276 
PRINCIPLES OE GUNNERY. 
As before, suppose <0 bas a mean value <0, and suppose a — p expressed 
in degrees = D; then 
D 
— 18 °// /* j 
7T 
du 
180y Osee</> 
2 sec ^ 
w sec 0/(w sec 0) 
Put u — v cos <0, then the equation becomes 
D sec ^ = 180 ^ r 
K Jq% 
But by Bashfortffis law, 
so that - D= cos 0 . —i. C 
a s 
£ _- 18% /”\p sec $ (1000) 3 efo 
A©* 
^ q sec $ 
■(*) 
Niven has calculated the value of the integral — 
180y 
(iooo) 3 r 
.1700 ^ 
Iv 
(using Bashforthds value of A) for values of v down to 900. (Vide 
Table III.) Table III. connects degrees and velocity, and is denoted 
by pv 
If this integral is represented by D VJ equation (Ji) becomes 
d [3 
— J) = COS 0 {Dq sec ,/) —- Dp sec 
m 
There is no means of determining the exact value of the mean angle 
0 in this case. As, however, sec <0 varies very slowly for the greater 
part of the trajectory, it will suffice in this equation to put <0 = 
u 
for low angles of elevation; but for high angles of elevation <p should 
be more accurately determined from the formula 
tan 0 
__tan a + tan ft 
by means of a/table of natural tangents. With the aid of the D v Table 
the value of q sec <0 can be determined, and thence the value of q. 
X, Y, and T can be determined by means of the equations (d), (B), 
and (C). 
In the ascending branch, for the distance integrals, I —~~ (a — J3) 
° ’ ' * jp + q K 
must be added to the value of q> so found, and for the time integrals, 
p — q 
-J-' -- - -- (a — p) must be added. 
In the descending branch, from the value of <0 found as above, for 
