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How to calculate trajectory of projectile?

How would I calculate the trajectory of a projectile with only the weight of the projectile, the velocity of the projectile at barrel, and angle of the barrel?  How did the artillery operators during WWII know how much powder to put or the angle of the barrel?

Thanks for any help

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Jetpack53 years ago
Wikipedia does have an article on this, called External Ballistics. I'll assume that you found that one too dry, so let's see if we can simplify.

So, we'll start with the theory. We will also start simple and go from there. I'll warn you, we will use trigonometry and algebra.

Theory
Looking at the images, the cannon will shoot at a certain angle, a. It has a muzzle velocity of V, shown as the arrow at an angle.

Now, the velocity is a vector, meaning it has a speed and a direction. The velocity can be broken down into components, Vh and Vv, which are the vertical and horizontal arrows. The vertical and horizontal components are related to the velocity.

The components are orthogonal (at right angles), which means that a force acting in the direction of one component will not affect the other. Gravity is only vertical and so it does not affect the horizontal component.

So, let's start with the vertical component. We will focus only on the effect of gravity, since that is the main force acting on the shell. Gravity provides a constant acceleration, g, (the mass of the shell does not affect the acceleration). The relation between a constant acceleration and change in velocity (which we will call dv) are given as:
dv=g*t   or   t=dv/g

The bullet will reach the top of it's arc when the vertical velocity reaches zero, which is a change in velocity of Vv, and the time it takes is t=Vv/g.

Since the bullet is at the point only halfway to the target, the time to reach the ground is twice that, or t=2*Vv/g.

Looking now at the horizontal velocity, it doesn't have any forces on it in our simple example (this assumes no air resistance, etc...). So the acceleration is zero and the velocity doesn't change. For a constant velocity, the relation between position and time is:
d=v*t

Now, importantly, the time is the same time, so we can say:
d=Vh*t=Vh*2*Vv/g=2*Vv*Vh/g

So, if we know the vertical and horizontal components, we can calculate the distance. This is where trigonometry come in. Looking at the triangle, we can see that the components are related to the initial muzzle velocity and elevation angle:
Vh=V*cos(a)
Vv=V*sin(a)

This leaves us with the equation:
d=2*V*cos(a)*V*sin(a)/g

Combining terms, we get:
d=2*V^2/a*cos(a)*sin(a)

Using this equation, we can calculate the expected landing point based on the muzzle velocity and elevation angle...as long as our assumptions are valid, which they aren't.

In real life the bullet has air resistance, which increases as the shell is bigger and moves faster, which tends to make the shell fall short. Wind may not seem like much to a chunk of lead, but it can add up. If the shell tumbles, that will also change the way it flies. The rotation of the earth also has an impact in the form of the Coriolis Effect, which will tend to skew a shell towards the equator and is seen when they are fired over long distances.

We are also assuming that the earth is flat. If your target is on a hill or in a ditch you have to take that into account as well.

Even worse, the Earth isn't a flat plane either. It curves away with distance. The further you shoot, the more it curves. In fact, if it curves away faster than the bullet falls...you will just come around the Earth in an orbit...and hit yourself...

All of these factors make the equation very complicated and hard to make a simple solution like above. What to do? Well, I can take all of those forces on the shell and see where teh shell will go after a tenth of a second. At that point, I recalculate the forces and move it another tenth of a second. I can repeat this until the shell hits. Sounds boring? It is! They had guys that literally would sit in an office doing these sorts of calculations (actually, they were usually women). Because they were doing computations all day they were called "computers". The job was boring and tedious and one mistake would be a disaster.

So the eggheads got together to make an artificial computer, and using these methods developed tables that, in principle, could be used in the field. Certain types of slide rules could also be used.

More modern guns incorporate computers on-board, but these weren't available in WWII.

They also calculated the muzzle velocities based on the number of standardized powder charges that were loaded into the gun in the case of larger guns. Smaller guns in WWII had a fixed charge in a metal case and so a fixed muzzle velocity.

Practice
In a practical sense, these were really more of a guide. Wind, which could have a big impact on the motion of the shell, was a partial unknown. You could know the wind speed and direction at your height and location, but the wind often shifted at higher elevations. So, in practice the gunners for larger guns would use their tables to get an approximate target elevation and would simply fire off a shot to see where it would hit. Trained observers known as spotters would then relay back to the gunner where the shell hit, and based on this the gunner could adjust his aim. Once he was zeroed in, he would start firing "for effect" and hit the target repeatedly. He could at that point hit nearby targets fairly accurately as well. One of the earliest uses of balloons in warfare was to provide an elevated lookout post for artillery spotters. If they were in a defensive position and had time they could use practice shots to mark a number of predetermined targets for greater flexibility.

For smaller guns, they usually didn't have the luxury of fire tables. They would aim the gun roughly and fire, adjusting as needed for accuracy.

They had one final thing going for them. Large shells flying over large distances didn't have to be perfectly accurate. If the shell was big enough, missing by a few meters didn't matter, the explosion was enough! Hence the expression that "almost only counts in horse shoes and high explosives." (horseshoes and hand grenades is more common).

I hope that answers your question. If not, let us know where you need a hand.
cannon0001.jpgcannon0002.jpg
electronicz (author)  Jetpack53 years ago

WOW, thanks for tons everyone for the great info everybody, but I am sorry the Best Answer can only goes to one person :( Anyways, sorry it took so long to get back to you all. Thanks again everybody! :)

Thanks!

electronicz (author)  Jetpack53 years ago

Your welcome.

ANDY!3 years ago
split the initial velocity into two separate axis using trigonometry.

to find the approximate height it has traveled at any moment of time, use the equation d= (initial speed in the y axis in meters per second)(time) + 1/2(-9.8)(time in seconds)^2

to find the distance traveled parallel to the ground aka range: d= (initial velocity in the x axis in meters per second)(time in seconds)

This should give you some rough theoretical numbers for your trajectory. Look at kinematics for more info
Its drop is 9.80 meters a second. called G

It's air speed is initial air speed times ballistic coefficiency and time. called K

Ballistic coefficiency is the efficiency the projectile travels through the air it is a combination of airodynamics and weight.

They guessed the distance at first however they have better means now.

So you have distance and air speed you have flight time. called T

For every second to target you aim 9.8 meters above the target. this gives you angle.

Truth be told they guessed at first that is why the first guns had no sights.

Joe
kelseymh3 years ago
Look up calculus and ballistics. Start with the Wikiepedia articles to give you an overview, then move on to a first-year physics textbook.
Have you done any searching of your own? A simple google search will give you the answers you need.

Solders in WWII didn't do any calculating. Artillery men in the army had quick reference charts on the guns. The amount of powder that went with the projectile never changed only the angle of attack changed to put the projectile on target. No the Navy was different. The battleships had complex mechanical computers to help the gunners work out how many charges where needed, angle of attack and dealing with the movement of the ship so the projectile would hit it's target.