
283Trajectories
We now have the differential equations of motion but need initial conditions to solve them.
Let us examine the projectile at the instant of muzzle exit without worrying about how it
attained its state of motion there (this is the job of the interior ballistician). We shall dene
the initial tube angle in azimuth and elevation as θ
0
and ϕ
0
, respectively. Then our initial
velocity vector can be dened as
Veee
0
1
2
3
1230
00
0
=
=
Λ
Λ
Λ
[]
coscos
cos sin
sin
0
V
φθ
φθ
φφ
0
eee
123
(8.294)
And, if we also take the wind into account, we have
vVWeee
000
1
2
3
123
11
22
0
0
0
00
00
=− =
=
−
−
v
v
v
VW
VW
V
[]
33
0
−
W
3
123
0
eee