ON LAGRANGE’S BALLISTIC PROBLEM. 
219 
the base of the projectile is known, and also its velocity and final position. For other 
points <p and S have to be found to make both x 0 and t correct : the adjustment is facili¬ 
tated by the fact that uniform division of x 0 corresponds approximately to uniform 
division of <p. At the junction (x Q = \c) cp = 0 , and at the base of the projectile ( x () = 0) 
<p = —0-019578. Taking four values of </> equally spaced between these, and finding 
S to give the correct t, we have four points which correspond nearly to 10, 20, 30 and 
40 per cent, division of the initial gas, and are easily adjusted to exact value by inter¬ 
polation. The pressure falls from 3304-3 kg./cm . 2 at the junction to 2970-3 kg./cm . 2 
at the base of the projectile. The projectile is displaced 75-4 cm. from its seat, and 
has velocity 466-2 m./sec. 
(Plate 1 , curve 6 ).—The first reflected wave reaches the breech at time t = 0-003859 
sec., where the pressure is 2610-5 kg./cm . 2 Other points are found as in the last para¬ 
graph. The pressure at the base of the projectile is 2161-6 kg./cm. 2 , the displacement 
of the projectile 124-3 cm., and its velocity 550-4 m./sec. 
(Plate 1 , curve 7) (Articles 30-31).—The second middle wave begins at the above 
epoch t = 0-003859, pushing back the first reflected wave along a junction s = R,. 
This junction reaches the middle particles at time l = 0-005154. In the part of the 
first reflected wave that still remains the pressure falls from 1708-2 kg./cm . 2 at the 
junction to 1535-2 kg./cm . 2 behind the projectile. The displacement of the projectile 
is 202-1 cm. and its velocity 632-5 m./sec. The second middle wave differs from the 
first reflected wave by the presence of four additional terms with coefficients given by 
log {- m {tJZU,) 6 } = 3fl4397, log {>, 7 {%,/Zll,) 1 } = 3‘52835, 
log { - m {to/ZRy) 8 } = 3-60452, log { m (2 2 /2K x ) 9 } = 3’34841, 
where = 814,358-3 cm./sec. ; also — 466-85. At the breech u = 0 or 
cp — S = (S x —Rj )f% 1} leaving 0 to be found by trial. The pressure at the breech is 
1728-0 kg./cm. 2 . For intermediate points we take a number of values of 0 , find a by 
trial to give the correct t, then calculate x 0 and interpolate. 
(Plate 1, curve 8).—The second middle wave reaches the base of the projectile where 
s = S 2 = R x , r = R 2 = 371,266-2, giving a = t 2 , t = T 2 == 0-007137 sec. This point 
is found without trial. The pressure at the base of the projectile is 1030-2 kg./cm. 2 , 
its displacement — X 2 = 335-6 cm., and its velocity — U 2 = 71,827 cm./sec. — 
718-3 m./sec. The value of Z is Z., = —177-0. At the breech we have a pressure of 
1085-7 kg./cm. 2 . Other points are calculated as in the last paragraph. We have 
F 7 (2R a ) 
S 2 10 
0-00034563, 
F, (1) (2R a ) 
V 9 
^2 
= 0-0000015953, 
F^ (2R 2 ) _ 
V 8 
■‘"'2 
- - 0-000020614, 
= 0'00017489, = - 
0-0005648, 
F/ 5) (2R,) 
V 5 
— 9 
= - 0-004338. 
