224 
MESSRS. A. E. H. LOVE AND F. B. PIDDUCK 
The pressure ratio between the two ends of the gas is R = {f'[0)/f'(b)} y , where 
if (W (O)K 1 = a 2 /(a 2 —c 2 ). Hence 
Writing c = a sin 0 we find It = cos m 0, where m = 2y/(y— l) and 9 is found by trial 
from the equation 
sin 9 f m _, 
—cos 
COS @ J o 
In the case of y = ll/9, m — 11, we find, after some analytical reduction, the 
expansion 
R — 1—1^4- 7.5 2 1491 3 3 , 
11 — 1 + STTVD 6 + 
valid for small values of e. Either method gives R = 0-894 for e = 12/50, the corre¬ 
sponding value of 0 being 8° 9' 6". It will be noticed that although R is not equal to 
(l+de) -1 to the second order, the approximation is still a remarkably good one. The 
present theory will appear more satisfactory, as it is based on an exact solution valid 
for all values of e. 
As regards the energy factor, the previous definition in terms of the work done from 
an initial state of uniform density is not convenient, as this state is not one of the previous 
states of the gas. We may, however, define the energy factor in such a way that the 
kinetic energy of the gas is ae times that of the projectile.* 
Corresponding to the initial distance x from the breech we have in general w = a sin <p, 
where 
1 cos” 1-1 cj)d(p = K^,. 
Jo 1 b 
and 
e cos m 6 
m sin 0 
The corresponding velocity is V sin </>/sin 9. If x + dx corresponds to <p + d(p , 
K dxjb = cos” 1-1 (pdcp. The kinetic energy of the gas is 
! CV 2 
2 K sin 2 6> 
cos m 1 cj) sin 2 
(j>d<p, 
and that of the projectile |-MV 2 . Hence by definition 
r» 
cos” 1 1 (p sin 2 <pd(p , 
Jo 
a — 
e cos”‘ 9 sin 9 
* This was not done above because the problem would naturally present itself in the other form in 
practical calculations, where we should seek a factor which will make the kinetic energy of the projectile 
equal to the work of an assumed massless propellant. 
