Dude, why's your barrel 2feet long?

[nippinout]

I was trying to figure out on paper what the perfect barrel length would be under ideal conditions (zero friction, perfect seal, 100% efficiency...) to see how that compares to the 8-10" optimal effective barrel length that TK has found. What I found was quite surprising.

Let's say that cylindrical volume behind the ball is VC (Volume cylinder), and the volume between the rear face of the ball and the cylindrical volume VR (Volume rear of ball).

We can find VR by finding the volume of the disc shape bound by the barrel from the most rear point of the paintball up to hemisphere that is formed from that point to the barrel/ball interface- Then subtract the hemispherical volume from that disc volume.

Let's let the bore diameter be .692" -> r = (.692/2)


VR = [disc volume] - [hemisphere volume]

VR = [(pi)(r^2)(r)] - .5[(4/3)(pi)(r^3)]
= 0.04337674501

The cylindrical volume is found the the radius, pi, and the length. We are trying to find that length. By multiplying the cylindrical volume by a pressure, we get a potential energy of that pressurized air.

If we then set that energy to the kinetic energy of a paintball, we can find the length of barrel needed.

But this isn't all the information we need yet. The compressed air in the barrel will accelerate the paintball until the pressure in front of the paintball and behind the paintball are the same. Let's say the pressure in front of the paintball is 14.7psi.

[cylindrical volume][14.7psi] = 111.0052597in-lb

[(pi)(r^2)(L) + VR][14.7psi] = 111.0052597in-lb

(pi)(r^2)(L) + VR = 111.0052597/14.7

L = [(111.0052597/14.7) - VR]/[(pi)(r^2)]

L = 19.96"

19.96" wtf?!?

That's around twice as long as the optimal effective barrel length that TK has found.

Let's take a Viking's efficiency: 1000 shots from a 68ci 3000psi tank. That comes to 204in-lb per shot.

If we subtract the kinetic energy of a paintball from 204in-lb we get 92.995in-lb. (we are neglecting the use of energy in operating pneumatics and again the same ideal assumptions)

The Viking is one of the more efficient guns and close to half of the energy of getting that paintball to 300fps is going into overcoming friction, poor barrel/paint sealing, gas cooling, etc...

I'm wondering how much more efficient can we get?

A paintball does not deform much, so pressure plays no big role in how well it does or doesn't seal with the barrel. I'd like to get my hands on some Perfect Circle balls for some testing. It would be interesting to see how significant sealing has to do with efficiency.

I have a hard time seeing the majority of 90in-lb going to overcoming friction. I'm leaning towards the paintball/barrel seal as being the culprit in the 20" barrel length not adding up.

Or maybe friction is the painball's greatest enemy.

[athomas]

You are calculating the value as if the perfect length is when the internal pressure is 14.7psi which is atmospheric pressure. In fact, the measured starting pressure is equal to the (measured value + 14.7psi) the 14.7psi in the air and in the compressed source cancel each other out. In an ideal situation with no friction or external forces, the air pressure will always be more than atmospheric pressure out to infinity, so there will be some degree of acceleration out to infinity. Of course that won't happen but lets speculate on what would kind of acceleration would happen if it was true.

A paint ball mass is about .116oz
given a barrel id of .692"
Ideal efficiency = 111in-lb of air usage to propel a ball


So:
Chamber pressure > chamber size > barrel length

@30psi - air chamber = 3.70ci - barrel = 19.7in
@40psi - air chamber = 2.78ci - barrel = 14.8in
@50psi - air chamber = 2.22ci - barrel = 11.8in
@60psi - air chamber = 1.85ci - barrel = 9.8in
@70psi - air chamber = 1.59ci - barrel = 8.4in
@80psi - air chamber = 1.39ci - barrel = 7.4in
@90psi - air chamber = 1.23ci - barrel = 6.6in
@100psi - air chamber = 1.11ci - barrel = 5.9in
@110psi - air chamber = 1.01ci - barrel = 5.4in
@120psi - air chamber = 0.93ci - barrel = 4.9in
@130psi - air chamber = 0.85ci - barrel = 4.5in
@140psi - air chamber = 0.79ci - barrel = 4.2in
@150psi - air chamber = 0.74ci - barrel = 3.9in
@160psi - air chamber = 0.69ci - barrel = 3.7in
@170psi - air chamber = 0.65ci - barrel = 3.5in
@180psi - air chamber = 0.62ci - barrel = 3.3in
@190psi - air chamber = 0.58ci - barrel = 3.1in

The calculations were done using no friction and all energies were ideal. The air source in the chamber was instantly available so there were no losses due to flow restrictions.

Hopefully I haven't made any errors that will throw the data way off.

[bjjb99]

I like your treatment of barrel lengths from an energy standpoint, but I am having trouble with the following statement:

Originally posted by athomas
In an ideal situation with no friction or external forces, the air pressure will always be more than atmospheric pressure out to infinity, so there will be some degree of acceleration out to infinity.

We can treat the ideal case as a frictionless, airtight piston, behind which we release a fixed volume of gas at some elevated pressure. Since we're talking ideal cases here, let's assume we've got an ideal gas, too. At this point, you can apply the a standard relationship for ideal gases:

p1 * v1 = p2 * v2

This is also where the in-lb values come from, in an ideal gas, energy is conserved.

Let's set our initial absolute pressure (p1) at some value P+A, where A is the atmospheric pressure; I suppose this would make P the gauge pressure, which is what folks usually refer to when they're talking pressure. Note that you can't use gauge pressures directly in the above equation and have the results come out meaningful... only absolute pressures are suitable. I suspect this may be where your conclusion came from about always having greater than atmospheric pressure inside the barrel, regardless of length.

There is some final volume (v2) for which the final absolute pressure (p2) is less than the atmospheric pressure (A). If I did my algebra right, I believe the volume at which the internal pressure equals atmospheric pressure is given by:

v2 = (P+A)*v1 / A

At this point, our "piston" will begin to decelerate. You won't accelerate out to infinity.

BJJB

[argnot]

My brain just exploded.

[athomas]

bjjb99, you are correct. Once I thought about it, the ball will eventually pass a point where the volume behind the ball contains pressure less than atmospheric. This would act like a vacuum and the ball would start to decelerate at this point.

[bjjb99]

Originally posted by argnot
My brain just exploded.

Well, they don't call this forum "Deep Blue" for nuthin'.

BJJB

[Brophog]

When it comes to actual functionality, the point seems to be that the performace plus or minus is generally so small, considering the pressure deviation and ball consistency, that there is no "perfect length".

On paper you may find one that measures out slightly better, but in application does not perform as it "should".

Its like shooting air through a really big hole. You can screw around with the pressure all day long, even going to relative extremes, but due to the size of the hole, the velocity of the object is gonna be relatively unchanged.

[RealBoy]

Originally posted by nippinout

The cylindrical volume is found the the radius, pi, and the length. We are trying to find that length. By multiplying the cylindrical volume by a pressure, we get a potential energy of that pressurized air.

If we then set that energy to the kinetic energy of a paintball, we can find the length of barrel needed.

But this isn't all the information we need yet. The compressed air in the barrel will accelerate the paintball until the pressure in front of the paintball and behind the paintball are the same. Let's say the pressure in front of the paintball is 14.7psi.


Or maybe friction is the painball's greatest enemy.

Where did you get 14.7 psi from. I don't quite get it.

[athomas]

14.7 psi is atmospheric pressure at sea level.

[hitech]

It would be interesting to see how significant sealing has to do with efficiency.

Just use a small bore barrel. The smaller the bore the better the seal. Yes, it will have increased friction, but I think you are going to find that the increased seal far out weighs the increased friction. It's easy enough to test.

[phyregod]

Unfortunately, the world we live in has friction, and balls which, by suggestion only, are round. Several real world tests have been conducted, and it seems to be the finding that if your barrel is 10 inches or above, there is no difference in accuracy. unless, of course, you tried to fire paint through a ten foot barrel.. but this is the real world, right? The barrels which are 8 inches and below are less accurate.

As far as efficiency, I have a low pressure modified Spyder E-99. It needs a barrel with 8 inches of non-ported area to perform correctly. I use a 14" barrel that has 8" of solid barrel, 6 inches of ported barrel. any less and my velocity drops.. any more (like 10" or 12" of solid barrel and it stays the same. Just food for thought.

Many swear by the longer barrels being more accurate, but I believe that this can be attributed to the length of the barrel making it easier to aim. It is much easier to aim accurately with a rifle vs. a pistol, because the sights are so far apart.

and i do know that rifles are more accurate than pistols, that was just an example...