( ) ( )( ) ( )( )
( )
2
22 11128
2fDfDfDDftc grain −++−++=
(12)
( ) ( ) ( ) ( )
( )
2
22222 112112128
2fDfDffDfDDftc grain −+−++−++=
(13)
( )
( )
2222222222 2221212128
2DffDDfDDDffDfDDftc grain +−+−++−++=
(14)
( )
( )
22222 231010
2DDDffDfDtc grain +++−+=
(15)
2
2++−+−++= DDffDDDDc graingrain
gas
(18)
Let’s manipulate this a bit
( ) ( )
( )
2
22222 31010 128
2
+−++−++= DDffDDDDc grain
gas
(19)
init
c
( )
( ) ( )
( )
( )
22
2
22222
128
2
31010 128
2
++
+−++−++
=
DD
DDffDDDD
t
grain
grain
(22)
( )
( ) ( )
( )
( )
22
2
22222
128
31010 128
++
+−++−++
=DD
DDffDDDD
t
(23)
( )
( ) ( )
( )
22
2222
128
10710103
++
+++−
=DD
DDfDDDf
t
(24)
( )
22
2
128
3
++
=DD
D
A
(26)
( )
22
2
128
1010
++
+
−= DD
DD
B
(27)
( )
22
2
128
107
++
+
=DD
DD
C
(28)
( )
27.0
177
48
48128
48
222
2
==
++
=
A
(29)
( )
13.1
177
200
48128
16040
222
22
−=−=
++
+
−=
B
(30)
( )
86.0
177
152
48128
40112
222
22
==
++
+
=
C
(31)
With these numbers we have
( )
86.013.127.0 2+−= fft
(32)
If we look at the form for φ we have
( ) ( )( ) ( )
1111 2+−+−=+−= fffft
(33)
Comparing equations (32) and (33), we see from the first term that θ should be around -0.27.
This is a problem because the second term requires θ to be -0.13. Based on this the relationship
of equation (32) would have to be used directly to model the propellant. The issue is the fact that
when f = 0 there is still some solid propellant remaining.
Alternate Solution: Following the homework examples we can define the weight of the material
as the product of the specific weight times the volume.
V
grain
c
=
(1)
Here
grain is the density of the propellant grain, V is the volume of the grain and c is the weight
of the propellant grain. If we define the cross-sectional area of the solid propellant grain as if it
were a short, right circular cylinder as
4
2
d
Ainit
=
and
4
V
2
d
Ainitinit ==
(2)
where, d is the diameter of the grain and
is the thickness (length). We define the web thickness
as D. Now we have to account for the seven perforations. Using the picture from the notes we
see that each perforation has an initial diameter which we shall call δ. So our cross sectional area
looks like this
seven perforations. Mathematically then, the volume is
( )
22
22
7
44
7
4
V
−=
−== d
d
Ainitinit
(4)
4
4
init
(6)
Since matter can neither be created nor destroyed, the weight of the gas generated by burning a
propellant grain is equal to the initial weight of the grain minus the weight of solid grain left.
This is described conveniently by the fraction of the web remaining (mass fraction), f. If the
surfaces regress at the same rate there must be a relationship between f D and δ. We see that this
is
( ) ( )
fDdtd −−= 1
(7)
( ) ( )
fDt −+= 1
(8)
D
D
D
δ
D
fD
½(D-fD)
( ) ( )
2222222 64202022
4
22416
4DfDfDfDDDDc graingrain
gas −++++−++=
(16)
Let’s manipulate this a bit
( )
6420202222416
4
2222222 DfDfDfDDDDc grain
gas +−−−−−++=
(17)
( )
620202014
4
2222 DffDfDDDc grain
gas +−−+=
(18)
( )
( )
( )
22
2222
22416
4
620202014
4
++
+−−+
=
DD
DffDfDDD
t
grain
grain
(20)
( )
( )
( )
22
2222
22416
620202014
++
+−−+
=DD
DffDfDDD
t
(21)
( )
( ) ( )
( )
22
2222
22416
6202014
++
++−+
=DD
DfDDfDD
t
(22)
( )
( )
22
2
22416
2014
++
+
=DD
DD
R
(24)
Now assume a solution of
( ) ( )( )
fft R
+−= 11
(25)
( )
DD
D
2014
6
2
2
+
−=
(33)
Inserting the relationship that δ = ¼ D we obtain
19
6
4
20
14
6
22
2
−=
+
−=
DD
D
(34)
Problem 11 – A Japanese 18.1 inch Type 94 gun was the largest weapon ever mounted on a
warship. The gun had a chamber volume of 41,496 in3. An estimated 8” of the projectile
protrude into the chamber after it seats. The length of travel for the projectile from shot start to
shot exit is 806.3 in. The weapon has a uniform twist of 1 in 28. The type 91 AP projectile
weighs 2,998 lbs. The propelling charge weighs 794 lbs. The propellant used is called “DC1”
and consists of 51.8% NC (11.85% nitrated), 41.0% NG with 4.5% Centralite (symmetrical
diethyl diphenylurea C17H20N2O), 2.0% ortho tolyl urethane (added as an improvement to the
centralite) and 0.7% mineral matter (salts for wear and flash reduction). Assume DC1 propellant
has the following properties:
Adiabatic flame temperature T0 = 3000 K
Specific heat ratio γ = 1.23
Co-volume b = 27.0 in3/lbm
Density of solid propellant δ = 0.059 lbm/in3
Propellant burn rate
= 0.000300 (in/s)/(psi)
Web thickness D = 0.184 in
Propellant force
= 284,000 ft-lbf/lbm
a.) Determine the central ballistic parameter for this gun/projectile combination.
b.) Using the above data determine the projectile base pressure, velocity and distance down
the bore of the weapon for both peak pressure and charge burnout assuming the grain is
single perforated propellant
c.) Determine the muzzle velocity of the weapon and the pressure acting on the projectile at
muzzle exit.
Since we were not given any other information we shall neglect the compression of air ahead of
the bullet, and bore friction so that the bullet mass is equal to the effective mass or w1 = w. We
then write.
lbm998,2
1== ww
(2)
lbm794=c
(3)
We now write the empty chamber volume as
baseprojectilechamberempty VVU −=
(IB-36)
We shall assume that the base of the projectile is at the full bourrelet diameter since we have no
other information available. Thus we have
( )
( )
3
2
2
3in438,39in8
4
in1.18
in496,41U=−=
(4)
The initial volume available to evolve gas into is then
c
i−= UV
(IB-58)
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
in
lbf
s
in
000300.0
ftlbm
slbf
2.32
1
lbm
lbfft
000,284lbm794lbm998,2
998,22
794
1
998,23
794
1
in184.0in3.257
M
(6)
007.1=M
(7)
The method I recommend is to first find the fraction of the web remaining at maximum pressure
=2
in
lbf
606,39
max
B
p
(11)
And at “all burnt” the pressure at the breech is
( ) ( ) ( )
( )
( )( )
998,22
794
1
lbm794
ft
in
12
lbm
lbfft
000,284
+
−
w
c
p
pB
s
2
1+
=
(14)
So
( )
( )( )
=
+
=2
2
in
lbf
974,34
998,22
794
1
in
lbf
606,39
max
s
p
(15)
( )
( )( )
=
+
=2
2
in
lbf
974,34
998,22
794
1
in
lbf
605,39
c
s
p
(16)
Let’s find the position of the projectile at both times. First let’s find the artificial chamber
length.
A
li
V
=
(17)
With our numbers we have
in101
in3.257
in980,25
3
3
==l
(18)
Keep in mind that this number is the velocity at charge burnout NOT the muzzle velocity. The
( )
−
+
+
=lx
lL
pLp
c
c
(32)
Then
( ) ( )
=
+
+
=
−
2
23.1
2in
lbf
818,8
1013.175
1013.806
in
lbf
061,38Lp
(33)
The muzzle exit pressure is actually that acting on the projectile base which is obtained by
converting the space-mean pressure to a base pressure. We use equation (LG-66)
+= w
c
pp S3
1
(LG-66)
( )
( )( )
+
in
998,23
794
1
muzzle
S
Problem 12 – A British “2 Pounder” (so-called because the projectile weighed about two
pounds) was their main anti-tank weapon for the first two and a half years of the Second World
War. It fired a 40mm projectile that weighed 1.94 lbm. The gun had a chamber volume of 23.0
in3. Since the shot had a flat base, when crimped to the cartridge case none of the projectile
protrudes into the chamber. The length of travel for the projectile from shot start to shot exit is
70 in. The weapon has a uniform right hand twist of 1 in 30. The propelling charge weighs
0.583 lbs. The propellant used was single perforated cordite Assume the cordite propellant has
the following properties:
Adiabatic flame temperature T0 = 2,442 K
Specific heat ratio γ = 1.21
Co-volume b = 31.32 in3/lbm
Density of solid propellant δ = 0.059 lbm/in3
Propellant burn rate
= 0.00024 (in/s)/(psi)
Web thickness D = 0.0197 in
Propellant force
= 318,000 ft-lbf/lbm
a.) Determine the central ballistic parameter for this gun/projectile combination.
b.) Using the above data determine the projectile base pressure, velocity and distance down
the bore of the weapon for both peak pressure and charge burnout.
c.) Determine the muzzle velocity of the weapon and the pressure acting on the projectile at
muzzle exit.
d.) Create a pressure-travel and velocity-travel curve for this system. Annotate the location
of charge burnout on the pressure-travel curve.
lbm94.1
1== ww
(2)
lbm583.0=c
(3)
We now write the empty chamber volume as
baseprojectilechamberempty VVU −=
(IB-36)
Since we were told that the base of the projectile does not protrude into the cartridge case we
have
33 in0.230in0.23U=−=
(4)
The initial volume available to evolve gas into is then
c
i−= UV
(IB-58)
Putting the numbers in we have
( )
( )
3
3
3in13
in
lbm
059.0
lbm583.0
i
Now we can calculate the central ballistic parameter.
+
+
=2
1
1
2
1
22
2
1
3
1
w
c
w
c
cw
DA
M
(LG-110)
( )
( )
( )
( )( )
( )
( )( )
−
−
+
+
=
2
2
2
2
2
2
2
4
2
s
in
slbf
1
lbfft
94.12
583.0
1
94.13
583.0
1
in0197.0in948.1
M
(6)
903.1=M
(7)
The method I recommend is to first find the fraction of the web remaining at maximum pressure
w
c
p
pB
s
2
1+
=
(14)
( )
( )( )
=
+
=2
2
in
lbf
800,29
94.12
583.0
1
in
lbf
277,34
max
s
p
(15)
( )
( )( )
=
+
=2
2
in
lbf
988,22
94.12
583.0
1
in
lbf
442,26
c
s
p
(16)
Let’s find the position of the projectile at both times. First let’s find the artificial chamber
length.
A
li
V
=
(17)
With our numbers we have
in735.6
in948.1
in13
3
3
==l
(18)
The projectile position at peak pressure and charge burnout is given by
lelxm=+
(LG-155)
M
clelx =+
(LG-170)
Inserting our values we have
( ) ( )
( )
in573.111in735.61 =−=−= eelxm
(19)
c
