Problem 1 – An 155mm M549A1 Projectile has the following properties and initial conditions
876.0
26
024.0
28.4
13.0
3.0
=
−=+
−=
=
=
=
p
q
p
M
MM
l
M
L
D
C
CC
C
C
C
C
=
=
=
s
ft
000,3
mm155
ft
lbm
0751.0 3
muzzle
V
d
lbm96
in–lbm610,6
in–lbm5.505
2
2
=
=
=
m
I
I
T
P
At an instant in time after launch when
=
==
=
s
ft
764,1
4
Hz220
V
p
Determine
R
b.) The precessional frequency in Hz.
2=
dt
d
c.) The nutational frequency in Hz.
1=
d
DD C
m
( ) ( )
( )
( )( )
( )
5
2
3
10206.13.0
lbm962
in
ft
12
1
mm
in
4.25
1
mm155ft202.0
ft
lbm
0751.0
− =
=
D
C
LL C
m
Sd
C2
=
T
yI
md
k
2
2
1=
( )
( )
( )
2
2
2
2
2
2
2inlbm610,6
mm
in
4.25
1
mm155lbm96
1
−
=
y
k
539.0
1
2=
y
k
P
xI
md
k
2
2
1=
( )
( )
( )
2
2
2
2
2
2
2inlbm5.505
mm
in
4.25
1
mm155lbm96
1
−
=
x
k
058.7
1
2=
x
k
After finding the inertial properties we can now calculate the coefficients in the stability equation
( )
=
M
y
C
k
M2
1
( )
( )
54 10273.91072.1539.0 −− ==M
( )
−=
p
M
x
LC
k
CT 2
1
( )
( )
( )
456 10534.210521.3058.710225.5 −−− =−=T
2
cos
V
gd
G
=
( ) ( )
( )
( )
6
2
2
2
2
10241.5
s
ft
764,1
4cos
in
ft
12
1
mm
in
5.25
1
mm155
s
ft
2.32
−
=
=
G
( )
( )
( )
495.2
10273.94
03.0
5
2
=
=−
g
S
Since this number is greater than 1 the projectile is gyroscopically stable. We have another
( )
( )
( )
911.0
10565.5
10534.22
4
4
=
=−
−
d
S
( ) ( )( )
992.0911.02911.0401.0
495.2
1=−=
( )
( )
( )
rad00172.0
10273.9
10241.503.0
5
6
=
=−
−
R
To determine the precession and nutation frequencies we need to obtain the
turning rates of the slow arm (precession) and fast arm (nutation), invert them and adjust the
units. The precessional rate is given by
2=
if we recall how this quantity was derived, it was the turning rate of the precessional vector with
respect to dimensionless arc length. To convert it to an actual rate we must invoke
rad
2
s
dt
Now we perform the same procedure with the nutational rate. The nutational rate is given by
1=
Again, this was the turning rate of the nutational vector with respect to dimensionless arc length.
To convert it to an actual rate we must use
( )
Hz9.14
rad
rev
2
1
s
rad
76.93
1=
=
dt
d
Problem 2 – For the projectile given in Chapter 8, problem 37, determine the precessional and
nutational frequencies in Hertz.
Solution: To determine the precession and nutation frequencies we need to obtain the turning
rates of the slow arm (precession) and fast arm (nutation), invert them and adjust the units. To
T
yI
md
k
2
2
1=
(FS-58)
( )
( )
( )
2
2
2
2
2
2
2
2
2
2ftlbm337.5
in
ft
12
1
mm
in
4.25
1
mm105lbm1.32
1
−
=
y
k
714.0
1
2=
y
k
Now we need to use equation (FS-59) to find CMα* as follows
Sd
V
d
as an anti-tank weapon if the situation required it and its high rate of fire made it quite successful
as an anti-personnel weapon. Assume the properties of the system are given below
a) Calculate the gyroscopic stability factor at the beginning and the end of the flight
assuming a terminal velocity of 2450 ft/s and the spin rate is 10% lower than the
g
b) Is the projectile stable throughout the flight?
c) Assuming that this is the longest time of flight for the projectile at what spin rate will
the projectile become unstable?
rad
unstable
d) Where will the instability occur?
Projectile and Weapon Information
011.0
10.3
−=
=
p
l
M
C
C
=
=
=
s
ft
2850
mm40
ft
lbm
067.0 3
muzzle
V
d
=
=
=
=
cal
rev
30
1
lbm985.1
in–lbm263.6
in–lbm231.1
2
2
n
m
I
I
T
P
( )
2
2
2
2
2
2
2
2
2ft0135.0
in
ft
144
1
mm
in
4.25
1
mm40
4
1
4
1=
==
dS
The spin rate at the muzzle is given by
( ) ( ) ( ) ( )
=
== s
rad
4548
s
ft
2850
ft
in
12
in
mm
4.25
mm
cal
40
1
rev
rad
2
cal
rev
30
1
nVpmuzzle
From the problem statement we know that
( )( )
=
== s
rad
4093
s
rad
45489.09.0 muzzleterminal pp
At the muzzle the gyroscopic stability factor is
( )
( ) ( )
( ) ( )
( )
( )
( ) ( ) ( )
10.3
ft
in
12
s
ft
2850mm40ft0135.0inlbm263.6
ft
lbm
067.02
in
mm
4.25
s
rad
4548inlbm231.1
22
2
22
3
2
2
2
2
2
2
22
−
−
==
MT
P
gCSdVI
pI
S
814.5=
g
S
At the end of the flight the Sg is
( )
( ) ( )
( ) ( )
( )
( )
( ) ( ) ( )
10.3
ft
in
12
s
ft
2450mm40ft0135.0inlbm263.6
ft
lbm
067.02
in
mm
4.25
s
rad
3638inlbm231.1
22
2
22
3
2
2
2
2
2
2
22
−
−
==
MT
P
gCSdVI
pI
S
372.6=
g
S
Since the gyroscopic stability factor is greater than 1 throughout the flight the projectile is stable.
( )
( ) ( )
( ) ( )
( )
( )
( ) ( ) ( )
10.3
ft
in
12
s
ft
2450mm40ft0135.0inlbm263.6
ft
lbm
067.02
in
mm
4.25
s
rad
1698inlbm231.1
22
2
22
3
2
2
2
2
2
2
22
−
−
==
MT
P
gCSdVI
pI
S
097.1=
g
S
Without going into any great detail the projectile will go unstable at the muzzle first.
Problem 4 – For the projectile described in problem 35 of chapter 8:
a) Determine the precessional damping exponent.
2−=
b) Determine the nutational damping exponent.
1−=
c) Based on a) and b) above which of these modes will damp out?
d
e) Consider a cargo projectile with identical properties to our projectile in problem 2.
The designer did not secure the cargo well enough so that the cargo fails to spin up
completely during gun launch in a worn tube. When this happens, immediately after
muzzle exit the round spins down (and the cargo spins up a little more) so that the
projectile finally reaches a spin rate of 100 Hz. The velocity is unaffected.
a. Determine the gyroscopic stability factor for each of the two situations
g
g
b. Will both projectiles fly properly? Why or why not.
Solution: We need to define our parameters T, H, M, and P but we have some preliminary work
to do as follows
( ) ( ) ( )
ftlbm377.5
ft
in
144
in
mm
25.4
2
2
2
2
2
2
2=
−
T
yI
k
( ) ( )
( )
( )( )
( ) ( )
0000569.09.1
ft
mm
8.304lbm1.322
mm105ft093.0
ft
lbm
060.0
2
2
3=
==
LL C
m
Sd
C
( ) ( )
( )
( )( )
( ) ( )
0000123.041.0
ft
mm
8.304lbm1.322
mm105ft093.0
ft
lbm
060.0
2
2
3=
==
DD C
m
Sd
C
( )
( ) ( )
( )
( )( )
( ) ( )
0001946.05.6
ft
mm
8.304lbm1.322
mm105ft093.0
ft
lbm
060.0
2
2
3−=−
=+=+
MMMM CC
m
Sd
CC qq
( ) ( )
( )
( )( )
( ) ( )
0001138.08.3
ft
mm
8.304lbm1.322
mm105ft093.0
ft
lbm
060.0
2
2
3=
==
MM C
m
Sd
C
( ) ( )
( )
( )( )
( ) ( )
0000015.005.0
ft
mm
8.304lbm1.322
mm105ft093.0
ft
lbm
060.0
2
2
3=
==
pp MM C
m
Sd
C
Now we will calculate define our parameters T, H, M, and P
( ) ( )( )
0000673.00000015.0964.60000569.0
1
2=+=+=
p
M
x
LC
k
CT
( )
( ) ( ) ( )( )
0001824.00001946.0708.00000123.00000569.0
1
2=−−−=+−−=
MM
y
DL CC
k
CCH q
( )( )
0000806.00001138.0708.0
1
2===
M
y
C
k
M
( ) ( ) ( )
( ) ( )
03208.0
s
ft
1510
ft
mm
304.8
mm105
rev
rad
2
s
rev
220
377.5
547.0 =
=
=
V
pd
I
I
P
T
P
Now we can begin answering the questions.
The precessional mode damping exponent is found from
The nutational mode damping exponent is found from
Since both of these exponents are negative, both modes will damp.
The dynamic stability factor is obtained through
( )
0001824.0
Since this value is between 0 and 2 the projectile is dynamically stable. The gyroscopic stability
( )
( )( )
192.3
0000806.04
03208.0
4
2
2
=== M
P
Sg
( ) ( ) ( )
( ) ( )
014582.0
s
ft
1510
ft
mm
304.8
mm105
rev
rad
2
s
rev
100
377.5
547.0 =
=
=
V
pd
I
I
P
T
P
Then the gyroscopic stability factor is
( )
( )( )
659.0
0000806.04
014582.0
4
2
2
=== M
P
Sg
This projectile is unstable. The second projectile will not be stable and tumble some distance out
of the gun
Problem 5 – A 155 mm HE projectile is fired from a cannon. The muzzle velocity of the
projectile is 800 m/s and the twist of the rifling is 1:20. The projectile and filler properties are
given below. Assuming the aerodynamic forces and moments are negligible and that the
projectile is dynamically stable :
muzzle
b) The spin rate of the projectile in flight assuming the fill does not spin up in the bore
and both shell and fill come into dynamic equilibrium.
rad
total
c) Determine the gyroscopic stability factors for a) and b).
Projectile and Weapon Information
012.0
07.1
−=
=
p
l
M
C
C
=
ft
lbm
067.0 3
=
=
in–lbm335,3
in–lbm555
2
2
I
I
total
total
T
P
( )
2
2
2
2
2
2
2
2
2ft203.0
in
ft
144
1
mm
in
4.25
1
mm155
4
1
4
1=
==
dS
The spin rate at the muzzle is given by
( ) ( ) ( )
=
== s
rad
5.621,1
s
m
800
m
mm
000,1
mm
cal
155
1
rev
rad
2
cal
rev
20
1
nVpmuzzle
From the problem statement we know that, if we neglect aerodynamic forces and moments, the
projectile will neither drag down nor spin down. Therefore if the fill is not spun up in the
weapon and catches up with the shell body (which was spun up by the weapon) conservation of
momentum will tell us what the final spin rate will be. We can write that the angular momentum
of the shell only will be equal to the angular momentum of the shell plus the fill after they are in
dynamic equilibrium. We can write
totalPshellP pIpI totalshell =
or
shell
P
P
total p
I
I
p
total
shell
=
( )
( )
( )
−
−
=s
rad
5.621,1
inlbm555
inlbm431
2
2
total
p
=s
rad
2.259,1
total
p
For part a.) the gyroscopic stability factor is
MT
P
gCSdVI
pI
S2
22
2
=
( )
( ) ( )
( ) ( )
( )
( )
( ) ( ) ( ) ( )
07.1
ft
in
12
m
ft
281.3
s
m
800mm155ft203.0inlbm335,3
ft
lbm
067.02
in
mm
4.25
s
rad
5.1621inlbm555
2
2
2
2
22
3
2
2
2
2
2
−
−
=
g
S
54.16=
g
S
For part b.) the gyroscopic stability factor is
( )
( ) ( )
( ) ( )
( )
( )
( ) ( ) ( ) ( )
07.1
ft
in
12
m
ft
281.3
s
m
800mm155ft203.0inlbm335,3
ft
lbm
067.02
in
mm
4.25
s
rad
2.1259inlbm555
2
2
2
2
22
3
2
2
2
2
2
−
−
=
g
S
97.9=
g
S
Since the gyroscopic stability factor is greater than 1 in both cases the projectile is stable.
Problem 6 – For the projectile given in problem 5, above, determine the precession and nutation
frequencies in Hertz.
dt
dt
We need some intermediate steps to find M. First we determine 1/ky2 as follows:
T
yI
md
k
2
2
1=
( )
( )
( )
2
2
2
2
2
2
2inlbm335,3
mm
in
4.25
1
mm155lbm106
1
−
=
y
k
183.1
1
2=
y
k
Now we need to use equation (FS-59) to find CMα* as follows
M
=d
V
dt
d
( )
( )
( )
=
=s
rad
18.400081.0
m0.155
s
m
800
2
dt
d
( )
Hz67.0
rad
rev
2
1
s
rad
18.4
2=
=
dt
d
Now we perform the same procedure with the nutational rate. The nutational rate is given by
()
MPP 4
2
12
1−+=
( ) ( )
( )
−+=
−5
2
11013.44052.0052.0
2
1
0512.0
1=
( )
( )
( ) ( ) ( )
( )
−
−
=
s
ft
764,1
in
ft
12
1
mm
in
4.25
1
mm155
rev
rad
2
s
rev
130
inlbm610,6
inlbm5.505
2
2
P
018.0=P
Based on this the stability factors are found through
1
4
2
= M
P
Sg
( )
( )
( )
871.0
10273.94
018.0
5
2
=
=−
g
S
Since this number is less than 1 the projectile is NOT gyroscopically stable. We are done but
( )
( )
( )
911.0
10565.5
10534.22
4
4
=
=−
−
d
S
( )( )
911.02911.0
min =
−
=d
V
I
I
Pp
P
T
minmin
=d
V
I
I
Pp
P
T
minmin