364 
MINUTES OF PROCEEDINGS OF 
and from this expression may be deduced 
p _ 2p 3 { Gz + Mo 3 j 
(/cV 2 + 4p 3 z 3 ) sin § 2p*/fe (p 2 — r 2 ) 
-s/4^ 2 (sin S) 3 H- X; 2 \/ 4^ 3 + £ 3 
(14) 
18. Equation (14) gives the pressure acting between the studs or 
rib of the projectile and the driving-surface of the groove at any point 
of the bore, and for any inclination of the driving-surface; but, as 
before stated, in the Woolwich guns the normal to the driving-surface 
(that is, the line of action of R) may, without material error, be con¬ 
sidered as perpendicular to the radius. 
If in (14) 8 be put = 90°, the equation is simplified; and the resulting 
expression gives the total pressure on the studs for the Woolwich guns. 
Putting, then 8 = 90°, (14) becomes 
R _ 2p 3 \/4z 2 + /g 3 {Gz + Mv*) .q 5 n 
hr 3 (h — 2p,^) + 2p 3 ? (2z + pj&) 
19. Compare now (14) and (15), the equations giving the pressure 
on the studs for parabolic rifling, with the equations subsisting where a 
uniform twist is used. 
For a uniform twist, we have, as I formerly showed. 
R = 
27rp 2 
p x (27rp 3 /*; — rh) 
(27rp 2 + rhh) sin 8 
Vjfi + (sin 8) 2 
G, 
•( 16 ) 
where U is the pitch of the rifling, h the tangent of the angle which the 
groove makes with the plane of xy , the other constants bearing the 
meaning I have already assigned to them in this investigation. 
20. In the Woolwich guns, where S = 90°, (16) becomes 
R = 
27rp 3 V 1 + £ 3 
hr (h — fXj) + 27rp 3 (p, 1 /6- -j- ]) 
G, 
■(17) 
21. I proceed to apply these formulae, and propose to examine what 
are the pressures actually required to give rotation to a 400-lb. pro¬ 
jectile, fired from a 10-in. gun with battering charges, under the 
following conditions:—1st. If the gun be rifled with an increasing- 
twist, as at present. 2nd. If it be rifled with a uniform pitch; the 
projectile in both cases being supposed to have the same angular velo¬ 
city on quitting the gun. As the calculations for the uniform pitch 
are the simpler, I shall take this case first. 
22. I have before remarked that with a uniform twist the pressure 
on the studs of the projectile is a constant fraction of that on the base 
of the shot, and represents, so to speak, on a reduced scale, the pressure 
existing at any point in the bore of the gun. Calling the fraction in 
equation (17) (7, we have 
where 
R=C .G, 
2ttp 2 s/I + /fc 3 
hr {h — p-j) -Jr 27Tp 3 (pj/t + 1) 
...( 18 ) 
= •04426, .(19) 
