1
( )
( )
( )
962.0
10498.1
10204.72
4
5
=
=−
−
d
S
Then
( )( )
g
S
1
999.0962.02962.0 =−
So we actually need
001.1
4
2
== M
P
Sg
and
MP 004.4=
Thus the answer is not appreciably changed and
( )
( )
016.010046.6004.4 5== −
P
( ) ( )
( )
( )
( )
Hz699,1
s
rad
674,10
in
ft
12
1
in5.0
s
ft
013,3
inlbm10677.2
inlbm025.0
016.0 23
2
=
=
=
=−
d
V
I
I
Pp
P
T
Problem 14 – For the projectile given in chapter 8, problem 27, determine the precessional and
nutational frequencies in Hertz. Determine the minimum spin rate for the projectile to be stable.
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
pd
I
P
The precessional rate is given by equation (FS-172) as
()
MPP
S4
2
12−−=
(FS-172)
( ) ( )
( )
−−=
−4
210144.14032.0032.0
2
1
S
3
10108.4 −
=
S
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 equation (FS-
138) where
=d
V
dt
d
(FS-138)
( )
rad
s
ft
022,1
3
dS
( ) ( )
( )
( )
( )
=
=s
rad
83028.0
in
ft
12
1
in4.134
s
ft
022,1
dt
dF
( )
Hz2.13
rad
rev
2
1
s
rad
83 =
=
dt
dF
Now we will check for gyroscopic stability. If the projectile is just gyroscopically stable then
1
4
2
== M
P
Sg
and
MP 4=
Thus
( )
( )
021.010144.14 4== −
P
Then the spin rate would have to be based on
=V
pd
I
I
P
T
P
( )
( )
( )
2
2
2
2inlbm768.78
in134.4lbm1.321
=
x
k
964.6
1
2=
x
k
pd
I
P
The precessional rate is given by equation (FS-172) as
()
MPP
S4
2
12−−=
(FS-172)
( ) ( )
( )
−−=
−4
210144.14032.0032.0
2
1
S
3
10108.4 −
=
S
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 equation (FS-
138) where
=d
V
dt
d
(FS-138)
( ) ( )
( )
( )
( )
=
=s
rad
83028.0
in
ft
12
1
in4.134
s
ft
022,1
dt
dF
( )
Hz2.13
rad
rev
2
1
s
rad
83 =
=
dt
dF
Now we will check for gyroscopic stability. If the projectile is just gyroscopically stable then
1
4
2
== M
P
Sg
and
MP 4=
Thus
( )
( )
021.010144.14 4== −
P
Then the spin rate would have to be based on
=V
pd
I
I
P
T
P
H
Thus we need to calculate the rest of the “starred” coefficients
( )( )
lbm1.322
D
LL C
m
Sd
C2
=
( )
( )
( )
2
2
2
2inlbm768.78
in134.4lbm1.321
=
x
k
964.6
1
2=
x
k
Off the left side of the aircraft we have a different drag coefficient. So we write
( )( )
lbm1.322
D
Now H will be a little different
Then
Then the projectile is dynamically stable if
608.0
608.0
The spin rate would then be
( ) ( )
( )
( )
( )
Hz126
s
rad
800
in
ft
12
1
in134.4
s
ft
022,1
inlbm768.78
inlbm288.774
027.0 2
2
=
=
=
=d
V
I
I
Pp
P
T
LSLS
So the spin rate required to stabilize the projectile is slightly higher off the right side of the plane.
10.4 Roll Resonance
Problem 16 – Roll resonance of a projectile occurs when the spin rate approaches a “forcing”
frequency of the projectile. This occurs more frequently in fin stabilized projectiles than spin
stabilized projectiles because, in the latter, the spin rate is usually quite high in order to maintain
stability, the overturning moment is positive and these forcing functions unless they are
intentional – like thrusters are usually due to asymmetries (like bent fins) usually are small. If
we examine the projectile of Chapter 8, problem 37 instead as a fin stabilized projectile we can
write the equation for the pointing direction as
( )
0
21
321
+
++= i
ii eKeKeK
Here the subscripts 1 and 2 represent the fast and slow modes, respectively. The third term is the
forcing function where we can define
V
pd
=
(1)
and, after inserting the initial conditions, say, of
0 = 0 we obtain
( )
PTHiMP
iA
K−
−+
−
−
=
2
3
3
(2)
Based on what you know about the behavior of imaginary numbers and expressions (1) and (2)
above, determine the spin rate at which catastrophic yaw will occur. Use the aerodynamic
properties of the projectile from chapter 8, problem 37 but assume that the overturning moment
is the negative of what was provided (we are essentially faking a fin-stabilized version). You
may assume the velocity stays constant at 750 ft/s.
Solution: First, we will put everything in terms of the spin rate, p. The forcing frequency is
V
pd
( )
( )
( )
p
p
00289.0
s
ft
750
rev
rad
2
in
ft
12
1
in134.4
s
rev
=
=
The non-dimensional spin rate and overturning moments are is
=V
pd
I
I
P
T
P
( ) ( )
( )
−
−
=
s
ft
750
in
ft
12
1
in134.4
rev
rad
2
s
rev
ftlbm337.5
ftlbm547.0
2
2
p
P
pP 00030.0=
From problem 1.) we have, directly,
( )
( )
54 1014.81014.1714.0 −− −=−=M
To find the other parameters we need the “starred” coefficients.
LL C
m
Sd
C2
=
( ) ( )
MMMM CC
m
Sd
CC qq +=+
2
( )
( ) ( )
( )
( )( )
( )
5.6
lbm1.322
in
ft
12
1
mm
in
4.25
1
mm105ft0932.0
ft
lbm
060.0 2
3−
=+
MM CC q
964.6
1
2=
x
k
( )
( )
( )
65 1050.1964.61070.5 −− +=T
5
1074.6 −
=T
Let’s combine all this good stuff then into
54522
21074.600030.01084.100289.01014.800030.000289.000289.0 −−− −−−− ppipp
To make the denominator real we must multiply by the complex conjugate so we have