PRINCIPLES OP GUNNERY. 
267 
Now, to find tlie time tlie projectile would be at P, substitute 
T — 2*2 andy = 8*05 in equation, (8), or 
= - + a / ~ - = i-i + Vvn — 
2 T / V 4 32-2 
'o; 
or 1*94 secs. 
The remaining velocity at P at the end of 1*94 secs, can be found 
thus: 
T,- d2 T+T v -( n '^ x 1-94+ 2’ 
f v ,--1+ 1 V -gQg- X 1 94 + J wo 
= -3731 + *8057 = 1*1788 secs.; 
or v' = 1319 f.s. 
The distance travelled by the projectile in 1*94 secs, can be found 
thus : 
_ S v , — $ v __ ^319 - ^1420 „ 1762*4 - 1251*1 
P “ *1923 ” *1923 , 5 
w 
or S' = ON = 2659 ft. 
But from figure, 
NB — OB — ON = 3000 - 2659 = 341 ft .; 
or the position short of 3000 ft. in which the side of the ship would be 
hit is 341 ft. 
Example 3.— Find the height of the trajectory of the Martini-Henry 
rifle for a range of 500 yds. at intervals of 100 yds. Weight of bullet 
480 
= 480 grs. = lbs/* = *06857 lbs., diameter of bullet = *45 ins., 
muzzle velocity = 1353 f.s. 
The remaining velocity at 500 yds. = 1500 ft. can be found by 
Table I. thus: 
S 0 = ,Si 863 + x 1500 = 1584 + 2-953 X 1500 = 6013-5 ; 
UOoD I 
or remaining velocity at 500 yds. = 889 f.s. 
Then time over 500 yds. can be found by Table II. thus : 
2*953P = T m - T im = 5*2762 - 1*0461 = 4-2301 j 
or time over 500yds; =T’432 secs. 
Method of 
finding 
height of 
trajectory 
at given 
distances. 
* 7000 grs. Troy make 11b. Avoirdupois. 
34 
