19.2 Taylor Angles
Problem 1 – A “Bangalore Torpedo” was a device built by the United States during the Second
World War to clear beach (or any other) obstacles. It consisted of a long tube filled with
explosive that was detonated on the end. Assume that we have a similar device made of steel
and filled with Composition B. The device is 3 feet long. The I.D. is constant at 2 inches. The
O.D. varies with length. The first foot of length is 2-1/4 inches in diameter, the next foot of
length is 2-3/4 inches in diameter and the last foot of length is 3 inches in diameter. Assuming
that we detonate the device at the 2-1/4 inch end:
a) Draw a graph of the fragment velocities vs. length in ft and ft/s.
b) Draw a graph of the Taylor angles in ft and degrees from the device axis.
Assume that the tube is steel with a density of 0.283 lbm/in3. Assume that the filler density is
1.70 g/cc. Assume that the detonation velocity is 7.89 mm/s and the Gurney constant is 2.7
mm/s.
( )
( )
 
=
−
== in
lbm
111.1in
4
00.23.00
in
lbm
283.0 2
22
3
33
tube
AM steel
( )
( )
 
=
== in
lbm
179.0in
4
00.2
in
lbm
057.0 2
2
3
3
fillCompB AC
( )
( ) ( ) ( )
8.3
2
066.0
m
ft
281.3
km
m
1000
s
km
7.892
s
ft
421,3
22
sin 333 =→=
==
D
V
This Taylor angle would tend to tilt the fragments at the above angles in the direction of the
detonation wave (away from the initiation point). These angles are measured from the normal to
the axis. To get them into the angles measured from the charge axis we need to subtract then
from 90 degrees. Thus we have
2
3=−=−=
Problem 2 – Assume that we used the Paris gun so often that it finally blew up. We want to
determine the velocity of the fragments and their Taylor angles. Assume the section where the
explosion took place is centered over a jacket transition. Therefore the analysis consists of two
sections each 4 feet long. The I.D. of the weapon is 210 mm. the O.D. of the forward section is
constant at 350 mm. The O.D. of the jacketed section is also constant at 420 mm. Assume the
2
1
3
explosion begins at the projectile and propagates rearward. Assume that the Gurney constant for
the filler/propellant combination is 1.8 km/s.
a) Draw a graph of the fragment velocities vs. length in ft and ft/s.
b) Draw a graph of the Taylor angles in ft and degrees from the bore axis.
Assume that the tube is steel with a density of 0.283 lbm/in3. Assume that the filler/propellant
density averages to about 0.6 g/cc. Assume that the detonation velocity is 16,500 ft/s.
Solution: Let’s get everything in consistent units. The density of explosive mixture first.
( ) ( ) ( )
=
=33
3
3
3in
lbm
020.0
g
lbm
)1000(
046.2
in
cm
54.2
cm
g
6.0
mix
The I.D. of the tube is
( )
 
in27.8
mm
in
4.25
1
mm210 =
=
i
d
The two O.D.’s are
( )
 
in78.13
mm
in
4.25
1
mm350
1=
=
o
d
( )
 
in54.16
mm
in
4.25
1
mm420
2=
=
o
d
The next step is to get the sectional densities calculated for the explosive mixture and the tube.
We only need two stations. For the forward cross-section we have
( )
( )
 
=
−
== in
lbm
00.27in
4
27.813.78
in
lbm
283.0 2
22
3
11
tube
AM steel
( )
( )
 
=
== in
lbm
074.1in
4
27.8
in
lbm
02.0 2
2
3
1
fillmix AC
4
in
3
22
tube
( )
( )
 
=
=== in
lbm
074.1in
4
27.8
in
lbm
02.0 2
2
3
12
fillmix ACC
Problem 3 – A projectile is to be fabricated from steel and filled with TNT as depicted below.
For a detonation of the fill, graph the fragment velocities in m/s and Taylor angles in degrees
versus distance from the nose of the projectile. The required properties for this calculation are
given as follows:
2
1
Solution: The first thing we need to do is get everything in consistent units. The density of TNT
first.
For cross section 1 we have
( )
( ) ( )
 
=
−+−
=
+
=in
lbm
383.0in
2
431.00.720773.01.062
in
lbm
283.0
2
2
2222
3
12
1
casecase AA
Msteel
( )
( ) ( )
 
=
+
=
+
=in
lbm
073.0in
2
431.0773.0
in
lbm
059.0
2
2
22
3
12
1
fillfill AA
CTNT
( ) ( )
=
+
=
−
s
m
851
2
1
073.0
383.0
km
m
1000
s
km
039.2 2
1
1
V
For the Taylor angle we first need to find the angle
/2 from our formula
( )
( )
61.3
2
063.0
s
m
67302
s
m
848
22
sin =→=
==
D
V
This Taylor angle would tend to tilt the fragment at 3.61 in the direction of the detonation wave
(towards the base) but at this point our ogive is canted 30 towards the projectile axis so the
actual angle is 3.61- 30 or -26.39 (see below)
V1
26.39
6
2.739
1.549
0.406
982
4.184
-5.816
7
3.745
1.615
0.455
1013
4.317
4.317
8
4.702
1.257
0.455
1129
4.812
4.812
9
6.702
1.615
0.455
1013
4.317
4.317
8.702
The graphs of the data are as follows
This is the fragment velocity
0
200
400
600
800
1000
1200
0.0001.0002.0003.0004.0005.0006.0007.0008.000
Fragment Velocity (m/s)
Distance From Nose (in)
Fragment Velocity (m/s)
Fragment Velocity (m/s)
-30.000
-25.000
-20.000
-15.000
-10.000
-5.000
0.000
5.000
10.000
0.0001.0002.0003.0004.0005.0006.0007.0008.000
Taylor angle (deg)
Distance from nose (in.)
Taylor angle (a ) (deg)
Taylor angle (a) (deg)
19.1 Shaped Charge Jet Formation
Problem 1 – A conical shaped charge liner is to be fabricated from copper and filled with
Composition B as the explosive. The thickness of the liner is to be 0.1 inches and the half-angle,
is to be 45°. The length of the liner is 3 inches and the charge O.D. is 7 inches. Determine the
following using the Birkhoff et al. theory:
a.) Mass of the jet Answer:
 
lbm302.0=
j
m
b.) Mass of the slug Answer:
 
lbm990.0=
s
m
c.) Velocity of the jet Answer:
=s
m
752,4
j
V
d.) Velocity of the slug Answer:
=s
m
086,1
s
V
Note that depending how you discretize the problem you may get a somewhat (but not too)
different answer.
The required properties for this calculation are given as follows:
Comp B Gurney Velocity (2E)1/2 = 2.35 km/s
Comp B Detonation Velocity (D) = 7,890 m/s
Comp B density = 1.717 g/cc
Copper density = 0.323 lbm/in3
( ) ( )
 
 
lbm292.1in999.3
in
lbm
323.0V 3
3=
== linerliner
M
( )
 
 
lbm280.3in549.56
in
lbm
058.0V 3
3=
== CompBCompB
C
Now the liner velocity follows directly from
( ) ( )
=
+
=
−
s
m
485,2
2
1
280.3
292.1
km
m
1000
s
km
350.2 2
1
0
V
We can now find the angle,
.
( )
( )
−
−
=sin
2
cos
cos
0
V
UD
(4)
I solved this using iteration. Once we have all of these results, we can determine the jet mass
and the slug mass using the angle
. As you can see from our previous results, using this method
this angle does not vary too much. Our average is 57.790° so we have
( )
cos1
2
1−= mmj
( )
cos1
2
1+= mms
The overall liner mass is 1.292 lbm. So we have
( )
 
( )( )
 
lbm302.0790.57cos1lbm292.1
2
1=−=
j
m
 
lbm990.0790.57cos1lbm292.1
m
( )
−
++−= 2
tancotcscsin
cos
D
U
V
( )
( ) ( )
−
++−
=2
45
tancotcsc45sin
45cos
s
m
7890
V
=s
m
752,4
j
V
( )
−
−−−= 2
tancotcscsin
cos
D
s
U
V
( )
( ) ( )
−
−−−
=2
45
tancotcsc45sin
45cos
s
m
7890
s
V
=s
m
086,1
s
V
Problem 2 – A conical shaped charge liner is to be fabricated from copper and filled with
composition B as the explosive. The thickness of the liner is to be 0.15 inches and the half-
angle,
is to be 30°. The length of the liner is 5 inches and the charge O.D. is 8 inches.
Determine the following using the Birkhoff et al. theory:
The required properties for this calculation are given as follows:
Comp B Gurney Velocity (2E)1/2 = 2.79 km/s
Comp B Detonation Velocity (D) = 7910 m/s
Comp B density = 1.717 g/cc
Copper density = 0.323 lbm/in3
( ) ( ) ( )
=
=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°
( ) ( )
=
+
=
−
s
m
829,3
2
1
850.2
088.0
km
m
1000
s
km
79.2 2
1
01
V
We can now find the angle,
.
( )
( )
−
−
=sin
2
cos
cos
0
V
UD
Solve this using iteration in Excel. Once we have all of these results, we can determine the jet
mass and the slug mass using an average of the angle
. Our average is 51.200° so we have
( )
cos1
2
1−= mmj
( )
cos1
2
1+= mms
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
cos
The answers are shown in the table below. We could also have taken an average as well.
cos
s
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
We need to obtain the density ratio,
from
j
t
=
936.0
in
lbm
323.0
in
lbm
283.0
3
3=
=
( )
   
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.