Paintball

[xkpxfatassxx]

How many lbs of pressure or force does it take to shoot a .68 paintball approximately 300fps. I was assuming 50lbs because the arrow bow requires a 50lb pull sting, but I can see that it's most likely different. So does anyone know?

[aqua_scummm]

Depends on barrel length, etc.

More force over less distance, or less force over greater distance... What you're looking for is not force/pressure, but energy and work.

I could figure it out, but I just left the engineering dept at my school, so no thanks

[Walking_Target]

and the formula will be for the delta change in velocity, regarding the coefficient of friction of barrel material to gelatin... i have the formula around here somewhere, but i'm to lazy to dig it up.

[cockerpunk]

peak breach pressure is between 60 and 100 PSI.

force depends on accelleration, so the faster you want to get it moving, the larger the force.

[Surestick]

Originally Posted by xkpxfatassxx

How many lbs of pressure or force does it take to shoot a .68 paintball approximately 300fps. I was assuming 50lbs because the arrow bow requires a 50lb pull sting, but I can see that it's most likely different. So does anyone know?

Your question isn't clearly defined enough.
F=ma or force equals mass times acceleration.
Assuming you know the mass of the paintball and the speed you want it to reach to know the force needed to accelerate it to a given speed you need to know the acceleration (otherwise any force starting at the drag on a paintball at 300 fps and up will eventually accelerate the paintball to 300fps).
Without specifying barrel length to determine acceleration your question has no real answer.

The pull strength of the airbow isn't directly related to the force on the ball, the distance the string of the bow moves is just as important (50lbs over 1 inch is significantly less energy than 50lbs over a couple of feet). Contrast the 50lb pull strength of the airbow with the 800+psi input of a marker that operates on unregulated CO2 (say a lower end Spyder) and then contrast that with an Airstar Nova that fires the ball using only 90 psi.

What you really want is the curve of the force the bow string applies as it goes through its travel. Graph that over the distance it travels & the area under the graph gives you the amount of energy needed to accelerate the paintball to 300fps plus the energy needed to overcome friction & other losses in the airbow itself.

[Mayvik]

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:

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



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.

[Surestick]

^^^ Can you do that again in metric for those of us in Canada?

[Mayvik]

Yes!


But I won't

[Old Man]

That made my head hurt!!!!

[cockerpunk]

Originally Posted by Mayvik

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.

your first equation assumes uniform acceleration.

thats not a true in a paintball gun.

what you need to do is find the impulse of the air. bascially impulse simply mean the work done on the ball by the air as a function of time. meaning, the area under the curve of the pressure function.

if TKs pics had an accurate time scale at the bottom, thenw e could easically appoximate it.