9.1 Vacuum Trajectory
Problem 1 – A target is located at 20 km. A projectile muzzle velocity is 800 m/s, assuming a
vacuum trajectory, at what QE should one set the weapon to hit the target?
x
y
V0 = 800 m/s
V
mg
0 = ?
R = 20,000 m
=→= −
2
0
1
0
2
0
0sin
2
1
2sin V
gx
V
gx
( ) ( )
( )
( )
mil7.158
deg
mil
360
6400
93.8307.0sin
2
1
s
m
800
m000,20
s
m
81.9
2sin 1
0
2
2
2
2
0=
==→
=−
Problem 2 – The enemy in the above problem is very smart and has located his unit on the
reverse slope of a hill that is 3000 m in height with its peak located 18,000 m from your firing
position. Assuming that the target is at the same level as you (just behind the hill) determine a
firing solution (Q.E., if there is one) to hit him assuming a vacuum trajectory.
problem 1, above we only need to determine the altitude of the round at 18,000 m.
x
y
V0 = 800 m/s
V
mg
0 = 158.7mils
Rh = 18,000 m
R = 20,000 m
h = 3,000 m
0
22
0
2
0cos2
tan
V
gx
xy −=
Let’s look at our final equation to determine the trajectory in problem 8.1.1, above again.
Here we said that
If we recall the trigonometric identity that stated
Alright! We get the range, but can we lob it over the hill? Let’s check.
Problem 3 – The U.S. pattern 1917 (M1917) “Enfield” rifle was the most numerous rifle used
by our troops in the First World War. It was an easier rifle to manufacture than the M1903
“Springfield” (even though the Springfield was officially the U.S. Army’s service rifle) and the
troops liked its accuracy better. In fact, the famous Sergeant Alvin York was actually armed
( ) ( )
( )
( )
mil7.158
deg
mil
360
6400
93.8307.0sin
2
1
s
m
800
m000,20
s
m
81.9
2sin 1
0
2
2
2
2
0=
==→
=−
( )
( ) ( )
( )
m000,20m943,191.812sin
s
m
81.9
s
m
800
2sin
2
2
2
0
2
0=
==
g
V
x
( )
( )
( ) ( )
( )( )
( )
1.81cos
s
m
8002
m000,18
s
m
81.9
1.81tanm000,18
2
2
2
2
2
2
000,18
2
−=
=x
y
with an Enfield, not a Springfield as is commonly believed, when he single handedly captured
over 100 German soldiers in the Argonne Forest in 1918. The pattern 1917 used the standard
M1 30-’06 cartridge in U.S. service. The bullet had a mass of 174 grains (a grain is a common
unit of measure in small arms ammunition and is defined as 1/7000 of a lbm) and a diameter of
0.308 inches. This cartridge-rifle combination has a muzzle velocity of 2800 ft/s. Assuming a
vacuum trajectory:
a) Determine the angle in degrees to set the sights on the rifle (i.e. the Q.E.) if the
target is level with the firer and at 200 yards range.
b) If the target is at the same horizontal range but 20 yards higher, and the firer does
not adjust the sights, how much higher or lower will the bullet strike?
=→= −
2
0
1
0
2
0
0sin
2
1
2sin V
gx
V
gx
Problem 4 – You are asked to create a rough safety fan for a maximum range test at Yuma
Proving Ground. The test consists of a US M198 155 mm howitzer firing an M549 projectile at
maximum charge with rocket off. The projectile weighs 96 lbm. The muzzle velocity is 880
m/s.
b.) Determine the longest time of flight of the projectile.
I
Solution:
0
g
=
( ) ( )
( )
452sin
s
m
81.9
s
m
)880(
2
2
2
2
=R
m940,78=R
This is A LOT longer than the actual range so its VERY conservative. The time of flight of the
projectile is given by
g
V
tI
00 sin2
=
( )( )
( )
( )
=
2
s
m
81.9
90sin
s
m
8802
I
t
Problem 5 – It is desired to use a 105mm battery to set up a line of illumination candles over an
enemy defensive position. The enemy is positioned in depth of about 2 km. Assuming that 4
weapons are available and the candle array is to be placed such that the closest candle is 9 km
away from the gun position (assume all guns occupy the same point in space) and there is one
candle every 500m (i.e. 4 candles total at 9.0, 9.5, 10.0 and 10.5 km). Determine the Q.E. for
each weapon, the time set on each fuze and the time and order of fire of each gun such that all of
the expulsion events occur simultaneously at an altitude of 750m over the heads of the enemy.
Assume a muzzle velocity of 600 m/s and a projectile mass of 30 lbm. Use a vacuum trajectory
for the purposes of this problem (Note that this would NEVER be acceptable in actual practice).
Could the same result be achieved with less than 4 guns? Show why or why not with a
calculation.
Problem 6 – The fire control problem. During the age of battleships it was essential that a fire
control computer be installed on the ships. From 1915 onwards these were mechanical devices
by which one could put in their own ships course and speed as well as the target ships estimated
course and speed. This data and the expected muzzle velocity of the projectile (as well as time
of flight) allowed the guns to be aimed where the enemy ship would be when the shells landed.
To get a feel for the magnitude of the problem we are going to examine it in a greatly simplified
form assuming a vacuum trajectory. Given the data below, and assuming a vacuum trajectory,
provide a firing angle off the bow, elevation and timing to fire each of 4 guns so that a pattern is
created to hit if the target veers 10 degrees to the port or starboard of its present course. A hit
can be assumed to occur if the shell lands on the point where the enemy ship will actually be or if
it lands in it’s “danger space”. Because of the trajectory of the shells a hit will occur if the
trajectory passes over the target and lands within 100 yards behind it – the ship creates a
“shadow” or danger space. An example pattern might look like the one drawn in the figure but
feel free to create your own if it meets the above criteria. The shells should all be fired at the
same time. Assume your speed estimate of the target is exact. Also assume the four guns are
mounted in 2 pairs so that you only have two azimuths to work with but the elevations can be
varied independently. The target is about 400 feet long so some error in azimuth is acceptable –
but you need to tell me what that is. Plot the impact points and target position at the time of
impact. Remember that your ship is moving!
20 knots
The weapons are British 12” mark IX naval guns with a muzzle velocity of 2800 ft/s and a
maximum elevation of 20°.
yd
initial
To find the firing elevation we use
( )
s
ft
800,2
2
2
2
2
2
2
0
0=
V
So the approximate time of flight will be found from equation (VT-22)
( )
s
ft
2.32
2
I
We always want the projectile to land behind the enemy ship so we want to err on the long TOF
side. Let’s see how far the enemy ship has moved in roughly 14 seconds
( )
( ) ( ) ( )
ft467
s
hr
600,3
1
yd
ft
3
nm
yd
000,2
hr
nm
20s14 =
=D
(4)
Let’s see what this translates to along the three courses. We will break it down into an x and a y
component then subtract our ships speed from the y component to account for the velocity that
the ship imparted to the projectile. Let’s write the target speed in feet per second
( ) ( ) ( )
=
=s
ft
33.33
s
hr
600,3
1
yd
ft
3
nm
yd
000,2
hr
nm
20
tgt
V
(5)
We can find the x and y coordinate by using the sine and cosine of the absolute target course. I
shall put them in a table.
Target course (deg)
Target velocity x
(ft/s)
Target velocity y
(ft/s)
Relative target
velocity y (ft/s)
35
19.1
27.3
2.3
25
14.1
30.2
5.2
15
8.6
32.2
7.2
My strategy will be to use the first turret to fire at the 35° (near) course and the base course while
the second turret fires another shot at the base course and the 15° (far) course. I need to iterate
now. Assuming roughly a 14 second time of flight, the adjusted ranges to the target on the three
courses will be
Target course
(deg)
Target adjusted
range x (ft)
Target adjusted
range y (ft)
Total adjusted
range (ft)
Angle of fire
off the bow
(degrees)
35
35,732
12,002
37,694
71.4
25
35,803
12,005
37,762
71.5
15
35,879
12,007
37,835
71.5
The adjusted ranges were calculated through
Itgtinitx tVxR x
−=
(6)
Based on the above table it looks as though the two turrets will be pointing almost in the same
direction. Let’s deal with turret 1 first which will handle the longest range. To start we shall set
( ) ( )
( )
478.4
s
ft
800,2
ft000,38
s
ft
2.32
sin
2
1
2
2
2
2
1
01=
=−
(10)
( )( )
( )
( )
s577.13
s
ft
2.32
478.4sin
s
ft
800,22
2
1=
=
I
t
(11)
If we do these initial calculations for each gun we will get the following data
Now we determine the absolute x an y position of the impact points assuming the guns were all
fired simultaneously and (0,0) is at the firing point. We need to keep in mind that the firing ship
has a velocity of 25 ft/s so a constant velocity of 25 ft/s will be added to the shell in the positive y
direction.
Itgttgt tVy
We can tabulate all this
The values for each shell would then be
Gun
number
x
impact
(ft)
Adjusted
y impact
(ft)
Target x coordinate (ft)
Target y coordinate (ft)
1
36,015
12,461
35,740
35,808
35,883
12,372
12,411
12,438
2
35,920
12,428
35,740
35,809
35,883
12,371
12,410
12,437
3
35,847
12,333
35,741
35,809
35,883
12,370
12,409
12,436
4
35,752
12,300
35,742
35,810
35,883
12,369
12,408
12,435
Target Course (degrees)
35
25
15
35
25
15
Since the shells will land where predicted (at least in this problem), the only uncertainty is the
target position. We now plot these values in Excel to determine how well we did. Here I have
plotted the targets position and length to show that the calculations effectively resulted in a
“straddle”. This was the most desirable outcome of a naval engagement (for the firer at least)
because it nearly guaranteed a hit on the target with at least one shell.
35,600
35,700
35,800
35,900
36,000
36,100
Range (ft)
Comparison of impact points to target
Shell Impacts
35 degree target course
25 degree target course
15 degree target course
35 degree ship
25 degree ship
15 degree ship
Linear (35 degree ship)
Linear (25 degree ship)
Linear (15 degree ship)
Problem 7 – The French Infantry Rifle Model 1886 called the Lebel was their standard weapon
from 1886 into World War I and even saw limited use in the Second World War. You can see
this 51” long monster in any movie involving the French foreign legion. It used an 8 mm
cartridge called the “balle D” with a bullet mass of 198 grains and a diameter of 0.319 inches.
This cartridge-rifle combination has a muzzle velocity of 2296 ft/s. Assuming flat fire with K3 =
0.5 and using standard sea level met data (
= 0.0751 lbm/ft3, a = 1120 ft/s)
a) 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 200
yard increments out to 1000 yards.
b) If an infantryman is looking at a target at 2000 yards, What angle will the sight
have relative to the tube assuming they used standard met in the design?
Answer: About 10.3º
c) Comment on the validity of this method with respect to part b) above
We shall first calculate the projectile reference area and mass in consistent units
Now that we have k3 defined, we can create the table in order. Note that since we have assumed
that K3 is constant in this development, k3 is also constant. We will first call on
Vf
−
sft
2
s
2
3
0xxxkVV xx
Our table now appears like this
Range (yards)
V (ft/s)
0
2296
200
1953
400
1638
600
1351
800
1091
1000
859
Now that we have the impact velocity we can obtain the time of flight from
( )
s0209.0
s
ft
2296
0
0
0x
x
xx
x
x
xV
x
V
x
VV
x
V
V
V
x
t=
===
The table is now
Range
(yards)
V
(ft/s)
t
(seconds)
0
2296
0.000
200
1953
0.284
400
1638
0.620
600
1351
1.023
800
1091
1.518
1000
859
2.139
The elevation angle of the weapon is determined by solving for
0 with y and y0 set equal to zero.
Thus we have
+=→
+−+=
00
21
3
1
2
1
tan21
3
1
2
1
tan
2
0
2
00
x
x
x
x
V
V
x
gt
V
V
gtxyy
( )
( ) ( ) ( )
+=
+
=2296
21367.5
2296
21
3
1
ft
s
s
ft
2.32
2
1
tan
2
22
2
0
xx V
x
t
V
x
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
2296
0.000
–
200
1953
0.284
7.04
400
1638
0.620
15.88
600
1351
1.023
27.21
800
1091
1.518
42.16
1000
859
2.139
62.55
For the final task of part a) we calculate the angle of fall from
++−=
x
x
x
x
xV
V
V
V
V
gt 00
0
1
3
1
tantan 0
inserting the data from the table yields the final result
Range
(yards)
V
(ft/s)
t
(seconds)
0
(minutes)
(minutes)
0
2296
0.000
–
–
200
1953
0.284
7.04
-7.8208
400
1638
0.620
15.88
-19.825
600
1351
1.023
27.21
-38.631
800
1091
1.518
42.16
-68.953
1000
859
2.139
62.55
-119.76
For part b) we would have to follow the same procedure as above but at 2000 yards the sights
Problem 8 – British 0.303” ball ammunition is to be fired in a Mk.1 Maxim machine gun. The
bullets mass is 175 grains. When used in this weapon it has a muzzle velocity of 1820 ft/s.
Assuming flat fire with K3 = 0.5 and using standard sea level met data (
= 0.0751 lbm/ft3, a =
1120 ft/s)
a) 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 200
yard increments out to 1000 yards.
b) The weapon was used by British units assigned to bolster the Italians in the Alps
during the First World War (Italy came in on the Allied side because they wanted
the Tyrol region from Austria more than they wanted the Nice Region from
France). At an altitude of 3000 ft, how much higher or lower will a bullet fired
from this weapon impact a level target if the sights are set using the sea level
conditions above and the target is at 600 yards? At this altitude assume the
density and temperature of the atmosphere are
= 0.0551 lbm/ft3 and T = 20F.
00
3
2
3
2
0
00
x
x
V
x
V
( )
( ) ( ) ( )
+=
+
=1820
21367.5
1820
21
3
1
ft
s
s
ft
2.32
2
1
tan
2
22
2
0
xx V
x
t
V
x
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
1820
0.000
–
200
1512
0.362
11.36
400
1232
0.801
26.12
600
981
1.347
45.91
0
1820
200
1512
400
1232
600
800
0
1820
0.000
200
1512
0.362
400
1232
0.801
600
1.347
800
2.043
2.959
800
759
2.043
73.48
1000
565
2.959
113.83
For the final task of part a) we calculate the angle of fall from
++−=
x
x
x
x
xV
V
V
V
V
gt 00
0
1
3
1
tantan 0
inserting the data from the table yields the final result
Range
(yards)
V
(ft/s)
t
(seconds)
0
(minutes)
(minutes)
0
1820
0.000
–
–
200
1512
0.362
11.36
-12.85
400
1232
0.801
26.12
-33.87
600
981
1.347
45.91
-69.26
800
759
2.043
73.48
-131.4
1000
565
2.959
113.83
-246.8
For part b) we need to first calculate the speed of sound in air
RTa
=
( )( ) ( )
=+
−
−
=s
ft
1074R46020
Rslug
lbfft
17164.1a
We now need to re-calculate k3 from
( ) ( )
( )( )
( ) ( )
−
=
==
sft
1
0090.0
s
ft
10745.0
lbm0250.02
ft0005.0
ft
lbm
0551.0
2
2
3
33 aK
m
S
k
We need to calculate Vx, and t just like we did in the earlier problem using the modified value of
k3 above.
( ) ( )
( )
−=
−
−
=
−= s
ft
0045.0661.42ft
sft
1
0090.0
2
1
s
ft
1820
2
12
2
2
3
0xxxkVV xx