# Air Cannon Theory Help (Internal Ballistics)

1. Oct 14, 2007

### wannabe_genius

I have built a pneumatic tennis ball lancher (with compressed air). I need to develop theory (i.e. equations and formula) to determine the velocity as the ball leaves the barrel. I have the projectile motion after the ball leaves the barrel down pretty well. I'm having issues acounting for pressure drop among other things.

As seen in the attached picture, there is a PVC resevoir with has compressed air. The valve opens by a current from the battery and the air shoots the ball out. Note I need physics forumlas and not a link to a projectile motion calculator site.

Thank you.

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2. Oct 15, 2007

### PiratePhysicist

My suggestion is to take the experimental route for this. Measure values of muzzle velocity against pressure and then find the relation between the two so you have a method to predict the muzzle velocity based on the pressure. Ideally you should use something that can directly measure the muzzle velocity, if not you'll need to do calculations to take air resistance into account in the values you do measure (like how far the projectile goes).

However, for fun (cause Diff Eq is fun), here's what I can come up with off the top of my head for the formulas:
Initially you'll have:
$$F(t=0)=P_{diff}(x=x_0)*2\pi r^2$$ Which is the force is equal to the difference of pressure (the pressure inside the tube minus the pressure outside) times half of the surface area of a sphere.
Now you'll need to take into account the fact that as the ball is ejected from the tube the pressure behind it will decrease and so the force exerted on the ball with change, so in comes (I'm going to leave the subscript diff off of the pressure here for ease of typing, but realize I'm still referring to the differential pressure):
$$\frac{P_1}{V_1}=\frac{P_2}{V_1}$$
$$\frac{P_1}{2\pi r^2 x_0} = \frac{P_2}{2\pi r^2 (x_0+d)}$$
$$P_2=P_1\frac{2\pi r^2(d+x_0)}{2\pi r^2(x_0)}$$ at this point I'm going to substitute d for $$x-x_0$$.
$$P_2=P_1\frac{x}{x_0}$$
So taking that original force equation we had
$$F=P(x)2\pi r^2=\frac{P_1*2\pi r^2 x}{x_0 }$$ for now on I'm going to replace the explicit surface area there with an A (because this would be obnoxious to type otherwise). Also realize P_1 is just the pressure at "x=x0".
Up to this point I've been avoiding Calculus, because I don't know your level of education, however now it's a differential equation so, I have to use Calculus... kinda, it's Diff Eq so it's basically guess and check anyway.
$$a=x''=\frac{P_1*2\pi r^2 x}{x_0 m}$$
$$x''-\frac{P_1*2\pi r^2 x}{x_0 m}=0$$
This is very similar to the equation for a simple harmonic oscillator, just with the sign of the x term switched, this means we can just get rid of the imaginary portion of it giving us the two possible solutions:
$$x=e^{\sqrt{\frac{P_1 A}{mx_0}}t}+x_0$$ and $$x=e^{-\sqrt{\frac{P_1 A}{mx_0}}t}+x_0$$
So now consider which is more likely, the displacement exponentially increasing, or it exponentially decaying? (increasing is more likely because the pressure is positive, it seems unrealistic, but remember that the P is the pressure difference, so with a long enough tube and no friction the ball would oscillate in the middle of the tube)
So if you use that to calculate when the ball would leave the tube, then plug that time into
$$v=\sqrt{\frac{P_1 A}{m x_0}}e^{\sqrt{\frac{P_1 A}{m x_0}}t}$$
Now, if you want that to be valid you'll need to grease the inside of the tube and the ball *a lot* to reduce friction to being negligible. I left friction out intentially because it would depend on a few constants that would be hard for you to obtain (coefficient of friction between ball and PVC, the spring constant of the tennis ball and finally the compression of the tennis ball)

Last edited: Oct 15, 2007
3. Oct 15, 2007

### RKT

Piratephysicist,

I think you are assuming static pressure all along the barrel. I did a calc. long ago to try and determine muzzle velocity and the tech. I adopted is similar to yours. The results were way off from expected values.

I have not retried my calc. but I feel that there's a prob. in using the (PV = constant) to try and determine instantaneous pressure on the projectile at any point in the barrel. The reason I think is because the gas law eqn. applies for static pressure (like gas compressed in a vessel). At a point in the barrel far away from the valve, I doubt if the pressure can be calc. by substituting V2 = (V1 + b) in the eqn : P1/V1 = P2/V2. Here 'b' is the swept barrel volume and V1 is the reservoir vol. Here P1, V1 and V2 are known. This is what I did and as I said, my results were off.

At a point in the barrel away from the valve, the air is moving very fast. However, because the barrel vol. is pretty small compared to that of the reservoir, the air will not be fully decompressed. It will still be in a state of compression (although lesser than in the reservoir) and it will also be moving. IN SUCH A STATE, HOW DOES ONE CALC. PRESSURE ? If the air is fully decompressed (at 1 atm) and only moving along fast, the pressure exerted is simply the dynamic pr. = 0.5 x (density) x (vel)^2 (and thats pr. on a FLAT surface, not a curved one)

This thing is real tricky, thats for sure ................. lemme know what you think.

4. Oct 15, 2007

### PiratePhysicist

Oh.... it's PV=const? Whoops, that changes things ^_^ I hadn't slept in a day when I started writing all those up equations. However you said you did the calc right and still go off answers. Well, you could try correcting using van deer Waals gases instead of Ideal. Also, did you minimize the friction in your tests because friction is gonna play a *huge* part. That's also why I said this one should be handled by getting an empirical equation.

5. Oct 17, 2007

### RKT

Piratephysicist,

I didnt go for Vanderwalls corrections as I know that its not gonna help a lot. After all, just how much difference is it gonna make ? 10 % ? 20 % ? Not more than that I'm sure. I neglected friction as well, however that does not make up for the FACT that the calc. tech. is still lacking somewhere. I tried calc. for a model for which I already knew the muzzle velocity. I came up with a number that was abt 75 - 80 % short. A diff. THAT large just cannot be accounted for by friction and Van der Waal correction alone.

6. Oct 17, 2007

### RKT

Also, I forgot to mention, since I neglected friction, my numbers should have been somewhat higher than observed values, not 75 % LESSER. It was deep **** earlier, this makes it deeper. Tell you what, why dont we all decide on some input values (reservoir vol, pressure, barrel length, caliber etc.) and do our own calc. and post the results here ? The actual / observed val. will also be there for all to see. Any inadequacies with the calc. methods can be brought out depending on the values we get.

7. Oct 17, 2007

### PiratePhysicist

Um, okay, calm down. Also if you find my calculation in adequate, please make some constructive suggestions. And, 10%-20% is pretty big.

Now, what were the equations you derived that came out to being wrong and what data disproved it?

8. Oct 18, 2007

### Shooting Star

I have given a simple and rough treatment for another situation much like this in another thread. It's nothing but work done during adiabatic expansion. Then we can equate work done to mv^2/2. Perhaps you can have a look at it.

9. Oct 19, 2007

### clustro

You could always just set the thing at a known angle and pressure, and fire it. Use a stopwatch to track the time, and...something to check to the distance (one of those little rolly wheel thingies?). Then use the kinematic equations to derive velocity. Repeat as many times as you want. Although, I doubt firing angle will matter, since the ball's time in the tube is too short for the acceleration of the earth's gravity to begin slowing it.

Then just plot them to a least-squares line. If you have enough trials, you'll get decent results.

Also, someone above mentioned having actually known velocity values for their launcher. If the ball launchers are of similar design, then they may also be dynamically similar, and you could use the buckingham pi theorem to figure out your velocity that way using dimensionless groups.
They would just have to post their ballistics data and launcher specs for you to analyze.

For the hell of it, heres some stuff using that method:

The variables I chose to analyze are:
Pressure, P
Tube Length, L
Tube Diameter, D
Ball mass, m
Velocity (duh), v
time, t (I suppose the time period between firing and impact. You can measure time between any two points, as long as you're consistent between trials.)

The pi's come out to be:

pi1 = (mv^2)/PD^3

pi2 = L/D

pi3 = vt/D

The first pi is probably the most important one. Correct me if I'm wrong, but I am interpreting this as the ratio of the ball's kinetic energy to the work done by the gas. This is similar to what Shooting star said. Ideally, this ratio should be constant between trials for varying pressures, but in practice, it will probably change due to the system's dynamics.

The 2nd pi is simply a sizing factor. Obviously if you're tube were infinitely long, things would be different. It doesn't matter much though because I'm assuming you're not going to try and make correlations based upon tube sizes, but it allows us to generalize the results to any pneumatic ball launcher of similar design.

The 3rd pi is of course, relating the distance traveled by the ball to the diameter or length of the tube. (probably length is more appropriate, but that's what the math gave.)

From the theorem, each pi is a function of the other two. After a bit of hard work, you could make an empirical chart or equation to give you the velocity as a function of pressure by correlating the 3 pi values. It will just be time-consuming.

Better though I suppose than trying to figure it out analytically!

Last edited: Oct 20, 2007
10. Oct 20, 2007

### Shooting Star

I think the OP had wanted a theoretical model against which he could measure his practical data. Ultimately, he will have to resort to experiments as you have described to know exactly how his machine behaves, but without a theoretical baseline to compare against, how would one know how efficient a machine is, or to what limits he should be able to push it to?

“There is nothing so practical as a good theory.” – Almost Everybody.

11. Oct 20, 2007

### clustro

i would assume the limits he can push it to are limited by the structural integrity of pvc and/or the maximum pressure on his air compressor
but if he wants analytic solution, well, so be it.

Perhaps maybe pressure does not push the ball all along its path up the tube, and the ball only travels from the initial impulse of the mass of air smacking it, just like a baseball player hitting a baseball.

This would be less powerful than if the air pushed the ball up the entire tube, so if the formula given by pirate (well, okay, he goofed, but whatever) is overshooting the actual data, then this could be the case.

Last edited: Oct 20, 2007
12. Oct 20, 2007

### Shooting Star

I am so glad that you realised after your initial post that the air tube also plays a vital part. I have neglected friction between the ball and the air tube, also the angle of the tube, which can be easily included in the energy equation. As I have said, its a rough and simple first approximation.

It's unlikely that the ball will only get an initial smack from the air, in which case I think the air has to recoil like a solid object at that point. But certainly the efficiency will be less if the tube is too long; so there must be an optimum length of the tube depending on the pressure and friction.

In cannons, a piece of wood or metal is put around the shell, so that the gas does not waste energy passing around it. (Maybe not in very modern guns, but even just a few decades ago.)Other factors will come into mind as the experiment is explored further -- whether in theory or in practice.

13. Oct 20, 2007

### clustro

?
I don't understand

The dimensionless groupings definitely take into account the tube's dimensions...I don't know what you mean.

14. Oct 21, 2007

### Shooting Star

A simple bit of misunderstanding on my part, nothing else. Of course, the tube size was always there in your dimensionless grouping. But it seemed to me that to you it was not very important.

At one point you had written "Obviously if you're tube were infinitely long, things would be different. It doesn't matter much though because I'm assuming you're not going to try and make correlations based upon tube sizes,…”. Well, I thought it did matter.

After that, in your last post, you have considered how the compressed gas may interact with the ball, which I considered to be a more constructive and realistic analysis.

15. Oct 21, 2007

### clustro

As for what you said earlier about there being an optimum barrel length, I may have found it, but I don't know if I did it correctly.

I really wish I had a scanner to put my work in easier (since reading text for math formulas is a bear), but I'll try to keep this as clear possible.

First, I used shooting star's equation for adiabatic expansion of the gas.

Then, I inserted values for V1 and V2.

For V1, I am picturing the ball as resting on the bottom of a cylinder.
V1 = (pi*r^3)/3
For V2, I am picturing the ball halfway out of the pipe.
V2 = (L*pi*r^2) - (2*pi*r^3)/3

Inserting V1 and V2 into the formula gives us the energy transferred to the ball as a function of several variables, one of which is the barrel length, L.

Taking first and second partial derivatives dW/dL and d^2W/dL^2 (those are supposed to be the fancy squiggle-d's) and setting the first equal to zero gives us our extrema. The second derivative is negative when the computed extreme value of L is inserted, proving W to be a maximum at the discovered value of L.

The maximum value of L is therefore discovered to be:

L = [(Pa/K)^(-1/g)](1/pi*r^2) + 2r/3

Where
r is the radius of the tennis ball (approximately 2.86 cm)
Pa is atmospheric pressure (101325 Pascals @ sea level)
g = gamma

I assumed an initial pressure of 150 psig, which is 164 psia, which converts to 1.13 MPa
After calculating K and inserting all other known values yields:

Seems kind of short, so maybe I goofed somewhere, but I don't think I did...
I suppose in practice longer barrels are used for artillery pieces and guns because they give more accuracy.

Anyone else care to verify?

16. Oct 22, 2007

### Shooting Star

Great going!

To clear up some minor points.

V1 is simply the volume of the compression chamber. Nothing to do with where the ball sits. So, why have you given it as (pi*r^3)/3. A factor of 4 is missing, if it's spherical.

V2 should be taken as the volume of the compression chamber plus everything including the barrel upto the point where the ball is ejected, or upto the point where in practice the gas is doing no more work on the ball.

If the height difference between the mouth of the barrel and where the ball sits is h, then work done on the ball is mgh+mv^2/2. Here, the ball is being ejected at some angle to the horizontal.

I suggest for the first approximation, you consider a horizontal barrel. Then mgh=0. We will put in friction later.