Air cannon mathematics

[bishnu]

Im not posting this in the physics forum because i feel its more of a amth question. My friends and i have bulit an air cannon and i wish to analysis it mathematically...after a couple of hours i came up with an equation i think would do. I was wondering if people could look at my work and see of they see anything wrong with it.

 

First off my aircannon has a simple chamber then a valve and then a barrel. You pump air in to the camber and then let relase the valve and then the projectile is shot(ps i have never used LaTex before so we will see how this goes)

first off i started with the equation

[math] p=\frac{F}{A}[/math]

In this problem i am going to assume the barrel fits perfectly in the barrel so there is no room inbetween the barrel and projectile.

[math]pA_b=F[/math]

[math] p_iv_i=p_fv_f[/math]

[math]\frac{p_i*v_i}{v^i+A_b*x}[/math]ab= area of the barrel and x=distance traveled in the barrel

[math]f=\frac{p_iv_iaf_b}{v_i+a_bx}[/math]

this equation tells the force on the projectile at anytime druring the barrel

friction=kv^2 k=constant of proptionality

[math]f=\frac{p_iv_ia_b}{v_i+a_bx}-kv^2[/math]

now intergrate both sides from 0 to the length of the barrel xb to get the energy

now im getting lazy of typing in all that stuff so ill just tell you what i did from here on i substituted 1/2mv^2 for energy now i just sepertated the v and keep the v on the other side in the friction equation. Now i just subsitituted this eqaution into the v in the friction part and did that infintly many times which gives you 2 infinite series they converege and i did a little algebra to reduce it and i got

[math]v=(\frac{2}{m_p+2kx_b}(p_iv_i\ln({\frac{a_bx_b}{v_i}+1})+\frac{2k^2x^2_b}{m_p}))^(1/2)[/math]

i want to know if anyone can spot any errors

[Dave]

Surely an easier way to do this would be to estimate the time the force of the air acts on the projectile after you release the valve (i.e. the order of milliseconds) and then you know that:

 

Ft = m(v-u) = mv

 

(Impulse = change in linear momentum - that's ignoring gravity, which doesn't really matter).

[bishnu]

I would but i want to take in to account that the force acting on the projectile isnt constant and i want my equation to use pressure and i also want friction

[bishnu]

i know it is wrong becasue it doesnt output corrcect numbers

[Dave]

You need to take some values for the friction then.

[Dave]

Okay, this is what I've come up with.

 

Let's assume we have a perfectly cylindrical barrel with our rather snug fitting spherical slug sitting in it, of internal radius r (we don't really care about the external radius). Now lets assume we're going to start at some x position, so we might as well say 0 and the end of the barrel is x_f.

 

We also know the force exerted by the pressure of the air over this distance is going to be constant, so let's call that [math]F_p[/math]. You've said your resistive force is [math]kx^2[/math] - fair enough.

 

So we'll just ignore gravity and have it lying flat for the time being.

 

You can set up an equation by looking at Newton's second law:

 

[math]m\ddot{x} = F_p - kx^2[/math]

 

This is fairly simple to solve as long as you know that [math]\ddot{x} = v\frac{dv}{dx}[/math].

 

Then you just integrate both sides, noticing that your slug starts at rest. Also, to work out [math]F_p[/math], assuming your slug is spherical, you'll have to use half of the surface area of the sphere as your value for A. You need to find out your value of p also.

 

There's also a fairly good chance I don't know what I'm talking about and/or have made a mistake since I'm very tired. I've also just noticed that the pressure in the tube obviously isn't constant, so you can ignore most of this :| Hope it helps a bit.

[Guest SniperMax]

Hello.

 

I have also made air cannons in the past.

 

I've worked out an equation for s(t), where s ist the distance travelled down the barrel, and t is the time. You can take its derivatives for velocity and acceleration.

 

And that's what I did:

 

[math]p0*V1=p1 (V1 + A*s(t))[/math]

 

that's beasically the same thing you have done, so you were going in the right direction. p0 is the pressure in the reservoir before pulling the trigger. V1 is the reservoir's volume, and p1 is the muzzle pressure. A is the cross sectional area of the barrel.s ist the distance as a function of time.

 

The pressure at any point s is described by the following formula

 

[math]p=p0 + \frac{\frac{p0*V1}{V1 + A*s1} - p0}{s1}s(t)[/math]

 

,where s1 ist the length of the barrel, also a constant.

 

using [math]a=\frac{F}{m}[/math] and [math]F=pA[/math], we get

 

[math]a=(p0 + \frac{\frac{p0*V1}{V1 + A*s1} - p0}{s1}s(t)) \frac{A}{m}[/math]

 

,where m is the projectile mass.

 

You can set up the differential equation and solve for s(t).

The initial conditions are s'(0)=s(0)=0.

 

The solution:

 

[math]s(t)=-\frac{(A s1 + v1) (-1 + Cosh(\frac{A \sqrt{p0} t}{\sqrt{-m (A s1 + V1)}}))}{A}[/math]

 

You will find out that friction doesn't slow down the slug that much. That is because your barrel is most likely smooth, as opposed to a rifled one.

 

This calculation was pretty accurate in my experiments as long as the seal between projectile and barrel was perfect. I used steel darts with film cones for this. The resistance you feel when trying to push the slug down the barrel is just what makes the pressure behind it possible, and thus it doesn't slow down the projectile.

 

You can prove it mathematically if you like. The frictional force acts in the opposite direction of the pressure that comes from the air reservoir, thus increasing air pressure behind the projectile. The increased pressure yields a higher force, and it would mean a higher acceleration, but there's the friction again, slowing it down. In the end, frictional effects are cancelled out. At least that's the conclusion I've come to after measuring the muzzle velocity several times with varying amounts of grease and comparing with the formula. The values fit quite well together, with an average error of about 5m/s (off the top of my head).

 

See if it contradicts your own experiments.

 

I hope this helps, otherwise feel free to ask.