194 
ON THE MOTION OF ELONGATED PROJECTILES. 
The first and third give on eliminating 0", 
(pc'" cos 0 + 2 '" sin 0) [P+(R - P) (cos 2 0 + sin 2 0)] = 0. 
Hence the last equation is 
0" £4 0" + g cos 0 J= '0. 
In forming the energy equation, the Lagrangian equation 
d_ /dT\ _ dT_ dU 
dt\do) dO dO 
had been multiplied by 20, so that we expect a value of 0" zero given 
by the energy equation in addition to the true value of 0". Thus 
we take 
and then 
2 MN 
<9"=--2p # c °s0 
/ in o CMN 
^"=-2-jjp&n cos 0. 
In order to find the value of y IY (y f , y", y" all vanishing) we 
substitute in the second equation of linear momentum to third powers 
of t : we get 
0 = Py”£+(K-P) Ucospr'~ 
so that 
yl v = P^_R kpi cos/3 
a CMN (P - R) 
A 2p2 
Ung cos 2 0. 
For an ordinary elongated shot 
P>R 
and C y M, (P — R), U, n are all positive, (an ordinary shot having spin 
put on it so that it describes a right-handed screw): hence the signs 
of yjr'" and y lY are opposite to that of N, and the same as that of 0". 
2 MN 
It will be noticed that the value of 0", namely- g cos /3, is 
independent of the velocity of projection, and of the spin: if then 
the shot be held with its axis horizontal and let fall, and the pointed 
or fore-end drop the quicker, then 0" is positive and N is negative, and 
if it hit the ground with the hinder end N is positive: in the former 
case the shot will drift to the left and in the latter to the right. 
As a matter of fact a shot from a rifled gun always drifts to the 
right (n being positive) and a homogeneous shot with a pointed end 
would fall point first (very slightly) though a shell (with a lighter 
fore-end) might remain horizontal when dropped. We conclude that 
