ will be still the superior one.
$\mathrm {V}$
$v$
$\mathrm {V}$
​If in the same locality, we are enabled to observe two different bodies, both projected, and to measure the vertical and horizontal distances to the point of fall, we can determine both the angle of emergence of the wave-path ($e$) and the maximum velocity of the wave. Thus, for example, let both the bodies, be projected by the second semiphase of the wave, and let $ab$ and a $a'b'$ denote the coordinates in $x$ and $y$, of the two trajectories; then by Eq. XL. we have
* $-b=a\tan {e}-{\frac {a^{2}}{4\mathrm {H} \cos ^{2}{e}}}$
* $-b'=a'\tan {e}-{\frac {a'^{2}}{4\mathrm {H} \cos ^{2}{e}}}$
| $\mathrm {Tan\,} {e}={\frac {a^{2}b'-a'^{2}b}{aa'(a'-a)}}$ | (XLIII.) |
| $\mathrm {H} \cos ^{2}{e}={\frac {aa'(a'-a)}{4(ab'-a'b)}}$ | (XLIV.) |
$\mathrm {H}$
${\frac {\mathrm {V} ^{2}}{2g}}$
| $\mathrm {V} ^{2}=g\times {\frac {aa'(a'-a)}{2\cos ^{2}{e}(ab'-a'b)}}$ | (XLV.) |