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( ) ( ) ( ) ( ) ( ) ( )( )
( )
−
+
−=
2
2
2
1
rad
ft
ft
1
s
ft
0cos293245802458
ft
1
0017.0
s
ft
293
s
1
516.0
dt
dV
( ) ( ) ( )( )
=
−
−+
2
s
rad
85.39
s
rad
012870
s
1
25.21
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
( ) ( ) ( )( )
−=
−
−+
−
+
−
=
2
2
s
rad
922,35
s
rad
012870
s
1
25.21
s
ft
0cos024580
s-ft
rad
136.0
s
ft
129302440
s-ft
rad
6.122
dt
dh
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
−
+
−
=
3
s
ft
0cos124582440
s-ft
rad
136.0
s
ft
000293
s-ft
rad
6.122
dt
dh
vector. Given the problem statement we can draw the velocity vector in our C.S.
From the above diagram it is obvious that in the 1 direction we have
=s
ft
0
1
V
=s
ft
0
2
V
−= s
ft
440,2
3
V
Or, in vector notation
1 (AC axis)
2
3 (Right)
V
−= s
ft
440,2
3
eV
We again assume that the flight of the aircraft causes a wind in the opposite direction to the
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
2440293
s
ft
293
s
ft
2440 3113 eeeeWVv
The scalar magnitude of the relative velocity is unchanged
( )
=
+= s
ft
2458
s
ft
2440293 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 −=
1 (AC axis)
2
3 (Right)
i
( ) ( ) ( ) ( ) ( ) ( )( )
( )
−
+
−=
2
2
2
2
rad
ft
ft
1
s
ft
0cos0245802458
ft
1
0017.0
s
ft
0
s
1
516.0
dt
dV
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
( ) ( ) ( )( )
=
−−
−+
−
+
−−−
=
2
2
s
rad
922,35
s
rad
012870
s
1
25.21
s
ft
0cos024580
s-ft
rad
136.0
s
ft
129302440
s-ft
rad
6.122
dt
dh
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
−−−
+
−
=
3
s
ft
0cos124582440
s-ft
rad
136.0
s
ft
000293
s-ft
rad
6.122
dt
dh
Right side
−+−= 2
321 s
ft
184,189.5376,1 eeea
Left side
+−−= 2
321 s
ft
184,129.70376,1 eeea
Right side
−−= 2
321 s
rad
45.2922,3585.39 eee
h
dt
d
Problem 35 - One of the interesting aspects of the forces acting on a projectile occurs as the
projectile leaves an aircraft sideways. This problem is encountered all the time in the AC-130
gunship. Let’s examine a 105 mm HE projectile being fired into a city from both the top of a
building and from the AC 130 in flight. The velocity of the projectile is 1510 ft/s. With the
information provided, a) calculate the total acceleration vector for both cases. b) Comment on
the differences.
Positional Information:
north the toMPH 300at flying isaircraft The
rad/s 0.5at theofright the torotating is nose projectile The
downrange) lookingleft the to(Nose 5.1, up) (Nose2
calm is Wind
10 : Horizontal Vector toVelocity of Angle
True 80:VectorVelocity ofAzimuth
LattitudeNorth 5.33
vectorvelocity
−=+=
−
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
Please supply all answers in an inertial coordinate system labeled 1,2,3 with 1 being due north
and 3 being due east.
From the above diagram it is obvious that in the 1 direction we have
1 (North)
V
( ) ( )
( )
( ) ( )
=
== s
ft
5.1464
s
ft
80sin10cos151080sin10cos
3
VV
Or, in vector notation
+−= s
ft
5.14642.2622.258 321 eeeV
The wind is stated as being calm but we can assume that the flight of the aircraft causes a wind in
the opposite direction to the airplanes velocity vector, thus we can write for the case of the AC-
130:
( )
−=
−= s
ft
440
mi
ft
5280
s
h
3600
1
h
mi
300 11 eeW
Then the relative velocity vector is
s
s
The scalar magnitude of the relative velocity is
( )
=
+−+= s
ft
5.1643
s
ft
5.14642.2622.698 2
2
2
v
We shall 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 given
and
thus we can add these
to the velocity vector per our earlier convention, thus we have
2−=−=
( )
( ) ( )
970.05.78sin8cos1
3==
i
2
3 (East)
ω
1 (North)
V
( )
( ) ( )
( )
321
2
2
970.0139.0197.0
ftlbm377.5
rev
rad
2
s
rev
220ftlbm547.0
eeei +−
−
−
=
T
p
I
pI
+−= s
rad
4.1365.197.27 321 eeei
T
p
I
pI
The second term is calculated through
−−−=
−
−=
s
rad
067.0492.0015.0
099.0001.0487.0
970.0139.0197.0 321
321
eee
eee
i
idt
d
So then we can sum the two terms to obtain
+−= s
rad
1362028 321 eeeh
The gravitational acceleration vector is simply
−= 2
2s
ft
2.32 eg
( ) ( ) ( ) ( ) ( )( )
+
= −
s
ft
80cos5.33cos5.146480sin5.33cos2.698
s
rad
1029.72 5
2
( )
AZLVLV coscossin2 213 −=
s
ft
s
rad
3
++−= 2
321 s
ft
062.0115.0086.0 eeeΛ
Now we need to calculate the reference area, the drag coefficient and the rest of the particular
coefficients
( )
2
2
2
2
2
2
2
2
2ft093.0
in
ft
144
1
mm
in
4.25
1
mm105
4
1
4
1=
==
dS
The total yaw angle is
( ) ( ) ( )
5.20.2sin5.1cos5.1sinsinsincossinsin 22212221 =−+−=+= −−
t
22
( ) ( ) ( )
( )
( )
=
== s
1
059.0
lbm1.322
41.0ft093.0
s
ft
5.1643
ft
lbm
060.0
2
~
2
3
m
vSC
CD
D
( ) ( )
( )
( )
=
== ft
1
00017.0
lbm1.322
9.1ft093.0
ft
lbm
060.0
2
~
2
3
m
SC
CL
L
( ) ( )
( )
( ) ( ) ( )
( )
( ) ( )
−=
−
== s
1
0004.0
ft
in
12
in
mm
4.25lbm1.322
01.0
rev
rad
2
s
rev
220mm105ft093.0
ft
lbm
060.0
2
~
2
3
m
SdpC
Cp
p
N
N
( )
( ) ( ) ( )
( )
( )
( )
( ) ( )
=
=
+
=s
ft
0002.0
ft
in
12
in
mm
4.25lbm1.322
005.0mm105ft093.0
s
ft
5.1643
ft
lbm
060.0
2
~
2
3
m
CCvSd
CNN
N
q
q
( )
( ) ( ) ( )
( )
( )
( )
( ) ( )
−=
−
=
+
=s
1
658.0
ft
in
12
in
mm
4.25ft-lbm377.52
5.6mm105ft093.0
s
ft
5.1643
ft
lbm
060.0
2
~
2
2
2
2
2
2
2
2
2
2
3
2
T
MM
MI
CCvSd
Cq
q
( ) ( ) ( )
( )
( )
( )
( ) ( )
=
== s-ft
rad
12.1
ft
in
12
in
mm
4.25ft-lbm377.52
8.3mm105ft093.0
s
ft
5.1643
ft
lbm
060.0
2
~
2
2
3
T
M
MI
vSdC
C
( ) ( ) ( ) ( ) ( ) ( )( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )
( )
−=
+
−
−
+
−
−−
−−−
+
−
−=
222
2
2
2
2
s
ft
10
s
ft
115.0
s
ft
2.32
s
rad
970.028197.0136
s
ft
0002.0
s
ft
197.05.1464970.02.698
s
1
0004.0
s
ft
5.2cos2.2625.1643139.05.1643
ft
1
00017.0
s
ft
2.262
s
1
059.0
dt
dV
( ) ( ) ( ) ( ) ( ) ( )( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )
( )
−=
++
−−−
+
−−−
−−
−
+
−=
22
2
2
2
3
s
ft
38
s
ft
062.00
s
rad
197.020139.028
s
ft
0002.0
s
ft
139.02.698197.02.262
s
1
0004.0
s
ft
5.2cos5.14645.1643970.05.1643
ft
1
00017.0
s
ft
5.1464
s
1
059.0
dt
dV
Thus the linear acceleration vector at this instant in time is
−−−= 2
321 s
ft
3810137 eeea
( ) ( ) ( )
−=
−
−+
2
s
rad
55
s
rad
2828
s
1
658.0
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
( ) ( ) ( )
−=
−−−
−+
−−−
+
−
=
2
2
s
rad
435
s
rad
2020
s
1
658.0
s
ft
5.2cos139.05.16432.262
s-ft
rad
004.0
s
ft
970.02.698197.05.1464
s-ft
rad
12.1
dt
dh
( ) ( )
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
( ) ( )( ) ( )( ) ( ) ( ) ( )( )
( )
−
+
−−−
=
3
s
ft
5.2cos970.05.16435.1464
s-ft
rad
004.0
s
ft
197.02.262139.02.698
s-ft
rad
12.1
dt
dh
at the case where we are firing off a building. First we note that the vector diagrams are the
same.
Now let’s repeat our earlier work. Our projectile velocity vector is the same so
+−= s
ft
5.14642.2622.258 321 eeeV
2
3 (East)
80
10
2
3 (East)
78.5
8
0.5 rad/s
ω
1 (North)
2
3 (East)
V
10
321 970.0139.0197.0 eeei +−=
( ) ( )
( )
( )
=
== ft
1
00017.0
lbm1.322
9.1ft093.0
ft
lbm
060.0
2
~
2
3
m
SC
CL
L
( ) ( )
( )
( ) ( ) ( )
( )
( ) ( )
−=
−
== s
1
0004.0
ft
in
12
in
mm
4.25lbm1.322
01.0
rev
rad
2
s
rev
220mm105ft093.0
ft
lbm
060.0
2
~
2
3
m
SdpC
Cp
p
N
N
Pitch damping is different (but it’s in the noise)
( )
( ) ( ) ( )
( )
( )
( )
( ) ( )
=
=
+
=s
ft
0002.0
ft
in
12
in
mm
4.25lbm1.322
005.0mm105ft093.0
s
ft
0.1510
ft
lbm
060.0
2
~
2
3
m
CCvSd
CNN
N
q
q
( )
( ) ( ) ( )
( )
( )
( )
( ) ( )
−=
−
=
+
=s
1
604.0
ft
in
12
in
mm
4.25ft-lbm377.52
5.6mm105ft093.0
s
ft
0.1510
ft
lbm
060.0
2
~
2
2
2
2
2
2
2
2
2
2
3
2
T
MM
MI
CCvSd
Cq
q
The overturning moment is different but the Magnus moment is identical
( ) ( ) ( )
( )
( )
( )
( ) ( )
=
== s-ft
rad
02.1
ft
in
12
in
mm
4.25ft-lbm377.52
8.3mm105ft093.0
s
ft
0.1510
ft
lbm
060.0
2
~
2
2
3
T
M
MI
vSdC
C
( ) ( )
( )
( ) ( ) ( )
rad
05.0
rev
rad
2
s
rev
220mm105ft093.0
ft
lbm
060.0
~
2
2
2
3
2
M
pCSd
( ) ( ) ( ) ( ) ( ) ( )( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )
( )
−=
−++
−−−
+
−−−
−−
−
+
−=
22
2
2
2
1
s
ft
4
s
ft
086.00
s
rad
139.0136970.020
s
ft
0002.0
s
ft
970.02.262139.05.1464
s
1
0004.0
s
ft
5.2cos2.2580.1510197.00.1510
ft
1
00017.0
s
ft
2.258
s
1
054.0
dt
dV
( ) ( ) ( ) ( ) ( ) ( )( )
( )
−−−
+
−
−=
2
2
2
2
rad
ft
ft
1
s
ft
5.2cos2.2620.1510139.00.1510
ft
1
00017.0
s
ft
2.262
s
1
054.0
dt
dV
rad
h
d
rad/s 2at up rotating is nose projectile The
5.1,1
horizontal andsouth duehour per miles 20at blowing is Wind
10 : Horizontal Vector toVelocity of Angle
True 300:VectorVelocity ofAzimuth
LattitudeNorth 75.48
==
+
Projectile Information
02.0
50.2
50.3
28.0
−=
=
=
=
p
N
L
M
D
C
C
C
C
( )
( )
=
=+
−=+
3
ft
lbm
060.0
005.0
5.16
p
q
q
M
NN
MM
CC
CC
=
=
=
s
rev
150
ft-lbm40.66
ft-lbm13.19
2
2
p
I
I
T
P
Note that the above numbers are guesses at the projectiles characteristics, they do not represent
the real projectile’s performance as no data is available from any source researched.
Please supply all answers in an inertial coordinate system labeled 1,2,3 with 1 being due west
and 3 being due north.
( ) ( ) ( ) ( ) ( )
=
== s
ft
2132
s
ft
30cos10cos250030cos10cos
1
VV
( ) ( ) ( )
=
== s
ft
434
s
ft
10sin250010sin
2
VV
( ) ( ) ( ) ( ) ( )
=
== s
ft
1231
s
ft
30sin10cos250030sin10cos
3
VV
Or, in vector notation
++= s
ft
12314342132 321 eeeV
Let’s look at the wind vector which (by inspection) is
( )
−=
−= s
ft
33.29
mi
ft
5280
s
h
3600
1
h
mi
20 33 eeW
The next vector we require is the relative velocity vector defined as
++=
−−
++=−= s
ft
12604342132
s
ft
33.29
s
ft
12314342132 3213321 eeeeeeeWVv
The scalar magnitude of the relative velocity is
=
++= s
ft
2514
s
ft
12604342132 222
v
Now we shall 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 given
and
thus we can add these
to the velocity vector per our earlier convention, thus we have
2==
1
2
3 (North)
i
31.5
11
