Calculating Force of a Small Potato Cannon

[jameslat]

Hello,

I built a small potato cannon out of PVC pipe/steel. It's 15.5 cm long, and about 7.6 cm wide on the inside. It's made of 2 parts the barrel and combustion chamber (or as i refer to, "the death chamber"). the "death chamber" is cylinder about 7.6 cm long and 7.6 cm thick.
and the barrel is about 7.6 cm long but it's only about 2.5 cm thick so as to allow a great deal of pressure to be focused on the projectile. I use hairspray as my fuel.

Is there a maximum power point in which after a certain amount of hairspray, it won't matter how much more is added because it will still give the same results?

(assuming I know the Velocity initial when it comes out of the barrel) How can I calculate the force that the the fuel is causing?
(would it be a momentum equation like p=(mass of projectile)*velocity , but how would i incorporate the length of the barrel?)



Please help put me on the right track and,
Thanks so much for your time,

-James

 

[Stephen Tashi]

Hello,

I built a small potato cannon out of PVC pipe/steel. It's 15.5 cm long, and about 7.6 cm wide on the inside. It's made of 2 parts the barrel and combustion chamber (or as i refer to, "the death chamber"). the "death chamber" is cylinder about 7.6 cm long and 7.6 cm thick.
and the barrel is about 7.6 cm long but it's only about 2.5 cm thick so as to allow a great deal of pressure to be focused on the projectile. I use hairspray as my fuel.

You'll shoot your eye out, kid.

Is there a maximum power point in which after a certain amount of hairspray, it won't matter how much more is added because it will still give the same results?

I'm not a combustion chemist, but I would guess yes since there is only so much oxygen in there.

(assuming I know the Velocity initial when it comes out of the barrel) How can I calculate the force that the the fuel is causing?
(would it be a momentum equation like p=(mass of projectile)*velocity , but how would i incorporate the length of the barrel?)

Force must vary as potato goes down the barrel. So it isn't really correct to say "the force". If you can write force as a changing function of time or distance, calculus can be used. Otherwise you can talk about the "average force". Then question arises "average over what? distance or time?". I think (Force)(Time) = (Mass)(Velocity) would apply if Force was the average force over time. However, you're more likely to know the distance down the barrel that the time it took the potato to travel that distance.

So you might use Work = (Force)(Distance) = Kinetic Energy = 1/2 (Mass)(Velocity squared). The drawback is that energy equations are less reliable than momentum equations since energy gets lost as friction, heat etc.