Predicting force and velocity of a pneumatic projectile

1. Jan 15, 2006

Crucible Guardian

Here's a problem I've been working on for a while, and have made some breakthroughs on, but have recently gotten stuck. It stemmed out of my intrest in pneumatic cannons. Consider a pneumatic cannon with a known inner surface area, a known barrel lenght, and with a projectile with a known mass. This cannon is pressurized to a known pressure. How can you predict the velocity of the projectile at any given moment?
Here's what I've got so far:

let P equal the original pressure.
Since pressure is equal to a force over an area, the force (F) acting on any given area is P*A.

The inside surface area of the pneumatic cannon changes as a function of x, which is the distance travelled down the barrel by the projectile. This fuction will be called A(x)

A(x) is equal to the original chamber surface area plus the ammount of surface area 'revealed' by the distance x down the barrel by the projectile. The barrel is assumed to be cylindrical.

Let (Ao) equal the original chamber area
Let (Rb) equal the radius of the barrel

A(x)=(Ao+2*pi*Rb*x)

Therefore

F=P*(Ao+2*pi*Rb*x)

Since force is equal to mass times acceleration, acceleration is equal to force over mass

Let M equal the mass of the projectile
Let a equal the acceleration acting on the projectile

(P*(Ao+2*pi*Rb*x))/M=a
Acceleration is equal to the second-order derivative of x (the change in the change in distance with respect to time with respect to time)
so

Let t equal time

d2 x/d2 t =(P*(Ao+2*pi*Rb*x))/M
Since this will have to be integrated with respect to time, all x-values must be on the side of the derivative

multiply by M and divide by P
(d2 x/d2 t)*(M/P)=Ao+2*pi*Rb*x

subtract 2*pi*Rb*x and take each side to the e power. Using exponent rules, the left side can be broken up into a product of two exponents
(e^((d2 x/d2 t)*(M/P)))*(e^(-2*pi*Rb*x))=e^(Ao)