−+
=
0
1
21
x
xVR
t
W
t
t
going to hit the ground (y = 0). We’ll use.
ttVyttVyy xRyR
R−+−+=
tan
as stated above for our case
ttVy x
R−
tan
Problem 30 – A British 12” projectile has a K3 of 0.8 and a weight of 850 lbm. If it is fired at an
initial QE of 130 mils with a muzzle velocity of 2,800 ft/s:
e.) Create a table of range (yards), altitude (yards), velocity (ft/s), time of flight
(s), inclination angle (degrees), and drift (yards) if the projectile is fired with
no wind.
f.) Repeat part a.) if the projectile is fired with a headwind of 25 ft/s for the first
3000 yards of flight and a crosswind (left or right – your choice) of 35 ft/s for
the remainder of the flight. Tabulate every 1000 yards with the impact
location as the last entry in the table.
Solution: As usual we will draw the problem
Vf
( )
( )
2
4
2
3ft
sft
1
10289.9
2
1
s
ft
777,2
2
1
0
−
−
=
−= −xxkVV xx
Now that we have the velocity we can obtain the time of flight from
( )
x
xx
x
x
xV
x
VV
x
V
V
V
x
t
===
s
ft
777,2
0
0
0
The elevation angle of the weapon was given in this problem.
+−=
0
21
3
1
2
1
tan 2
0
x
x
V
V
gtxy
( )
( )
( )
+
+= 777,2
21
3
1
ft
s
s
ft
2.32
2
1
ft313.7tan
22
2
x
V
x
t
xy
we calculate the inclination angle from
++−=
x
x
x
x
xV
V
V
V
V
gt 00
0
1
3
1
tantan 0
( )
( )
( ) ( )
+
−= 777,2
21
3
1
s
ft
777,2
s
s
ft
2.32
2
1
ft313.7tantan
2
x
V
t
For the final task of part a) we note that with no cross wind there will be no drift so we can write
our table as
Range
(yards)
Vx
(ft/s)
t
(seconds)
y
(yards)
z
(yards)
(degrees)
0
2777.0
0.000
–
–
7.313
1000
2632.3
1.110
121.8
0
6.566
2000
2491.3
2.281
229.7
0
5.732
3000
2354.1
3.520
322.0
0
4.796
4000
2220.8
4.832
396.8
0
3.743
5000
2091.5
6.224
452.1
0
2.554
6000
1966.0
7.703
485.2
0
1.209
7000
1844.3
9.279
493.2
0
-0.317
8000
1726.6
10.960
472.9
0
-2.055
9000
1612.7
12.758
420.0
0
-4.038
10000
1502.8
14.685
329.9
0
-6.304
11000
1396.7
16.756
197.0
0
-8.895
12000
1294.5
18.987
14.5
0
-11.858
12068
1288.0
19.146
0.0
0
-12.075
The maximum values were determined by using the spreadsheet I created and iterating until y = 0
(actually I got tired when y = -0.079 ft). Once the range to impact is determined, the data was
used in each equation above to obtain the results in the table.
For the wind effects things are a bit tricky. We will take the problem in two parts. First we shall
−+
=
0
1
21
x
xVR
t
W
t
t
( )( ) ( )
−
−+
=
s
ft
777,2
1
s
ft
2521
s
x
t
t
t
the altitude can be obtained by substituting with a fixed
0 and using the values of time and
velocity calculated above.
+−=
0
21
3
1
2
1
tan 2
0
x
x
V
V
tgxy
( )
( )
( )
+
+= 777,2
21
3
1
ft
s
s
ft
2.32
2
1
ft313.7tan
2
2
2
x
V
x
t
xy
Once all of these values are known we then find the maximum range values by iteration and
( )
( )
( )( )
( ) ( )
−
=
==
−
sft
1
015.0
s
ft
120,1491.0
lbm021.02
ft10174.5
ft
lbm
0751.0
2
24
3
33 aK
m
S
k
−+=
0
12
x
x
xxx V
V
WVV
(WT-51)
( )( ) ( )
−
−+
=780,2
1
s
ft
202
s
ft x
xx
V
VV
Now the time of flight can be obtained through equation (WT-72) which again relies on the
values from part a.) of the solution noting that we shall substitute our values of x in for R.
the altitude can be obtained by substituting into equation (FF-111) with a fixed
0 and using the
values of time and velocity calculated above.
0
3
2
0
x
V
We can complete our table as follows:
Range
(yards)
[Vx]
(ft/s)
[t]
(seconds)
[y]
(inches)
[z]
(inches)
0
2780.0
0.000
0
–
100
2546.0
0.113
-0.002
0
200
2322.2
0.236
-0.0179
0
300
2108.5
0.372
-0.0673
0
400
1904.9
0.521
-0.1789
0
500
1711.4
0.687
-0.3944
0
600
1528.1
0.873
-0.7747
0
Now we can evaluate the cross wind, noting that a cross wind will not affect the down range
velocity. The deflection comes about through equation (WT-38)
−=
0
x
zV
x
tWz
(WT-38)
Which for our case is
( )
−
=
s
ft
780,2
ft
s
s
ft
20 x
tz
Range
(yards)
Vx
(ft/s)
t
(seconds)
[y]
(inches)
[z]
(inches)
0
2780.0
0.000
–
–
100
2549.3
0.113
0
1.146
200
2328.7
0.236
0
4.798
300
2118.0
0.371
0
11.319
400
1917.3
0.520
0
21.149
500
1726.6
0.685
0
34.823
600
1545.8
0.868
0
52.995
Problem 32 – Precision shooters are always in search of “tight groupings”. That is, the grouping
of the impact points of the projectiles at a given range. We shall examine the rifle and projectile
combination given in problem 31 and determine the effect each of several parameters has on the
precision. Vary each of the following parameters individually by 1% (up and down) from
problem 31 and determine the miss distance in inches for a 200 yard range. Please carry answers
to 5 significant figures as a baseline of comparison.
a.) Muzzle velocity
b.) Projectile mass
c.) Drag coefficient (CD)
d.) Air density
e.) Air temperature (in Rankine)
f.) Launch angle of departure (we’ll assume this is due to weapon or shooter motion)
g.) Headwind (0 ± 0.2 ft/s)
h.) Crosswind (0 ± 0.2 ft/s)
i.) Choose any two of the above and vary them together – what is the result? Is the
answer simply a linear superposition of the two individual errors? Why or Why not?
Is this true for all of the parameters? Can you draw any conclusions from looking at
all of the results?
+−=
0
21
3
1
2
1
tan 2
0
x
x
V
V
gtxy
2
3
2
1
0
−= xkVV xx
0
0
0xx
x
x
xVV
x
V
V
V
x
t==
−+=
0
12
x
x
xxx V
V
WVV
(WT-51)
0
x
zV
+−=
0
21
3
1
2
1
tan 2
0
x
x
V
V
tgxy
Converting the mass we obtain
( )( )
lbm02164.0lbm02143.001.1 ==
u
m
lbm02121.0lbm02143.099.0 ==
l
m
( )( )
=
=33 ft
lbm
075851.0
ft
lbm
0751.001.1
u
( )( )
=
=33 ft
lbm
074349.0
ft
lbm
0751.099.0
l
We again use
aK
m
S
k33 2
=
The results are as follows
0
(minutes)
Range
(yards)
ρ
(lbm/ft3)
k3
Vx
(ft/s)
t (seconds)
[y]
(inches)
[z]
(inches)
4.84
200
0.075851
0.01505
2,324.4
0.23604
-0.01272
0.00000
4.84
200
0.074349
0.01475
2,333.0
0.23560
+0.01269
0.00000
These results are identical to the drag coefficient changing because k3 is linear in both variables.
The air temperature affects the speed of sound through
RTa
=
(D-5)
Based on this we can determine the standard temperature using
( )
( )( )
R14.522
Rslug
lbfft
17164.1
s
ft
120,1 2
2
2
2
=
−
−
== R
a
T
Calculating the upper and lower limits and inserting them into our table we obtain
0
(minutes)
Range
(yards)
T
(R)
a
(ft/s)
k3
Vx
(ft/s)
t
(seconds)
[y]
(inches)
[z]
(inches)
4.84
200
527.37
1,125.6
0.01494
2,327.6
0.23587
-0.00317
0.00000
4.84
200
516.92
1,114.4
0.01486
2,329.7
0.23576
+0.00319
0.00000
For the launch angle variation we use the above formulas to calculate the error in the vertical
0
9.4 Generalized Point Mass Trajectory
Problem 33 – A projectile fired from a British 12” mark IX naval gun had a muzzle velocity of
2800 ft/s and was fired at a QE of 130 mils. Assuming a vacuum trajectory, at what range and
deflection would the shot hit the ground?
Assume the firing is taking place at 50 South Latitude and the round is being fired due north.
x
y
V0 = 2800 ft/s
V
mg
0 = 130 mils
R = ?, z = ?
b.) Fired to the left
ft
321 s
dt
c.) Discuss the effect of the angular momentum on the projectile nose (which way does it
tip?)
Please ignore the Coriolis acceleration, assume there is no yaw at muzzle exit and assume a
muzzle velocity of 2,440 ft/s, the weapon has a right hand twist.
Projectile Information
67.0
81.2
46.3
35.0
2
0
−=
=
=
=
p
N
L
M
D
D
C
C
C
C
( )
( )
=
=+
−=+
3
ft
lbm
060.0
003.0
2.16
p
q
q
M
NN
MM
CC
CC
=
=
=
s
rev
033,2
in–lbm00258.0
in–lbm00026.0
2
2
p
I
I
T
P
Please supply all answers in an inertial coordinate system labeled 1,2,3 with 1 being along the
aircraft flight path and 3 being off the right side of the plane. Treat all missing coefficients as
equal to zero.
=s
ft
0
2
V
=s
ft
440,2
3
V
Which can also be written as
3
ei =
We need to determine the angular momentum vector so first we need the instantaneous angular
The initial total yaw angle is zero so the drag coefficient is
( )( )
35.0046.335.0
2
20 =+=+=
DDD CCC