You also have to account for drag on the paintball as it accelerates through the air ahead of the ball (adjusted to account for any effects due to blow-by), and friction against the barrel (adjusted for the weight of the paintball due to gravity on the bottom half of the barrel), ball not being a homogeneous sphere (liquid filled with a seam and possibly dents) and any rotation of the ball caused by uneven air impingement (such as from a low-blow bolt), boundary layer effects on the ball, etc etc.
Not a simple question...but we'll just approximate here with some basic Newtonian physics.
Simple Newton Stuff: http://upload.wikimedia.org/math/a/c...beac1fcdff.png
Where vf = final velocity (300fps), vi = initial velocity (0), a = acceleration and d = final location (length of the barrel, since velocity is at the muzzle).
So you've got
(300ft/s)^2 = 0 + 2 * a * (14in = 1.17 ft) = a * 2.33 ft
Gives you a = (90000 ft2/s2 ) / 2.33 ft = 38626 ft/s2
Plug into Newton F=ma, using m of a paintball as 3.2 grams (~.0071 lb), you get
F = 274.2 lb ft/s2
or divide by 32.174 to convert to lbf (
Force Converter)
F = 8.522 lbf
Your .68 paintball has an exposed surface area As = 1/2 * (4*Pi*r^2), r being 0.34 in
As = .726 sq in
Pressure is force over area, or P = F/As and you get a pressure of
P = 11.738 lbf/sq in
So in a perfectly frictionless 14" barrel, perfectly sized to the diameter of the ball, firing into a vacuum without the ball spinning, you'd need around 12psi to get the ball up to 300 fps.
Drag:
Obviously this is much lower than cockerpunk's range above, but you're not firing in an ideal environment, either.
Now the ball isn't immediately going from 0 to 300, so this back pressure from drag is only in effect exactly at the muzzle. From initial firing until that point, the drag is increasing directly with the velocity as the ball accelerates. For simplicity sake and to get an order of magnitude, we'll calculate the drag force on the ball right at the muzzle, when it's moving at 300fps, and assume we need to add roughly that number to the pressure in order to overcome drag as the ball is accelerating.
Coefficient of drag on a sphere is about 0.5 (varies depending on smoothness), and the equation for drag is
http://upload.wikimedia.org/math/4/1...9bdb306ae8.png
Where Fd =s your drag force, rho (little p) is the density of the fluid (air), v is your speed, Cd is about 0.5 as defined above, and A is our As from before (converted to feet at .0605 ft). At 300 fps using a density of air as 0.075 lb/ft3, we get
Fd = -1/2 * .075 lb/ft3 * (300ft/s)^2 * .5 * .0605 ft^2
Fd = -1020.9375 lb ft / s2
or converted again..
Fd = -31.732 lbf over our .726 in2 area gives us a "back pressure" of
43.71 lbf/in2
That's as complicated as I feel like getting without busting out the differential equations and such to account for the other aspects I mentioned above. In conclusion, I would be very surprised if you could get a ball moving with less than 50psi, and probably closer to 75psi.