...the force to a little paintball

[Manuel_FZR]

But:

[bjjb99]

Some of the equations you are using assume constant acceleration. A paintball's acceleration down the barrel is by no means constant. In general, the initial acceleration is quite high and then drops off sharply as the ball travels down the barrel.

The cross-sectional area of a paintball exposed to pressure is simply the area of a disc equal in diameter to the paintball (i.e. pi * r^2) or right around 0.000234 m^2. I'm not quite sure how you arrived at your result of 0.0000962 m^2. In any case, using the pressure formula you provided, I calculate the initial force on the paintball to be right around 97 N.

F = P * A
P = (60 psi / 14.7 psi/bar) * 101325 N/m^2/bar = 4.14x10^5 N/m^2
F = 4.14x10^5 N/m^2 * 0.000234 m^2 = 96.8 N

Going back to the acceleration calculations you provided and assuming a constant acceleration through the barrel (which we know is not really the case), we can use the following:

x = (1/2) * a * t^2
v = a * t
solve for t to get t = v/a and substitute it into the first equation

x = v^2 / (2*a)
solve for a

a = v^2 / (2*x)

Using the numbers you provided (v = 280 fps = 85.3 m/s, x = 15 cm = 0.15 m), we get an acceleration of 2.42x10^4 m/s^2. Applying this acceleration and the mass you listed (m = 3.248 g = 0.003248 kg), I calculate the force to be right around 79 N.

So the average force is around 79 N with an initial force of around 97 N. This implies that the force on the paintball as it exits the barrel must be less than the average force, which physically makes sense. If we assume a linear decrease in pressure as the ball travels down the barrel (again, not completely accurate), then the final pressure would end up around around 38 psi. I recently read a post somewhere on automags.org in which the residual pressure in an automag barrel was stated to be between 25 and 50 psi (someone please correct me if I got those two numbers wrong, as I'm working from memory).

Frictional effects should be quite small with a good paint to barrel match. If you can blow a paintball out of the barrel with little effort, there can't be that much friction holding it in place.

BJJB

[Manuel_FZR]

[bjjb99]

By computing the pressure the way you did, using the surface area of the ball rather than the cross-sectional area, you are determining the "inward" pressure on the ball rather than the "forward" pressure. The pressure used to push the ball forward is the only portion of the pressure vector we're interested in, since the remaining vector components cancel out or are constrained by the barrel itself. Since we want the forward pressure only, you can treat the ball as a piston with a circular flat surface exposed to the pressure.

Thus, for the forward-direction force exerted on the ball by pressure, you should use an area of pi * r^2, where r is the radius of the paintball, measured in meters.

For your question regarding equations:

a = v / t
t = x / (v/2) = 2*x/v

substituting for t in the first equation, we get

a = v / (2*x/v) = v^2 / (2*x)

which is exactly the equation I used, so we're in agreement here. :)

A velocity of 280 fps is 85.3 m/s.
The distance you mentioned was 15 cm, which is 0.15 m
Thus our acceleration is (85.3)^2 / (2*0.15) = 2.42x10^4 m/s^2

F = m * a

The mass you listed was 3.248 g, which is 0.003248 kg

F = 0.003248 * 2.42x10^4 = 78.8 N

I think you may have miscalculated somewhere. Since you're off by a factor of ten, it's most likely a decimal point position error... a mistake I make far too often when entering values into a calculator.

BJJB

[Manuel_FZR]

[lonsch]

where did you guys get 60 psi from. when we dino'ed the mag at the tech class it was 90 psi????

[Manuel_FZR]

Thanks for the information

lonsch: Someone said, that it is 60 psi; but if there are 90psi - thanks for the info! What exactly measured you at the tech class? What pressure is this exactly? The pressure of the bolt to the ball? ...
Have you messured othe guns?

[AGD]

60 psi for the mag and the impulse
95 pis for angel
45 for matrix
110 for autococker

AGD

[Manuel_FZR]

[AGD]

It is the air pressure behind the ball that comes out of the front of the bolt.

AGD

[Manuel_FZR]

Originally posted by AGD
It is the air pressure behind the ball that comes out of the front of the bolt.

AGD

[lonsch]

"But how could this be? They all shoot the same fps (300fps ?)... how could the pressure, witch accelerates the ball be different? "

easy, different volume

Tom at the tech class you showed us graphs of the mag peak pressure and it was at 90 psi. did i miss something? is the peak pressure diierent witht the super bolt?

[Manuel_FZR]

[Wat]

its not really the bolt pressure but the momentum transfer. Pressure is more of a static, continuous force while a collision is more of a dynamic event.

Anyways, you would want to find the momentum of the bolt. Find the bolt velocity at point of contact and its weight to determine momentum.

Everyone says low pressure stop chops. Do they imply that in case of a partial feed the bolt hits the ball and stops or bounces off without chopping? With paint as fragile as it is these days, does this ever really happen? I would have thought high rate of fire = high bolt speed = high momentum = chops regardless how low of a pressure you have behind the bolt.

[kilaueakid]

I think it is bolt speed you are looking for. Different guns have different bolt speeds. Some travel at a higher feet per second than others. I think most blow forward designs are anywhere from 10 to 25 feet per second. Although not proven, it would seem that a slower bolt forward design would be "more gentle" to fragile paint.

I'm sure we have all seen markers that instead of chopping a ball, it just stops the bolt and wedges the ball between the feed neck and face of the bolt. I think it was a cocker I saw this in and possibly pbjosh's shiva he designed. I am not sure how this works, but it may be related to the pressure operating the cocking mechanism, not the pressure used to accelerate the ball down the barrel.

kila