PRINCIPLES OP GUNNERY. 
41 
It must be remembered that tbe velocity of translation is given 
by tbe pressure of tbe powder gas on tbe base of tbe projectile, and that 
the velocity of rotation is given by tbe pressure of tbe sides of tbe 
grooves against tbe driving edges of tbe studs of tbe projectile, as it 
travels through tbe bore. 
Tbe pressure on tbe driving edges of tbe studs depends on tbe angle 
of spiral in tbe gun, and also on tbe employment either of uniform or 
increasing twist, in order to give the requisite rotation to tbe projectile. 
To compute moment of inertia of the core :— 
(1) Head. m X K\* = tt p (2*84) 5 x *01519 = T0657rp, 
(2) Body. 
m 2 K 2 * 1 2 = tt p x 6*04 =5= 5'659ttp ; 
.*. total moment of inertia of core = m x K x 2 + m 2 K 2 2 * * = 6*72 4nrp; .....(2) 
.*. moment of inertia of empty shell = 42*3l7rp — 6*724 ttp = 35*597 rp .(3) 
If the shell is tilled with water (the density of which does not differ greatly from that of the 
powder in a filled shell), then for the core, the density p — 1, and for the cast-iron shell p — 7*11. 
Therefore from (3), 
moment of inertia of empty shell = 35*59 x 7‘llTr = 253*07r, 
and from (2), „ core filled with water = 6*7tt, 
.*. total moment of inertia of filled shell — 259*77r.(4) 
To find the radms of gyration of filled 16-pr. common shell :— 
1st. Suppose solid. 
(1) Ogival head. 
(2) Body. 
The volume of ogival head, or ^ = tt<P x *1573 1 
•. M x = Tvpd* x *1573 = tt p (3*54)3 x -2573 = 6*977Trp. 
M 2 — wp x length = 7 rp * 6*9 = 21*627rp. 
.*. Mi + 3£ 2 = 28*607rp. 
2nd. To find m x + m 2 of core. 
(1) Head. % = tt p (2*34)3 x *1573 = 2*0157rp. 
(2) Body, tt p (1*17) 2 * x 6*04 = 8*2687rp. 
m 1 + m2—l0'28‘irp. 
mass of empty shell = (28*60 —10*28) tt p — 18*327rp 
= 18*32 x 7 -IItt = 130*3?r (since p = 7*11), 
mass of filled core = 10*287r (since p = 1), 
•\ /, filled shell = (130*3 +10*28) tt = 140*58tt, 
. 7 2 _ moment o f inertia of filled shell _ 259 * 77 r 
mass of filled shell. “ 140*6?r ~ ^ 8 ^ms., 
_ .*. 7 e = a/F848 = 1*35 ins. ,* 
where k is the radius of gyration of the filled shell. 
Now, if the muzzle velocity of a filled 16-pr. common shell, fired with a service charge of 3 lbs, 
R.L.G. powder, is 1358 ft. per second, and weight of filled shell = 15 lbs. 13 ozs., the total energy at 
muzzle, or total work done by 3 lbs. of powder 
- WV ' 2 $ 1 , ^ttx 2 ( 15-81 x (1358) 2 f . /2 x 3*1416\ 2 7 
2 g i 1 + \nd) k ) 2 x 32*19 x 2240 C 1 _r \ 30 x 3*54 ) 1,b4:8 ) ft 
2 9 
202 * 2(1 
*19x2240 
*001894) ft, tons = 202*2 x 1*0019 = 
30x3 
202*6ft. tons. 
202*6 
3 
tons 
.*. work done by each lb. of powder in the charge = — 67-5 ft. tons, 
O 
. s . iuce the work done in rotating the projectile is such a small per-centagc of the work clone 
giving velocity, the former is usually neglected in calculations for penetration of iron plates, &c. 
0 
Pressure 
required 
to give 
rotation to 
projectiles. 
