Air Cannon Ballistics

[bout0046]

For the longest time I have been trying to figure out how far a projectile will go after being launched from an air cannon. I'm trying to figure out how big of an air chamber I will need to build to shoot a certain object so far.

 

I was looking around and found someone saying that PV=1/2mv^2 gives a good estimate, but I wasn't too sure and confused about when solving for v, you're only left with a length(I think).

 

So I am wondering if someone could help me figure out either what the velocity of an object would be as it's leaving the barrel, or how far it could go. The pressure of the air chamber, volume of air chamber, mass of object, area of object, length of barrel, will all be known. I'm not looking for anything super accurate, so anything like wind, friction, etc. can be ignored.

 

I have been trying to do this for a while now and can't seem to come up with anything. Years of physics and I can't think about what to do, haha. I don't know if maybe I'm over thinking or it's harder than it looks. Any help will be greatly appreciated. Thanks

[DrP]

If you can work out the force on the object from the air pressure then you are away. You can use F=Ma to get accelaration. Then use the standard equations of motion to work out velocities, distances etc.. Type "equations of motion" into google.

[swansont]

I was looking around and found someone saying that PV=1/2mv^2 gives a good estimate, but I wasn't too sure and confused about when solving for v, you're only left with a length(I think).

 

That will give you a reasonable upper bound as long as the temperature doesn't change much. It ignores drag, so the actual numbers you achieve will be smaller. If you solve for v you should end up with a speed — if you don't you have made a mistake. Then you assume a launch angle of 45º and apply kinematics.

[John Cuthber]

There are two different things in that equation symbolised by "v" and "V" which probably isn't helping anyone.

[swansont]

There are two different things in that equation symbolised by "v" and "V" which probably isn't helping anyone.

 

Ah, yes. I was assuming it was recognized that they are different terms: on the left, V is Volume and on the right, v is speed

[bout0046]

So by using PV=1/2mv^2 where v is velocity, and V is volume of air chamber, P is pressure of air chamber, and m is mass, I can get the initial velocity of the object as it's leaving the barrel? And from this I can just use equations of motion?

 

So solving for v, I get v= Sqrt(2PV/m), but this leaves units in a length I think. But this doesn't make sense to me.

[swansont]

So by using PV=1/2mv^2 where v is velocity, and V is volume of air chamber, P is pressure of air chamber, and m is mass, I can get the initial velocity of the object as it's leaving the barrel? And from this I can just use equations of motion?

 

So solving for v, I get v= Sqrt(2PV/m), but this leaves units in a length I think. But this doesn't make sense to me.

 

 

P is N/m^2, V is m^3, m is kg. N is kg-m/s^2

 

[math]\frac{kg*m*m^3}{m^2*s^2} = \frac{m^2}{s^2}[/math]

 

so you get m/s, just as you'd expect

[bout0046]

Oh ok, that makes sense. I was trying to do it in English units; I was being lazy . I think I understand it now. Thanks everyone for your help.

[dirtyamerica]

What if you aimed the canon straight up, shot it and timed how long the projectile is in the air? Divide that number (T) by two so that represents the object travelling from the cannon to the top of its travel OR from the top to the ground (t). The object velocity out of the cannon OR velocity when it hits the ground should be nearly the same. Then v at m/s =1/2att.....

 

or is it v = at ?

 

I'm having a brainfart at 5 in the morning..

Edited June 4, 2008 by dirtyamerica

[swansont]

That works if there's no air resistance. If there is, the trip down can take longer.

[John Cuthber]

There are other complications. The molecules in air are traveling at about the speed of sound (on average). Once the projectile is traveling near the speed of sound very few of the air molecules "pushing" it can catch up with it.. Since they don't reach it they can't push - the force drops and so does the acceleration.

That puts a limit on the top speed.

hot light gases work better but there's still a limit.