ON LAGRANGE’S BALLISTIC PROBLEM. 
223 
A solution of the type z — f(x) </> ( t ) is possible if and only if 
f" (x) = A fix) {/' (x)}y+\ {0 (t) } y <p" (i t) = B, 
where A and B are constants connected by the equation 
B 
= ypo 
po 
(l — /°o) 7 A. 
If S is the area of the cross-section, the equation of motion of the projectile, which is 
supposed to be at x = b, is p S = M f(b) q>" ( t ). Now in general 
P = Po {l-po) y {/' [x] cp 
Hence the equation of motion of the projectile is satisfied if 
MB/(6) = S P ,(l- P „)M/'(&)}-\ 
or 
A = 
ybf(b){f{b)Y 
where e = C/M = SbpJM is the ratio of the mass of the propellant to that of the projec¬ 
tile. Writing w = / (x) and q — dwjdx, the first integral of the differential equation 
for w is 
7 _x 2 1 
^ A (y— l) a 2 —w 2 ’ 
where a is a constant. Since f(x) vanishes with x, the final integral is 
1 
2 
1 ' (a 2 —w 2 ) 
Jo 
y- 1 dw — 
U(v-i) 
X . 
Writing c — f (b) for the length of the column of gas at the instant considered, we have 
therefore 
(a 2 —w 2 )y~ l div — 
[A (y-l) 
Substituting for A we have 
f (b) = (y- 1 ) e 
Y-l 
b, {f(b)y-' = 
i 
A (y — 1) a 2 —c 2 
O 
°'-bc ' A (y-l) 
- sb ^r 
so that 
j■ (a 2 —w 2 )y~ l dw = ———(a 2 — c 2 )y~ l . 
Jo 2y C 
This equation determines cja, and when it is known w is given by 
1 {a 2 —w 2 )y~ l dw — ^ -—— (« 2 —c 2 )?- 1 -• 
Jo 2yC 0 
VOL. CCXXII.—A. 
