60 
MINUTES OF PROCEEDINGS OF 
will have been found. The angle of elevation (height of the tangent 
scale) and the deflection are then to be either taken directly from the 
other columns of the tables or interpolated between, the next smaller 
and next larger distances. 
Fig. 6. 
As regards the angles of elevation, it must, however, be borne in 
mind that the angles of descent y t and y 2 (Fig. 6) are always reckoned 
from the straight lines ML and MT, and that therefore also the cor¬ 
responding angles of elevation of the practice tables must likewise 
be reckoned from these lines, and can only be employed when the 
points D or T happen to lie in the same level as the muzzle of the gun. 
If these points lie higher, the trajectory must be raised by the ground 
angle of the point to be struck; if they lie lower, the trajectory must 
be lowered by the corresponding ground angle. Thus, if for the angle 
y x and the distance ML the angle of elevation is found to be fa the 
angle to be used will be fa + n x . 
If the angle of elevation fa is determined from the angle y 2 and the 
distance MT the angle <j) 2 + n 2 must be employed. In both cases must 
the same result ensue. If one of the points L or T lies below the 
level of the muzzle, the ground angles n x or % are to be subtracted. 
A few examples will illustrate more clearly the method of proceeding 
to be observed :— 
Ex. 1.—An escarp wall has to be breached with the short 15 c.m. 
breech-loading gun; but, owing to the distance of the battery from 
which it is to be done, systematic breaching is not contemplated. The 
distance of the top of the crest of the protecting glacis is 1200 metres, 
and the ground angle of this crest is found by measurement to be + 41'. 
Moreover, after considering all the circumstances of the case, it has 
been deemed advisable to fix the lowed point of impact at 5*84 metres 
below the crest of the glacis. The horizontal distance of the point of 
impact from the crest is 28 metres. 
We have now, 
x 1 = 1200 metres, a = 28 metres, 
# 2 = 1228 „ n 1 — + 41', u = 5-84 metres, 
y x = 1200 tan 41' = 14-31 metres, 
y 2 = y 1 — u — 14-31 — 5'84 metres = 8 47 metres, 
tan n 2 = 
8*47 
1228 
= 0-0069, andw 2 =24', 
tan <* = - = ^ = 0-2087. 
Cl 2 o 
