( )
xVgV
V
++−=
p
ND C
V
d
SVmVSC
dt
d
m
2
1
2
1
In this case x is the ball axis which will have to point up to cause the ball to move left. With that
p
N
x
xzxD
zC
V
d
SVVVSC
dt
dV
m
−−= 2
2
1
2
1
We can divide these by the mass and write them in terms of distance to yield
2
2
xDxx VC
m
S
VV
−=
gVVC
m
S
VV yxDyx −−=
2
−−=
x
xNzxDzx V
d
VC
m
S
VVC
m
S
VV p
2
2
2
Dividing by the axial velocity we have
xDx VC
m
S
V2
−=
(1)
x
yDy V
g
VC
m
S
V−−=
2
(2)
(3)
Because the coefficients are constants we define
11
2
2
ˆkK
m
S
C
m
S
CDD ===
(4)
( ) ( )
 
( )( )
 
( )
=
=−
ft
1
109.252.0
oz
lbm
16
1
oz52
in
ft
12
1
in9.2
4
ft
lbm
076.0
3
2
2
2
2
2
3
1
k
We have already solved equations (1) and (2) with the following results:
xk
xx eVV 1
0
−
=
(FF-51)
( )
1
11
01
−= xk
x
e
kV
t
(FF-53)
( )
Bxp =
( )
Axq −=
It follows that we can write
+
−= −− BdxBdxBdx
zCeAeeV
(7)
BxBxBx
zCeAeeV −− +−=
B
A
CeVBx
z−= −
The initial conditions are at x = 0, Vz = Vz0 thus
B
A
VC z+= 0
Then
( )
Bx
z
BxBx
zz eVe
B
A
B
A
e
B
A
VV −−− +−=−
+= 00 1
(8)
Rewriting we have
( )
Bx
z
Bx eVe
B
A
dt
dz −− +−= 0
1
(9)
Then the drift is
( )
0
0
1zteVet
B
A
zBx
z
Bx ++−= −−
(10)
We need to solve these for the spin rate. Let’s put in some numbers
( ) ( )
=
=s
ft
7.102
mi
ft
280,5
s
hr
600,3
1
h
mi
70
0
x
V
s
mi
s
600,3
h
( )
( )
( )
( )
 
 
s64.01
ft
1
109.2
s
ft
7.102
1ft5.60
ft
1
109.2
3
3
=
−
=
−
−
−
et
( )
( )
( ) ( )
 
( )
( )
( )
( )
 
−=
+
−
=
−
s
ft
0.13
2
s64.0
ft
1
109.2
s
ft
7.102
1
s
ft
7.102
s64.0
s
ft
2.32
4tan
s
ft
2.86
3
2
y
V
( )
 
( )
( ) ( )
 
( )
( )
( )
 
( )
( )
( )
 
( )
( )
( )
 
+
−
+
−+=
−
−
−2
3
3
3
2
2
2
s64.0
ft
1
109.2
s
ft
7.102
s64.0
ft
1
109.2
s
ft
7.1021ln
s64.0
ft
1
109.2
s
ft
7.102
1
2
1
s64.0
s
ft
2.32
2
1
4tanft5.604
y
 
ft928.1=y
5.12
2.86
1.19
tantan 11 =
=
=−−
x
y
V
V
1.2
2.86
11.3
tantan 11 =
=
=−−
x
z
V
V
So we are off a bit in the vertical but fine in deflection
9.6 Modified Point Mass Trajectory
Problem 40 – The Paris gun was built by Germany in the First World War to shell Paris from 75
miles away. The weapon was a 210 mm diameter bore with the shells pre-engraved to account
for wear of the tube. During firing of this weapon all things such as wind effects, Coriolis etc.
had to be accounted for. When the United States entered the war, the doughboys (the nickname
for American troops) were to take the St. Mihiel salient where the gun was located. We shall
assume that the Germans have turned the gun to fire on the Americans. The projectile is at some
point in space defined below. To demonstrate your knowledge of the Modified Point Mass
equations please
a) DRAW the situation
b) Calculate the vector yaw of repose for this projectile: use equation (8.6.16)
R
c) Write the acceleration vector for this projectile: use equation (8.6.29)
at the instant in its trajectory when the velocity (relative to the ground) is 2100 ft/s and the
conditions below apply.
Note: You don’t need all of the information below It is provided to you so you can
compare the differences in formulations with the 6-DOF model.
d) Why don’t we need to obtain the angular acceleration vector dh/dt?
Positional Information:
rad/s 1at down rotating is nose projectile The
25.0,0.5
horizontal andsouth duehour per miles 15at blowing is Wind
1 : Horizontal Vector toVelocity of Angle
True 190:VectorVelocity ofAzimuth
LattitudeNorth 48
==
+
Projectile Information
01.0
02.0
50.3
28.0
−=
−=
=
=
p
p
l
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
130
ft–lbm40.66
ft–lbm13.19
2
2
p
I
I
T
P
Please supply all answers in an inertial coordinate system labeled 1,2,3 with 1 being due south
and 3 being due west.
( ) ( )
( )
( ) ( )
=
== s
ft
2068
s
ft
10cos1cos210010cos1cos
1
VV
( )
( )
( )
=
== s
ft
37
s
ft
1sin21001sin
2
VV
( ) ( )
( )
( ) ( )
=
== s
ft
365
s
ft
10sin1cos210010sin1cos
3
VV
Or, in vector notation
 
++= s
ft
365372068 321 eeeV
Let’s look at the wind vector which (by inspection) is
( )
+=
+= s
ft
22
mi
ft
5280
s
h
3600
1
h
mi
15 11 eeW
The next vector we require is the relative velocity vector defined as
 
++=
−
++=−= s
ft
365372046
s
ft
22
s
ft
365372068 3211321 eeeeeeeWVv
The scalar magnitude of the relative velocity is
=
++= s
ft
2079
s
ft
365372046 222
v
We now need to determine the unit vector, l along the axis that the projectile is pointing. Keep
2
V
1
3 (West)
 
321321 176.0018.0984.0365372046
s
ft
2079
1eeeeee
v
l++=++
== v
3 (West)
1==
( )
( )
026.05.1sin1
2==
i
( )
( ) ( )
178.025.10sin5.1cos1
3==
i
1
2
10.25
1.5
33
22
11
VV
VV
VV
−=
=
−=
33
22
11
−=
=
−=
( )
LVAZLV sinsincos2 321 +−=−
( ) ( ) ( ) ( )( )
+−
−=−
s
ft
48sin365190sin48cos37
s
rad
1029.72 5
1
( )
AZLVAZLV coscossincos2 312 −−=
( ) ( ) ( ) ( ) ( )( )
−−
=−
s
ft
190cos48cos365190sin48cos2068
s
rad
1029.72 5
2
( )
AZLVLV coscossin2 213 −−=−
( ) ( ) ( ) ( )( )
−−
−=−
s
ft
190cos48cos3748sin2068
s
rad
1029.72 5
3
 
++−= 2
321 s
ft
221.0070.0040.0 eeeΛ
Now we need to calculate the reference area and some of the 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
( )
Λgαvαv
V++−+−= RNRLD p
CvCC
dt
d
~~~ 2
( ) ( ) ( )
 
( )
( )
 
=
== s
1
030.0
lbm2202
28.0ft373.0
s
ft
2079
ft
lbm
060.0
2
~
2
3
m
vSC
CD
D
( ) ( )
 
( )
( )
 
=
== ft
1
00013.0
lbm2202
5.2ft373.0
ft
lbm
060.0
2
~
2
3
m
SC
CL
L
( ) ( )
 
( )
 
( ) ( ) ( )
( )
 
( ) ( )
−=
−
== s
1
0006.0
ft
in
12
in
mm
4.25lbm2202
02.0
rev
rad
2
s
rev
130mm210ft373.0
ft
lbm
060.0
2
~
2
3
m
SdpC
Cp
p
N
N
We now need the cross product below
 
+−−==s
ft
368.16092.4846.2
002.0008.0000.0
365372046 321
321
eee
eee
αvR
Now we can solve component by component
( ) ( ) ( ) ( ) ( )
+
−=
2
2
2
1
s
ft
02079
ft
1
00013.0
s
ft
2046
s
1
030.0
dt
dV
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
−=
+
−
−
−−
+
−=
222
2
2
2
2
s
ft
29
s
ft
070.0
s
ft
2.32
s
ft
092.4
s
1
0006.0
s
ft
008.02079
ft
1
00013.0
s
ft
37
s
1
030.0
dt
dV
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
−=
++
−−
+
−=
22
2
2
2
3
s
ft
10
s
ft
221.00
s
ft
368.16
s
1
0006.0
s
ft
002.02079
ft
1
00013.0
s
ft
365
s
1
030.0
dt
dV
Thus the acceleration vector is
 
−−−= 2
321 s
ft
102961 eee
V
dt
d
Problem 41 – If we were to use a modified point mass assumption for both of the cases sited in
problem P 8.5.2
a) Calculate the vector yaw of repose for both cases
Answer:
 
321 0009.00060.00003.0 eeeα−+=
R
for the building
 
321 0000.00007.00009.0 eeeα−−−=
R
for the gunship
b) Draw and explain what this vector represents
c) Comment on whether this model is applicable for each case and why
Solution: For a spin stabilized projectile, the vector yaw of repose was given as follows:
In either case we need to calculate v × dV/dt for each case. We shall start with the round fired
from the AC-130. In this case we have
31
s
02.320
So we can write
321 s
The first term in the numerator is
( )( )
 
( )( ) ( )
+−−
−−=
− 3
2
321
2
s
ft
429,23667,178633,2
rev
rad
2
s
rev
2209.1ftlbm547.022 eee
V
v
dt
d
pCI LP
 
−+=
− 4
4
3
6
2
6
1
6
s
ft–lbm
103.67103.51310565.72 eee
V
vdt
d
pCI LP
For the second term we have
( )( )
 
( )
 
( ) ( ) ( ) ( )
+−−
−=
−− 3
2
321
2
2
2
2
2
2
s
ft
911,45667,178790,4905.0
rev
rad
2
s
rev
220
ft
mm
8.304
mm105
lbm1.3222 eeeg
V
v
dt
d
pCmd p
M
 
−
−+=
−− 4
4
3
6
2
6
1
62
s
ftlbm
102.24101.941022.262eeeg
V
vdt
d
pCmd p
M
Now for the denominator we have
 
pp
NMML
CCdpCCvSdv
2222
=+
( ) ( )( ) ( ) ( )
( )
( )( )
−
+
01.005.0
ft
mm
8.304
s
22208.39.1
s
5.1643
s
192,5
2
2
2
 
2
2222
ft–lbm
192,5 +
=+
 
=+ 4
4
62222
ft–lbm
103.256,101
pp NMML CCdpCCvSdv
 
−
−
−++
−
−+
=
4
4
6
4
4
6
321
4
4
6
321
s
ftlbm
103.256,101
s
ftlbm
102.241.9422.26
s
ftlbm
103.673.513565.7 eeeeee
αR
 
321 0009.00060.00003.0 eeeα−+=
R
Or
 
 
321
4
4
6
4
4
6
321
0007.00051.00001.0
s
ftlbm
101.293,53
s
ftlbm
104.352.270982.3
eee
eee
α−+=
−
−
−+
=
R
Which, you can see in the dominant term is accurate to within about 15% We proceed along the
same lines to obtain the answers for the ground based weapon.
( )( )
 
( )( ) ( )
−+
−−=
− 3
2
321
2
s
ft
184,2729,14345,27
rev
rad
2
s
rev
2209.1ftlbm547.022 eee
V
v
dt
d
pCI LP
 
+−−=
− 4
4
3
6
2
6
1
6
s
ft–lbm
1028.6103.42106.782eee
V
vdt
d
pCI LP
For the second term we have
( )( )
 
( )
 
( ) ( ) ( ) ( )
++−
−=
−− 3
2
321
2
2
2
2
2
2
s
ft
130,6729,14812,1905.0
rev
rad
2
s
rev
220
ft
mm
8.304
mm105
lbm1.3222 eeeg
V
v
dt
d
pCmd p
M
 
−
−−=
−− 4
4
3
6
2
6
1
62
s
ftlbm
1023.31076.7104.102eeeg
V
vdt
d
pCmd p
M
Now for the denominator we have
 
( ) ( )
CCdpCCvSdv
2222
ft
8.304
=+
 
( ) ( )( ) ( ) ( )
( )
( )( )
−
+
=+
01.005.0
ft
mm
8.304
s
22208.39.1
s
0.1510
s
383,4
2
2
2
2222
 
pppp NMMLNMML CCdpCCvCCdpCCvSdv 222
2
2
2222
s
ft–lbm
383,4 +
=+
4
62222
ft–lbm
If we used
 
 
321
4
4
6
4
4
6
321
0001.00006.00011.0
s
ftlbm
100.976,37
s
ftlbm
1031.33.224.41
eee
eee
α+−−=
−
−
+−−
=
R
The answer would be much closer in this case because the projectile has a much smaller effective
yaw to it.
We need to draw this situation to understand it. In the case where we fired out of the AC-130 the
V
i
t
R
Unit length
 
321
321
028.0237.0027.0
891.0156.0425.0
970.0139.0197.0 eee
eee
li ++=
−
−=
then
( )
321
321
105.0012.0215.0
028.0237.0027.0
891.0156.0425.0 eee
eee
lilα++−=−==
R
And for the case of firing from the top of the building we get
 
321
321
970.0174.0171.0
s
ft
0.1510
s
ft
5.14642.2622.258
eee
eee
l+−=
+−
=
321 970.0139.0197.0 eeei +−=
 
321
321
011.0025.0034.0
970.0174.0171.0
970.0139.0197.0 eee
eee
li ++=
−
−=
( )
321
321
010.0031.0026.0
011.0025.0034.0
970.0174.0171.0 eee
eee
lilα++−=−==
R