16.1 Penetration and Perforation of Metals
Problem 1 – A German 280 mm armor piercing projectile weighs 666 lbm and is about 34 inches
in length. It strikes a British warship in the 1/2” thick vertical side plating at an angle of 12°
from horizontal along the path depicted below. The initial impact velocity is 2,000 ft/s.
Determine the residual velocity of the shell after passing through each compartment and how far
through the ship it will go (i.e. in which compartment will it stop)
Assume the density of the armor plate to be
= 0.283 lbm/in3.
Solution: We shall use the limit velocity concept to determine shell penetration and once again
we need to convert EVERYTHING to cgs units. Thus we can write:
( )
( )
cm36.86
in
cm
54.2in34 =
=l
( )
in03.11cm0.28
mm
cm
10
1
mm280 ==
=d
( )
( )
g727,302
kg
g
1000
lbm
kg
20.2
1
lbm666 =
=m
( ) ( )
=
=
=s
cm
960,60
s
m
60.609
ft
m
3048.0
s
ft
000,2
s
V
( )
( )
cm27.1
in
cm
54.2in50.0
1=
=t
Path of shell
12
0.50” thick
0.25” thick
7.00” thick
4.00” thick
(assume normal to shell path)
1.25” thick
We shall now determine the limit velocity for part a) from
s
l
( )
cm
0.28
d
Thus the model is not really valid but we shall use it anyway throughout the problem.
4
in
3=
Now we can calculate a from
( )
−+
=−
s
m
1sec000,4
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
112sec
0.28
27.1
727,302
0.28
000,4
0.28
36.86 12sec
0.28
27.1
75.0
3
15.0 75.0
eVl
( )
993.0
lbm75.13
3
1
lbm666
lbm666 =
+
=a
We also need
( )( )
in03.113
So the residual velocity is then
r
For the next bulkhead in the line of travel
in
We shall now determine the limit velocity for the ¼” thick plate from
Again we have perforation of the plate. We calculate residual velocity as before.
( )
−+
=−
s
m
1sec000,4
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
112sec
0.28
635.0
727,302
0.28
000,4
0.28
36.86 12sec
0.28
635.0
75.0
3
15.0 75.0
eVl
( ) ( )
( )
( )
lbm873.612secin25.0in03.11
4
in
lbm
283.0 75.02
2
3=
=
m
( )
997.0
lbm873.6
3
1
lbm666
lbm666 =
+
=a
( )
( )
cm78.17
in
cm
54.2in00.7 =
=t
We shall now determine the limit velocity
=
=s
ft
725,1
s
m
8.525
l
V
Here we again have perforation of the plate. We calculate residual velocity again.
( ) ( )
( )
( )
lbm45.19212secin00.7in03.11
4
in
lbm
283.0 75.02
2
3=
=
m
( )
912.0
lbm45.192
3
1
lbm666
lbm666 =
+
=a
( ) ( ) ( )
( )
−= s
ft
725,1960,1912.0 215.2
1
215.2215.2
r
V
=s
ft
950
r
V
Now our projectile encounters a deck plate of 1.25”. The penetration model requires that we
reorient ourselves and the impact angle is 78. Let’s calculate some more.
( ) ( )
=
=
== s
cm
956,28
s
m
56.289
ft
m
3048.0
s
ft
950
rs VV
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
112sec
0.28
78.17
727,302
0.28
000,4
0.28
36.86 12sec
0.28
78.17
75.0
3
15.0 75.0
eVl
( )
( )
cm175.3
in
cm
54.2in25.1 =
=t
=
=s
ft
027,1
s
m
93.312
l
V
Thus the shell will NOT penetrate the deck plate. This is exactly what happened in this situation.
Problem 2 – An explosively formed penetrator impacts a 4 inch thick rolled homogeneous armor
(RHA) plate at a velocity of 1500 m/s. The penetrator parameters are given below. Determine if
the penetrator will perforate the target using the Lambert/Zukas model given
a) A normal impact
m
l
Penetrator Information
mm22
mm95
=
=
d
l
=
=
s
m
1500
lbm25.1
s
V
m
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
178sec
0.28
175.3
727,302
0.28
000,4
0.28
36.86 78sec
0.28
175.3
75.0
3
15.0 75.0
eVl
( )
( )
cm16.10
in
cm
54.2in4=
=t
We shall now determine the limit velocity for part a) from
=s
m
299,1
l
V
Since the striking velocity is above the limit velocity the penetrator should penetrate. For part b)
we use the same formula but a different angle.
=s
m
389,1
l
V
This is a little closer to not penetrating but it will still go through. We shall do a quick
applicability check by recalling that the model is valid for t/d > 1.5.
( )
cm
2.2
d
Thus the model is valid.
Problem 3 – A German 7.5 cm Gr 34A1 projectile is fired at a 2 inch thick armor plate at a 30
degree obliquity. The impact velocity is 400 m/s. The penetrator parameters are given below.
a) Determine whether the penetration mode will be plugging or piercing through use of
the Jacobson model for a normal impact.
b) Determine if the penetrator will perforate the armor though use of the Lambert model.
c) Comment on the validity of the model.
Penetrator Information
( )
−+
=−
s
m
1sec4000
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
−+
=−
s
m
10sec
2.2
16.10
568
2.2
4000
2.2
5.9 0sec
2.2
16.10
75.0
3
15.0 75.0
eVl
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
−+
=−
s
m
130sec
2.2
16.10
568
2.2
4000
2.2
5.9 30sec
2.2
16.10
75.0
3
15.0 75.0
eVl
cm5.7
cm39
=
=
d
l
=
=
s
m
400
kg75.5
s
V
m
crit
d
To apply this equation to the oblique impact we need to use line of sight thickness so the armor
=s
m
658
l
V
Problem 4 – A Japanese 20mm projectile with the properties below impacts the ½” thick
aluminum armor plate on a U.S. plane’s rear gun mount at 30° obliquity. If the projectile and the
armor have the properties below:
a.) Determine how deep the projectile will penetrate into the armor (assume
= 0.03)
b.) If the projectile perforates the armor, determine its residual velocity
m
Estimated Penetrator Information
mm60
mm20
mm40
=
=
=
L
d
s
=
=
s
m
500
g128
s
V
m
=3
in
lbm
283.0
p
Estimated Armor Information
068.1
418.4
=
=
B
A
psi000,39=Y
=3
in
lbm
098.0
t
mm20
( )
rad848.075.0sin
mm40
2
mm20
mm40
sin 11
0==
−
=−−
( )( ) ( )( ) ( )( ) ( )( ) ( )( )( ) ( )( )
( )( )
−
−−−
+−= −
22
124
sin12224124
3
1
3
24
24 1
22
k
467.1=k
Now we need to determine the coefficients α and
through
( )
−−−
−+= 1412
2
41 0
2
A
( )
( )
−−+−
−
−+
−
=2
2
0
2
224
1414612
2
24
18
B
Making the proper substitutions yields
( ) ( )( )( ) ( ) ( )( )( ) ( )( )
−−−
−+= 12412203.0848.0
2
203.041418.4 2
899.4=
( ) ( )( )
( )( ) ( )( ) ( ) ( )( )( ) ( )( ) ( )( )
( )
( )( )
( )( )
−−+−
−
−+
−
=2
2
2
2224
1241242612203.0
848.0
2
203.0
224
128
068.1
177.0=
Now we are ready to determine the penetration depth, keeping in mind that at a 30º obliquity the
target plate is effectively 0.577 inches thick. From the notes we have
+
+
=Y
V
kd
LP t
t
p
2
0
1ln
22
1
We have to be careful with units here
( )( ) ( )
( )( )
( ) ( ) ( )
( ) ( ) ( )
−
−
+
+
=
ft
in
12
slbf
ftlbm
2.32
in
lbf
000,39
m
mm
1000
mm
in
4.25
1
s
m
500
in
lbm
098.0
899.4
177.0
1ln
2
mm20467.1
mm60
in
lbm
098.0
in
lbm
283.0
177.02
1
22
2
2
2
2
2
22
2
2
3
3
3
P
in09.2mm1.53 ==P
=3
in
lbm
283.0
s
available in 2” thick plates
( )
( )
cm08.5
in
cm
54.2in2=
=
B
t
( )
( )
cm5.15
mm
cm
1.0mm155 =
=d
( )
( )
cm6.101
in
cm
54.2in40 =
=l
The equation for the limit velocity is given by
( )
−+
=−
s
m
1sec4000
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
So, as a starting point, for one plate at 30º we have
( )
( )
( )
( )
( )
( )
−+
s
m
5.15
08.5
10818.4
5.15
5.15
6.101 30sec
5.15
08.5
75.0
4
3
15.0 75.0
eVlB
=s
m
359
lB
V
Thus the projectile will perforate. This is not good. The limit velocity is a bit over ¼ of where
we want it to be so let’s try 4 plates
( ) ( )
( )
( )( )
( )
( )
( )( )
( )
( )
−+
=−
s
m
130sec
5.15
08.54
10818.4
5.15
4000
5.15
6.101 30sec
5.15
08.54
75.0
4
3
15.0 75.0
eVlB
=s
m
227,1
lB
V
Solution: We shall use the limit velocity concept to determine shell penetration and once again
we need to convert EVERYTHING to cgs units. Thus we can write:
( )
( )
cm41.89
in
cm
54.2in2.35 =
=l
( )
in12cm5.30
mm
cm
10
1
mm305 ==
=d
( )
( )
g000,406
kg
g
1000
lbm
kg
20.2
1
lbm894 =
=m
( )
=
=s
m
6.548
ft
m
3048.0
s
ft
800,1
s
V
( )
( )
cm25.8
in
cm
54.2in25.3
1=
=t
We shall now determine the limit velocity for part a) from equation (TB-34)
s
l
( )
cm
5.30
d
Thus the model is not quite valid but we’ll use it anyway. Let’s calculate the residual velocity
( )
−+
=−
s
m
1sec000,4
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( )
p
p
l
p
sr VVaV
1
−=
(TB-44)
Now we find our coefficient, a. We start with calculating the mass of the hole using equation
(TB-37).
75.02 sec
4tdm =
(TB-37)
( ) ( )
( )
( )
lbm8.15755secin25.3in12
4
in
lbm
283.0 75.02
2
3=
=
m
Now we can calculate a from equation (TB-49) which is
+
=
mm
m
a
3
1
(TB-49)
( )
944.0
lbm8.157
3
1
lbm894
lbm894 =
+
=a
75.0
sec
3
2d
t
p+=
(TB-57)
( ) ( ) ( )
( )
−= s
ft
109,1800,1944.0 137.2
1
137.2137.2
r
V
=s
ft
384,1
r
V
For the next bulkhead in the line of travel (actually a turret barbette)
( ) ( )
=
== s
m
422
ft
m
3048.0
s
ft
384,1
rs VV
( )
( )
cm86.22
in
cm
54.2in00.9 =
=t
We shall now determine the limit velocity for the 9” thick barbette from equation (TB-34)
( )
−+
=−
s
m
1sec000,4
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
120sec
5.30
86.22
000,406
5.30
000,4
5.30
41.89 20sec
5.30
86.22
75.0
3
15.0 75.0
eVl
=
=s
ft
003,2
s
m
0.611
l
V
So we don’t penetrate the barbette wall.
Problem 7 – A German 280 mm Armor Piercing projectile weighs 666 lbm and is about 34
inches in length. It strikes a British warship in the 0.43” thick vertical side plating at an angle of
10° from horizontal along the path depicted below. The initial impact velocity is 1,900 ft/s.
Determine the residual velocity of the shell after passing through each compartment and how far
through the ship it will go (i.e. in which compartment will it stop)
Assume the density of the armor plate to be
= 0.283 lbm/in3.
1.50” thick
3.25” thick
9.00” thick
8.00” thick
15
0.43” thick
Solution: We shall use the limit velocity concept to determine shell penetration and once again
we need to convert EVERYTHING to cgs units. Thus we can write:
We again determine the limit velocity for part a) from equation (TB-34)
s
l
( )
cm
0.28
d
The model isn’t truly valid. For the residual velocity we have to invoke equation (TB-44).
Path of shell
10
6.00” thick
1.00” thick
( )
−+
=−
s
m
1sec000,4
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
110sec
0.28
09.1
727,302
0.28
000,4
0.28
36.86 10sec
0.28
09.1
75.0
3
15.0 75.0
eVl
( )
p
p
l
p
sr VVaV
1
−=
(TB-44)
Now we find our coefficient, a. We start with calculating the mass of the hole using equation
(TB-37).
75.02 sec
4tdm =
(TB-37)
( ) ( )
( )
( )
lbm76.1110secin43.0in03.11
4
in
lbm
283.0 75.02
2
3=
=
m
Now we can calculate a from equation (TB-49) which is
( )
lbm76.11
3
1
lbm666
+
We also need equation (TB-57)
( )( )
in03.113
So the residual velocity is then
r
For the next bulkhead in the line of travel (actually a turret barbette)
( ) ( )
=
== s
m
5.574
ft
m
3048.0
s
ft
885,1
rs VV
( )
( )
cm32.20
in
cm
54.2in00.8 =
=t
We shall now determine the limit velocity for the 8” thick plate from equation (TB-34)
( )
−+
=−
s
m
1sec000,4
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
−+
=−
s
m
110sec
0.28
32.20
727,302
0.28
000,4
0.28
36.86 10sec
0.28
32.20
75.0
3
15.0 75.0
eVl
=
=s
ft
936,1
s
m
04.590
l
V
So no penetration of the barbette occurs –this is what actually happened.
Problem 8 – A British Short Magazine Lee Enfield (SMLE) is fired at a sniper plate across no-
man’s land in WWI (because of the static nature of the fighting, snipers in the opposing lines
fired through small holes in thick metal plates to minimize exposure). The projectile has a mass
of 175 grains and the projectile is 1.1 inches long and made of lead (ignoring the copper jacket).
The diameter is 0.310 inches. The range to the target is 200 yards so we can assume an impact
velocity of 1,512 ft/s. The angle of impact is 12.85 minutes from the normal. Using the
Lambert-Zukas model, determine the thickness of armor plate that the projectile will penetrate
(i.e. obtain V50).
kg
lbm
20.2
grains
000,7