ON THE MOTION OF ELONGATED PROJECTILES. 
195 
for these cases AF is small compared with the other quantities and 
proceed to find the initial motion when IV is zero. 
5. In this case, 0" = 0, A t r"' = 0, y lY = 0, initially and on expanding 
the variables in powers of r, as before, we get from the third powers 
of T 
.0 =Px iy + (R-P)(x' cos /3 + z iy sin/3)cos (3 + (R — P) Uff" sin/3 
0 = Pz iy + (R-P) (x iy cos (3 + z lv sin /3) sin (3 - (R - P) U&" cos /3.(xiii) 
0= Ap y cos /3-Cn 6"' .(xiv) 
0=RU (x iy cos /3 + z iy sin /3)- Mgz'" 
Hence as we have shown that z'" = 0, 
& ,Iv cos/3 + 2 Iv sin/3 = 0 .(xv) 
In order to determine O'" we must consider the coefficients of t 4 
in the equations of motion. The first three and the last equations 
of motion give 
/ 0= Px y + (R — P) [ U6 iy sin (3 + 4#'" (x sin 2/3 — z cos 2/3) + ( x y cos /3 + z y sin /3) cos /3] 
jo =Py y + (R-P) U\js iy cos /3.(xvi) 
j 0 = Pz y + (R - P) [ - U6 iy cos /3 - 4(9'" (x cos 2/3 + z sin 2/3) + ( x y cos (3+z y sin /3) sin /3] 
' 0 = R U (x y cos /3 + z y sin (3) - ^MgzP + 6 A <9'" 2 
The first and third of these last equations give that 
0 = R(x y cos /3 + z y sin /3) + 46"' (R - P) (x sin (3-z cos /3) 
and by the last, remembering that we have proved that (xiii), (xv) 
Pz iy — (R-P) U&" cos j8, 
0= -4 Uff" (R- P) (x sin (3-z cos /3) - 
Also 
(R-P) UO'"cos 13 + 6A6'" 2 
. Mg 
x sin /3 — z cos /3=-p cos /3, 
0 = 2 0 fl 
Mg U p P ^ (2 +1) cos /3 - 3 A6"'~J 
As before we neglect the value O'" = 0 obtained from the kinetic 
energy equation and take the root 
.... , r tt P — R 
6 =-MgU-pj- cos/3 
.(xvii) 
We write - (P — R) instead of + (R — P) because for an elongated 
shot P > R. 
Our former equation (xiv) gives us 
r— .(s™i) 
whence by (xv) 
v (P-RfMG TT „ _ 
y =- p2 -j 2 9 n u cos z 3 
...(xix). 
26 A 
