2 
MINUTES OF PKOCEEDINGS OF 
Suppose that 2000$ —= = N } the tabular number corresponding to the 
1 cl* 
mean value of V and v. Then the above equations take the form 
d 2 d 2 
Thus from the above equations we find that s — and t — are functions 
w w 
of V, v, and of N, which depends upon the external form but not on the 
weight or diameter of the projectile. Taking equation (2), suppose that 
V is the initial velocity, and that v 1} v 2 , v 3j &c., are the velocities of the 
1 + 4 
+ 4* 
&c.. 
1000, 
( 1 
W 
( 
^i 3 
d* 
1000 
r_L 
! w 
2 A 2 1 
\ 
d * 
1000 
(1 
' W 
' SAT 
w 
&c.. &c. 
In the calculation of the General Time Table for ogival-headed shot, 
F=1700 f.s., z? 1 =1690 f.s., ?; 2 =1680 f.s., v 3 =167Q f.s., &c., and N x cor¬ 
responds to a velocity 1695 f.s., iV 2 to 1685 f.s., N 3 to 1675 f.s., &c. 
Thus a table was formed showing the times in which the projectile 
successively lost 10 feet in velocity. Afterwards the table was com¬ 
pleted by interpolation. 
(1) Suppose it was required to find by the help of the General Table 
in what time the velocity of a 700-lb. elongated shot would be reduced 
f] 2 
from 1344 to 1129 f.s. Here <^=11‘52 inches and — = *1896. By 
w J 
Table 23 we find 1"*0806 corresponding to a velocity 1344 f.s., and 
/72 
2”'1464 to a velocity 1129 f.s. Hence (time required) x — = 2"*1464 
—1"‘0806= 1"*0658, which gives the required time =l' /# 0658-^*1896 
= 5"*621. 
(2) Suppose that the initial velocity of a 250-lb. elongated shot was 
1310 f.s., and that it was asked in what time its velocity would be 
d 2 
reduced to 1022 f.s. Here d— 8’92 inches, and — = *3183, and the time 
w 
required=(3 ,/ -0335-l"*2163)^-3183 = l"-8172^*3183 = 5"-709. 
