Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
( ) ( )
( )
( )
( )
Hz.130
s
rad
813
in
ft
12
1
mm
in
4.25
1
mm105
s
ft
1500
ftlbm547.0
ftlbm377.5
019.0 2
2
min =
=
−
−
=p
Problem 10 - A right circular cylinder is to be fired horizontally for an impact test. If the
cylinder is made of steel (ρ = 0.283 lbm/in3) and it is 0.5 inches in diameter and 0.75 inches long.
Determine:
a) The spin rate required to stabilize the projectile (if it can be stabilized)
b) Comment on the answer above – what is dominant in the problem?
c) The precessional and nutational frequencies of the projectile at this spin rate in Hz.
The projectile properties are provided below
92.0
29
02.0
0.8
18.0
4.0
=
−=+
=
=
=
=
p
q
p
M
MM
l
M
L
D
C
CC
C
C
C
C
=
=
=
=
s
ft
000,6
in75.0
in5.0
ft
lbm
0751.0 3
muzzle
air
V
l
d
Solution: To calculate the requested information we need to calculate the reference area,
moments of inertia and mass in addition to the information provided above.
2
in
144
8
8=
( )
( )( )
( )
25
2
2
2
2
2
2
2
2
ftlbm10809.1
in
ft
144
1
in75.0
4
in5.03
lbm042.0
12
1
4
3
12
1=
+=
+= −
l
d
mIT
( )
23
2
2
2
2
2ft10364.1
in
ft
144
1
in5.0
4
1
4
1−
=
==
dS
( )
( )
( )
( )( )
( )
5
23
3
1071.492.0
lbm042.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0
−
−
=
=
p
M
C
Now we find the remaining inertial properties
T
yI
md
k
2
2
1=
( )
+=
p
M
x
LC
k
CT 2
1
( )
( )
( )
456 1086.31071.4810214.9 −−− =+=T
2
cos
V
gd
G
=
( ) ( )
( )
( )
8
2
2
2
2
10718.3
s
ft
000,6
0cos
in
ft
12
1
in5.0
s
ft
2.32
−
=
=
G
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
( )
( )
081.010638.14 3== −
P
Then the spin rate would have to be based on
=V
pd
I
I
P
T
P
( )
( )
( )
( )
( )
Hz700,3
s
rad
300,23
in
ft
12
1
in5.0
s
ft
000,6
ftlbm10045.9
ftlbm10809.1
081.0 26
25
=
=
−
−
=
=−
−
d
V
I
I
Pp
P
T
We have another condition, however, for dynamic stability where we need
( )
dd
g
SS
S− 2
1
( )
( )
( )
128.0
10029.6
1086.32
3
4
=
=−
−
d
S
Then
( )( )
1240.0128.02128.0 =−
Thus the projectile may be spin stabilized with a spin rate of
( )
dd SS
M
P−
=2
4
165.0=P
and
Hz600,7=p
We can see that this is dominated by the large overturning moment and high velocity.
2
S
011.0=
S
138) where
=d
V
dt
d
(FS-138)
( )
( )
( )
=
=s
rad
524,1011.0
in
ft
12
1
in0.5
s
ft
000,6
dt
dS
2
F
155.0=
F
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 equation (FS-138).
( )
( )
( )
=
=s
rad
300,22155.0
in
ft
12
1
in0.5
s
ft
000,6
dt
dF
Hz550,3
rad
rev
2
1
s
rad
300,22 =
=
dt
dF
Problem 11 – Modifications are made to a 155mm M483A1 projectile so that it has the following
properties and initial conditions
20.1
2.15
0285.0
573.4
975.1
2.0
=
−=+
−=
=
=
=
p
q
p
M
MM
l
M
L
D
C
CC
C
C
C
C
=
=
s
ft
900,2
ft
lbm
0751.0 3
muzzle
V
lbm103
in-lbm1.537
2
2
=
=
m
I
T
P
At an instant in time after launch when
=
==
=
s
ft
000,1
2
Hz100
V
p
Determine
a.) If the projectile is stable
b.) The precessional frequency in Hz.
c.) The nutational frequency in Hz.
( ) ( )
( )
( )( )
( )
6
2
3
105.72.0
lbm1032
in
ft
12
1
mm
in
4.25
1
mm155ft202.0
ft
lbm
0751.0
− =
=
D
C
LL C
m
Sd
C2
=
( ) ( )
( )
( )( )
( )
5
2
3
104.7975.1
lbm1032
in
ft
12
1
mm
in
4.25
1
mm155ft202.0
ft
lbm
0751.0
− =
=
L
C
MM C
m
Sd
C2
=
667.0
1
2=
y
k
P
xI
md
k
2
2
1=
( )
( )
( )
2
2
2
2
2
2
2inlbm1.537
mm
in
4.25
1
mm155lbm103
1
−
=
x
k
141.7
1
2=
x
k
After finding the inertial properties we can now calculate the coefficients in the stability equation
( )
( )
( )
455 1047.21049.4141.7104.7 −−− −=−=T
2
cos
V
gd
G
=
( ) ( )
( )
( )
5
2
2
2
2
1064.1
s
ft
000,1
2cos
in
ft
12
1
mm
in
4.25
1
mm155
s
ft
2.32
−
=
=
G
( )
( )
( )
974.1
1014.14
03.0
4
2
=
=−
g
S
Since this number is greater than 1 the projectile is gyroscopically stable. We have another
condition, however, for dynamic stability where we need
( )
dd
g
SS
S− 2
1
Here we find the dynamic stability factor through
H
T
Sd
2
=
( )
( )
( )
107.1
1046.4
1047.22
4
4
−=
−
=−
−
d
S
( ) ( ) ( )( )
439.3107.12107.1575.0
974.1
1−=−−−=
So the projectile is not stable! This is actually a Magnus instability brought about by the fact that
( )
( )
( )
=
=s
rad
2.500255.0
in
ft
12
1
mm
in
4.25
1
mm155
s
ft
000,1
dt
dF
rad
rev
2
1
s
rad
dt
dF
Problem 12 - What is the minimum spin (Hz.) required to stabilize the projectile in problem 11 ?
Problem 13 - For the projectile given in chapter 8, problem 25, determine the precessional and
nutational frequencies in Hertz. Determine the minimum spin rate for the projectile to be stable.
We need some intermediate steps to find M. First we use equation (FS-58) to determine 1/ky2 as
follows:
T
yI
md
k
2
2
1=
(FS-58)
( )
( )
( )
2
2
2
2inlbm025.0
in5.0lbm092.01
=
y
k
910.0
1
2=
y
k
( )
( )
( )
( )( )
( )
88.2
lbm092.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0 23
3
=
−
M
C
( )
rad
s
ft
013,3
3
dS
2
F
019.0=
F
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 equation (FS-138).
=d
V
dt
d
(FS-138)
( )
( )
( )
=
=s
rad
397,1019.0
in
ft
12
1
in0.5
s
ft
013,3
dt
dF
( )
Hz3.222
rad
rev
2
1
s
rad
397,1 =
=
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=
Then the spin rate would have to be based on
=V
pd
I
I
P
T
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
We have another condition, however, for dynamic stability where we need
( )
dd
g
SS
S− 2
1
Here we find the dynamic stability factor through
H
T
Sd
2
=
Thus we need to calculate the rest of the “starred” coefficients
DD C
m
Sd
C2
=
( )
( )
( )
( )( )
( )
6
23
3
1078.62938.0
lbm092.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0
−
−
=
=
D
C
Sd
( )
( )
( )
( )( )
( )
5
23
3
10208.669.2
lbm092.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0
−
−
=
=
L
C
Sd
( )
( )
( )
( )( )
( )
00
lbm092.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0 23
3=
=
−
p
l
C
( )
MMMM CC
m
Sd
CC qq +=+
2
( )
( )
( )
( )( )
( )
4
23
3
10038.15.5
lbm092.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0
−
−
−=−
=+
MM CC q
( )
( )
( )
( )( )
( )
6
23
3
10154.105.0
lbm092.02
in
ft
12
1
in5.0ft10364.1
ft
lbm
0751.0
−
−
=
=
p
M
C
We also need
P
xI
md
k
2
2
1=
( )
( )
( )
23
2
2
2nlbm10677.2
in5.0lbm092.01
ikx
=−
634.8
1
2=
x
k
Now we find H and T as follows
( )
+
−−=
MM
y
DL CC
k
CCH q
2
1
( ) ( )
( )
( )
4465 10498.110038.1910.01078.610208.6 −−−− =−−−=H
