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( )
( )
( )( )
( )
( )( )
=
+
+
=22 in
lbf
290,25
94.12
583.0
1
94.13
583.0
1
in
lbf
442,26p
(31)
Now we use equation (LG-177) with this number to obtain
( )
−
+
+
=lx
lL
pLp
c
c
(32)
Then
( ) ( )
=
+
+
=
−
2
21.1
2in
lbf
317,13
735.6428.38
735.670
in
lbf
290,25Lp
(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)
010 20 30 40 50 60 70
0
4375
8750
13125
17500
21875
26250
30625
35000
Pressure-Distance Curve for 2 pounder
Travel distance [in]
Breech Pressure [psi]
pBx( )
x
010 20 30 40 50 60 70
0
375
750
1125
1500
1875
2250
2625
3000
Projectile Velocity [ft/s]
V x( )
x
Charge
Burnout
in bulletized form, how you would solve these equations numerically. Hint: Start with equations
(3.80) and (3.77) through (3.79) making sure that you alter equation (3.78) to the new burn rate.
Rearranging we have
( )
+
+
+
−=
1
1
3
1
2
1
w
c
w
c
DlxA
c
dt
df
(5)
Because we need to solve these equations numerically, we can leave them in this form. The
proper algorithm would then be as follows:
a.) Set the initial values and time step
b.) Solve equation (5) for df/dt
c.) Solve equation (3) for dV/dt
d.) Solve equation (LG-93) for the pressure
e.) Determine the velocity by multiplying the result of step c.) by the time step and adding it
to the previous velocity
f.) Determine the distance travelled by multiplying the velocity obtained in e.) by the time
step and adding it to the previous distance
g.) Determine the fraction of the web remaining by multiplying the result of step b.) by the
time step and adding the result to the previous value of f
Problem 14 - Write a code using any software you want to solve problem 13 Only write the
code so it solves the interior ballistics problem up to charge burnout. Check the code by setting
α = 1 and show that you get answers close to that obtained in problem 12
Problem 15 - Use your code developed in problem 14 to obtain a solution to the same problem
assuming black powder is the propellant. Only use your code to take the problem to the all-burnt
point. Assume the properties of black powder are as follows:
Adiabatic flame temperature T0 ~ 2,300 K
Specific heat ratio γ ~ 1.22
Co-volume b ~ 31 in3/lbm
Density of solid propellant δ = 0.060 lbm/in3
Propellant burn rate
= 0.04044 (in/s)/(psi0.511)
Web thickness D = 0.0197 in
Propellant force
= 105,000 ft-lbf/lbm
in
max
B
=2
in
lbf
000,12
c
B
p
in8=
m
x
in26=
c
x
=s
ft
645
max
p
V
=s
ft
250,1
c
V
Problem 16 - A Japanese (designed by the British firm of Vickers) 14”/45 cannon is to be
examined. It fired a 14” (36cm) projectile that weighed 1,485 lbm. The gun had a chamber
volume of 17,996 in3 [6]. Assume 4 inches of the projectile protrudes into the chamber. The
length of travel for the projectile from shot start to shot exit is 540.8 in [6]. The weapon has a
uniform right hand twist of 1 in 28. The propelling charge has 4 increments, where each weighs
78.45 lbs. The propellant used was DC which consisted of 64.8% NC, 30% NG, 4.5% centralite
and 0.7% mineral matter [6]. Assume the propellant geometry is such that θ = 0.1. Assume the
DC propellant has the following properties:
Adiabatic flame temperature T0 = 3200 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
= 0.000298 (in/s)/(psi)
Web thickness D = 0.165 in
Propellant force
= 365,000 ft-lbf/lbm
The weapon was “zoned” to fire using 2, 3 and 4 bags which were called, “weak”, “reduced” and
“full” [7]. For each of these charge configurations
a.) Determine the central ballistic parameter for this gun/projectile/propellant combination.
b.) Using the above data determine the projectile breech pressure for both peak pressure and
charge burnout.
c.) 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.
d.) Determine the muzzle velocity of the weapon and the pressure acting on the projectile at
muzzle exit.
e.) Plot the pressure vs. distance, based on the above results at the instant of peak pressure
and muzzle exit
+
+
=2
1
1
2
1
22
2
1
3
1
w
c
w
c
cw
DA
M
(LG-110)
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
in
lbf
s
in
000298.0
ftlbm
slbf
2.32
1
lbm
lbfft
000,365lbm8.313lbm485,1
485,12
8.313
1
485,13
8.313
1
in165.0in9.153
M
(6)
204.1=M
(7)
in
max
B
And at “all burnt” the pressure at the breech is
in
c
B
These were breech pressure values we need to see what the values are at the base of the projectile
so we use the Lagrange gradient
w
c
p
pB
s
2
1+
=
(16)
So
( )
( )( )
+
in
485,12
8.313
1
c
s
Let’s find the position of the projectile at both times. First let’s find the artificial chamber
length.
A
li
V
=
(19)
With our numbers we have
in354.78
in9.153
in062,12
3
3
==l
(20)
The projectile position at peak pressure and charge burnout is given by
( )
( )
M
mM
M
llx
+
+
=+ 2
(LG-151)
( )
M
cllx +=+ 1
(LG-169)
Inserting our values we have
( )
( ) ( )
( )( )( )
( ) ( )
1.0204.1
1.02204.1
21.0
204.1
+
+
+
+
M
M
M
m
(21)
( ) ( )
( ) ( )
in56.168in354.781.01in354.7811.0
204.1 =−+=−+= llx
M
c
(22)
Now let’s look at velocities
( )
+
−
=
1
12
1
1
w
c
w
fAD
V
(23)
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
485,12
8.313
1
ftlbm
slbf
2.32
1
lbm485,1
in
lbf
s
in
000298.0
01in165.0in9.153
2
2
2
c
V
(26)
=s
ft
672,1
c
V
(27)
Keep in mind that this number is the velocity at charge burnout NOT the muzzle velocity. The
( )
−
+
+
−
=
−
1
354.7856.168
354.788.540
23.11
223.0
(28)
657.1=
(29)
The muzzle velocity is then obtained as
( ) ( )
c
M
c
muzzle xV
c
wl
elxc
V2
13
+
+
+
=
−
(30)
( ) ( ) ( )
( )
( )
( ) ( )
+
+
+
=
−
2
2
2
204.1
2
s
ft
672,1657.1
lbm
3
8.313
485,1in354.78
in354.7856.168lbm8.313
slbf
ftlbm
2.32
lbm
lbfft
000,365 e
Vmuzzle
(31)
=s
ft
535,2
muzzle
V
(32)
The muzzle exit pressure is determined from equation (LG-177)
( ) ( )
( )
−−
+
+
=
=lx
lx
xv
xv
p
xp
cc
c
(LG-177)
In this case the space mean pressure at burnout is found first from equation (IB-55)
+
+
=
w
c
w
c
pp B
2
1
3
1
(IB-55)
Inserting our numbers we obtain
( )
( )
( )( )
( )
( )( )
=
+
+
=22 in
lbf
160,36
485,12
8.313
1
485,13
8.313
1
in
lbf
350,37p
(33)
Now we use equation (LG-177) with this number to obtain
( )
−
+
+
=lx
lL
pLp
c
c
(34)
Then
( ) ( )
=
+
+
=
−
2
23.1
2in
lbf
672,11
354.7856.168
354.788.540
in
lbf
160,36Lp
(35)
The muzzle exit pressure is actually that acting on the projectile base which is obtained by
