( )
=
=s
m
461
ft
m
3048.0
s
ft
512,1
s
V
The angle needs to be put into degrees so we write
( )
deg214.0
MOA
deg
60
1
MOA85.12 =
=
We shall now determine the thickness assuming the impact velocity is the limit velocity 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
The simplest way to do this is to iterate and we obtain
Problem 9 – Using the same information in problem 8, above, determine how deep a projectile
will penetrate into a 1 inch thick steel sniper plate assuming the impact is normal this time and
the impact velocity is 1,800 ft/s. Use the Tate formula. The additional target and bullet
properties are as follows:
=2
in
lbf
000,10
p
=2
in
lbf
000,36
t
=2
in
lbf
000,000,29
t
E
The Tate formula shows the energy balance between the projectile and the target. When solved
for the penetration depth, the following equation results
( ) ( )
dvvlvu
Y
P
i
c
V
v
p
p
=
(1)
In this formula u(v) is the instantaneous velocity of the base of the hole, v is the instantaneous
velocity of the projectile and l(v) is the instantaneous projectile length. The limits of integration
are between the impact velocity, Vi, and a cutoff velocity, vc, that depends on whether the
projectile is stronger than the target or not. For this case we can use
p
pt
c
YR
v
−
=2
for
pt YR
(2)
−
2
1
p
t
=
(4)
( )
( )
t
pt YR
A
2
12 −−
=
(5)
( )
( )
−+−−+
−
++
++
−
2222
2
2
2
12
p
p
Y
YR
ii
Y
AVV
Avv
L
vl p
pt
(6)
Where L is the initial length of the projectile. Comment on the results.
ttpip RuYuV +=+− 2
2
2
pp
Y
7.1=
(TB-59)
+=
t
t
tt
E
R
57.0ln
3
2
(TB-60)
We determine the values for the bullet and target resistances as follows
( )( )
=
=22 in
lbf
000,17
in
lbf
000,107.1
p
Y
(7)
( )
( )
( )
=
+
=2
6
2in
lbf
649,244
000,36
1029
57.0ln
3
2
in
lbf
000,36
t
R
(8)
Let’s get the units correct on the densities
( ) ( )
( )
=
=2
2
2
2
3
ft
s
in
lbf
152.0
slbf
ftlbm
2.32
ft
in
12
in
lbm
407.0
p
(9)
( ) ( )
( )
=
=2
2
2
2
3
ft
s
in
lbf
105.0
slbf
ftlbm
2.32
ft
in
12
in
lbm
283.0
t
(10)
Now we can begin calculating the values for the solution. We start with equation (2) where,
inserting the values we obtain
( )
ft
s
in
lbf
152.0
2
2
2
c
The density ratio is
( )
ft
s
in
lbf
152.0
2
2
2
The term A is
( )
2
2
2
s
ft
s
in
lbf
105.0
( )
( ) ( ) ( )
=s
ft
in
s
ft
in
lbf
000,17
ft
s
in
lbf
152.0
in
2
2
2
2
dvvlvuP
i
c
V
v
(15)
So the Tate model predicts that the soft lead bullet will simply mush up and make an 0.016 inch
dent. This is due to the fact that I assumed the lead in the bullet was so soft. In reality lead does
not have a yield strength. It is common to use spall strengths in place of Yp and Rt which in this
case gives us
Which is a bit deeper.
Problem 10 – At the end of World War I the German navy surrendered to the British at Scapa
Flow (A large anchorage in northern Scotland). When it appeared that surrender negotiations
were breaking down the German sailors opened the sea cocks (valves in the bottom of the ships)
and sank most of their ships within sight of astounded British onlookers. In the 1920’s the
British raised what ships they could and used them as targets to assess the penetration and
bursting characteristic of their heavy shell [5]. One such test was against the side armor of
S.M.S. Baden, the largest warship built by Germany during the war. We would like to examine
two impacts against the armor of this vessel. In each case the projectile data is given below:
Projectile Information [6]
=
=
=
=
3
in
lbm
283.0
in90
in15
in66
p
s
d
l
=
=
s
ft
550,1
lbm938,1
s
V
m
=
3
in
lbm
283.0
t
=
3
in
lbm
283.0
t
kg
lbm
2.2
and for the thickness, diameter and length we have
( )
( )
cm015.3
in
cm
54.2in187.1 =
=
GD
t
( )
( )
cm1.38
in
cm
54.2in15 =
=d
( )
( )
cm4.167
in
cm
54.2in66 =
=l
We should also put our striking velocity in m/s for comparison purposes.
( ) ( )
=
=s
m
44.472
ft
m
3048.0
s
ft
550,1
s
V
The limit velocity for part a) is then
( )
−+
=−
s
m
1sec4000
75.0
sec
75.0
3
15.0
d
t
le
d
t
m
d
d
l
V
( ) ( )
( )
( )
( )
( )
( )
( )
( )
−+
=−
s
m
15.16sec
1.38
015.3
10809.8
1.38
4000
1.38
4.167 5.16sec
1.38
015.3
75.0
5
3
15.0 75.0
eVlGD
=s
m
71
lGD
V
75.02 sec
4tdm =
(TB-37)
( ) ( )
( )
( )
lbm26.615.16secin187.1in15
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)
We also need equation (TB-57)
75.0
sec
3
2d
t
p+=
(TB-57)
( )
( )( )
( )
027.25.16sec
in153
in187.1
275.0 =+=
p
So the residual velocity is then
r
For part b) we need to start again.
lB
16.2 Penetration and Perforation of Concrete
Problem 11 – A 0.50 caliber projectile is fired at an extremely thick concrete wall of 2,100 psi
unconfined compressive strength and density of 0.084 lbm/in3. It strikes with no obliquity and a
2000 ft/s velocity. How far does it penetrate?
Projectile Information
mm70.12
mm50.63
=
=
d
s
=
=
s
ft
2000
grains662
s
V
m
( )
MPa48.14
in
lbf
MPa
05.145
1
in
lbf
100,2
2
2=
=
c
f
Now we use our curve-fitted equation
5603.0
48.93 −
=c
fS
( )
2
1
2
4VV
d
m
cs−=
=−
s
V
V
c
m
t1
1
1cos
( )
( )
s000044.0
610
9.518
cos
s
kg
486,854,6
kg043.0 1
2
1=
=−
t
( )( )
( )
( )( )
( )( ) ( )
( )
( )
( )( )
m0127.02
m
N
1048.14909.20
s
m
9.518
m
kg
327,2065.0
1ln
m
kg
327,2065.0m0127.0
kg043.02
2
6
2
2
2
3
3
2
2
+
+
=
P
Problem 12 – Areal density is a measure of the mass of one square inch (or sometimes a square
foot or meter) of a material used as armor. Given a threat projectile with the properties below,
find the armor solution with the lowest areal density that prevents perforation if your choices are
steel with density 0.283 lbm/in3, concrete of 2,500 psi unconfined compressive strength and
density of 0.084 lbm/in3, and aluminum with the following properties:
068.1
418.4
=
=
B
A
psi000,39=Y
=3
in
lbm
098.0
t
Assume the projectile strikes with15 degrees of obliquity.
Penetrator Information
mm15
mm12
mm35
=
=
=
L
d
s
=
=
s
m
600
kg03.0
s
V
m
=3
in
lbm
283.0
p
mm12
The length of the ogive is given by equation (AP-2) as
We again determine the limit velocity for part from equation (TB-34) but in this case we know
the limit velocity and need to find the thickness required to stop the projectile. This is easily
achieved through a spreadsheet
( ) ( )
s
2.130
2.1
s
The thickness turns out to be 1.01 cm or about 0.398 inches. The areal density is then
Now for the concrete we need to determine the parameter, N. We shall assume that B = 1 as was
stated in the notes.
( )
917.224
2=
Now we need the empirical parameter S but to get it we need to convert our unconfined
compressive stress, fc’ to MPa.
in
lbf
05.145
in
2
2=
c
Now we use our curve-fitted equation
Now we need to find V1, c and t1. Let’s get all of our units consistent first.
mm
1000
The rest of the units are consistent so we can write
Ndm
fSdmV
Vcs
3
32
12
2
+
−
=
(CP-36)
( )( )
( ) ( )
( )
( )
( )( )
( )
( )( )
+
−
=
3
3
3
2
63
3
2
2
2
1
m
kg
327,2109.0m012.0kg03.02
m
N
1024.1796.18m012.0
s
m
600kg03.02
V
=s
m
568
1
V
( )
2
1
2
4VV
d
m
cs−=
(CP-33)
( )
( )( )
( ) ( )
( )
=
−= 22
2
22
2
2s
kg
000,947,1
s
m
568600
m012.04
kg03.0
c
Finally we can get t1 from
=−
s
V
V
c
m
t1
1
1cos
(CP-37)
( )
( )
s00004.0
600
568
cos
s
kg
000,947,1
kg03.0 1
2
1=
=−
t
Now we have everything we need to calculate the penetration depth
d
fS
VN
Nd
m
P
c
21ln
22
1
2+
+=
(CP-58)
( )( )
( )
( )( )
( )( ) ( )
( )
( )
( )( )
m012.02
m
N
1024.1796.18
s
m
568
m
kg
327,2109.0
1ln
m
kg
327,2109.0m012.0
kg03.02
2
6
2
2
2
3
3
2
2
+
+
=
P
m14.0=P
So we need 5.5 inches of concrete at normal obliquity. This problem, however stated that the
obliquity was 15 degrees. Thus we need to reduce our thickness by the line of sight distance or
The areal density is then
( )
( )
=
=23
concrete in
lbm
446.0
in
lbm
084.0in31.5AD
004.5=
( ) ( )( )
( )( ) ( )( ) ( ) ( )( )( ) ( )( ) ( )( )
( )
( )( )
( )( )
−−+−
−
−+
−
=2
2
2
2917.224
1917.241917.24917.261917.2203.0
977.0
2
917.203.0
917.224
1917.28
068.1
126.0=
Now we are ready to determine the penetration depth, keeping in mind that at a 15º obliquity.
From the notes we have
We need to reduce our thickness by the line of sight distance or
aluminum in
in
Comparing the three areal densities we see that the aluminum makes the best armor here.
Problem 13 – We would like to compare defenses against the projectile in problem 10.
Assuming this projectile impacts a steel plate at zero degrees obliquity
a.) The thickness of the armor required to prevent penetration
b.) The thickness of 1500 psi unconfined compressive strength concrete required to do
the same assuming that the concrete does not spall– assume a density of 0.084 lbm/in3 for the
concrete – note that this particular shell has a secant ogive so as an estimate for the purposes of
this problem – divide the resultant ogive length by 2 – please note that this is simply a guess and
not based on physics.
c.) Comment on the validity of the answer to part b.
( )
( )
s
1.38
10809.8
1.38
s
5
Solving this for the minimum thickness we obtain
min =t
( )
in68.45
2
in361.91 ==
oeff
l
( )( )
( ) ( )
( )
( )
( )( )
( )
( )( )
+
−
=
3
3
3
2
63
3
2
2
2
1
m
kg
327,2054.0m381.0kg8812
m
N
1034.1025.25m381.0
s
m
44.472kg8812
V
=s
m
6.441
1
V
( )
44.472
s
kg
000,000,2
2
1=
Now we have everything we need to calculate the penetration depth
d
fS
VN
Nd
m
P
c
21ln
22
1
2+
+=
(CP-58)
( )( )
( )
( )( )
( )( ) ( )
( )
( )
( )( )
m381.02
m
N
1034.1025.25
s
m
5.469
m
kg
327,2054.0
1ln
m
kg
327,2054.0m381.0
kg8812
2
6
2
2
2
3
3
2
2
+
+
=
P
m51.3=P
This is a reasonable answer.
16.3 Penetration and Perforation of Soils
Problem 14 – A 0.50 caliber projectile is fired at a soil berm with properties established below.
How far does it penetrate?
( )
52
0=
125.0
m
kg
125,2
m
kg
860,1
1
3
3=
−=
( )
3
3
1
2
1
−−
+=
E
c
( )
( )
( )
500.0125.01
in
lbf
1022
in
lbf
500,1
1
3
3
2
7
2=−−
+=
( )
( )
+−−
−+
+
33
2
2
3212
3
3
2
1
1
3
E
cc
c
( ) ( )( )( )
−−−
( )
( )
( )
( )
( )
( ) ( )
( )
( ) ( )( ) ( )
( ) ( )( )( )
( )
+−−−
+−−
−
−+
+
4
7
33
2
77
271.02
7
2
1022
500,1
1271.02271.021125.01
500.03271.02125.012500.0
1022
500,13
1125.0
102
500,13
500.0
1022
500,1
1
500.0
1
486.1=B
With A and B determined we have everything we need to calculate the retarding force parameters
αs and
s as follows
( )
−−−
−+= 1412
2
41
40
2
2
A
d
cs
( )
( ) ( ) ( )( ) ( ) ( )( ) ( )
−−−
−+
=1541521.0120.1
2
51.041343.2
in
lbf
500,1
4
m0127.0 2
2
2
2
s
( )
lbf7.093,1
m
in
37.39
in
lbfm
7056.0 2
2
2
2
2
=
−
=
s
( )
( )
−−+−
−
−+
−
=2
2
0
2
2
0
2
24
1414612
2
24
18
4
B
d
s
( )
( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
( )
( )
( )
−−+−
−
−+
−
=2
2
2
23
2
2
524
154154561521.0
120.1
2
51.0
524
1)5(8
486.1
m
kg
860,1
4
m0127.0
s
=m
kg
0305.0
s
Let’s change the units of m, Vs and αs to consistent units
s
lbf
s
( )
kg043.0
lbm
kg
20.2
1
grain
lbm
7000
1
grains662 =
=m
( ) ( ) ( )
=
=
=s
m
610
s
cm
960,60
in
cm
54.2
ft
in
12
s
ft
2000
s
V
Now we have
+=
s
s
s
V
m
P
2
0
1ln
2
( )
( )
( ) ( )
( )
−
+
=
2
2
2
2
s
mkg
8.864,4
s
m
610
m
kg
0305.0
1ln
m
kg
0305.02
kg043.0
P
cm9.84m849.0 ==P