ON LAGRANGE’S BALLISTIC PROBLEM. 
199 
R x for v and s x for s in the formulae belonging to the first middle wave. When they 
meet, the first middle wave becomes obliterated, and the. second middle wave begins 
to be generated at the time and place in question, and encroaches upon the two first 
reflected waves. 
To determine the second middle wave we have the conditions that at its advancing 
front, where r = R l5 the Z belonging to it is equal to that belonging to the first reflected 
wave from the right, and at its receding front, where s — s x , the Z belonging to it is 
equal to that belonging to the first reflected wave from the left. Riemaxn’s method 
may be applied in exactly the same way as in Article 15. If P is the point (/, s'), A 
the point (R l5 s'), B the point (R l5 s,) and C the point (/, Sj), we have 
Z(r', s') = [VZ] 1 -[VZ] B + [VZ] C + Z 
Jab 
/3Y _ _5V N 
\ds r + s 
W[ z(AA^ 
/ Jbc \or r + s 
dr, 
At A we have 
r = R,, s = s', £ = 0, V = 
- /Rm+3' 
/\5 
r' + s' 
and Z is the result of substituting R, for r and s' for s in the formula 
Z = i 1 + Z 1 „+(il V ‘ 
<T OCT, 
At B we have 
1/ 
ccr 
10 
<x 
l945a 
Vfx{(r-u) 
r = Rj, s = s u g = 
v = &py<i-*o{+9or-uor + Ton 
and Z is the result of substituting R t for r and s, for s in the formulae for the first middle 
wave, or in those for either of the first reflected waves. For the present we shall denote 
it by Z B , and observe that it is independent of r' and s'. 
At C we have 
r = r', s = Si, 
f = 0, 
V = 
f r' + Sj 
\r' + s' 
5 
and Z is the result of substituting r' for r and s x for s in the formula 
Z = K I + L 1 «+ (-A)‘j F| {a + u ) \- 
\cr dcrj l c r ) 
Along AB, where r = Rx and s increases from s' to s l3 we have 
T 3 \ 4 f 1 [ ccr,, 
Z — ^ + 1^ (Rj— s) + 
cr 3cr 
1U 
crl945a ’) 
= (Pi + g/ )( I ^-^)( R i + s ) :< (2Q-180^ + 420g 3 -280f) > 
cs r + s (r' + s) b 
(r' + s') 
(R,—F) (s—s') m 
^ (r' + s')(-R 1 + sy 
2 F 
VOL. CCXXII.-A. 
