PRINCIPLES OP GUNNERY. 
281 
Similarly, 
Similarly, 
.F= ’3443 sin 1° 52' 
= 112-2 ft. 
T — § {T im — 2 7 1680 ) 
= 2-48 secs. 
since F 1168 = 1-9081 
^ 80 = -0493 
1-8588 
t 
H 
To explain: the arc OII (where II is the vertex of the trajectory) 
lias been computed, and 
00= X=. 3441 ft. 
HQ = F = 112 ft,, 
or the height which the shot would have to fall from the vertex to 
reach a horizontal line through the muzzle of the gun. But if the 
range is measured to a lower level, G, the distance UG must be added 
to the height the shot has to fall—suppose in this case 15 ft. ; then 
the height the shot has to fall from the vertex is 112 + 15 = 127 ft. 
In computing the descending branch, it is necessary to decide upon 
some angle at the end of the arc and as near as possible to the actual 
angle of descent. In this case compute the descending branch from 
0° to 5°. 
Then, proceeding as before, since D = 5°, 
d 2 
D 0 = D v + — D = Z ) 1168 + f x 5 = 6-3 ; since D 1168 = 2 550 
w + 3-75 
.-. v' = 953-5 f.s., 6/300 
and 
p = 2° 30' - 
1168_ — 953 
1168 + 953 
= 2° 30' — 10' = 2° 20'; 
then 
X = 4- cos 2° 20' (# 953.5 — #n 68 ) 
= 2917 cos 2 ° 20' = 2911 ft., 
T = 2917 sin 2 ° 20' = 119 ft. nearly. 
i.e., there is yet 127-119 ft. == 8 ft. to fall. 
since # 953 . 5 = 4 8 5 3 
^iiG8 = 2665 
2188 
4 
,3)8752 
2917 
