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20.2 Shaped Charge Jet Penetration
Problem 3 - A conical liner is to be fabricated from copper and filled with Composition A-3 as
the explosive. The thickness of the liner is to be 0.10 inches. The length of the 20º conical liner
is 4 inches and the charge O.D. is 4 inches. The case is fabricated from steel and is an 8 inch
long cylinder, 0.12 inches thick. Determine the following using the Birkhoff et al. theory
ignoring effects of confinement (If the region over the liner is discretized into 4 segments that
should be sufficient):
a.) Masses of the jet and slug
j
s
b.) Velocities of the jet and slug
c.) The velocities of the case material (plot this as fragment velocity vs. case length)
d.) The direction in which the case material fragments will be projected (plot this as
departure angle vs. case length)
e.) If the standoff is 8 inches determine the maximum penetration into rolled homogeneous
armor plate at a 15º angle of obliquity using the formula of Dipersio and Simon and
The required properties for this calculation are given as follows:
Composition A-3 Gurney Velocity (2E)1/2 = 2.63 km/s
Composition A-3 Detonation Velocity (D) = 8.14 km/s
Composition A-3 Density = 1.59 g/cc
Copper density = 0.323 lbm/in3
Steel density = 0.283 lbm/in3
( ) ( ) ( )
=
=33
3
3
3
3in
lbm
053.0
g
lbm
)1000(
046.2
in
cm
54.2
cm
g
59.1
CompA
2.000
20º
8.000
4.000
( ) ( )
=
+
=
−
s
m
558,3
2
1
749.0
035.0
km
m
1000
s
km
63.2 2
1
01
V
We can now find the angle,
.
( )
−
2
cos
0
V
UD
( )
−
−−−= 2
tancotcscsin
cos
D
s
U
V
( )
( ) ( )
−
−−−
=2
20
tancotcsc20sin
20cos
s
m
140,8
s
V
( )
( )( )
7.8
s
m
140,82
s
m
462,2
sin
2
1=
=
−
Now the only thing that changes as the detonation wave passes over the liner (from the case
standpoint) is the fact that there is less explosive to accelerate the case. We handle this by
subtracting the cone volume from the explosive just like when we analyzed the liner. Let’s start
with the apex of the cone.
2
in
2
3
31
Now the case segment velocity follows directly from
s
2
661.0
km
s
01
with the Taylor angle determined as above. The data is tabulated below.
Position
M = rV/L
C = rV/L
Segment
Velocity
(m/s)
θ/2 (deg)
0
0.427
0.666
2462
8.7
1
0.427
0.661
2457
8.6
2
0.427
0.622
2415
8.5
3
0.427
0.544
2321
8.2
4
0.427
0.427
2149
7.6
The plots of the data are as follows
V0 (m/s)
2100
2150
2200
2250
2300
2350
2400
2450
2500
0 1 2 3 4 5 6 7 8
Distance From Base (in)
Fragment Velocity (m/s)
V0 (m/s)
j
t
=
936.0
in
lbm
323.0
in
lbm
283.0
3
3=
=
( ) ( ) ( )
=
=33
3
3
3in
lbm
057.0
g
lbm
)1000(
046.2
in
cm
54.2
cm
g
717.1
CompB
We need to determine, for each section, the liner mass to charge mass ratio in order to determine
our velocity, V0 for our later calculations. With our truncated cones we will simply assume each
section is a cylinder at the average radius of the section. Then we can write
1
2
3
4
5
2.887
4.000
30°
( ) ( ) ( )
=
+
=
+
=in
lbm
088.0in
2
0577.
in15.02
in
lbm
323.0
2
23
01
1
t
rr
MCu
( ) ( )
=
+
−
=
+
−= in
lbm
850.2in
2
0577.0
4
in
lbm
057.0
2
2
2
2
3
2
01
2
1
rr
rC cCompB
Now the liner segment velocity follows directly from
s
2
850.2
km
s
01
We can now find the angle,
.
( )
−
2
cos
0
V
UD
2
The overall liner mass is the sum of all our individual masses (or it could be calculated directly
from the geometry). It is 2.197 lbm. So we have
( )
( )( )
lbm410.0200.51cos1lbm197.2
2
1=−=
j
m
( )
( )( )
lbm787.1200.51cos1lbm197.2
2
1=+=
s
m
The jet and slug velocities are obtained for a conical liner from
( )
−
++−= 2
tancotcscsin
cos
D
U
V
The answers are shown in the table below. We could also have taken an average as well.
For the Slug velocity we have
( )
−
−−−= 2
tancotcscsin
cos
D
s
U
V
If we take all of our data and put it in a table we get
Position
M = rV/L
C = rV/L
Segment
Velocity
(m/s)
(deg)
Vj (m/s)
Vs(m/s)
1
0.088
2.850
3829
54.200
8119.292
1113.340
2
0.264
2.731
3612
52.809
7845.654
1043.828
3
0.439
2.492
3393
51.406
7555.759
974.483
4
0.615
2.134
3142
49.811
7207.738
896.713
5
0.791
1.656
2822
47.772
6731.217
798.788
Total
2.197
11.863
Average
51.200
7491.932
965.431
If the standoff is 1 meter the length of the jet will be the difference in velocities of the tip and the
slug (assuming the values we calculate above are correct and constant) by the time it takes the tip
to arrive at the target. First let’s find how long it takes the tip to reach the target.
s000134.0
s
m
4927
m1 =
==
tip
V
s
t
We’ll use the average slug velocity to get a reasonable answer
( )
( )
m875.0s000134.
s
m
965
s
m
7492 =
−
=− tVVL slugtip
Since this is a fairly large shaped charge this length makes sense.
We can determine the penetration into a steel plate from the DePersio and Simon Formula for no
particulation. This is
−
=1
min
0
V
V
sP
j
t
=
936.0
in
lbm
323.0
in
lbm
283.0
3
3=
=
Since no other information was given we shall use 2 km/s as the cutoff velocity giving us
( )
m98.21
s
m
000,2
s
m
119,8
m1.1
936.0
=
−
=P
That’s an awful long hole! Let’s look now what the density rule would give us. The Density
rule is given by
t
j
LP
=
Inserting our numbers from above we get
( )
m935.0
in
lbm
283.0
in
lbm
323.0
m875.0
3
3=
=P
Quite a difference! Normally to be conservative, the warhead guys will use the density rule as a
rough calculation and the armor guys will use some other model.
Problem 5 - A trumpet liner is to be fabricated from copper and filled with Composition A-3 as
the explosive. The thickness of the liner is to be 0.12 inches. We shall approximate the trumpet
liner as indicated below where the half-angle,
is to be variable. The length of the liner is 4
inches and the charge O.D. is 4 inches. Determine the following using the Birkhoff et al. theory:
a.) Masses of the jet and slug
c.) If the standoff is 8 inches determine if the jet will perforate 5 inches of Rolled
Homogeneous Armor plate at a 70º angle of obliquity using the formula of Dipersio and
Simon and assuming no particulation
The required properties for this calculation are given as follows:
Composition A-3 Gurney Velocity (2E)1/2 = 2.63 km/s
Composition A-3 Detonation Velocity (UD) = 8.14 km/s
Composition A-3 density = 1.59 g/cc
Copper density = 0.323 lbm/in3
Steel density = 0.283 lbm/in3
( ) ( ) ( )
=
=33
3
3
3
3in
lbm
053.0
g
lbm
)1000(
046.2
in
cm
54.2
cm
g
59.1
CompA
1
2
3
4
2.000
10º
20º
30º
25º
1.000
1.000
1.000
1.000
( )
( )( )
lbm545.0008.32cos1lbm590.0
2
1=+=
s
m
The jet and slug velocities are obtained for a conical liner from
( )
−
++−= 2
tancotcscsin
cos
D
U
V
( )
( ) ( )
−
++−
=2
10
tancotcsc10sin
10cos
s
m
140,8
V
The answers are shown in the table below. In this case we can take an average since we need
this later in the problem. For the Slug velocity we have
( )
−
−−−= 2
tancotcscsin
cos
D
s
U
V
( )
( ) ( )
−
−−−
=2
10
tancotcsc10sin
10cos
s
m
140,8
s
V
Position
M = rV/L
C = rV/L
Segment
Velocity
(m/s)
(deg)
Vj (m/s)
Vs(m/s)
1
0.021
0.685
3610
35.225
11884.351
330.079
2
0.084
0.734
3356
33.423
6616.681
350.288
3
0.186
0.790
3066
31.375
708.590
53.325
4
0.299
0.560
2587
28.007
1749.983
96.758
Total
0.590
2.771
Average
32.008
5239.901
207.612
j
t
=
The penetration is then given by
