Pneumatic Golf Ball Cannon Distance HELP

[skuzzy21]

Alright, So for my Grade 11 class our project was to design and build something that shows scientific characteristics. I decided to make a golfball cannon

It uses a regulator with a schraeder valve to fill with air. In one shot we used 80 Psi to shoot a golfball with a diameter of 1.68 inches. Our chamber uses 2'' ABS pipe, which I calculated to have a volume of approximatly 0.618 L per sq foot. Our pressure chamber is 5' long in total, where it leads to a quick release valve. We use very tight wadding for our golfball so it fits the barrel, which has an inside diameter of 1.7'', very nicely. We shot on a 45 degree angle, and our golfball has a mass of 1.620 ounces (45.93 g)

Is there any way someone could help me with this, perhaps there is an equation that would be used to find projectile distance or atleast something that I could show, like velocity or acceleration when shot. Any other variables copuld be estimated, or ask me and I may be able to tell you.

youtube.com /watch?v=ylbWorbZeOM

 

[mfb]

Maybe you can re-use some equations from the water rocket thread. Instead of water, you have a solid golf ball, which has a single exit velocity, therefore you don't have the issue with differential equations.
With the exit velocity v, you can calculate the horizontal and vertical velocity (with your angle of 45°). Neglecting air resistance, the horizontal component will stay constant and the vertical component feels an acceleration of g=9.81m/s^2 downwards. This can be used to calculate the distance of the golf ball.

 

[sophiecentaur]

If your compressed air reservoir is pretty large compared with the volume of the cannon barrel, you can assume the pressure is constant during the launch. You can then work out the energy (kinetic) imparted to the ball by Volume times Pressure. That will give you a good idea of the speed of the thing. If the reservoir is not large, you could still work out the average pressure as the ball travels along the barrel and then get a (slightly lower) value for its final K.E..
This assumes (correctly) that the mass of air moving along the barrel is negligible.

You can do all those sums yourself, I'm sure.

 

[Bob S]

Trajectoware (at http://www.trajectoware.com/) is a program to calculate the range of a dimpled golf ball with backspin, once the intital conditions (velocity, launch angle, backspin) are known. It is surprising how different the trajectory is than a naive parabolic trajectory with quadratic (turbulent) drag.

 

[haruspex]

If your compressed air reservoir is pretty large compared with the volume of the cannon barrel, you can assume the pressure is constant during the launch. You can then work out the energy (kinetic) imparted to the ball by Volume times Pressure.

I don't think that works. If the reservoir has a much larger volume than the barrel then the pressure will still be quite high when the ball exits the barrel. That will waste a lot of the energy.
Assuming the barrel volume is significant, need to assume adiabatic expansion. Let
V_r = reservoir volume
V_b = barrel volume
V_f = V_b+V_r
P_o = initial pressure
P_a = atmospheric pressure
γ = heat capacity ratio, C_p/C_v = 1.4 for air
The useful work done on the ball is
P_o(V_r-V_r^γV_f^1-γ)/(γ-1) - P_a(V_f-V_r)
This is maximised wrt V_b at V_f = V_r(P_o/P_a)^1/γ

2'' ABS pipe, which I calculated to have a volume of approximatly 0.618 L per sq foot

Per linear foot.

I assume the 80psi is excess over atmospheric, so in MKS units we have:
P_o = 80*6895+100000 = 651600 Pascals
P_a = 100000 Pascals
V_r = .0031 m³
Optimum V_f = .0031(6.516)^1/γ = 11.4 litres
making Vb = 8.3 litres.
That's over 13 feet long at 2" internal diameter. But there will be losses, so a somewhat shorter pipe would be better.
Backspin would certainly help. Anyone have suggestions for how to impart that?

 

[sophiecentaur]

I don't think that works. If the reservoir has a much larger volume than the barrel then the pressure will still be quite high when the ball exits the barrel. That will waste a lot of the energy.

Wasting energy doesn't count in my calculation. If that bothers you then you would just shut off the valve (which has to be there, in any case) when the ball has emerged. Either way, you waste the energy of the gas that's in the barrel at the end. I am a great believer in ball-park figures and my simple, initial calculation tells you something about the general situation. If you want to include the drop in pressure in a real system then that's fine but that's another layer of complication and would require more detailed knowledge of the design. My main point was that it's easy to get preliminary results on the basis of Energy, rather than thinking of Forces and acceleration etc.

 

[haruspex]

Wasting energy doesn't count in my calculation. If that bothers you then you would just shut off the valve (which has to be there, in any case) when the ball has emerged. Either way, you waste the energy of the gas that's in the barrel at the end. I am a great believer in ball-park figures and my simple, initial calculation tells you something about the general situation.

Let me be clearer: in your pressure * volume calculation, what is the volume?
I believe it is the volume swept out by the ball as it travels up the barrel.
So a very short barrel means very little velocity.

 

[sophiecentaur]

Yes. The swept volume of the 'piston' times the pressure, is the work done. Having a large reservoir will just ensure that the pressure is almost constant over the whole of the event. A short barrel will reduce the energy delivered, whatever.

 

[rcgldr]

A word of caution, I'm not sure about 80 psi, but from my college days, I recall a high pressure golf ball cannon with a range of about 1/2 mile.

 

[sophiecentaur]

Just a thought. If this could be classed as an 'airgun' then the UK regs restrict the energy to 12 lb ft (16J), I believe. This is all a bit more powerful than that!