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Thread: Things to ponder with how much pressure you need.

  1. #1
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    Post Things to ponder with how much pressure you need.

    In my learning I questioned what is the lowest pressure need to propell a paintball to 280 fps in 10 inches? Now granted this is the inital baby-step in coming up with a more complex answer. So imagine a system that is open at one end for the paintball to leave and a the PSI level that is constant through acceleration. I haven't figured in friction of gelatin on aluminum/steel/brass. If you happen to know the frction coefficient (u) of gelatin on one of these combinations I would appreciate it if you let me know. I just don't happen to have a block of gelatin and aluminum handy. If I did it's a simple test to gather the friction number.

    Onto the fun stuff..... *Note: all numbers are in metric.

    Time
    t = 0.254000508001 / 85.3444 = 0.002976196 seconds for the ball to travel 10 inches and peak at 280 fps.

    Acceleration
    a = 85.344 / 0.002976196 = 28675.53078 m/s^2 acceleration for 10 inches

    Force
    F = 0.003188 * 28675.53078 = 91.41759212 newtons

    Area
    A = 4(PI)(r^2) / 2 = 0.000468603 m^2 ; this is area (standard .68 ball)

    Pressure
    P = 91.41759212 / 0.000468603 = 195085.3753 pascals

    195085.3753 pascals = 28.294743 PSI

    If this is correct, then this is extremely interesting. If I'm correct we can't supply a constant flow of ~28 PSI for 10 inches against an area of 0.000468603 m^2 for 0.002976196 seconds. So the next thing to figure in the PSI volume expansion ratio. That should bump this up to ~100 psi.

  2. #2
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    Nice job! Looks about right too! Yes somewhere around 28 psi will get the ball going in 10 inches. The problem is you waste the rest of the pressure out the end of the barrel after the ball leaves. empty barrel + 28 psi = wasted energy and low efficiency.

    To get best efficiency you need to start with "x" psi and use it all in 10 inches but still get the ball to 280 fps.

    For example start with 80 end up with 10 when the ball leaves gives you an average of 35 psi accelerating the ball.

    Do some math, I'll monitor the post.

    AGD

  3. #3
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    Wow (you can take this post out if you feel it doesn't belong) That is really impressive. I thought algebra was tough. Keep up the good work.
    I have no Sig.

    Good Traders
    ________________________________________
    billybob_81067, Jon/XPM

  4. #4
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    Well, you cant have porting anywhere in that barrel, you know that. And it also depends on hte surface of the barrel, and like you said you couldn't find the friction coefficient. But, problem is, if hte user of this "super-low pressure gun" changes barrels, say, from a creamic to a hoed aluminum, then the pressure will have to go up/down, depending on the switch. Paint travels over certain sufaces much easier than others. But you see this on all guns, so no biggie.

    Another I dea is, using an ultra-low pres. gun with a lubed barrel/ ball. I would think you could use lower pressure if hte ball or barrel was lubed to an extent. You barrel would have to be coated evenly (or turbulance could result on the ball), and be about 0.002 bigger than the ball to allow for the thickness of the bit of oil. Then you should be able to drop that psi from your already low 100 or so PSI to about 60. Im not sure how these barrels would be lubed, either by oil, which is rather primitive, or by new, ultraslick materials. Problem with oils is that it rubs off, and onto the ball, cuasing air turbulance, which cuases inaccuracy.

    Ultra low psi would mean great things for paintball. more shots per tank.. less money. Of course, fill stations would raise prices most likely becuase of the loss of business due to ppl not coming in as often.

    Just some thoughts...

  5. #5
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    Integretion

    I wouldn't say integretion is my strongest subject, but I do know how to. Integrating over time, PSI, and distance isn't a small task for me. I guess I should have paid more attention in the later parts of calculus II class.

    Luckily I picked up some "light" reading. "Fluid Mechanics" by Merle Potter and John Foss. Right now I'm still on chapter 1 and I feel like the Lawn Mower Man in the amount of information each page holds. Eventually I'll hit paydirt in later chapters that talk about compressible fuilds.

    Baby steps...and building blocks...no task is to difficult. Eventually, I will hopefully be able to create the building blocks of a Java API paintball simulator; where all paintball markers can be modeled and tested under virtually.

  6. #6
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    lunacy

    I haven't seem any specs on this yet, but the new 'Lunacy' from splatmaxx section8 is quoted at 25psi operating pressure. Seeing as the original Nova did 90psi no problem this could be interesting.

  7. #7
    Well,

    I have gone off about this topic more than once on other forums, but here it goes.

    What is being looked for is not pressure/velocity but ENERGY/velocity.

    A mag has about .55 ci chamber with a 400+ input pressure.

    So the energy usage is about 220 in3*lbs of energy.

    A shocker uses a chamber of 1.8 ci with about 180+ psi input.

    So the energy usage is about 324 in3*lbs or better.

    Now a tank of 68 ci in size, at 3000 psi is storing about 204,000 in3*lbs of energy. If the gun runs down to 400 psi, then the last 68ci at 400psi can't be used.

    Get the picture?

    Here is more. If you take the total area used by a dump chamber gun, like a 'Mag or Shocker, from the point that the ball is 10" down the barrel, and include the volume for the dump chamber and the bolt/powertube/valving arrangement, you get a total static pressure of about 40-45 psi.

    A Shocker and Matrix are both very in-efficient. Both have very large volumes between the dump chamber and the bolt head. Both have huge dump chambers.

    A Mag has a small dump chamber and small bolt/powertube volume.

    Hence the effecientcy.

    Enjoy,

    Josh
    "If you build it they will run" - pbjosh
    MM006610 bought new in '94. One owner.
    http://itspaintball.com For Pneu Ideas

  8. #8
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    Hey pbjosh afraid that I'm going to have to throw a monkey
    wrench into your theory, the matrix is inefficient in stock
    form because it wastes gas with every shot. The aarvdvark
    bolt kit removes this waste and allows the matrix to use
    the gas effectively. most users are experiencing about
    800 shoots per 68/3000, much the same as a mag.
    Oh well back to the drawing board.

  9. #9
    I have never really had a chance to measure the internals of a Matrix, so I don't know the volume.

    And, I have a design that is a dump chamber, and right now I am running at les than 200 ci*lbs of energy each shot, putting it below a stock angel.

    Here are a couple of things I posted on another site:


    The basics for the gun and tank are this:

    We are looking for VOLUME * PRESSURE = ENERGY.

    Energy is the amount of air used, at a certain pressure, and a certain volume. The more Energy used, the less efficient the gun. Easy.

    Now look at this. A tank has a fixed amount of air, or volume, and that volume is charged to a certain pressure. Say we have a 68/3000 psi tank, charged all the way. We multiply the volume with the pressure to get 204,000 in3*lbs of potential energy in the tank. If a gun runs the tank down to 400 psi, then they don’t use the last 27,200 in3*lbs of energy in the tank. So you get 176,800 in3*lbs of energy used. If you figure out the guns number of shots from 3000 psi down to 400, then divide 176,800 by the number of shots fired, and you get the amount of energy used in each shot.

    So, just figure out how many shots per tank, and the tank size, you get the energy used by each gun.

    I did this to figure out the volume needed to build the Shiva. I also figured out some other stuff, but that will come later.

    I did figure out the that fixed volume guns, like the ‘Mag, could give a good number for the amount of air used because they have the volume and pressure measurable for each shot. A ‘Mag has about .55 ci of volume in the dump chamber, which is ran at about 400+ psi. I figured the gun uses about 210-225 in3*lbs of energy in each shot. A Shocker, which has about 1.8 ci of volume, and runs at 180 psi+, we can figure out that the Shocker uses about 324 in3*lbs of energy. And, from what I have back figured from amount of shots fired per tank, the Shocker can use over 365 in3*lbs per shot on a consistent basis. A Matrix uses about 310 in3*lbs of energy, and the average blowback uses about 270 in3*lbs of energy.

    Now, I need to get new good number figured out for the Shiva, but I am running less than 200 in3*lbs of energy used. And that is without the best paint to barrel match.

    If a person takes an Autococker (or whatever), and takes an accurate pressure reading, then fires a couple of hundred shots, with their paint and barrel, and record the average velocity, and the pressure drop, they would get a VERY accurate reading of the energy used by that gun, with that barrel, and that paint. The only hard part is finding a high quality pressure gauge. So, if you want to figure out how efficient your gun is, there is the guide.

    Fill up your tank, get a GOOD gauge, fire a couple of hundred rounds, and you have the amount of energy used. SIMPLE.

    All the “pressure to get what velocity” questions are just a crock of dookie. What they are wanting to ask is the amount of energy used. And now you have a way to find it.


    I consider the wrench removed.

    Your turn

    Your home work is to tell me what tank you have, what gun you have (barrel and all that) and what ever else. THEN give me a number for amount of you get and the amount of air you use. I can then average the amount of air used per shot for you guys, and we can figure out how truly efficient each gun is. Just as long as everybody is honest, and does try to count the amount of shots taken, then we can get good numbers. Accurate counts, and honest pressure.

    Full report due this weekend

    Josh

  10. #10
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    OUCH you throw a mean gauntlet there pb
    I will supply the dump chamber volume for a (my) matrix soon

    as for accurate gauges try here http://www.winters.ca/pre_process.html

    I bought a 0-300psi 1 1/4" gauge from one of their dealers
    for C$ 11.00

  11. #11
    oh, I already have the gauges, and they are 0-200 and industrial, for use with the 'Shiva'.

    And for everybody reading this post, please attach your numbers, for tank size and pressure, plus type of gun. A chart of some such that tells the number of shots per gun and the size of the tank can be produced from that. Let me make some other posts for this so I can get a average number of shots per tank per gun, and then I can post some good numbers for the energy usage of several guns

    Later,

    Josh

  12. #12
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    Back again The volume of a matrix dump chamber is
    21.5 ml (measured with graduated syringe) or 1.312 cubic in

    A handy link for the conversion challenged http://www.onlineconversion.com/

  13. #13
    VERY NICE! you figured it out with a syringe! smart.

    Okay, the matrix does still dump some air, right?

    and run at 140 psi?

    the 140psi * 1.3ci = 182 in3*lbs which is VERY GOOD.

    So, where does the rest of the air go. Some one had built a paintball gun that shot as much air out the back of the gun as shot air out the front of the gun! HAHAHAHAHAHAHA! oh boy, that seems really silly.

    LOL- that still gets me.....

    Josh

    hehehehe

    how many shots from a tank do you still get?

    Humrph....hehe...giggle....

  14. #14
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    actually it takes about 180 psi for 280 fps I have to confirm this as It's been about a month since I've last
    played so 180psi * 1.3ci = 234 in3*lbs

    still not bad

    further 204000 in3*lbs/234 in3*lbs =872 shots

    That number is of course "ideal" so what happens in real life is different.

    As for shots per tank I confess that I haven't had the bolt kit long enough to personally confirm or deny 800
    shots, 68/3000 It is what people at pbnation are claiming that they are experiencing. plus I have to use a
    scuba tank so I never get a real 3000psi fill.

    None the less my original post was to warn you of an incorrect
    assumption as to why the stock matrix got lousy
    gas mileage. There are no elves involved in the arrdvark bolt
    kit merely the correction of a design mistake that allowed
    air to be wasted on the return stroke of the bolt
    (by chance did you already know this?)

    OH and how about some links to those "other" sites

  15. #15
    As to other sites,

    Actully, I have written almost the same thing to most of the other sites out there. I usually hit the Tinker's guild and the pbnation. I just put a survey up in pbnation paintball talk asking for shots per tank adn all that. chaeck it out. And on pbnation in the paintball news section is the 'Shiva' project.

    The 'Shiva' is mine.

    Josh

  16. #16
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    Question "I'm not dead yet!" Monty Python

    Wow, I read all the above posts. Interesting.

    I think I'm on the right track in figuring out pressure loss in piping, but I need to figure out how to calculate SCFM.

    Most problems explain what SCFM is and give it a value to calculate ICFM, but never states how one derives SCFM.

    Anyone know how to do this?

    UPDATE: Never mind I figured it out and have a completed formula. All I need is a couple of reference tables for relating density to temperature and some others.
    Last edited by CHK6; 01-15-2002 at 08:22 AM.

  17. #17
    okay everybody,

    I did a quick survey on pbnation, and i guess I will post one here soon.

    Here is an energy used chart, per gun.

    Now, these are not tested numbers, but from a list of peoples postings about which gun, how many shots, and what tank. So, error is included. Just for now, this is just a beginning. I want to fill this up to the point that we can figure out valves and barrel ratios also.

    So, if you read this, post something like....

    I have a SFL Snagler ACE with an 68/4500 tank and I get 900 shots per tank @ 225 psi

    Thanks,

    Josh
    Attached Images Attached Images

  18. #18
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    To: PBJosh

    PBJosh,

    The energy the tank holds also heavily depends on local barometric pressure, temperature, and altitude above sea level.

    I have also found some other interesting tidbit relating to PSI levels. When talking "shop talk" I found that you either have to say PSIA or PSIG and just saying PSI will get you funny looks from individuals that work with compressed air. And if you say what the PSIG is you have to give the local barometric pressure.

    I also found out that the barrel id has an effect of how many shots you get with the energy in a bottle.

    Also when calculating pressure temperature you have to convert Farenheit to Rankine. Something I wasn't doing and really scewed my results.

    Just sharing the information I have found in better hopes to truely understand it all myself.

    Later gator.

  19. #19
    Well, it all gets down to averages. A person can tell me how much pressure is in their tank, but the gauges on a tank is not to accurate. Also I will get basic averages for the number a shots per tank, generally rounded off to 50 or 100.

    Barrel has alot to do with it, but the general average for the gun will be a bigger factor. And part of this is to educate. The amount of energy used per gun on average is a number nobdoy uses to figure out efficinct, but should be incorporated in everybodies ideals for how a gun works. How many people think that low pressure means high efficiency? With a basic graph, this should help people realize what gun they want, but also what tank. If a person thinks that one valve is better just because of its working pressure, then they are not taking into account the amount of energy used per shot.

    A different bolt, or even operating pressure can affect all of this also.

    Again with the accuratecy of a post to a survey, and the gauges, and everything else, the difference for the local barometric pressure temp, and PSIG and all that, it doesn't really matter. I am dealing with larger numbers and averages than that. And the reason for a large # of averages is to sort that information outta the mix. I don't want an effecientcy number based on a certain air pressure, certain paint, certain barrel. I want an average for that gun, in any local barometric pressure, at any temp, and etc.

    So, tell us, how many shots from what gun and the tank you use at what pressure and what velocity.?

    Josh

  20. #20
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    Pbj,

    Nice data table!! Glad to see someone wants the facts. I agree with you that a large sample size will average out the variables. I would like to see the formulas you used to get the energy in the tank and the per shot energy.

    Thanks

    AGD

  21. #21
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    Post An inward view

    To answer the question I get 250 to 300 shots on any given day with 3.5 oz CO2 on my Phantom. Pressure ranges from 800 to 1000 on the guage.

  22. #22
    AGD! proud to have you on this!

    Okay, here is all I did to figure out the numbers.

    I originally posted over a 2 years ago on the tinker's guild to a question about the pressure it takes to get the ball to 300 fps, like on this post. I was posting on how the number that they are looking for is a value of energy, not pressure.

    Then I collected the numbers for the volume of a couple of guns with fixed chamber dump valves. Since energy is a measure of Volume x Pressure, I figured I could get the right numbers from multiplying the Volume of the dump valve with the pressure the gun ran. And, then I smacked my head and said "Hey, all the guns have a fixed amount of energy attached to them! The HPA tank!"

    So in the basic 68/3000 tank, multiplied, you get 204,000 ci*lbs of energy in the tank! And if a gun runs till 400 psi, a person would take the volume of the tank, minus the amount of air unused, like 68 ci x 400 psi = 27,200. So, 204,000 - 27,200 = 176,800. And if a gun gets 800 shots, then divide the energy that can be used, 176,800, buy the number of shots, 800, you get the energy used, so:

    176,800 / 800 = 221 ci*lbs of energy on each shot.

    I just asked about the gun, the tank size and the number of shots @ which pressure. That gave me the total energy used per system. So, even if it doesn't have the exact amount of air used down the barrel, it does give a relative number for the energy used per gun. And I must say, the 'Mags are the best dump shamber out there right now, with 200-225 ci*lbs of energy used. A gun like the Shocker is horrid, with 365-400 ci*lbs common.

    Well, that about it, just whiped up an Excel file and that was that.

    Now go my children, and spread the word.

    Josh

  23. #23
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    Lets get more elaborate here. I'm no expert at math anymore, its been a few years since I've picked up a math book, but I am interested in some of the dynamics here.

    for ease of calculation:

    Mag chamber pressure = 400 psi
    translates to 220 in-lb of work force

    Actual force on the paintball at start is 150lbs when properly vectored, as it is the cross sectional area of the ball that is important, not half of the surface area. 150lbs represents the pressure in one direction(out the barrel).

    Chamber = .55 ci
    cu area of 1 inch of barrel (id = .690) = .3739 ci

    Lets plot the acceleration force of the expanded gas.

    distance = force
    start(0) = 150 lbs
    1 in = 89 lbs
    2 in = 63.4 lbs
    3 in = 49.2 lbs
    4 in = 40.2 lbs
    5 in = 34 lbs
    6 in = 29.4 lbs
    7 in = 26 lbs
    8 in = 23.2 lbs
    9 in = 21 lbs
    10 in = 19.2 lbs

    As you can see, the amount of force on the ball decreases as the gas expands and the ball accelerates down the barrel. Therefore the rate of acceleration is not going to be standard. Most acceleration is calculated using a constant force.


    Now, I can't find my books with all my formulas, so I need your help here. Can anyone figure out a formula for calculating the actual rate of acceleration of a paintball down the barrel of a paintball gun. If we can come up with a standard formula taking into account the changing volume of the chamber-barrel as the ball accelerates it will help us realize the benefits of changing pressures, volumes and barrel lengths.

    I did some quick calculations and noted by eye that the effect of doubling the chamber size and decreasing the chamber pressure doesn't work. It is more like double the chamber size, decrease the pressure by 25%. The amount of in-lbs of useable gas increased. In the other direction we could greatly increase the efficiency by increasing the pressure and decreasing the chamber size. The starting force would be increased and would probably cause the balls to explode in the gun though.

    I know my data has gaps in it but the fundamental concept is basically sound.

    Hope this helps.

    We can do it for the betterment of mankind on the painball field.
    Except for the Automag in front, its usually the man behind the equipment that counts.

  24. #24
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    Warning, a whole lot of theory and physics going on...

    Fair warning, I'm going metric on you guys. That being said, here
    we go...

    First off, some basic formulas.

    1: F = m a
    2: F = P A
    3: P1 * V1 = P0 * V0

    Notes:

    Formula 1 is your basic "force equals mass times acceleration"
    equation. Formula 2 relates force, pressure, and area. Formula 3
    describes how pressure and volume interrelate. The "*" symbols in
    Formula 3 are multiplication symbols, for those who are used to
    seeing "x" instead.

    Since we're interested in the acceleration of the ball at any
    point in a barrel, we really want to solve Formula 1 in terms
    of the variable "a".

    a = F / m

    Now, we have a pretty good idea what the mass of a paintball is
    (around 0.116 ounces, or around 0.0033 kilograms). What we need
    to know is how the force on the ball varies as the ball travels
    down the barrel. This is where the second formula comes into
    play.

    The force on the ball as a result of the gas behind it is
    dependent on the pressure of that gas and the cross-sectional
    area of the paintball.

    F = P A

    We can calculate the cross-sectional area easily enough. A
    paintball is right around 0.68 inches in diameter (0.017 meters).
    Area is simply Pi R squared, which gives a final area of
    0.00023 square meters. Why am I using square meters for something
    so tiny? Because the pressure values I'm going to be using are
    expressed in units of force per square meter of area. So now
    we have the following for Formula 2:

    F = 0.00023 P

    We can plug this into our acceleration solution for Formula 1 to
    get the following:

    a = 0.00023 P / 0.0033 (I plugged in the mass as well)

    a = 0.0697 P (simply divided 0.00023 by 0.0033)

    Ok. Now we have acceleration expressed as a function of pressure.
    Now we have to answer how the pressure changes as the ball travels
    down the barrel. I'm going to assume that the barrel we're using
    has zero porting in it, just to make my life easier.

    This brings us to Formula 3:

    P1 * V1 = P0 * V0

    This basically says that the energy held in a gas essentially
    remains constant. Of course, this also assumes that the gas
    temperature does not change. We'll make that assumption for
    this exercise.

    Our initial pressure (P0) is going to be 400 psi
    (2.757 megapascals). The initial volume (V0) is 0.55 ci
    (0.00000901 cubic meters). We can plug these two values into
    Formula 3 and get the following:

    P1 * V1 = 24.85 Newton-meters (i.e. Joules)

    The volume of space behind the ball as it travels down the barrel
    is going to be the volume of the dump chamber (0.55 ci) plus the
    area of the barrel times the distance the ball has traveled.

    V1 = V0 + A * x,

    where x is the distance the ball has traveled, measured in meters.

    We can then get the following expression for our pressure:

    P1 = 24.85 / (V0 + A * x).

    P = 24.85 / (0.00000901 + 0.00023 * x)

    Ok. We now know how the pressure should change as a function of
    ball position. We can plug this back into our acceleration
    formula:

    a = 0.0697 P

    a = 0.0697 * 24.85 / (0.00000901 + 0.00023 * x)

    a = 1.732 / (0.00000901 + 0.00023 * x)

    or, if you want to keep all the variables as variables rather than
    having the numbers plugged in...

    a = P0 V0 A / (m * (V0 + A x) )

    P0 measured in pascals (newtons per square meter)
    V0 measured in cubic meters
    m measured in kilograms
    A measured in square meters
    x measured in meters

    Our acceleration should be in meters per second squared.

    And there you have it... one acceleration formula. Keep in mind
    that this does not take into account frictional forces experienced
    by the ball as it slides against the barrel wall. These forces
    are not always insignificant, though if you have a good paint to
    barrel match and can blow a paintball out the barrel with ease,
    you may well be able to ignore frictional effects.

    I've got a semi-handy Microsoft Excel spreadsheet which calculates
    a paintball's acceleration down a nonported barrel and determines
    what the theoretical exit velocity should be. I'm working on
    tuning this equation so that the ball exits the barrel at the
    desired velocity and when the pressure in front and behind it are
    equal. I can vary the pressure, initial volume, barrel length,
    and frictional forces so far.

    BJJB

  25. #25
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    Excellent. That's what we're talking about.

    If it isn't too much trouble, could I have a copy of the excel worksheet. It would save me doing the same work all over.

    thanks

  26. #26
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    athomas wrote:
    >
    > If it isn't too much trouble, could I
    > have a copy of the excel worksheet. It
    > would save me doing the same work all
    > over.

    I'll have to make sure it's reasonably presentable... after all, I created it to be "me friendly" rather than "user friendly". When I get it ready I'll post an attachment in .zip format, since it doesn't look like this forum supports .xls or .wbk. It's going to be an ancient version of excel (4.0), but the recent versions should still be able to read it.

    BJJB

  27. #27
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    Excel Spreadsheet

    Well, I've made the spreadsheet as friendly as I could, and have zipped it up so I can attach it. Man, it's just under the filesize limit. I hope this works...

    There are a couple of things to keep in mind when playing around with this spreadsheet. First, it does not necessarily represent reality, as it is only a theoretical treatment of the problem at hand. Second, it handles things a bit differently than what has been discussed... basing everything on very small increments of time (calculus a-la excel) rather than ball position.

    I have yet to figure out why, but the spreadsheet fails to give even remotely close velocity numbers for an automag's dump chamber design (500+ fps from a 10 inch barrel at 400 psi input?!). I think this has much to do with my inaccurate guesses for the volumes of the power tube and bolt, both of which will cause an immediate pressure drop from the dump chamber initial pressure at the moment of gas release. Perhaps more accurate numbers for these two volumes would correct the spreadsheet results. There are also the matters of non-instantaneous gas relase from the dump chamber and gas leakage around the bolt and power tube (likely small?) to contend with.

    Have fun playing around with it. Maybe you can improve some things here and there, though you may want to fiddle around with a copy in case things get out of hand.

    BJJB
    Attached Files Attached Files

  28. #28
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    May 2001
    Location
    Austin, TX, USA
    Posts
    36

    Baby steps taken a bit further....

    Let me first say I have really enjoyed this thread, I have learned a lot. I am 99.999% almost done in figuring out pressure drop in a pipe with compressible gasses. The following is a massive culmination of research I have gathered from the last three weeks and was able to cross reference verify with other sources.

    Here is the generic formula that needs to be solved.

    delta(P) = F (L/D) * [(p * u^2)/(2 * g)]

    delta(P) is the absolute pressure drop, not guaged.
    F is friction factor of material.
    L is the length of the pipe.
    D is the pipe diameter
    p is avg. density at flow conditions.
    u is mean velocity
    g is gravity

    To find (u):
    u = Q/A
    A = cross section of a circle.
    Q = volume flow in entrance conditions. (ICFM)

    To find (Q):
    Q = (Z * R * T) / (60 * M * Ps)
    Z is the compressibility of the gas.
    R is universal gas constant
    T is pressure temperature
    M is molecular weight of the gas
    Ps is pressure (PSIA)

    To find (p):
    p = (P1 / 14.7) * d
    d is the density of air at amospheric temperature.

    To find (F):
    You need a materials chart

    To find out if the flow of the gas behind the ball is turbulent calculate the Reynold's Number:
    Re = (p * u * D) / y
    y is absolute viscosity of air at atmosphereic pressure.
    If the result of Re is greater than 4000 the flow of the gas behind the ball is considered turbulent.

    Now all I have left is to plug in the numbers. Which I'll do this weekend when I have some time.

  29. #29
    Join Date
    May 2001
    Location
    Austin, TX, USA
    Posts
    36

    Angry Grrrr....

    This doesn't sound right. After some calculations I figured that with a 10 inch SS pipe with a mean velocity of 140 fps using Nitrogen with 80 degree weather the pressure drop is 2.142318 PSIG.

    Now that's with a constant flow of gas for the length of the pipe. So if we started with ~28 PSIG we would end up with a pressure guage reading of ~26 PSIG at the end of the pipe. Giving use slightly less than 280 fps. This gives us the solution of show much max volume you need at the lowest pressure possible. Volume being the volume of gas the 10 inch barrel consumes.

    So the flip side is how little volume can we get away with at the highest pressure?

    Since we can't provide a constant flow of gas for the entire length of the barrel, (well we can, but why stop there and be in-efficient; aka gas piggy), we need to figure out what fixed amount of volume of Nitrogen is needed at under what preesure will accelerate the ball.

  30. #30
    Join Date
    May 2001
    Location
    saint john nb canada
    Posts
    460
    Looking at pbjosh's table and concluding that 200in3*lb
    is a close enough for our purposes number, and that the
    typical maximum output pressure for a fixed air tank is
    850psi

    200in3*lb/850psi=.235 in3 not suprisingly about half
    the volume of the mag at 400psi.

    Given the anecdotal nature of pb's table, it is remarkably
    constant from marker to marker

    so what makes a shocker so wasteful and an impulse so
    efficient ?

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