PRINCIPLES OE GUNNERY. 
Let Ofj — li , and </> the angle of spiral at p. Then from equation (1) 
tan <p = tan a + 2ch 
\m n / 
TT 
Therefore twist of rifling at the pointy is 
one turn in 
1 + 
!(;->) 
calibres, . (A) 
For example : in the 12-in. gun of 25 tons, the twist at the muzzle 
is one turn in 50 calibres ; that at the breech is one turn in 100 calibres. 
The length of rifling is 127 ins. It is required to find the twist in 
calibres at 8 ins. from the muzzle. 
Here n = 50, 
m = 100 , 
l — 127 ins. 
h = 8 ins. 
Then, substituting in the above formula (A), the twist at 8 ins. from 
the muzzle is 
50 
one turn in 
1 + 
8/50 
(~ - 1 ) 
127 \100 ) 
= 51*626 calibres. 
Next, to find the linear distance the exterior of the projectile has 
rotated while passing down the bore. 
Suppose Oq — h (the distance passed down the bore by the projectile); 
then 
jpq = Hi + clb® = ]i tan a + It 3 
qi-iy 
\m n / 
tan a — tan a 
21 
■nil ^ 
n 
21 
(B) 
For example : it is required to find the linear distance which the 
exterior of the projectile has rotated in the 12-in. gun of 25 tons, when 
it has passed down 8 ins. of the bore from the muzzle. 
In equation ( B ), 
U — 8 ins. 
and substituting as before, the linear distance of rotation of the exte¬ 
rior of the projectile at 8 ins. from the muzzle is 
Sn 
50 
*(l00 5o) 
254 
X 64 = 
4000tt 
25400 
= *49 ins. 
Determin¬ 
ation of 
linear ro¬ 
tation of 
exterior of 
projectile, 
while pass¬ 
ing down 
the bore. 
