26 
SHORT NOTES ON PROEESSIONAL SUBJECTS. 
In calculating the loss of velocity due to the resistance of the air, the following 
approximate formula will be found to give very close results, when the velocities 
are as high as those given in the above table, and the range is moderate, viz :— 
V 
v =-— 
1 -\-cVx 
where v = required velocity in feet, 
V = initial velocity in feet, 
x = range in feet. 
e = 0-000062 —, 
W 
R = semi-diameter of shot in ft., 
W weight of projectile in lbs. 
Thus, determine the remaining velocity of a Palliser projectile fired from the 
9-in. rifled' M.L. gun, with a charge of 4Bibs., the initial velocity being 1840ft., 
and range 1000 yds. Here 
r= 1340 feet, 
x — 3000 feet, 
R = 446 inches or 0*375 ft., 
TF == 250 lbs., 
c = 0*000062 x------ 
250 
= 0-000000034875 ; 
1 + 0-000000034875 x 1340 x 3000 
_1340 
1*140 
= 1175*2 feet. 
In comparing the action of projectiles on striking an iron defence, such, for 
example, as an< iron-clad ship, it is usual to estimate the effect by the stored-up 
“ work ” or “ energy ” in the shot at the moment of impact. 
This is equal to the weight of the shot in lbs. multiplied by the square of its 
velocity in feet, and divided by twice the force of gravity. The result being a 
number of foot pounds, which, for convenience, are turned into foot tons, as a 
less number of figures are required to express the latter. Thus, 
Total foot tons = 
2g x 2240* 
where TF = weight of shot in lbs., 
v = velocity of shot on impact in feet, 
g = force of gravity = 32*19 feet, 
2240 = number of pounds in a ton. 
This equation gives the total “ energy ” in the shot. 
When, however, it is required to compare two projectiles as far as their power of 
penetration is concerned, it is usual to express the stored-up " energy ” in terms 
of their diameter or circumference. Thus a shot is said to have so many foot 
tons of “ energy ” per inch of its circumference—its total “ energy ” must therefore 
be divided by the number of inches in its circumference, and the last equation 
will become 
Toot tons per inch of shot’s circumference 
JFv 3 1 where 7r = 3*1416, 
2 g X 2240 X 2ttR 
R = semi-diameter of shot in inches . 
Thus, determine the total “ energy ” and “ energy ” per inch of circumference of 
a Palliser projectile fired with a charge of 43lbs. from the 9-in. rifled M.L. gun; 
range 1000 yds. 
and “ energy ” per inch of shot’s 
circumference 
2394 
Here we have already found 
v = 1175*2 feet, 
. . „ 250 xll75*2 2 
so total “ energy = 
64*38 x2240 
= 2394 foot tons; 
2x3*1416x4*45 
= 85*63 foot tons. 
