How long is the ball in the barrel when shot?

[SSMercury]

Originally posted by EchoX
Take note that the ball has to accelerate down the barrel, making the ball slow at hte beginning. Now calculus would have to play a part in this math.

Heh. Might make an effort...

Ok, made me a "pretty" (sarcasm) graph using Windows Paint to show ball slowing down, near the end, if it's loosed by a single puff of air, and forgot to add the graph of a slower, longer acceleration.

[athomas]

Originally posted by SlartyBartFast


Not so. Under discussion is the impossible scenario of two balls being in the barrel simultaneously. The second ball will have air pushing it for the whole length of the barrel but the first only up to the point at which the second supposedly enters the barrel.
The second ball then experiences more accelration and would hit the first.

Not quite. If the two balls were in the barrel at the same time and the first was moving on gained momentum while the following ball was accelerating, then the pressure built up between the balls as the 2nd approached the first would slow the 2nd and add additional acceleration to the 1st. Plus if the possibility that two balls could be in the barrel at the same time, then the bolt must reset itself while the ball is in the barrel. This would cause blowback and remove pressure from behind the traveling ball. This effect would be the same for each ball fired, not just the first ball in a two shot sequence.

[hitech]

Originally posted by J_Hoyt
How about people start putting down the math they're doing instead of just the answers they're getting.

Vi = Initial velocity of ball
Vf = Final velocity of ball
s = displacement of ball (as it applies here, the length of the barrel)
a = Acceleration of ball
t = time the ball takes to reach the end of the barrel

16" barrel = 16/12 feet = 1.333 feet = S value...
Ball reaches end of 16" barrel in .009 sec or 9 milliseconds.

Here is what Tom said in his tech tip:

Barrel Efficiency, Tech Tip #1

...The balls acceleration rate is approx. 50,000 feet per second to get to 300 feet per second in 10 inches. The entire barrel travel time is about 6 thousandths of a second... Although this may sound incredible if someone out there would like to do the math you will see that I'm close.

It seems that your numbers match Tom's pretty close. What to try it for a 10 inch barrel?

[petefol]

the acceleration in a 10" barrel would be about 54000fps if the velocity was 300fps, and the change in time would be .005 sec.
the calculations i did were the same as j_hoyt's but i used .8333ft for the displacement (10in/12ft). I'm guessing tom probably did his calculations with a different velocity, or just took an educated guess.

[J_Hoyt]

Yes, I think Tom either used an 11" barrel for his calculations or he used 270-280 FPS for his velocity. Either would have given him about 50 000 FPS/Second.

[Hoplon]

I've had multiple balls in my barrel lots of times.











It's called double feeding.

[hitech]

Although this may sound incredible if someone out there would like to do the math you will see that I'm close.

So, it would appear that this is in fact correct. They did the math and Tom is close.

[athomas]

The acceleration curve of a paintball indicates that it achieves most of its velocity in the first 3 - 4 inches of the barrel. It takes about 3ms to reach about 80% of its final velocity. The remaining travel speed works out to approximately 3.5inches per millisecond. This is the ball gradually reaching final velocity and traveling near final velocity. There isn't much additional increase in velocity because the friction and air pressure pushing against the ball prevents any great amount of additional acceleration.

The findings of mag efficiency using a 10 - 11 inch barrel are true but the ball actually reaches much of its final velocity much sooner than that. A barrel that is 16" long would still reach much of its final velocity in the first 3 - 4 " as well. In the remaining length of the barrel, the ball would only gradually increase and possibly even decrease in velocity near the very end.

For 300fps the ball travels 3600 inches per second. That equates to 3.6 inches per millisecond.

The ball reaches much of its velocity in about 4 inches of the barrel and it takes about 3 millisecond. For every 3.6 inches farther, it takes about 1 millisecond of travel time.

Note: I did a lot of rounding in the numbers to make it more readable and generically understanable. The actual numbers and acceleration profile is here in deep blue from a while ago for those wishing to see the relationship.

[Ryderdude15]

barrel vacuums

hey i'm not in calc or anything but i was thinking... would it be possible for two balls to be in the barrel at the same time if like the balls had a good enough seal that the second ball going down the barrel was causing pressure to push the first ball out of the barrel. i think someone earlier said that there would be no pressure in the barrel to propel the first ball once the bolt backed up or whatever but thats like two rubber stoppers in a plastic tube... if you have the right amount of friction the stopper will get pushed down the tube from the pressure between the two stoppers. does this make any sense? i'm real tired so i dunno. jus thought i would say what jus popped into my head. lemme know if what im talkin about is possible. thanks

[athomas]

What you are saying is sort of true. The first ball would have the same momentum and velocity if a second ball was in the barrel or not if the gun was setup so that it obtained maximum velocity at the end of the barrel under those conditions. The conditions being that the bolt opens before the ball exits the gun. The second ball could never catch the first in the barrel because of the situation you mentioned.

One problem with getting two balls in the barrel at the same time would be the acceleration that would have to be placed on the second ball to get it into the breach. The first ball would have to be at or above its maximum velocity before the bolt is opened. Once the bolt is opened the barrel loses pressure behind the ball and further acceleration is gone. The first ball will actually start to decrease in velocity at this point. In order to get the second ball into the breach before the first ball exits the barrel it must be done in less than 2 ms given that the first ball takes at least 4 ms to accelerate to 300fps and is only in the barrel about 6ms total. Even then you could only start firing the second ball as the first exits the barrel.

To actually fire the second ball and have it traveling down the barrel behind the first ball it would have to be in the barrel in about 1 ms. The travel distanc is about 1" to get the ball into the breach. The average velocity is about 83fps which means the actual final velocity is probably more than 160fps. At 160fps the ball will probably break as it hits the back side of the breach in the barrel.

[Joemafia69]

joel, your numbers look right, but not for the every day scenario. your equations are for a constant accleration in a vacuum. Thus elimanting friction, air resistance, and variable acceleration over the whole 16" barrel. what you still need to look at is the "fluid" flow of the air. For the sake of simplicity assume it is a lamanar flow over the whole time of travel, BUT you can't do it over the whole 16" (unless the barrel is not ported) then your equations would work (not counting friction and air resistance) over the first 6-8 inches of the barrel (er however long it is to the porting) because of conservation of mass (assuming no losses as well) try redoing the calculations not counting the barrel porting, because after that the ball does not accelerate constantly anymore. a good setup to try is if anyone has a freak, set the velocity of the gun so it is 300 fps with the tip on, then take the tip off and then fire the ball over the cronograph. this will tell you the velocity to use as your VF and give a better model to use. i will try crunching through it with the bernouli equation tomorrow inclass to see what i get as a theoretical Vf, if anyone can try the model i described and post the velocity they get over teh chrono i would greatly appreciate it