PRINCIPLES OP GUNNERY. 
40 
the number of ifiches in its circumference— i.e., by dividing the energy 
by the number of inches in the circumference of the projectile. Thus., 
JFv* 
enemy per inch of circumference = —- —— 7 ft. tons; 
1 4480 g X t rd 
where d is the diameter of the projectile in inches. 
E WV 2 
Also, enemy per inch of circumference at muzzle = —- = —---; 
' 1 Trd 4480 g X t rd 
thus, muzzle energy per inch of circumference of 10-in. Palliser shell 
5160 _ , ny.nr. . 
~ 3-1416 x 9-93 — 0 6 ' ° nS ' 
Energy This energy is energy due to the velocity of translation of the 
tninsiation.projectile. But the projectile has rotation round its longer axis; so 
that there is also energy due to the velocity of rotation, which, being 
small as compared to the energy due to translation, is usually neglected. 
Energy But in considering the rotation of projectiles, the energy due to 
rotation, rotation must also be taken into account. 
If o) represent the angular velocity of the projectile, K its radius of 
gyration in feet, then 
1 . , ,. W{Kidf 
enemy due to rotation = — — • 
%j 
energy The total energy of a rifled projectile in motion 
== energy due to translation 4- energy due to rotation 
=J* a+ 5 (Xra)2= ? K+ 
where JV = weight of projectile and v its velocity.* 
* If The the muzzle velocity, and co the muzzle angular velocity, then 
total energy of projectile at muzzle = ~ ( V' 2 + K-<sfi); 
i i. 24 tF. 1Q(y . 
hut a) = ( see P* 37), 
7c- 
and K = ^ ’ 
where Tc = radius of gyration in inches. 
WV 2 f /27T\ 2 7 
■4*. total energy at muzzle == —g— £ 1 + \fgj V 1 j 5 
Where n — twist of rifling in calibres, and d — diameter of projectile in incheSi 
fW \ 
It can be shown that the moment of inertia f — Tc 2 ) of the head of a solid projectile with ail 
ogival head of diameters & 
= 7 rpd 5 x *015189, . (A) 
or M7c“= 7r pd 5 x *015189 ; 
where p is the density of the projectile, 31 is the mass of the projectile, and d its diameter. 
Also, that the volume of the ogival head 
= 7 rcfi x *15733. . ... (B) 
Example .—Tojind moment of inertia of 16-jn*. common shell (vide Changesin War Stores, §2190):— 
Ogival head. 1st. Suppose solid. 
(1) \~^ 3f 1 kV== Trpd 3 x ‘01519 = 7rp (3*54) 5 x *01519 = 8*443 x Trp. 
Body. 
(2) () () 3f 2 Ic. 2 2 = 7rp p x length — irp - ° x 6*9 = 33*87 x vp. 
.*. total moment of inertia of solid projectile = M\7c^ + 31 2 7c^= 42‘3l7r p. ...(1) 
