July-August - Air Defense Artillery School

1 foot per second \\'ould be (d% H factor)~. To deter-
100
mine the difference in muzzle velocity between the assumed
and the deyeloped muzzle yelocity it is only necessary to determine
the horizontal range de\'iation TO of the burst, then
add to this the horizontal range effect OP of the vertical
de\'iation, and diyide the resultant length TP by (d% H
Rl
factor) 100 .
Then 6.i\ IV =
length of line TP
(d%H factor) Rl
100
Referring to figure 2 which is a horizontal projection of
the problem.
The horizontal range deviation due to a lateral deyiation
from the batten' and Rank is TO. CD = ER.
but TD = TC + CD = ::t
0202
l'
1000 sin
=+=
l')} 01
-----
1000 tan T
= :J
02 H
1000 sin £2sin T
01 H
+ ------~
1000 sin £1tan T
The horizontal range deviation due to a vertical deviation
from the battery is DP. Referring to figure 1:
01(11sin £1 (11H
D P = ::t 1000 ~~::t 1000
Then TP = TD + DP
ancl 6.i\IV =
(11H t'h H
:!: 1000 :!: 1000 sin £1 tan T
By the process of factoring.
61\IV = ::tC1(11::tC281 """C3()2
where,
H
CI = lORI (d%J-) factor)
C1
C2 = sin £1 tan T
C1
C3= .. T
SIn £2 SIn
(d%H factor~ RI
100
The sign to be used with these factors is given above. To I
simplify the computations, the yalues of Cl for the 90mm
ancl 3 inch AA guns are given in Tables I and II. ~
The vertical correction required to move the bursts onto
the muzzle velocity line is equal to minus (11corrected b\
the vertical angle from 01 which is sub tended by the length
1\IK. This correction is equal to 6MV multiplied by the
d.p factor gi\'en in the graphical firing tables. Then 4
d = (11+ 6.MV (cI factor) j
The lateral correction is equal to the minus ()1 converted
to the horizontal plane. 1
-01
orelA = --- ..
cos £1
..
02 H
1000 sin £2 sin f