( ) ( ) ( )
513.05.31sin11cos1
3==
i
Which gives us
We determine the direction of ω through the geometry
s
The vector di/dt is then given by
The angular momentum vector, h is then obtained through
I
T
The first term on the RHS is
( )
( ) ( )
( )
321
2
2
513.0191.0837.0
ftlbm40.66
rev
rad
2
s
rev
150ftlbm13.19
eeei ++
−
−
=
T
p
I
pI
++= s
rad
13952229 321 eeei
T
p
I
pI
The second term is calculated through
−−
196.0927.1319.0
So then we can sum the two terms to obtain
++= s
rad
14052228 321 eeeh
The gravitational acceleration vector is simply
−= 2
2s
ft
2.32 eg
13
22
VV
=
13
22
=
( )
LVAZLV sinsincos2 123 +−=−
( ) ( ) ( ) ( )( )
+−
−= −
s
ft
75.48sin2132300sin75.48cos434
s
rad
1029.72 5
3
( )
AZLVAZLV coscossincos2 132 −=
( ) ( ) ( ) ( ) ( )( )
−
= −
s
ft
300cos75.48cos2132300sin75.48cos1231
s
rad
1029.72 5
2
( )
AZLVLV coscossin2 231 +−=
( ) ( ) ( ) ( )( )
+−
= −
s
ft
300cos75.48cos43475.48sin1231
s
rad
1029.72 5
1
+−−= 2
321 s
ft
270.0205.0114.0 eeeΛ
Now we need to calculate the reference area and a whole slew of particular coefficients
( )
2
2
2
2
2
2
2
2
2ft373.0
in
ft
144
1
mm
in
4.25
1
mm210
4
1
4
1=
==
dS
( ) ( ) ( ) ( ) ( ) ( )( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )
( )
−=
++
−
+
−
−−
−
+
−=
22
2
2
2
3
s
ft
35
s
ft
270.00
s
rad
837.052191.0226
s
ft
0004.0
s
ft
191.02132837.0434
s
1
0007.0
s
ft
8.1cos12602514513.02514
ft
1
00013.0
s
ft
1260
s
1
036.0
dt
dV
Thus the linear acceleration vector at this instant in time is
−−−= 2
321 s
ft
353385 eeea
( ) ( ) ( )
−=
−
−+
2
s
rad
14
s
rad
227226
s
1
318.3
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
( ) ( ) ( )
−=
−
−+
−
+
−
=
2
2
s
rad
42
s
rad
5252
s
1
318.3
s
ft
8.1cos191.02514434
s–ft
rad
041.0
s
ft
513.02132837.01260
s–ft
rad
022.1
dt
dh
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
−
+
−
=
3
s
ft
8.1cos513.025141260
s–ft
rad
041.0
s
ft
837.0434191.02132
s–ft
rad
022.1
dt
dh
I
dt
T
MtMMll qpP
( )
( )
−+−+−++= 333312213
3~
cos
~~~~ i
pI
hCvivCivivCiCC
dh
P
MtMMll qpP
Stage 1 – with 1 due north
2
6
4
V1
V3
V2
( )
LVAZLV sinsincos2 326 −−=
Now let’s change the velocity components
4
1
3
V1
V3
V2
5
6
5
V6
( )
LVAZLV sinsincos2 456 +−=
( )
AZLVAZLV coscossincos2 465 −=
( )
AZLVLV coscossin2 564 +−=
Now we convert back where 4 = 1, 5 = 2 and 6 = 3
( )
LVAZLV sinsincos2 123 +−=
( )
AZLVAZLV coscossincos2 132 −=
( )
AZLVLV coscossin2 231 +−=
Problem 37 – An AC-130 is flying at a speed of 200 miles per hour in still air. If the 105mm
weapon is fired sideways, calculate the axial acceleration vector and the angular acceleration
vector acting on the projectile through use of the 6 DOF equations if the projectile is fired to the
left. Discuss the effect of the angular momentum on the projectile nose (which way does it want
to tip?)
Please ignore the Coriolis acceleration, assume there is no yaw at muzzle exit and assume a
muzzle velocity of 1,500 ft/s, the weapon has a right hand twist.
Projectile Information
01.0
9.1
0.8
39.0
2
0
−=
=
=
=
p
N
L
M
D
D
C
C
C
C
( )
( )
=
=+
−=+
3
ft
lbm
060.0
005.0
5.6
p
q
q
M
NN
MM
CC
CC
=
=
=
s
rev
220
ft–lbm377.5
ft–lbm547.0
2
2
p
I
I
T
P
3
1
2
V3
V1
V2
Please supply all answers in an inertial coordinate system labeled 1,2,3 with 1 being along the
aircraft flight path and 3 being off the right side of the plane. Treat all missing coefficients as
equal to zero.
airplanes velocity vector, thus we can write for the case of the plane:
( )
−=
−= s
ft
293
mi
ft
5280
s
h
3600
1
h
mi
200 11 eeW
Then the relative velocity vector is
−=
−−
−=−= s
ft
1550293
s
ft
293
s
ft
1550 3113 eeeeWVv
The scalar magnitude of the relative velocity is
2
3 (Right)
( )
=
+= s
ft
1528
s
ft
1500293 2
2
v
We shall again orient the projectile i–j–k triad in our coordinate system (we only need to concern
ourselves with i since the projectile is symmetric. We were told that
and
are zero and we had
no muzzle disturbance, thus we have
We can now define the unit vector along the projectile geometric axis as
0
1=i
0
2=i
1
3−=i
Which can also be written as
3
ei −=
We need to determine the angular momentum vector so first we need the instantaneous angular
velocity vector. We know that there is no muzzle disturbance. Thus di/dt is zero.
0=
dt
di
The angular momentum vector, h is then obtained through
+= dt
d
I
pI
T
pi
iih
Since di/dt is zero, we only need concern ourselves with the first term. The first term on the
RHS is
1 (AC axis)
2
3 (Right)
i
( )
( ) ( )
( )
3
2
2
ftlbm337.5
rev
rad
2
s
rev
220ftlbm547.0
ei −
−
−
=
T
p
I
pI
reference area, the drag coefficient and the rest of the particular coefficients
( )
2
2
2
2
2
2
2
2
2ft0932.0
in
ft
144
1
mm
in
4.25
1
mm105
4
1
4
1=
==
dS
The initial total yaw angle is zero so the drag coefficient is
( )( )
39.000.839.0
2
20 =+=+=
DDD CCC
Now we’ll calculate all of our coefficients
( ) ( ) ( )
( )
( )
=
== s
1
052.0
lbm1.322
39.0ft0932.0
s
ft
1528
ft
lbm
060.0
2
~
2
3
m
vSC
CD
D
( )
( ) ( ) ( )
( )
( )
( )
( )
=
=
+
=s
ft
00023.0
ft
in
12lbm1.322
005.0in134.4ft0932.0
s
ft
1528
ft
lbm
060.0
2
~
2
3
m
CCvSd
CNN
N
q
q
( )
( ) ( ) ( )
( )
( )
( )
( )
−=
−
=
+
=s
1
617.0
ft
in
12ft–lbm337.52
5.6in134.4ft0932.0
s
ft
1528
ft
lbm
060.0
2
~
2
2
2
2
2
2
2
3
2
T
MM
MI
CCvSd
Cq
q
( ) ( ) ( )
( )
( )
( )
( )
=
== s–ft
rad
048.1
ft
in
12ft–lbm337.52
80.3in134.4ft0932.0
s
ft
1528
ft
lbm
060.0
2
~
2
2
3
T
M
MI
vSdC
C
( ) ( ) ( ) ( ) ( ) ( )( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )
−=
+
−
−−−
+
−−−
−−
−
+
−=
222
2
2
2
2
s
ft
32.32
s
ft
0
s
ft
2.32
s
rad
10067.141
s
ft
00023.0
s
ft
015501293
s
1
00042.0
s
ft
0cos0152801528
ft
1
00017.0
s
ft
0
s
1
032.0
dt
dV
( ) ( ) ( ) ( ) ( ) ( )( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )
=
++
−
+
−
−−
−−−
+
−
−=
22
2
2
2
3
s
ft
73.40
s
ft
00
s
rad
0000
s
ft
00023.0
s
ft
029300
s
1
00042.0
s
ft
0cos1500152811528
ft
1
00017.0
s
ft
1500
s
1
032.0
dt
dV
Thus the linear acceleration vector at this instant in time is
+−−= 2
321 s
ft
7.703.324.89 eeea
( ) ( )
331221211233
2
3
3~~
cos
~~ ++−+−−−+−= gihihCivivCvvivCvC
dt
dV
qp NNtLD
I
dt
T
MtMMll qpP
( )
( )
−+−+−++= 333312213
3~
cos
~~~~ i
pI
hCvivCivivCiCC
dh
P
MtMMll qpP
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
( ) ( ) ( )( )
=
−
−+
−
+
−−−
=
2
2
s
rad
06.307
s
rad
067.1410
s
1
617.0
s
ft
0cos015280
s–ft
rad
0043.0
s
ft
129301500
s–ft
rad
048.1
dt
dh
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
−−−
+
−
=
3
s
ft
0cos115281500
s–ft
rad
0043.0
s
ft
000293
s–ft
rad
048.1
dt
dh
Problem 38 – An F-86 Saber jet flying with a speed of 600 mph. The aircraft is pulling up out of
a dive and the pilot is experiencing 2.5 g’s. The pilot fires his 0.50 caliber machine guns at a
ground target during the pull out. At the instant the plane makes an angle of 30 degrees to the
ground, a projectile leaves the muzzle of the weapon with a muzzle velocity of 1,800 ft/s, 3
degrees to the right of the bore axis and is rotating to the left at 2 rad/s in the plane of the gun,
relative to the gun. Also the projectile is pitched up 1.5 degrees and yawed left 1 degree to the
line of fire. Ignoring Coriolis effects, write the vectors that completely define the initial
conditions of the projectile so that you could use them in a 6-DOF model. DO NOT solve the
equations of motion – just write the initial conditions. Assume coordinate directions as follows:
1 – Horizontal direction in which the aircraft is flying
2 – Straight up
3 – To the right as viewed from the rear of the aircraft
Here we know an but neither of the other two terms. To find r we note that
rV
=
(2)
Substitution of equation (2) into equation (1) yields
V
an
=
(3)
Thus the projectile velocity vector is
( ) ( )
( )
( ) ( )
=
== s
ft
557,1
s
ft
3cos30cos800,13cos30cos
1
VV
( )
( )
( )
−=
=−= s
ft
900
s
ft
30sin800,130sin
2
VV
( ) ( )
( )
( ) ( )
=
== s
ft
82
s
ft
3sin30cos800,13sin30cos
3
VV
Or, in vector notation
+−= s
ft
82900557,1 321 eeeV
If the projectile is rotating to the left relative to the gun we can draw the situation as
1
2
3
30
3
ω
( )
( )
=
=s
rad
0.1
s
rad
30sin2
1
( )
( )
=
=s
rad
732.1
s
rad
30cos2
2
=s
rad
0
3
Or, in vector notation
++= s
rad
0732.1000.1 321 eeeω
But this has to be combined with the aircraft angular velocity to yield
++=
+
++= s
rad
091.0732.1000.1
s
rad
091.0
s
rad
0732.1000.1 3213321 eeeeeeeωtotal
iω
i=
dt
d
−−== s
rad
259.1437.0871.0
513.0191.0837.0
091.0732.1000.1 321
321
eee
eee
i
dt
d
The final piece of information we need is the projectile orientation
The projectile pointing vector is then
1
i
28.5
=
++= s
ft
679,2
s
ft
82340,1319,2 222
v
Problem 39 – In the study of ballistics we normally neglect Magnus forces as small and only the
moments generated by these forces are significant. In low velocity projectiles, however, Magnus
forces may be considerable. Consider a baseball thrown with spin towards a batter. The ball
weighs 5 ounces and is 2.9” in diameter. The distance from the pitcher’s mound to home plate is
60’ 6”. Assume the ball is thrown at 70 miles per hour and that there is no wind. The release
point is 4 feet above the ground with an upward angle of 4º. We need to determine how much
spin is required to move the ball 2 feet to the left as viewed from the pitcher. Please perform the
following calculations:
a.) Assuming constant drag and Magnus force coefficients develop the equations of motion for
the ball assuming flat fire is valid and the spin axis remains vertical for the entire flight.
b.) Determine the spin rate and direction to cause the desired motion.
c.) Determine the final velocity, time of flight and height of the ball.
d.) Determine if the assumption of flat fire really was valid.
Please list all of your assumptions. The following coefficients may be assumed:
52.0=
D
C
=3
ft
lbm
076.0
09.0=
p
N
C
5.60