( )
s02344.0
s
ft
1820
0
0
0x
x
xx
x
x
xV
x
V
x
VV
x
V
V
V
x
t=
===
( )
( )
ft
ft
1
0000762.0
s
ft
060,2
1
0
x
xk
xx eeVV
−
−
==
( )
( )
+
−+= 2
1
1
1
2
0
0
0
0
1ln
1
2
1
2
1
tan tkV
tkV
tkV
gt
xx
x
x
( )
( ) ( )
( )( )( )
( )( )( )
+
−
+
=
2
22
2
0
0000762.0060,2
0000762.0060,21ln
s
ft
1
0000762.0
s
ft
060,2
1
2
1
s
s
ft
2.32
ft2
1
tan t
t
t
t
x
( ) ( )
+
−+= 2
2
00246.0
1570.01ln3706.6
5.01.16tan t
t
tx
t
Using the above relation we obtain the angle multiply this by 60 to obtain the angle in minutes.
Range
(yards)
V
(ft/s)
t
(seconds)
0
(minutes)
0
2060
0.000
–
200
1968
0.298
8.07
400
1880
0.610
16.65
600
1796
0.937
25.78
800
1716
1.278
35.49
1000
1639
1.636
45.84
For the final task of part a) we calculate the angle of fall from
+−= tkV
V
gt
x
x
10 0
02
1
1tantan
( )
( ) ( ) ( )
+
−= s
ft
1
0000762.0
s
ft
060,2
2
1
1
s
ft
060,2
s
s
ft
2.32
tantan
2
0t
t
t
(seconds)
0
(minutes)
0
2060
0.000
–
–
200
1968
0.298
8.07
800
1716
1.278
35.49
-40.087
1000
1639
1.636
45.84
-53.365
One can see from these data that the difference between a vacuum trajectory and flat-fire,
constant drag is appreciable depending on your point of view. If you were worried about impact
velocities, it makes a huge difference. If you were worried about impact angle, not so much.
Problem 10 – A U.S. 37 mm projectile is fired with a muzzle velocity of 2,600 [ft/s]. The
projectile weighs 1.61 lbm. Assuming K2 = 0.841 [unitless] and using standard sea level met
data (
= 0.0751 lbm/ft3, a = 1,120 ft/s,
( )
−
−
=Rslug
lbfft
716,1R
)
a) Determine the drag coefficient CD and drag force on the projectile if the projectile
is fired in still air.
Answer:
lbf04.33=
D
F
b) Create a table containing range (yards), impact velocity (ft/s), time of flight (s),
initial quadrant elevation angle (minutes) and angle at impact (minutes) in 100
yard increments out to 800 yards.
c) If this weapon is used at an increased altitude assume the density and temperature
of the atmosphere are
= 0.060 lbm/ft3 and T = 30F. how much higher or lower
will the weapon have to be aimed to hit a target at 800 yards
Answer: The weapon must be aimed 0.28 mils or 0.98 minutes lower.
D
We can now calculate k2 from
s
s
2
0xxxkVV xx
Our table now appears like this
Range (yards)
V (ft/s)
0
2600
100
2524
200
2447
300
2371
400
2295
500
2219
600
2142
700
2066
800
1990
Now that we have the impact velocity we can obtain the time of flight from
( )
−
=
−
=
600,2
1
600,2
ln
s
ft
600,2
ft
1
ln
0
0
0x
x
x
x
x
x
xV
V
x
V
V
V
V
V
x
t
The table is now
Range
(yards)
V
(ft/s)
t
(seconds)
0
2600
0.000
100
2524
0.117
200
2447
0.238
300
2371
0.362
400
2295
0.491
500
2219
0.624
600
2142
0.761
700
2066
0.904
800
1990
1.052
The elevation angle of the weapon is determined by solving for
0 with y and y0 set equal to zero.
Thus we have
x
xV
V
x
600,2
ln
600,2
ln
ft
0
Using the above relation we obtain the angle multiply this by 60 to obtain the angle in minutes.
Range
(yards)
V
(ft/s)
t (seconds)
0
(minutes)
0
2600
0.000
–
100
2524
0.117
2.51
200
2447
0.238
5.11
300
2371
0.362
7.83
400
2295
0.491
10.67
500
2219
0.624
13.63
600
2142
0.761
16.73
700
2066
0.904
19.98
800
1990
1.052
23.39
( )
−
−
600,2
1
s
ft
600,2
1
2
2
2
0
2
0
0
0x
x
x
xV
V
V
V
inserting the data from the table yields the final result
Range
(yards)
V
(ft/s)
t (seconds)
0
(minutes)
(minutes)
0
2600
0.000
–
–
100
2524
0.117
2.51
-2.56
200
2447
0.238
5.11
-5.32
300
2371
0.362
7.83
-8.33
400
2295
0.491
10.67
-11.59
500
2219
0.624
13.63
-15.15
600
2142
0.761
16.73
-19.04
700
2066
0.904
19.98
-23.29
800
1990
1.052
23.39
-27.96
For part c) we need to first calculate the speed of sound in air using
We now need to re-calculate k2 from.
( ) ( )
( )( )
( )( )
=
== s
1
1967.0
s
ft
085,1841.0
lbm61.12
ft01157.0
ft
lbm
060.0
2
2
3
22 aK
m
S
k
We need to calculate Vx, and t just like we did in the earlier problem using the modified value of
k2 above.
( ) ( )
( )
−=
−
=−= s
ft
1967.0600,2ft
s
1
1967.0
s
ft
600,2
2
0xxxkVV xx
( )
−
=
−
=
600,2
1
600,2
ln
s
ft
600,2
ft
1
ln
0
0
0x
x
x
x
x
x
xV
V
x
V
V
V
V
V
x
t
Range
(yards)
V
(ft/s)
t
(seconds)
0
2600
0.000
100
2541
0.117
200
2482
0.236
300
2423
0.359
400
2364
0.484
500
2305
0.612
600
2246
0.744
700
2187
0.880
800
2128
1.019
We actually only needed the values for 800 yards but I just copied the last table. From our
earlier calculations the sight on the weapon should be set for a Q.E. of 23.39 minutes if the target
is at 800 yards. We thus need to invoke
Here we put in our expected Q.E. and our actual impact velocity and TOF from our table in part
c)
Problem 11 – You are asked to create a rough safety fan for at maximum range test at Yuma
Proving Ground. The test consists of an experimental 155 mm howitzer projectile at a severe
overpressure charge. The projectile weighs 106 lbm. The muzzle velocity is 980 m/s.
a.) Using a vacuum trajectory calculate and plot the trajectory envelope for the test
b.) Determine the longest time of flight of the projectile.
c.) If we assume an average constant drag coefficient (on the way up) of CD = 0.5/M
and the projectile were fired vertically what would the maximum ordinate be? assume
standard sea level met data (
= 0.0751 lbm/ft3, a = 1,120 ft/s,
( )
−
−
=Rslug
lbfft
716,1R
)
hint: look at the flat fire assumptions and re-derive the equation of motion assuming
there never is an x-velocity (30 points)
Solution:
( ) ( )
s
m
81.9
2
m400,97=R
This is A LOT longer than the actual range so it is VERY conservative. The time of flight of the
projectile is given by equation (VT-22)
g
V
tI
00 sin2
=
(VT-22)
( )( )
( )
( )
=
2
s
m
81.9
90sin
s
m
9902
I
t
s200=
I
t
gVC
dt
dV
yD
y−−= 2
ˆ
(1)
Since the drag is inversely proportional to Mach number we see that
M
K
CD
2
=
where we know from the problem statement that K2 = 0.5. We can then write
M
K
m
S
C
m
S
CDD
2
2
2
ˆ
==
As in the notes we can define
aK
m
S
k22 2
=
gVk
dt
dV
y
y−−= 2
(2)
gVk
dt
dV
y
y−=+ 2
(3)
This is of the form
dy =+
( )
+
−
=
−−
ttt
dtk
tdtkdtk
yCedtegeV 0
2
0
2
0
2
0
(4)
Since k2 is a constant we can evaluate all of the integrals directly
tktktk
g
−− +−=
22
2
0k
g
e
k
g
VV tk
yy −
+= −
(6)
( )
C
k
gt
k
ge
k
eV
ty
tk
tk
y+−−−=
−
−
222
2
2
0
(8)
Inserting the initial conditions that at t = 0, y = y0 we obtain
0
2
2
0
g
V
gt
ge
eV
tk
tk
y+++−−−=
−
−
( )
( ) ( )
0
222
2
2
01
1y
k
gt
k
eg
k
eV
ty
tk
tk
y+−
−
+
−
=
−
−
We know that at the maximum ordinate the velocity (dy/dt) equals zero so we can write
tk
y
e
k
g
V
k
g
2
0
2
2−
=
+
( )
2
2
2
2
2
2
2m019.0
mm
m
1000
1
mm155
44 =
==
dS
( ) ( )
=
=33
3
3
3m
kg
206.1
m
ft
281.3
lbm
kg
2.2
1
ft
lbm
0751.0
( )
( )
=
=s
m
3.341
m
ft
281.3
s
ft
120,1
a
( ) ( )
( )
( )( )
( )( )
=
=s
1
040.0
s
m
3.3415.0
lbm1062
kg
lbm
2.2m019.0
m
kg
206.1 2
3
2
k
The time of flight to maximum ordinate is then
s07.40=
s
t
The altitude is then
( )
( )
( ) ( )
( )
( )
( ) ( )
( )
( ) ( )
( )
m0
s
1
040.0
s07.40
s
m
81.9
s
1
040.0
1
s
m
81.9
s
1
040.0
1
s
m
980
2
s07.40
s
1
040.0
2
s07.40
s
1
040.0
+
−
−
+
−
=
−
−
ee
ty s
( )
m921,9=
s
ty
Which is more realistic than the vacuum trajectory!
Problem 12 – The drag of a sphere below Mach number of 0.5 is well approximated by CD =
AM2 + BM + C where A = 0.0262, B = 0.0456 and C = 0.4666 and M is the Mach number. If we
would like to analyze the motion of a cannon ball fired from a Demi-Culverin circa 1646 which
fired a spherical shot weighing 9 lbm and of 4.5 inch diameter, please develop the equation of
motion for Vx (only) as a function of downrange distance based on the flat fire assumptions we
have developed in class. Then using standard sea level met data (
= 0.0751 lbm/ft3, a = 1,120
ft/s,
( )
−
−
=Rslug
lbfft
716,1R
), determine Vx at a range of 1,800 yards (the reported maximum range
of the weapon). Assume a muzzle velocity of 550 ft/s.
Let us define k as
( )( )
( )
ft
ft
in
144lbm92
2
2
Then
( )
ft
s
ft
120,1
a
Then we can write
CEVDVkV
dx
dV
xxx
x++−= 2
Separating variables we have
−=
++
x
V
Vxxx
xdxk
CEVDVV
dV
x
x0
2
0
Integrating yields
1
2
1
2
2
2
4
2
tan
4
2
2
ln
2
1Ckx
ECD
EDx
ECD
C
E
CEVDV
V
Cxx
x+−=
−
+
−
−
++
−
If we insert the initial conditions that at x = 0 Vx = Vx0 we can determine C1 to be
182.14
1=C
Which, upon inserting our values we obtain
−−
−
++
=−
2
1
2
2
2
14
tan
4
ln
2
1
00
0
ECD
E
ECDC
E
CEVDV
V
C
C
xx
x
Using this value and iterating, the velocity at 1800 yards (5400 feet) we obtain
=s
ft
197)5400(
x
V
Problem 13 – A French 240 mm projectile is fired from a model 1873 cannon with a muzzle
velocity of 440 [m/s]. The projectile mass is 144 kg. Assuming K1 = 0.55 [unitless] and using
standard sea level met data (
= 0.0751 lbm/ft3, a = 1,120 ft/s,
( )
−
−
=Rslug
lbfft
716,1R
)
a) Create a table containing range (meters), impact velocity (m/s), time of flight (s),
initial quadrant elevation angle (degrees) and angle at impact (degrees) in 600
meter increments out to 7,800 meters.
b) Experiments were conducted in France in 1884 to determine the effect of rotating
and locations on the range of this projectile. These experiments were reported on
by Breger in Notes on the Construction of Ordnance No. 27, 10 June 1884. The
farther rearward the rotating band was placed, the greater the projectile yaw at
muzzle. The same Q.E. that provided a range of 7,800 meters actually only made
7,688 meters with a (assume constant) yaw angle of 4º. Assume that the 7,800
meter range was achieved with zero initial yaw. With this information calculate
the cubic drag coefficient,
2
D
C
.
Solution:
1== KCD
Note that this is unusually large, but it works in the drag model for the actual data pretty well.
We can now calculate k1 from equation (FF-50)
0s
xx eeVV
Our table now appears like this
Range (m)
V (m/s)
0
440.0
600
413.3
1200
388.3
1800
364.8
2400
342.7
3000
321.9
3600
302.4
4200
284.1
4800
266.9
5400
250.7
6000
235.5
6600
221.2
7200
207.8
7800
195.3
Now that we have the impact velocity we can obtain the time of flight from equation (FF-53).
( )
( )
−
m
1
10042.1
s
m
440
4
1
0
x
kV
The table is now
Range
(meters)
V
(m/s)
t
(seconds)
0
440.0
0.00
600
413.3
1.41
1200
388.3
2.91
1800
364.8
4.50
2400
342.7
6.20
3000
321.9
8.00
3600
302.4
9.93
4200
284.1
11.97
4800
266.9
14.15
5400
250.7
16.47
6000
235.5
18.94
6600
221.2
21.57
7200
207.8
24.37
7800
195.3
27.35
The elevation angle of the weapon is determined by solving equation (FF-57) for
0 with y and y0
set equal to zero. Thus we have
( )
( )
+
−+−+= 2
1
1
1
2
00
0
0
0
1ln
1
2
1
2
1
tan tkV
tkV
tkV
gtxyy
x
x
x
(FF-57)
( )
( )
+
−+= 2
1
1
1
2
0
0
0
0
1ln
1
2
1
2
1
tan tkV
tkV
tkV
gt
xx
x
x
( )
( )
( )
( )
( )
( )
( )
( )
+
−
+
=
−
−
−2
4
4
4
22
2
0
s
m
1
10042.1
s
m
440
s
m
1
10042.1
s
m
4401ln
s
m
1
10042.1
s
m
440
1
2
1
m2
s
s
m
81.9
tan
t
t
t
x
t
Range
(meters)
V
(m/s)
t (seconds)
0
(degrees)
0
440.0
0.00
–
600
413.3
1.41
0.91
1200
388.3
2.91
1.90
1800
364.8
4.50
2.97
2400
342.7
6.20
4.14
3000
321.9
8.00
5.41
3600
302.4
9.93
6.79
4200
284.1
11.97
8.29
4800
266.9
14.15
9.91
5400
250.7
16.47
11.67
6000
235.5
18.94
13.57
6600
221.2
21.57
15.62
7200
207.8
24.37
17.81
7800
195.3
27.35
20.16
( )
( )
( )
( )
+
−=
+−=
xx
xx
xVx
tt
V
V
x
tV
V
gt 440
ln
m2
s
s
m
440
1
s
m
440
s
s
m
81.9
tanln
1
2
1tantan
2
2
2
2
00
00
0
inserting the data from the table yields the final result
Range
(meters)
V
(m/s)
t (seconds)
0
(degrees)
(degrees)
0
440.0
0.00
–
–
600
413.3
1.41
0.91
-0.95
1200
388.3
2.91
1.90
-2.06
1800
364.8
4.50
2.97
-3.36
2400
342.7
6.20
4.14
-4.88
3000
321.9
8.00
5.41
-6.65
3600
302.4
9.93
6.79
-8.68
4200
284.1
11.97
8.29
-11.01
4800
266.9
14.15
9.91
-13.67
5400
250.7
16.47
11.67
-16.67
6000
235.5
18.94
13.57
-20.02
6600
221.2
21.57
15.62
-23.71
7200
207.8
24.37
17.81
-27.72
7800
195.3
27.35
20.16
-31.99
For part b) let’s recap what we know:
The linear drag coefficient was calculated above as CD =
0
D
C
= K1 = 0.55.
1) Insert equation (FF-51) into equation (FF-64) and solve for k1 and suffering all of the
messy algebra – then back calculate K1 to CD and the difference is our cubic term.
−
−+
−
−+=
x
x
x
x
x
x
x
x
xV
V
V
V
V
V
V
V
V
x
gxyy 000
0
0
ln11
2
1
ln
1
2
1
tan
2
2
00
(FF-64)
xk
xx eVV 1
0
−
=
(FF-51)
2) Iterate k1 in our spreadsheet or program to determine its value so the range comes out
to 7,688 meters and back calculate to find the cubic term.
K10.5699=
e0.0751=
lbm
ft3
e
3.2813
2.2
=
CDK1
=
m144=
kg
1.206=
kg
m3
k1
S
2m
K1
=
k11.079 10 4−
=
1
m
m
x7688=
V 191.9=
t1
e
k1x
1−
( )
=
t 27.223=
s
2x
gt2
2
Vx0 k1
t
Vx0 k1
t
( )
2
0r 0.352=
rad
360
=
degrees
r0.564−=
rad
ratan tan
0r
( )
g t
Vx0
1
2 x
ln
V
+
−
=