Air Cannon Physics: Projectile Motion Analysis

[carguy454]

I am a high school physics student and am building an air cannon to study projectile motion for our end-of-year project. I am trying to figure out what air pressure I need to propel a 1 lb projectile 100 ft/s. Here's what I have so far. Let me know if I'm right or wrong. Thanks.

Work=(1/2)(m)(v^2)
W=(1/2)(1)(10,000)
W= 5,000 ft-lbs

If I use a 5 foot barrel with a three inch diameter then
Force=1,000 lbf
Area=28.26 in^2
Pressure~36 psi

 

[PhanthomJay]

carguy454, welcome to PF!

I am a high school physics student and am building an air cannon to study projectile motion for our end-of-year project. I am trying to figure out what air pressure I need to propel a 1 lb projectile 100 ft/s. Here's what I have so far. Let me know if I'm right or wrong. Thanks.

Work=(1/2)(m)(v^2)
W=(1/2)(1)(10,000)
W= 5,000 ft-lbs

If I use a 5 foot barrel with a three inch diameter then
Force=1,000 lbf
Area=28.26 in^2
Pressure~36 psi

If the projectile weighs 1 pound, it's mass is not 1. Convert the weight (in pounds) to its mass (in slugs) using the appropriate formula and units. Also, the area you calculated used a radius of 3 inches, not a diameter of 3 inches.

 

[carguy454]

Ok.

W=(1/2)(0.3)(10,000)
W=1500 lbf * s^2

Assuming 5 ft barrel w/ 3 in diameter
F=300 lbf*s^2
A=7.065 in^2
P~43 psi

Is this right?

 

[PhanthomJay]

Ok.

W=(1/2)(0.3)(10,000)
W=1500 lbf * s^2

Assuming 5 ft barrel w/ 3 in diameter
F=300 lbf*s^2
A=7.065 in^2
P~43 psi

Is this right?

Ok.

W=(1/2)(0.3)(10,000))

the mass works out to 1/g =1/32 = about 0.03 slugs. Looks like you slipped a decimal point.

W=1500 lbf * s^2

Due to your decimal point error, the Work done is 150 ft-lbs. Your units are way off, work has the same units as energy (ft*lbs).

Assuming 5 ft barrel w/ 3 in diameter
F=300 lbf*s^2

that's 30 lbs. And watch your units!

A=7.065 in^2
P~43 psi

Is this right?

4.3 psi, which is the average pressure required to be exerted on the projectile, which assumes the projectile diameter is 3 inches. Note the 30 pound force required to accelerate the projectile to a speed of 100fps as it leaves the cannon, is an average force.

 

[carguy454]

Oops. Those decimals will get you get every time.

That seems like an awfully low pressure though. You're sure I'm supposed to use slugs instead of lbs?

 

[Topher925]

You're sure I'm supposed to use slugs instead of lbs?

Yep, Jay has it right. Remember that 4.3psi is the AVERAGE pressure along the 5ft tube and not the actual pressure of your air source. If you want to know what pressure to fill your air tank, then we will need more information.

 

[PhanthomJay]

Oops. Those decimals will get you get every time.

That seems like an awfully low pressure though. You're sure I'm supposed to use slugs instead of lbs?

In the SI system, a Newton is defined as the force that will accelerate a 1 kg mass at a rate of 1m/s^2, per Newton 2. In the Imperial system, used almost exclusively in the USA, a Slug is defined as the unit of mass that will accelerate at a rate of 1ft/s^2 when a force of 1 lb is applied to it. It follows from weight = mass time acceleration of gravity, that a 1 kg mass weighs 9.8 N, and a 1 slug mass weighs 32.17 lbs. Consequently, a 0.031 slug mass weighs 1 lb. But Topher925 makes an excellent point; while an average force of 30 lbs, or average 4.3 psi pressure acting on a 3 inch diameter flat faced projectile, is required to accelerate it to a velocity of 100ft/sec at the end of the 5 foot tube, your question as to what air pressure is required to produce that force, remains unanswered.

 

[carguy454]

What other information would I need to find the actual air pressure from the average air pressure?

 

[Dr.D]

There are a couple of problems that have not come up yet in this discussion as I see it.
1) You have to maintain this average air pressure throughout the full length of the stroke in order to accelerate the projectile to the launch speed you have calculated. This can be a problem in many cases if your air supply us unable to deliver sufficient air to maintain pressure behind the projectile as the chamber volume grows. Remember that you have assumed a constant air pressure back there, which assumes that there is enough air available at pressure to sustain this condition; this is often difficult to achieve without an explosive charge.
2) Maintaining the cylinder air pressure is highly dependent on the leakage around the projectile. If the projectile is a tight fit in the bore, then there is friction to be taken into account which was not included in the energy based calculation for the required pressure.
Overall, I think you can take the result you have at this point as nothing much more than a preliminary, low-side, estimate of the lowest possible air pressure that will do anything at all. For a better estimate, you will need a better, more comprehensive model of the situation.

 

[carguy454]

There are a couple of problems that have not come up yet in this discussion as I see it.
1) You have to maintain this average air pressure throughout the full length of the stroke in order to accelerate the projectile to the launch speed you have calculated. This can be a problem in many cases if your air supply us unable to deliver sufficient air to maintain pressure behind the projectile as the chamber volume grows. Remember that you have assumed a constant air pressure back there, which assumes that there is enough air available at pressure to sustain this condition; this is often difficult to achieve without an explosive charge.
2) Maintaining the cylinder air pressure is highly dependent on the leakage around the projectile. If the projectile is a tight fit in the bore, then there is friction to be taken into account which was not included in the energy based calculation for the required pressure.
Overall, I think you can take the result you have at this point as nothing much more than a preliminary, low-side, estimate of the lowest possible air pressure that will do anything at all. For a better estimate, you will need a better, more comprehensive model of the situation.

1. My air supply will be able to handle it
2. I'm trying to get a rough estimate of how much pressure is needed, not accounting for friction.

 

[Dr.D]

It sounds like you have what you asked for then.

 

[Dr.D]

You do have to be concerned about the fact that you must get this volume of air INTO the barrel, so you need to assure that you do not get into difficulties related to choked flow in your fittings that connect the air supply to the barrel. This may not (or may) be a problem, but it is something to look into before you commit to your design.

 

[nvn]

carguy454: Work is integral of force times incremental distance. Also, from Boyle's law, neglecting air leakage, p1*V1 = p(x)*V2, where p1 = initial air pressure in pressure reservoir tank, V1 = reservoir volume (not including barrel volume), p(x) = pressure in system when projectile is at location x, and V2 = system instantaneous new volume. Therefore, p(x) = p1*V1/(V1 + A2*x), where A2 = uniform barrel cross-sectional area. If I worked the integral correctly, solving for p1, and neglecting friction and head losses, gives

p1 = [0.5*m*v2^2 + m*g*L*sin(theta)]/{V1*ln[1 + (A2*L/V1)]},

where m = projectile mass, v2 = exit velocity, L = barrel length, and theta = barrel slope angle (measured from horizontal plane). E.g., if m = 0.453 59 kg, v2 = 30.480 m/s, A = 0.004 560 4 m^2, L = 1.5240 m, theta = 0 deg, and V1 = 0.008 000 m^3, then p1 = 42.121 kPa. Using these same example values except changing V1, notice as V1 approaches infinity, p1 approaches 30.316 kPa (4.397 psi).

 

[Dr.D]

I still think that this is all rather pointless. If friction is so small as to be ignored, then the blowby (leakage) will be substantial. If the blowby is negligible, then the friction is significant and the motion will be significantly affected by the friction. You cannot make realistic calculations ignoring both of them. Which way is it?

 

[Topher925]

I still think that this is all rather pointless. If friction is so small as to be ignored, then the blowby (leakage) will be substantial. If the blowby is negligible, then the friction is significant and the motion will be significantly affected by the friction. You cannot make realistic calculations ignoring both of them. Which way is it?

I concur. If this is just a homework assignment your turning in on paper then who cares. But if your actually trying to apply a model to a device you will build then friction or blow by must be taken into consideration.

 

[Dr.D]

The opening statement was, "I am a high school physics student and am building an air cannon to study projectile motion for our end-of-year project." I interpret that to mean that these are supposed to be design calculations, in which case all that has been done thus far is a waste of time.

There is just no point to spending time on calculations that have no relation to reality. In point of fact, both friction and blowby will be present in the real case, so both should be included in a realistic model. There was surprise expressed at the low air pressure required, but that really is only another indicator of the fact that the model is entirely unrealistic and the results meaningless.

 

[nvn]

One person on internet who has experimented with air cannons claims you don't get much blow by if you use a film cone on the projectile (?). He didn't specify how the film cone is made, so use your imagination. The frictional force of the film cone would be mu*p(x)*As, which would be subtracted from p(x)*A2 for the integration, giving work W = integral[(A2 - mu*As)*p(x)*dx], where mu = kinetic coefficient of friction, and As = film cone surface area in contact with barrel. You could also subtract from p(x) the pressure head loss due to sudden contractions, sudden expansions, and/or valves by assuming the air velocity is roughly equal to the projectile velocity as a function of x.

 

[minger]

I'm not 100% sure about the conical "film", but think about a dart gun. The darts have the feathers on the back, which helps seal the projectile with the barrel. What I assume the author was referring to is a flexible membrane which can seal against the side of the barrel while not exerting much force on it.