# Article about long range artillery

1. Dec 9, 2004

### Henk

I have to write an article about long range artillery and the physics involved.
I thought that perhaps I could tell about the coriolis effect or the perfect angle (at which the projectile covers the greatest distance).

I wondered if anyone has some other interesting ideas to write about.

(Sorry for my bad English, I'm from Holland)

2. Dec 9, 2004

### prasanna

One more effect is the MAGNUS EFFECT.
Since you are from Holland, you would know something about cricket.
Magnus effect takes place when a fast bowler swings the ball.

3. Dec 10, 2004

### CollectiveRocker

Write something about projectile motion and trajectory. Those are two principles behind long range artillery.

4. Dec 10, 2004

### Tjl

In addition to CollectiveRocker's post, considering that this is k-12 you most likely do not need to include any information on wind resistance and change in gravity. So include in your paper how the projectile motion of the artillery is parabolic. A more broad topic, would be to research Kinematics.

5. Dec 10, 2004

### NateTG

Hitting things with artillery is a hard problem

Depending on how involved you want to get, you should probably start with 'projectile motion' which assumes that the earth is flat, and that the acceleration of gravity is constant. This leads to nice clean parabolic motion that everyone has to deal with in physics class.

Even the slightly more painful problem of hitting a target at a different height with a gunshot is surprisingly difficult. (I wouldn't know how to do it w/o using calculus or numerical methods.)

The next step is probably compensating for the rotation of the earth. Keep in mind that how the rotation of the earth affects your shots also depends on where you are on the earth (compare shooting from the north pole to shooting along the equator).

From there, since this is long-distance artillery, you might consider adding considerations for the curvature of the earth.

If you're feeling a bit more adventerous you can start adding in calculations for the change in gravity as the projectile flies higher.

Dealing with air resistance is also a mathematically relatively hairy endeavor.

6. Dec 11, 2004

### arildno

Another line of options opens up if you want to explore a bit of the HISTORY of long-range artillery:
1) The development of gunpowder/force-generating mechanisms.
2) The need (and difficulties with) to standardize muzzles, bullet shapes and to develop DEPENDABLE weapons which were SAFE for the user (if not for the target..)

7. Dec 12, 2004

### Henk

actually I am in my first year of university (I study physics). I posted the topic here because frankly I didn't know what K-12 meant.

Those are some interesting ideas. I think I'll use the setup Nate described. Starting with a simple model and adding more factors along the way.

Last edited: Dec 12, 2004
8. Dec 12, 2004

### Henk

It really is harder than I thought it would be. Could someone check if I'm correct? First I assume there is no friction, the earth is flat and doesn't rotate.

A projectile is shot with an angle $$\theta$$ to the horizontale and a speed v so:

$$v_{0x} = v\cos(\theta)$$
and
$$v_{0y} = v\sin(\theta)$$

$$a_{x} = 0 \rightarrow \ v_{x} = v_{0x} = v\cos(\theta)$$
$$x = vt\cos(\theta)$$

$$a_{y} = -g \rightarrow \ v_{y} = -gt + v_{0y}$$
$$y = -gt^2 + v_{0y}t + h$$

if the projectile is shot from groundlevel h = 0.

If the projectile hits the ground: y = 0 so:

$$0 = -\frac{1}{2}gt^2 + v_{0y}t \ \rightarrow \ t = \frac{2v_{0y}}{g}$$
$$x = vt\cos(\theta) = \frac{2v^2\cos(\theta)\sin(\theta)}{g}$$

To calculate the angle for the furthest distance:
$$0 = \frac{dx}{d\theta} = \frac{2v^2}{g}(-(\sin(\theta))^2+(\cos(\theta))^2) \ \rightarrow \ \theta = \frac{\pi}{4}$$

Is this correct?
If it is then this is quite simple, however a minor change makes it a lot more complicating. If I choose $$y_{0} = h$$ then:
$$y = v_{0y}t - \frac{1}{2}gt^2 + h = 0 \ \rightarrow \ t = \frac{v_{0y} + \sqrt{v_{0y}^2 + 2gh}}{g}$$
$$x = v_{0x}t = \frac{v_{0x}v_{0y} + v_{0x}\sqrt{v_{0y}^2 + 2gh}}{g}$$

And I don't think I can solve this for $$\theta$$.

I also tried to make g not constant so:
$$g = g(y) \rightarrow v_{y}=\int{g(y)dt$$

I also don't think I can solve this.

Last edited: Dec 12, 2004
9. Dec 12, 2004

### Staff: Mentor

Yes it is.

10. Dec 13, 2004

### Henk

11. Dec 13, 2004

### NateTG

As a brief note, you can use the substitution
$$\cos(x)\sin(x)=\frac{1}{2}\sin(2x)$$
which will make your range formula slightly easier to understand.

Regarding articles about long range artillery:
I recall a reading a bunch of stuff about Gerard Bull (who was working on the Iraqi supergun) and gun-barrel based space launch , which is, I believe the most extreme example of long range artillery.
For your article it wouldn't hurt to point out that long range artillery trajectories are essentially orbits.

Regarding dealing with height offsets:
I'm sure that there are endless charts and formulae for calculating the angles for hitting things. I remember working out the general problem with $h\neq0$ in a second year calculus class once. In retrospect, it might be easier to try to work out the angle $\theta$ off the horizontal to go a distance $x$ along a slope of angle $\phi$, rather than dealing with a fixed height offset. (Finding the optimum angle on a slope is a standard physics problem.) However, I haven't really looked at it, so that's just a guess.

Showing that even simplistic attempts to do more realistic artillery calculations are hard isn't necessarily a bad paper. You can also do work to determine how large the error from various sources would be on a long-range artillery shot rather than trying to account for them in your calculation.

12. Dec 20, 2004

### Henk

I have found a lot of information for the article. There is just one thing that I can't find. At what scale are the Coriolis effect and "changes in gravity" important? At what speed, distance or heigth is it really necessary to adjust for them?

Anyone any ideas?

13. Dec 20, 2004

### NateTG

If you can approximate the magnitude of error you're making with an upper bound, and the amount of time the projectile spends in the air, you can find a maximum error due to it. For example:

http://www-das.uwyo.edu/~geerts/cwx/notes/chap11/artillery.html
Has some pretty reasonable analysis for coriolis forces.

It's possible to deal with changes in gravity in a similar fashion, but remeber that gravity can change direction as well as magnitude if you start looking at really long distances. (An extreme example might be shooting from the north to south poles where the error will be very large indeed.)

BTW, now that you've got all this stuff worked out, bend your mind around this: A navy warship has artillery that can hit a tenis-court sized target 35 kilometers away. What kind of angle precision is necessary to make that happen?