Problem 6 – Verify Equation (3.2.148) is valid for any θ.
Solution: We start from
( ) ( )
( )
−
==−
+
x
x
c
x
x
c
cc
dxrpAdxpAxVxV
w
c
w
22
1
1
3
1
2
M
i
c
c
p
+
=1
1
V
Also let’s assume that
Al
i=V
Then
M
cAl
c
p
+
=1
1
(b)
Also for θ 0
( )
M
cllx +=+ 1
(c)
Inserting (b) into (a) we get
( ) ( ) ( )
( )
−
+
+
−
+
+
+
=−
−
1
11
1
3
1
1
2
1
1
1
22
lx
lx
lx
w
c
lw
c
xVxV
c
c
M
c
(d)
Now inserting (c) into (d) we get
( ) ( ) ( )
−
+
+
−
+
=−
−
1
1
1
3
1
1
2
1
1
1
22
lx
lx
w
c
w
c
xVxV
c
c
( ) ( )
+
=−
1
1
22
3
1
1
w
c
w
c
xVxV c
+
1
1
1
3
1
w
c
w
w
We get our desired form of
( ) ( )
+
=−
3
1
1
22
c
w
cxVxV c
Problem 7 – You are asked to design a gun to propel a fragment for an explosive initiation test at
3,000 ft/s. The diameter of the chamber and bore is to be 0.50 inches. You have on hand some
M10 flake propellant with the ballistic properties given below. Because of space limitations the
device (chamber and bore) can not exceed 5 feet in length. The fragment plus sabot weighs 0.1
lbm. Assume the propellant has the following properties:
Adiabatic flame temperature T0 = 3,000 K
Specific heat ratio γ = 1.2342
Co-volume b = 27.76 in3/lbm
Density of solid propellant
= 0.0602 lbm/in3
Propellant burn rate coeff.
= 0.00002468 (in/s)/(psi)
Web thickness D = 0.011 in
Propellant force
= 339,000 ft-lbf/lbm
a.) Using the above data determine the bore and chamber length for the weapon as well as
the amount of propellant required – be careful to leave some air space in the chamber.
b.) Once the system is established in a.) determine the central ballistic parameter, value of
peak breech pressure and location of the projectile when peak pressure occurs
Solution: First we must calculate the bore area.
( )
22
2
2
in196.0in
4
5.0
4===
d
A
Now let’s make some assumptions as follows
in48ft4==L
let’s assume a chamber length of , say, 6 inches, then the empty chamber volume is
( )
( )
32 in178.1in196.0in6U ==
( ) ( )( ) ( )
3
3in178.1
in
lbm
0602.08.0U8.0
==
c
lbm057.0=c
The central ballistic parameter can be found directly by plugging in the given values from the
+
1
2
1
2
1
w
c
cw
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
in
lbf
s
in
00002468.0
ftlbm
slbf
2.32
1
lbm
lbfft
000,339lbm057.0lbm1.0
1.02
057.0
1
1.03
057.0
1
in011.0in196.0
M
925.0=M
Let’s first see where the projectile is when the charge burns out. Here we have to note that for a
flake propellant θ = 0. So our equation is
( )
llexfM −= −1
(LG-127)
We need to determine our faked chamber length from
Al
i=V
But
( )
( )
( )
3
3
3
base in236.00
in
lbm
0.0602
lbm057.0
in178.1VUV =−
−=−−=
c
i
then
( )
( )
in2.1
in196.0
in236.0
V
2
3
=== A
li
Then using equation (LG-127) we have
( )
( )( )
( )
in2.1in2.1 01937.0 −= −
exc
in827.1=
c
x
That’s not too far down the tube. Let’s see what the velocity is there – we use equation (LG-99)
( )
+
−
=
1
12
1
1
w
c
w
fAD
V
(LG-99)
Inserting our data yields
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
1.02
057.0
1
ftlbm
slbf
2.32
1
lbm1.0
in
lbf
s
in
00002468.0
01in011.0in196.0
2
2
2
c
V
=s
ft
195,2
c
V
Wow – that’s cookin! Looks like we will have to shorten the tube a bit. let’s see what length
tube we need based on this result.
( ) ( ) ( )
+
+
=−
−
3
1
22
c
wl
elxc
xVxV
M
c
c
(LG-187)
We rearrange this expression to yield
( ) ( )
( )
+
+
−
= −3
1
22 c
wl
elxc
xVxV
M
c
c
Inserting our values we have
( ) ( )
( ) ( )
( )
3
057.0
in2.1827.1lbm057.0
lbm
lbfft
000,339
s
ft
195,2
s
ft
000,3
925.0
2
2
2
2
2
2
+
+
−
−
−
e
= 2
s–lbf
ft–lbm
849.25
We can now use this value in equation (LG-186) to determine the length of the tube
( ) ( )
llxx c−+
+
−
=−
1
1
1
2
1
( ) ( )
( )( ) ( )
in2.1in2.1827.11
slbf
ft–lbm
2.322
2342.11
slbf
ft–lbm
849.25
2342.11
1
2
2−+
+
−
−
−
=
−
x
in414.3=x
So this stubby little gun will blast the fragment towards the target. The device is mostly
chamber!
So the tube is 9.414 inches long of which 6 inches is chamber the amount of propellant is 0.057
lbm. We already found the central ballistic parameter. at peak pressure we find that
( )
( )( )( )
02925.0
10925.0
2
1
+
−+
=
+
−+
=
M
M
fm
This yields a negative value. This is indicative of a system that achieves peak pressure as the
charge burns out! Based on this the location of peak pressure is the same as charge burnout and
the value is given by:
( )
( )
fM
i
Bef
w
c
w
c
c
p−−
−
+
+
=1
1
11
3
1
2
1
V
(LG-138)
At peak pressure the pressure at the breech is then
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( )
( )( )
01925.0
301
01.03
057.0
1
01.02
057.0
1
in0.236
lbm057.0
ft
in
12
lbm
lbfft
000,339
max
−−
−
+
+
−
=epB
=2
in
lbf
340,388
max
B
p
We’d better have a lot of steel to hold this – We really would not allow this to happen in a design
– I hope your designs are better!
Problem 8 – A test of an experimental propellant grain yields a parametric relation for the
fraction of web remaining of
( )
1
2
2
1++= tCtCtf
(1)
In equation (1), C1 and C2 are constants. The burn rate model for this propellant was determined
to be
B
p
dt
df
D−=
(2)
Determine the expression for the velocity of the projectile as a function of time for a given gun–
projectile system. Describe the assumptions used and why they are relevant.
Ap
dt
dVc
wB
=
+2
1
(LG-94)
We can rearrange equation (2) to yield
1
1
−= dt
dfD
pB
(3)
Inserting this relationship into equation (LG-94) we obtain
1
1
12
−=
+dt
dfD
A
dt
dVc
w
(4)
Rearranging we have
1
1
12
+
−= dt
dfD
c
w
A
dt
dV
(5)
We can differentiate the fraction of web remaining in equation (1) to yield
21
2CtC
dt
df +=
(6)
( )
dtCtC
D
c
w
A
dV
tV +
+
−=
0
1
21
1
1
0
2
2
(8)
( ) ( )
( )
bn
bxa
dxbxa
n
x
n
1
1
0+
+
=+
+
Then we have
1
1
21
1
2
CtC
D
A
+
+
( )
4
1
1
1
2
1
1
1
12
2
0C
C
C
D
c
w
A+
+
+
−=
+
(10)
a.) Develop the proper equations for the motion of the projectile.
b.) Size the charge so that the system will function as required. – make sure it fits too!
c.) Determine the peak pressure in psi and location of peak pressure in inches.
d.) Determine the location of charge burnout in inches and the velocity at burnout in ft/s. Please
note that if your design allows unburnt propellant to exit the tube please state that and calculate
muzzle exit pressure and velocity in this case.
e.) Determine the muzzle velocity in ft/s.
( ) ( )
lx
w
c
w
c
cw
fDA
df
dx +
+
+
−
−= 2
1
1
2
1
22
2
1
3
1
1
(LG-108)
At this point we have to insert equation (1) into the above to yield
( )
( )
( )
lx
w
c
w
c
fcw
fDA
df
dx +
+
+
−
−
−= 2
1
1
23
1
22
2
1
3
1
1
1
(2)
( )
3
1
f
df
−
We now separate the equation to yield
( ) ( )
( )
df
f
f
M
lx
dx
3
1
1
−
−
−=
+
(4)
( ) ( )
lx
lx
dx
x
+=
+
ln
0
(6)
The first term on the RHS is
( )
( )
( )
+
+
−
−
−−=
−
−−
3
12
tan
3
1
1
1
ln
6
1
1
1
3
3
0
3
f
f
f
M
f
df
M
f
(7)
( )
( )
( )
+
−
−
−
=
−
−−
3
12
tan
3
1
1
1
ln
6
1
1
1
3
3
0
3
f
f
f
M
f
df
M
f
(8)
The second term on the RHS is
( ) ( )
( )
+
−
−
−
=
−
−
3
12
tan
3
1
1
1
ln
6
1
1
1
3
3
0
3
f
f
f
M
f
fdf
M
f
(9)
( ) ( )
( )
f
f
−
−
3
3
1
6
3
3
1
6
3
3
(10)
( ) ( )
( ) ( )
( )
K
fM
f
f
f
fM
lx +
+
−
−
−
+
−
−
=+ −
3
12
tan
3
2
1
1
ln
1
1
ln
6
ln 1
3
3
3
3
(11)
( ) ( )
( ) ( )
( )
K
fM
f
f
f
fM
lx +
+
−
−
−
−
−
=+ −
3
12
tan
3
2
1
1
1
1
ln
6
ln 1
3
3
3
3
(12)
( )
+
−=+ −
3
12
tan
3
2
ln 1fM
Klx
(13)
Taking the anti-log of both sides yields
+
−=+ −
3
12
tan
3
2
exp 1fM
Klx
(14)
3
3
3
3
3
33
(17)
We insert this into equation (LG-93) to obtain
+
+
=
+
−−
1
1
3
12
tan
3
3
2
3
1
2
1
1
w
c
w
c
A
c
lep
fM
B
(18)
Now we insert equation (1) into equation (18) and rearrange to yield
( )
+
−− −
+
+
−
=3
12
tan
3
3
2
1
1
31
3
1
2
1
1fM
Be
w
c
w
c
Al
fc
p
(19)
The differentiation of this is governed by the product rule
( ) ( )
+
+
−
+
+
+
−
=
+
−−
+
−− −−
3
12
tan
3
3
2
1
1
3
3
12
tan
3
3
2
1
1
311
3
1
2
1
1
3
1
2
1
1fMfM
Be
df
d
w
c
w
c
Al
fc
e
w
c
w
c
Al
fc
df
d
df
dp
(20)
The first term yields
( )
+
−−
+
−− −−
+
+
−3
12
tan
3
3
2
1
2
3
12
tan
3
3
2
1
311
2
1
3
2
1
1fMfM
w
c
fc
w
c
fc
d
0
33
1234 =−+
++ M
ff
M
f
(25)
and (19) with f replaced by the value (fm) determined from equation (25)
−
+
−= −1
3
12
tan
3
3
2
exp 1m
m
f
M
lx
(26)
−= 1
33
M
celx
(26)
+
=
1
12
1w
c
w
AD
Vc
(27)
22
c
( )
( )
=
=psis
in
00067.0
psi329,3
s
in
22.2
At this point I set up all of the equations in MathCAD and iterated on the charge weight until I
2
1
1
2
2
1
+
w
c
cw
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
2
in
lbf
s
in
00067.0
ftlbm
slbf
2.32
1
lbm
lbfft
000,105lbm128.0lbm2
22
128.0
1
23
128.0
1
in02.0in3
4
M
6.51=M
c.) Pressure and location at peak pressure. Now that we have the central ballistic parameter we
0
33
1234 =−+
++ M
ff
M
f
This yields four roots: -30.776, -0.5 ± 0.866i and 0.969. The only root that has physical meaning
is the last one so we write
969.0=
m
f
Now we can insert this into equations (26) and (27) but we need to find the artificial chamber
length first. This was given by equation (LG-130) as
c
Al
i−== UV
(LG-130)
Rearranging we have
A
c
l
−
=
U
Putting the numbers in we have
( )
( )
( )
( )
( )
2
2
3
2
2
in3
4
in
lbm
06.0
lbm128.0
in6in3
4
−
=l
m
( )
( ) ( )
( )
( )
( )
( )
( )
( )( )
( )
( )( )
( )( ) ( )( )
+
−− −
+
+
−
1969.02
tan
3
3
6.512
2
2
3
1
23
128.0
1
22
128.0
1
in698.5in3
4
ft
in
12969.01
lbm
lbfft
000,105lbm128.0
B
=2
in
lbf
212
m
B
p
d.) The location, pressure acting on and velocity of the projectile at charge burnout are provided
by equations (26) through (28). It is always best to determine the location first. Here we use
equation (26) which, with numbers is
( )
( )
−= 1in698.5 33
6.51
exc
in1005.2 14
=
c
x
97.0=
muzzle
f
This value states that shot exit will occur before peak pressure – kind of a crappy design but it
works. The tube length is then
in93.3=
muzzle
x
A stubby little thing. It would be better to use less propellant and a longer tube – but let’s see
what you come up with. For the muzzle exit pressure we tailor equation (19) to
( )
+
−− −
+
+
−
=3
12
tan
3
3
2
1
1
31
3
1
2
1
1muzzle
muzzle
f
M
muzzle
Be
w
c
w
c
Al
fc
p
Inserting numbers we have
( )
( ) ( )
( )
( )
( )
( )
( )
( )( )
( )
( )( )
( )( ) ( )( )
+
−− −
+
+
−
=3
197.02
tan
3
3
6.512
2
2
3
1
23
128.0
1
22
128.0
1
in698.5in3
4
ft
in
1297.01
lbm
lbfft
000,105lbm128.0
ep muzzle
B
=2
in
lbf
209
muzzle
B
p
The muzzle velocity is given (for this case) by equation (LG-99)
( )
)(1
2
1
)(
1
1
tf
w
c
w
AD
tV −
+
=
(LG-99)
Putting in the values we have
( )
( )
( )
( ) ( )
( )
( )( )
( )
97.01
22
128.0
1lbm2
lbfs
in
00067.0
slbf
ftlbm
2.32in02.0in3
4
3
2
2
2
−
+
=
muzzle
V
Problem 10 – A 7-perf grain is a very common geometry in weapons. The geometry is such that
the web between the outer diameter of the grain and the I.D. of the outer perforations and the
web between the perforations themselves is equal. Determine, up to the point of slivering, the
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
We see that the diameter of the grain is given by
34 += Dd
(3)
The cross sectional area of the grain is then the area defined in equation (2) less the area of the
seven perforations. Mathematically then, the volume is
( )
−+=
−== 2
2
22
4
7
34
44
7
4
VD
d
Ainitinit
(4)
D
D
D
δ
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
( ) ( )
fDt −+= 1
(7)
Since all of the surfaces regress at the same rate we can write the weight of the unburnt
propellant grain at time t as
( ) ( ) ( ) ( )( ) ( )
−+=
−= ttfD
ttd
t2
2
22
4
7
34
44
7
4
V
(8)
( )
( )
−++= 2222
4
7
92416
4
VfDDft
(9)
( )
( )
222 22416
4++= fDDftc grain
(10)
D
fD
½(D-fD)