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Question for Neil Winston

Discussion in 'Uncategorized Threads' started by bodybuilder, Mar 12, 2007.

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  1. bodybuilder

    bodybuilder TS Member

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    Does porting make you Lose Velocity? And do you think Porting hurts your Patterns? Thanks
     
  2. Neil Winston

    Neil Winston Well-Known Member

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    No.

    No.

    But it does make it shoot lower. It's has to. *

    Neil

    *we're sure to hear about forces having to "overcome" movement, inertia, or the like before they start to work. But look at the formula: F=ma. There's nothing in there about "later" or "except when you don't want it to work this way;" the acceleratioin occurs right at the very same time as the force is applied. If porting pushes the end of the gun down to reduce muzzle rise, it does so when the gasses are there, that is, before the shot exits.

    I bought a brand new MX-8 o/u from Frenchy Frigon and tested the POI a lot, had it ported, and retested it. It shot lower. Not much - let's say 3 inches at 40 yards - but enough to easily measure.
     
  3. Pocatello

    Pocatello Active Member

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    Neil, would you shoot me an email please? I have a request to make of you.

    Thanks.

    Larry
     
  4. Bob Hawkes

    Bob Hawkes Well-Known Member

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    Neil, Very interesting, thank you.Problem,It seems that some days I can't stay within a foot of where I ought to be. Oh well, Thanks,Bob
     
  5. Neil Winston

    Neil Winston Well-Known Member

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    Ajax, I accept John Brindle's version, as presented in his book "Shotgun Shooting." In his explanation, there are two phases of recoil. (I know that Rollin is as least as up-to-date on this as I and may have more to add.)

    1) In Phase one, the gun rotates around its center of mass. The fact that the line of the bore is above the center of mass causes the muzzle to rise (and the butt to fall if it can) without regard to what the gun is pushing against.

    2) in Phase two, the gun lifts as a result of pushing against the shoulder.

    Brindle doesn't say this, but I think that if there is anything to the unsingle theory, it is all in the first phase, since, by my calculations (and one measusrement) the second phase is almost all long after the shot has left the barrel.

    In terms of whether porting moves the POI down or not, it makes no difference when all this happens. Imagine there were no first phase of recoil at all, it was all due to the shoulder pivot. Still - if the muzzle-jump is reduced by the porting, then it must have started when the shot (and compressed powder-gasses) were in the barrel, so it still shoots lower.

    Neil
     
  6. Neil Winston

    Neil Winston Well-Known Member

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    Jimmy, it's the gasses released which force the muzzle down. It's the gas retained, by adding (according to Hatcher 4%) to the rearward speed of the gun which raises the muzzle, pivoting around the point of contact with the shoulder. This upward movement reduction - a consequence of slower recoil - cuts down on face-slap, but is unrelated to the POI downward deflection, since it, by definition, occurs _after_ the shot exits.

    Neil
     
  7. old tex

    old tex TS Member

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    Neil

    Regardless of the laws of physics which cause the ported bbl. to shoot a bit lower it in fact does in my personal experience.

    It also follows that it has less cheek slap.

    Thanks for your explanation, I always enjoy your technical treatises even though most of them pass slightly over my head (maybe I was "ported" in my youth).

    Unc
     
  8. Neil Winston

    Neil Winston Well-Known Member

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    There are seveal ways to look at it, Jimmy, each useful in one way, less in another.

    Here's the unequal-pressure explanation. Near the end of the barrel, before the ports, the pressure, and so the force, is equal in all directions. Now let's jump forward and open all the ports. Now on the bottom part of the barrel all the material is still there for the gasses to "push against." But near the top, that's no longer true, some steel is gone leaving nothing for the gasses to act on. So the forces at the end of the barrel are unbalanced, bigger down, and unbalanced forces lead to accelerations.

    It's an interesting sidelight that the square-edged holes only pass about 70% as much gas as calculations lead you to expect.

    Yours in Sport,

    Neil
     
  9. Neil Winston

    Neil Winston Well-Known Member

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    PBB, please don't say anything about "overcoming stationary mass." That’s what I brought up in my first post. Nothing's overcoming mass. Forces can overcome other forces, for example, you tugging on a new case of titanium castings in your shop can overcome the force of friction between it and the floor and so it will move. But you are not overcoming it's mass. You are accelerating its mass.

    To put some numbers to the problem, the mass of the gun is not waiting around to be covercome. By the time the shot gets to the ports, the gun has moved about a third of an inch and is moving something like 15 feet per second.

    I have to doubt that 50% figure. If Hatcher is right there's only 4% to work with, and to get that you would

    1) have to evacuate _all_ the gas and

    2) somehow make all of it head straight back.

    Add to that the fact that the ports are at 45 degrees does not mean that even a substantial portion of the gasses exit at any angle (to the vertical) at all.

    I'm afraid I'd have to see some proof, Mike. It seems unlikely to work out, that's all.

    Yours in Sport,

    Neil
     
  10. Rollin Oswald

    Rollin Oswald Active Member

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    Neil,

    Your experimental results that ported barrels shoot lower than unported barrels puzzles me. Given that a shot and wad mass is moving down the barrel at about 1200 fps and the barrel has just begun its upward movement around the gun's center of mass in the first phase of recoil, the length of time between the shot and wad's passing the ports and their exiting the barrel cannot be more than a millisecond or two.

    It is that short period of time that is available to force the barrel downward as the gasses exit upward through the ports. In addition, that "jet" effect must slow the upward movement of the barrel - all in that minute period of time as the shot mass travels from the porting to the muzzle.

    How is that possible? I do not doubt your findings but simply fail to understand them. It is that nagging "why" question again. Can you help? Any idea about how far upward the muzzle would need to move to raise the POI 3" and therefore the barrel rise that is prevented due to porting? You math whizzes might be able to help, here. (My math competency significantly diminishes with numbers greater than 6.)

    Rollin
     
  11. Neil Winston

    Neil Winston Well-Known Member

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    Rollin, let's call the time for the force to work no more than 1/2500 second. But at the same time, we have to remember, it's at the end of the barrel, out where it has the best leverage to exert a torque. Look at all those photo comparing muzzle jump with and without ports. All that effect (the difference in the outcomes) had to be exerted while the shot was still on the barrel. It's not much time, but you only need about 1/10 inch difference in muzzle position to get that 3 inches at the target, and that - 1/10 inch - not much of a distance either.

    Neil
     
  12. Rollin Oswald

    Rollin Oswald Active Member

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    Neil,

    That makes sense; thank you.

    Rollin
     
  13. GunDr

    GunDr Well-Known Member

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    Neal,

    You didn't mention this...so I'll ask.

    Wouldn't you also get some effect on the muzzle, before the shot passes through the porting. Upon firing, and with the shot column moving 1200 fps, doesn't this compress the air ahead of it. And with this compression, only so much can exit the muzzle of the bbl. Allowing a balance to exit the porting.

    My thought on this is....the porting is exiting compressed air as the shot begins it's movement.... and as the shot passes the porting, it exits the expanding gas.

    Wouldn't this compression of the air ahead of the shot column explain why some bbl's "banana peel" with a muzzle obstruction?

    Doug Braker
     
  14. Neil Winston

    Neil Winston Well-Known Member

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    Doug, I doubt the effect is large but it may exist. I think the banana peel result is based on the compressed air you cite, but we have to consider that in that case, the muzzle is blocked. I've got a perfect example hanging in my garage as a warning; I'll try to photograph and post it here. It's a classic.

    Neil
     
  15. Neil Winston

    Neil Winston Well-Known Member

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    AA, by "square-edged" I meant holes just as might be drilled or eroded, that is, no one has gone in with a file to try to "break" the edges or make it look like a trumpet or the like. The above reference covers the general idea.

    I think we "know" that the gas which hits that front edge of the angled hole exits at a 45 degree angle, the metal at 45 degrees (relative to the gas flow) bends it back. And if the barrel were so thick that a port was so long you couldn't "see in" it would all bend back.

    But barrels are thin, and you can see straight in right to the other side. And what part you can see into is by far the majority of the hole. What's to make compressed gas, looking straight up and with nothing above it but the sky, bend back at all? It won't. of course. It just goes straight out.

    "But wait!" you say, "the gas from the 45 degree part, bends the rest back too, at least a little."

    "But wait", I say, "there's way more gas coming straight out, and that just straightens that other gas as well, so the effect is diminished, not increased."

    There's a bit of a complication, though. The gas looking up isn't really going to go straight out, it's going to go forward-out. It has velocity forwards and it's going to keep it until stopped by something. So if you take this "angle of exit" stuff to have any substantial effect, and I don't put much stock in it, the forward motion of the gas is going to reduce what you are after.

    And so on. I'm glad of the opportunity TS.com gives me to promote my ideas, and questions make me do my job better.

    Yours in Sport,

    Neil
     
  16. southpark

    southpark TS Member

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    Neil, when you did your before and after measurments did you note the temperature of the ammunition each time?
     
  17. Dove Commander

    Dove Commander TS Member

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    I think you all got "gas"
     
  18. Neil Winston

    Neil Winston Well-Known Member

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    Mike (PBB) I'll deal with your posts out of order, but the first to address is the .50 BMG observation.

    Needless to say, I was talking about trap loads. The effect of powder in rifles is completely different, in fact, on page 281 (Hatcher's Notebook, Stackpole Books, ISBN 8117-0614-1) he cites an experimental .50 gun as having 70 ft. lb. recoil without a muzzle brake, half that with one. In other words, Hatcher would not be surprised by your findings.

    Neil
     
  19. Neil Winston

    Neil Winston Well-Known Member

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    Your other three posts can be dealt with together.

    First a correction. In quoting Hatcher's 4% I left out a term. In his example, (1250 fps, 1 1/8 oz., fiber wads) it was more like 6%, not 4%. I regret, the error.

    Your plan to make the barrel really thick does address my post to AA, in that I take it you can't look straight down into the ports and see the inside of the barrel, and this will direct the gasses back in a way many times more effective than standard ports.

    The problem needs some numbers, since what we need is ratios: your claim of 50% or my counter-assertion that that's way, way too high.

    I went through the recoil calculation for the powder in a different way in early February's slow powder = more recoil thread but Hatcher's method does a better job of focusing in on the powder effect. It's better for answering this ratio question.

    The following analysis is adapted from Hatcher's chapter "The Theory of Recoil," extensions beyond the method are my own as are any errors or misunderstandings they may creep it.

    You can account for the momentum of the recoiling gun by adding up the momenta of 1) the shot/wad, 2) the powder, and 3) adding the jet effect of the expansion of the powder at muzzle exit.

    1) The momentum of the shot/wad is easy, just the muzzle velocity times the mass.

    (OK, here's where we are going to do only the math we need to, and all we need is ratio. To deal with mass accurately, in the above formula, I should include the gravitational constant but I'm not going to because how many would understand it? So please go forward with the understanding that the math "units" won't come out right, but the ratios will. I'm not defending this degree of simplification as a general rule, but here it will work periectly, and clear away some clutter which might otherwise obscure the "meaning" of the result.)

    back to 1) With the parenthetical warning in mind, we will call the mass 500 and the speed 1200, leading to a product of 600,000. "Six hundred thousand whats?" you ask, and I say "Just 600,000, that's all we need."

    2) The momentum of the powder is more interesting. The mass is easy, and for this example we'll call it 20. But the speed takes some thought. The powder at the muzzle, right at the back of the wad, is going the same speed as the wad, 1200 fps. But the powder (gas, now) still remaining in the shell isn't moving at all. So we'll take the average speed of the powder as half the muzzle velocity, taking into account both very different speeds, zero and full.

    So, this term of your momentum will be 20 times 600 = 12,000

    3) The jet effect. Experiments have made it possible to estimate that _in a shotgun_ treating the speed of the gas as 1.5 the muzzle velocity (at the outside, also 1.25 is suggested, page 289, for standard shotguns.) We'll use 1.5.

    Using the conventions above, we'll multiply 20, the mass of the powder, by 1200, the MV, and then by 1.5 to account for the jet effect. So we get 36,000.

    This gives us a sum of 12,000 and 36,000 or 48,000 and this is what we compare to the shot momentum, 600,00. Using these admittedly rough calculations, we can put the effect of the powder at 8% of that of the shot, and so about 7% of the total, which is 648,000

    Now imagine we were in a universe where guns worked differently, where the shot went one way and the powder gasses in exactly the opposite. Here the powder effect is subtracted from the shot effect, leaving a sum of 552,000, which is 15% less than our original 648,000. This is the absolute limit of what you can expect from any system, in that it's taken all the powder and turned it back the other way.

    If all the powder is turned at 45 degrees, not 90, then you have to at least multiply by 0.7, reducing the max to 10%. Add to that the problems of turning the gas without reducing its speed (unlikely), keeping it's temperature (and so its pressure) up through a passage a third of an inch long, getting a high percentage of the gas to turn at all, and you are down well under 10%.

    Shortening the barrel will increase this number, adding shot will lower it.

    I can't see it any other way. There's only so much gas can be expected to do. There's not much of it, relative to the mass of the shot, and there's not much pressure in it at the end, relative to rifles, and those are what set its limits.

    Neil
     
  20. Dutchboy

    Dutchboy TS Member

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    Rollin, Neil, I think it is easy to understand the effect of porting on POI if you look at the system beyond the muzzle. A shotgun gives both FORWARD momentum as well as UPWARD momentum to the charge. The upward movement of the muzzle "throws" the charge upward.

    By porting, you reduce the upward momentum. Even though this effect is slight, it reduces the upward movement over the whole trajectory of the charge to the target.

    In other words, it is not just that the gun aims lower, it is also that the charge is moving upward at a lower rate. If you throw a ball up with a slower motion, it doesn't go as high. FWIW, Dutch.
     
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