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									John Lehman                                   Phys Project # 2
Bill Ryder                                        11/18/2005
Thadeus Ololdapash                                     Fall 05
                     Call for Fire
                         Projectile Motion




                        M198/155mm Howitzer
         The oldest weapon still used by the US military is the Howitzer. Adopted during
the early 19th century for volley fire, the 155mm howitzer is the military’s primary
weapon in artillery support to this day. The 155 howitzer (referred to by the military as
the M198) is used in every branch of the US Military. Naval Vessels for the last 200
years have armed their decks with the howitzer originally designated for naval volley fire
between ships but now the howitzer is used to support the Marines during Amphibious
Assaults. The Air Force uses the 105mm howitzer on their C-130 gun ships. The
howitzer is mounted with other various weapons on one side of the aircraft and will circle
its target while annihilating it with endless cannon fire. Both the Marine Corps and the
Army use the 105mm and the 155mm howitzer for infantry support.
          The M198 has the capability of firing multiple projectiles. Projectiles for the
Howitzers can vary from a phosphorus round, to a rocket propelled round. A phosphorus
projectile can be used to mark a target for incoming bombers, or more commonly used to
burn out brush since phosphorus burns at 2000 degrees. A rocket assisted projectile gives
the 155mm the capability of assaulting targets up to 12-28 miles away from its station.
         To initiate artillery fire, infantry personnel requiring support would radio back the
CP (command post) where the 155mm howitzers would be stationed. Using a
topographical map and a protractor, the Marine would report the position of the target(s)
to be annihilated by giving the latitude and longitude of the on the same map held by the
howitzer crew. The howitzer crew uses adjusting knobs to move the barrel left, right, up,
or down. Using a 6 - 8 digit grid system the M198 can contact a target within a 10 meter
radius.
         Projectile motion refers to the motion of an object projected into the air at an
angle. It was Galileo who first accurately described projectile motion. He showed that it
could be understood by analyzing the horizontal and vertical components separately. No
one had done this before Galileo.
         Before Galileo the general opinion on projection followed the “impetus” theory
which stated that an object shot from a cannon, for example, followed a straight line until
it “lost its impetus” at which point it fell abruptly to the ground. Later on by more
observation, it became clear that projectiles actually follow a curved path. But no one
knew what the path was, until Galileo.
         He came up with this conclusion by reasoning that projectile is not only
influenced by one motion, but by two. The motion that acts vertically is the force of
gravity and this pulls an object toward the earth at 9.8 meters per second. But while
gravity is pulling the object down, the projectile is also moving forward, horizontally at
the same time. And this horizontal motion is uniform and constant according to Galileo’s
principle of inertia.
         He was able to show that a projectile is controlled by two independent motions,
and this work together to create a precise mathematical curve. He actually found that the
curve has an exact mathematical shape, a shape that the Greeks had already studied and
called the parabola. The conclusion that Galileo reached was that the path of any
projectile is a parabola.
                 After some information on the M198 Howitzer, and a brief introduction on
what exactly projectile motion is, one may now want to know exactly how projectile
motion applies to the Howitzer. One may want to know how a three man crew would go
about hitting an actual target and how the physics of projectile motion. In example, if a
target is 9,000 meters away, at what angle would the barrel of the gun need to make with
the horizontal ground? To start this problem we need to know some information about
the weapon itself. One key piece of information is the magnitude of the velocity of the
projectile as it leaves the barrel of the gun. Let’s say that this information was not
known, considering it is very hard to measure such a fast speed. However, the maximum
range of the projectile is known, which happens to be about 18,150 meters. This is about
11.3 miles. Quite a far range! The maximum range of a projectile is attained at a 45
degree angle with the horizontal. This shall be proved later. So now there is enough
information to begin the problem.
         To start we consider the fact that projectile motion is actually 2 motions combined
into one; first the vertical motion, which changes with time due to the force of gravity,
and secondly the horizontal motion, which remains constant. We will start first by
examining the vertical motion. Let us see how long it takes the projectile to reach its
maximum height. We will need a kinematical equation to describe this part of the motion
as follows:

                                        V = V0y + at

V = the final velocity at its maximum height, V0y = the initial velocity in the vertical
direction, a = the acceleration the projectile undergoes, and t = the time it takes to reach
its maximum height. Since the projectile momentarily stops at the maximum height we
can conclude that V = 0. It is also given that the acceleration that the projectile
undergoes is due to gravity which we shall call g. Since gravity is an acceleration
working against the projectile as it travels upward this is actually a deceleration denoted
as –g. Now by substituting in the new information and a little algebra the equation
becomes:

                                           t = V0y
                                                g

         This equation represents how long it takes the projectile to travel to its maximum
height upward. We all know that what goes up must come down, and since projectile
motion happens to be symmetrical we can conclude that it takes the same amount of time
to return to ground level. So now for an equation to describe the time it takes during its
entire flight we simply multiply the right side by 2. Also, since V0y is only the vertical
component of the motion it can be written in its component form as V0sin where
the angle made with the horizontal, and V0 = the initial resultant velocity with respect
to both the vertical and horizontal components. The equation now becomes:

                                        t = 2 V0sin
                                               g

Let us call this equation 1. Now we shall examine the horizontal component on the
motion. As stated earlier this motion remains constant, therefore, it does not have any
acceleration. The kinematical equation to describe this motion is:
                                                       R = V0x t

We will call this equation 2. Here, R = the range of the projectile (which is the maximum
distance it travels in the horizontal direction), V0x = the initial velocity in the horizontal
direction, and t = the time the projectile remains in the air which was described using
equation 1. V0x written in its componenent form is V0cosCombining equation 1 and 2
together creates the formula:

                                                        R = 2V02 (sincos
g

Using the trigonometric identity 2 (sincos= sin2the new equation now becomes:

                                                   R = V02 sin2
                                                       g

This equation shall be denoted as equation 3, which can be used to calculate the range of
a projectile given the initial velocity and the angle at which it is fired. In the case of the
155mm Howitzer, as mentioned earlier the range for a 45 degree angle is known, but the
initial velocity is not. Therefore all that is needed is to solve equation 3 for V0 and plug
in the known values to attain the initial velocity. The work is done as follows:

                                                (Rg/sin21/2 = V0

                                [(18,150m)(9.8m/ss)/sin2(45o)] ≈ 422 m/s

Now that we have the initial velocity of the projectile fired from the Howitzer, we can
now solve equation 3 for By doing this the angle required to hit a particular target at a
particular distance can be calculated. This equation is:

                                   Sin-1(Rg/V02) = 
                               
So then, we can now use this equation to calculate what angle the barrel of the Howitzer
would have to make with the ground to hit a target that is 9,000 meters away.
Substituting the known information into the above equation yields:

                                Sin-1((9,000m)(9.8m/ss)/(422m/s)2) = 15o
                                             
                                             
                                   Range vs. Angle

     Range (meters)   20000

                      15000

                      10000                                                             Range

                      5000

                         0
                              0   20       40         60         80        100
                                       Angle (degrees)



        The above chart was created using equation 3 for angles from 0 – 90 degrees.
Notice how for every angle there is another angle that will achieve the same range.
These angles are called complimentary angles, not merely because they compliment each
other but because they mathematically add up to 90 degrees; this is a must for every man
to know in a howitzer crew. For instance, if there is an obstacle in the way of a target, a
60 degree angle may be used instead of a 30 degree angle. Both the 30 and 60 degree
angles have the same range, but the projectile travels higher with a 60 degree angle which
allows artillery to fire over an obstacle such as trees.
        Although aircraft has contributed a great deal in Infantry support since WWII,
The Howitzer canon will always remain the muscle of the American infantry.
Advancements in technology have made it easier for the modern soldier by overriding the
howitzers WWII eras adjustment knobs with computer controlled hydraulics, but the need
for weapons crew will never cease due to the uncertainties of natural conditions.
Moreover, technology and weaponry do not always work in harmony. Weapons such as
the howitzer undergo severe shock from the recoil of the weapon after firing not to
mention the in climate weather they are designed to withstand, however computers are
not designed to withstand either of those conditions. It is also important to remember that
power is necessary to operate a computer and such power sources are not always
abundant in a war zone such as in the middle of a dessert in Iraq. Therefore soldiers will
always be needed to manually aim the 155mm howitzer in the event that the computer
shuts down. More importantly crews will always be needed to load, fire, and unload the
weapon. Regardless, the training each artillery soldier goes through has remained the
same with little variations since WWII. They learn the equations discussed above,
however the problem is that the above equations only work when conditions are perfect.
The world we live in is not exactly as perfect as one may want it to be. There are many
variables that affect the flight of a projectile apart from gravity. One of these is wind
resistance. To account for this Howitzer crews will usually adjust there sights by 3
degrees to the right or the left for every 500 meters of the anticipated target. This is when
the wind is blowing against the Howitzer from the sides. If the wind is blowing head on,
or with the line of fire no adjustments are made considering this wind doesn’t change the
flight much at all.

								
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