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The science of recoil

Discussion in 'Handguns: Autoloaders' started by tackleberry45, Apr 18, 2008.

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  1. tackleberry45

    tackleberry45 Member

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    I was asked why a .40 "snaps up" for recoil and why a .45 "pushes back" I really did not have a great answer other than the only thing I couod think of which was powder burn rate. I have a Smith M&P .45 and STI .45 that are tame on the recoil side for .45. I also shot a Beretta PX4 .45 which was not so tame, at least to me. I guessed becuase of a little higher bore axis. So THR folks what is the correct answer as to why a .40 tends to snap up and a .45 pushes back??
     
  2. easyg

    easyg Member

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    The question itself is flawed.
    I don't even agree with the notion that the .45 "pushes back", and the notion that the .40 "snaps up".
    How they react is largely dependent upon the specifications of the pistols and the strength and ability of the shooter.
     
  3. makarovnik

    makarovnik Member

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    A lot of it has to do with velocity, bullet weight, barrel length and weight of the gun. I agree that the .40 is very snappy. You really feel like you need to hold on tight. Problem is that a lot of .40's are packaged in a small light frame. This is a mistake in my IMHO. A .40 in a large frame is quite controllable.

    The .45acp doesn't seem as snappy when you are shooting it but if you video taped yourself shooting one you would see they have more muzzle rise than you expected.
     
  4. ccmdfd

    ccmdfd Member

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    I'm with easyg on this one.

    There's so much subjectivity when it comes to perceptions of recoil that it's almost laughable reading about them on the net.

    So much depends upon the type of gun, it's weight, its bore axis. How well the grip fits the particular shooter (a very big thing IMHO) as well as any torque or rotational forces also play a role.
     
  5. Gun Slinger

    Gun Slinger member

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    See for yourself:

    Just "plug in" the numbers...

    Where:

    M = bullet weight in grains
    V = muzzle velocity in feet per second
    E = weight of powder charge in grains and
    W = weight of the rifle or pistol in pounds

    ([MV + 4700(E)] ÷ 7000)² ÷ (64.34809711) ÷ (W) = "free" recoil energy of the pistol or rifle (in foot-pounds)

    Example:

    A Glock 17 fires a 147 grain bullet with a muzzle velocity of 1000 feet per second using a cartridge that uses 4.0 grains of propellant and the Glock 17 weighs ~1.57 pounds unloaded. What is it's "free" recoil energy in foot-pounds?

    ([147 gr. x 1000 f.p.s. + 4700 x 4.0 gr.] ÷ 7000)² ÷ (64.34809711) ÷ (1.57 lbs.) = 5.55312 ft.-lbs. of "recoil"

    After that, how you "percieve" the recoil is pretty subjective, but the gun does the same thing everytime regardless of who is firing it.

    Hope that this helps,
     
  6. presspuller

    presspuller Member

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    Can't argue a bit with what has been said here.
    The only thing I would add is that the different grip angles on different pistols makes a difference in the way recoil is felt.
     
  7. Rustynuts

    Rustynuts Member

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    It's all physics. F=MA

    The 40 bullet isn't all that lighter than a 45, yet has to be accelerated (A) to a much higher velocity in essentially the same distance (gun barrel). Slightly lower mass (weight, or "M"), but much higher A = higher F (force). That's the simple notion of it, then you have the different shapes of the brass/bullets, burn characteristics of the powder, etc., and it starts getting more complicated. Then you get into the grip and gun characteristics as well.
     
  8. WinchesterAA

    WinchesterAA Member

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    Wow.. After reading and doing much of the math in "Understanding Firearms Ballistics", I actually understood EASILY what Gun Slinger said..

    I'll go ahead and recommend that book to everyone then.

    I sucked at math, or so I thought, but when firearms are involved it's easy for some reason...
     
  9. makarovnik

    makarovnik Member

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    I would like to add that when someone asks a question about recoil they are usually asking about perceived recoil. I think that should be assumed. If it isn't "perceived" then it doesn't bother a person and shouldn't be an issue unless you're concerned about wear and tear on your gun.
     
  10. Walkalong

    Walkalong Moderator

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    Velocity my friend, velocity. ;)
     
  11. 1911Tuner

    1911Tuner Moderator Emeritus

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    Muzzle velocity has little to do with recoil energy. Most of the recoil momentum has hit your hand or shoulder long before the bullet reaches the muzzle.

    You can have high velocity and low recoil impulse...as with the .22-250 rifle...and you can have low velocity and high recoil...like with the .45-70/405.

    If we accept Newton's 3rd Law that force forward equals force backward, we can start to unravel the mystery.

    The pressures generated by the burning, expanding powder gasses create a force vector between the bullet and the breechblock. Whatever force is imposed on the bullet is imposed on the breechblock in like measure...and at the same instant...so the gun is in recoil as soon as the bullet moves.

    Force vector is defined as having magnitude and direction.


    Here's a clue to consider:

    If we ascribe an average of 30 fps per inch of barrel length...gained or lost...and the muzzle velocity from a 5-inch barrel is 850 fps...and roughly an inch of that barrel is chamber...we have 120 fps of the total coming from the acceleration in the barrel. That leaves 730 fps unaccounted for.

    When does that 730 fps occur?

    During the violent rise to peak pressure, and the initial punch of sudden acceleration...resisted by the bullet's entry into the rifling...and the resulting rise in force necessary to push it forward.

    Likewise, the initial punch of equal force backward against the breechblock creates the greatest percentage of what we call recoil.

    Accelerating the bullet from that point doesn't generate the level of equal and opposite force that occurs at the onset of the event.

    Here's more:

    http://www.thehighroad.org/showpost.php?p=4417450&postcount=8
     
  12. ccmdfd

    ccmdfd Member

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    I'd love to see more science put into recoil.

    I'm somewhat recoil sensitive due to some arthritis. But I've found there are certain .45's I can shoot with no trouble, while some 9mm's cause me to wince.

    As I stated above, I think there's more to the story than just magnitude of recoil, which is easily calculated. There's directional issues-does the gun go straight up and down, or straight back into the hand, and torque. There's probably issues in terms of the rapidity of the impulse-a fast, sharp peak vs a slow round hump on the scale. There's certainly other issues I haven't thought of yet.

    I'd love to see studies done which look into these different aspects. Thus if one is sensitive to muzzle climb, they can look and see which guns/calibers have the least amount of climb. If one is sensitive to torque, same story.

    It would take some of the subjectivity out of the discussions.
     
  13. 1911Tuner

    1911Tuner Moderator Emeritus

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    Basically, the science is no more complex than Newton's 3rd Law...but there are other factors to plug into the equation when we bring up the recoil of a firearm.

    Torque occurs because of another equal and opposite created by the direction and pitch of the rifling.

    Muzzle flip is a function of the bore axis above the centerline of the wrist and the grip angle itself.

    Then, there's that one point that so many people fail to consider...and one that I've argued for years.

    The delaying/slowing effect of the bullet as it passes through the bore.

    We understand that the bullet being forced through the bore under high frictional resistance. All that's required to understand how high those forces are is to try to push one through by hand.

    This resistance causes the bullet to exert a forward drag on the barrel...and that drag is present as long as the bullet is there and moving.

    This forward drag exists while the gun is recoiling backward...and since the barrel is moving backward at the same instant that the bullet is moving forward...the rearward acceleration is being resisted...buffered, if you will...by the bullet's influence.

    Imagine a cork attached to a rope...fitted tightly into a plastic pipe.

    Grasp the pipe in one hand and the rope in the other...and pull. The instant that the cork slips and starts to move in one direction...the pipe slips and starts to move in the other.

    Keep pulling. The cork is being resisted by the pipe...and the pipe is being resisted by the cork. It requires continuing equal and opposite force to pull them apart...all the while, each one under resistance by the other.

    The "Equal and Opposite" thing works in reverse, too. When you push on an object, you are immediately pushed BY that object through the vectored force provided by your arm.

    Likewise, when you pull on an object...you are immediately pulled BY that object.

    Whatever resistance is imposed on the bullet by the barrel is also imposed on the barrel by the bullet.

    This delaying effect is only present in locked or fixed-breech weapons. Straight blowback designs aren't affected because the breechblock and barrel aren't connected. In these, what we feel as recoil is a function of the recoil/action spring, and the slide striking the impact abutment in the frame.
     
  14. Gun Slinger

    Gun Slinger member

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    Since perception is neither objectively quantifiable (except in subjective terms and it has no standarized unitary value assigned to it) nor a constant from one person to another, it is not likely that someone will have the same perception to a given quantity of recoil that another person has so the best means to convey the recoil of a given gun/load combination remains that as expressed in terms of foot-pounds of KE. Unless one has an objective unitary measuement with which to refer to a level of recoil generated by a particular gun/load combination, you can only convey what you feel and the person to whom you are describing such a sensation to, will still need to "percieve" it for himself and will do so, most likely, with a different outcome than you experienced, rendering your "evaluation" of the effects 'moot' for him.
    If nothing is percieved then there is nothing to quantify in the first place and such a comparison would be irrelevant since even airguns and the gentlest of rimfires generate a level of recoil (albeit a miniscule level) that can still be percieved by anyone.
     
  15. Gun Slinger

    Gun Slinger member

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    Actually, muzzle velocity has everthing to do with recoil energy.

    Since momentum (ρ) is the product of mass (m) times velocity (v):

    ρ = mv therefore mv = mv

    then we can conclude that the mass of the bullet times the velocity of the bullet divided by the mass of the gun will equal the "recoil" velocity imparted of the gun:

    m(bullet) x v(bullet) / m(gun) = v(gun)

    Since there has been a resultant "recoil" velocity (v) imparted to the gun, the gun now also possesses momentum (ρ) as well as a certain amount of kinetic energy that is related to its momentum ρ = mv via the integration of the momentum equation (ρ = mv) with respect to v as below:

    mv δv = m v δv = m ½v² = ½mv² = (KE) Kinetic Energy

    Kinetic energy and momentum are undeniably related and neither momentum (ρ) nor kinetic energy (KE) is time dependent (Δt) until such momentum (ρ) or kinetic energy (KE) is applied to a second body. At that point, momentum (ρ) will be conserved and only KE of the second body will differ since only momentum is (mathematically) conserved in such a "Newtonian" system.

    Regards,
     
  16. 1911Tuner

    1911Tuner Moderator Emeritus

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    Nope. You didn't read anything except the one sentence...and ignored the rest.

    Velocity at the muzzle is the result of the acceleration AFTER the punch that gets the bullet up to the speed that allows the small percentage of added accelleration that occurs...and provides final velocity at bullet exit.

    That initial punch is where 90% of the velocity occurs...for pistols, it's within the first half-inch of bullet travel...and probably more than 90% of the total recoil impulse.

    Go back and re-read the whole thing.

    Don't believe it? Try this. Warning:
    It's illegal...but it will demonstrate the reality in neat fashion.

    Go buy a cheap single-shot 12 gauge shotgun. Cheap is good, because you'll be destroying the barrel shortly after you do this.

    Lop off the barrel flush with the end of a fired shell, so that no projectile acceleration will be gained after the initial punch. Add weight to the stock so that the gun's mass will be equal to the original.

    Go fire the gun.

    I don't recommend placing your face on the stock's comb unless you relish the thought of a black eye and/or a bloody cheek. Ask me how I know...

    Cheers!
     
  17. Gun Slinger

    Gun Slinger member

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    I read your entire post prior to making my first post (#15) above.

    The experiment that you propose (and no I won't try it 'cause I an awfully fond of my freedom :D ) simply changes the location of the muzzle so that it is at the same point that the shotgun shell ends. In other words, it is still the "muzzle" (a.k.a. the end of the barrel) and it is the point at which the highest velocity of the projectile will be achieved since the shot/wad is no longer being accelerated by the high pressure propellant gasses remianing trapped behind it. The only change effected will be that the bullet (or shot charge) will not exit the "muzzle" as fast as it would as if it had the entire length of the barrel in which to continue to accelerate from the continued force (F) of the expanding gasses produced by the burning propellant. While the shot and wad will most certainly leave the 'new muzzle' (created by cutting the barrel off so that it is now the same length as the shot shell) at high velocity that veocity will not be as high as it would have been had the shot and wad been allowed to accelerate the full length of the barrel. I could not have come up with a better example with which to illustrate the point that I have been trying to make.

    Arguably, there will be some recoil with the arrangement that you suggest because you still have a mass (m) namely the shot and wad leaving the 'new muzzle' that you have created at a lower velocity (v) that will still result in momentum as well as K.E. being generated in both directions.

    The fact is, that regardless of how short the barrel is made, there will always be velocity (v) since the mass (m) is moving out of the barrel regardless of its length which means there will always be momentum (ρ) and K.E.

    I even find you (quoted again for the sake of clarity) to be in agreement with what I just explained above:

    The "bullet exit" is the muzzle.

    The acceleration of a bullet (or shot and wad) is a single impulse of exceedly short duration and momentum (not to mention K.E.) is a result of the final velocity (v) obtained of the mass (m) as it exits the muzzle even if the "muzzle" coincides with the end of the cartridge casing or shotshell mouth. It also a fine example of Newton's Second Law of Motion:

    "The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."

    Expressed mathematically, Newton's Second Law of Motion is:

    F = ma

    where "F" equals force (this is the "PUNCH" to which you refer above in your quoted statement), "m" equals mass and "a" equals acceleration.

    Additionally, if you substitute the value 0 (fps) for the velocity variable (v) in any of the following equations you will see that since any number multiplied by zero equals zero that you cannot possibly have momentum or kinetic energy without having velocity.

    mv δv = m v δv = m ½v² = ½mv² = (KE)

    ρ = mv

    which proves the following statement false:

    because it would violate the Newton's Third Law of Motion:

    "For every action, there is an equal and opposite reaction."

    Mathematically expressed Newton's Third Law of Motion is:

    mv = mv

    which proves again that muzzle velocity has everything to do with momentum (ρ) as well as the kinetic energy (KE) of both the bullet (or shot/wad) and the recoil of the gun.

    Regards,
     
  18. 1911Tuner

    1911Tuner Moderator Emeritus

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    Then you understand the physics of it without understanding internal ballistics.

    Force forward equals force backward.
    The maximum level of recoil occurs at peak force. Peak force occurs with peak pressure...and peak pressure doesn't occur when the bullet is about to clear the muzzle. It comes within a very small amount of bullet movement...even with slow rifle powders With some extremely quick pistol powders, it may peak before the bullet has even left the case.

    See Newton 1A: Objects at rest tend to remain at rest. (Inertia)

    Any object requires more force to get it started than it does to keep it moving afterward. It's during this rapid climb to peak pressure that the greatest percentage of the total muzzle velocity occurs...and likewise for the recoil impulse.

    As the bullet accelerates ever faster, it requires less force to accelerate it...ever faster. Thus the rapid pressure drop...and hence less force forward/backward...results in only a small amount of recoil after the punch.

    Consider again: (And these figures are representative rather than absolute...but experimentation will prove them to be pretty close.)

    If the barrel is 5 inches long, and the chamber accounts for an inch of the total...and we can figure on the 30 fps of velocity per inch pf barrel gained or lost rule of thumb...and the muzzle velocity is 850 fps:

    120 fps comes from the barrel length.
    That leaves 730 fps unaccounted for.

    If the initial punch takes up a half-inch...that leaves 3.5 inches of barrel to accelerate the bullet to the final velocity. Now we're down to
    105 fps after the punch..and we have 745 fps unaccounted for. Since it can't come from the 3.5 inches of barrel...it has to occur early in the bullet's travel...during the violent acceleration as the pressure climbs toward peak.

    And, now we've returned to force forward and force backward.

    I understand that as long as the bullet is in the barrel and accelerating, that an equal and opposite force will be created. It's just not very much force compared to that generated at the outset.

    Finally...This isn't theory without substance. I've proven it with real guns and real ammunition in blind tests. Revolvers with barrels and with barrels removed, leaving nothing but the chambers, and barrels cut off, leaving the forcing cones and a half-inch of rifling in order to keep the bullets from tumbling and destroying my chronograph. Then...with the forcing cones and the small rifled portion reamed to leave just barely enough rifling to spin the bullet. You'd be surprised at how small the difference in velocity was with the same ammunition lot.

    And even more surprising was that nobody who participated in the tests could detect any real difference in recoil. One maintained that the recoil in the barrel-less gun was sharper...three times in a row...even with lower bullet velocity.
     
  19. Gun Slinger

    Gun Slinger member

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    Hey, you're the one arguing against Newton's Laws of Motion. I understand internal ballistics far better than you surmise above.

    This statement violates Newton's Second Law of Motion: F = ma ...

    ...because in order to require less force (F) to continue to accelerate (a) the bullet would then require the body (m) being accelerated to actually become lighter, an absolute physical impossiblity at best.

    I have, however, learned from this thread, that I cannot educate the ineducable. :banghead:

    I give up. :)
     
    Last edited: Apr 20, 2008
  20. 1SOW

    1SOW Member

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    Is this the resulting conclusion?
    View muzzle velocity as the 1/4 mile drag strip speed:
    At a drag strip the slower top speed car may beat the higher top speed car through the 1/4 mile due to faster early acceleration. That faster accelerating car would be harder to keep the front end on the ground.

    .45s are usually heavier steel guns with a bigger slower bullet that accelerates slower. Powder type has a big affect here.
    That slower accelerating bullet would have less percieved recoil given all the other factors mentioned are equal.
    The mass of the gun itself also follows the laws of inertia/how fast the muzzle will accelerate in an upward direction from a given force.
    Along with bore axis etc., different strength recoil springs in the same gun will change percieved recoil/flip by changing the rate at which the recoil is felt by the shooter.
     
  21. TMann

    TMann Member

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    Another factor that is difficult to measure is the relationship between the size of ones hand and the gun's grip. It's pretty obvious that a nice, wide, rubberized grip will feel better than a small, narrow, hard grip. However, I believe that the shape and size of the shooter's hand will play a part, too. I am a short guy with small hands, and I think that some small guns just fit my hand better than they would someone who has a big hand. For example, there are many people who hate shooting the KT P3AT, and will tell stories of having a bloody thumb by the end of their range session. And yet, I went out yesterday and shot 75 rounds through mine yesterday with no problems.

    If the grip doesn't fit your hand, you won't enjoy shooting it.

    TMann
     
  22. 1911Tuner

    1911Tuner Moderator Emeritus

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    No, I'm not. Actually you are...or at least you seem to be.

    You're still not grasping my point. I understand that recoil energy/momentum is a function of Mass x Velocity. It's that one phrase..."Muzzle Velocity"...that doesn't ring. Remove that from your statement, and we're in agreement. The speed that the bullet is moving when it exits the muzzle has very little to do with recoil impulse, because the gun is in hard recoil well before that point. By saying that muzzle velocity determines recoil, it sounds like you feel that recoil doesn't start until the bullet reaches the muzzle. Kunhausen made that same mistake in his early theory of function...which he amended in his later editions. If that is, in fact what you're saying...then you don't understand the physics of it, either.

    Plainly stated: Action and reaction begin at the same instant. No?

    Pressures peak, early...quickly...and with some powders...violently. Pressures don't continue to climb until the bullet exits the muzzle. The gun is recoiling well before the bullet exits, and whatever added recoil occurs after the pressure peak amounts to a nudge compared to the initial impulse. The only exception would be if the mass of the powder charge is equal to or greater than the mass of the bullet. Then, you would have a readily measureable recoil impulse after bullet exit, though it would still be relatively small because with such a powder charge weight/bullet mass combo...recoil impulse comes from the acceleration of the bullet AND the powder gasses, plus any unburned powder that remains after the onset. Essentially, double the bullet's mass is factored into the mass/velocity equation.

    Let's try a hypothetical:

    You've got a rifle with a 30-foot barrel, made of a new space-age metal that weighs the same as a standard 22-inch barrel. Let's assign .30-06 as the caliber, and use a slow powder...like IMR 4350.

    Even as slow as this powder is, it would burn completely up in the barrel before the bullet could hit the air. When you fire this long rifle...will it kick?

    Of course it will. Even though the bullet never reaches the muzzle...the rifle will kick, because whatever force is imposed on the bullet is imposed on the breechblock in equal measure.
     
  23. saltydog452

    saltydog452 Member

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    Can I cut in on this dance?

    I wish that I could find the photos, but I remember seeing high speed photos of the newest Federal HST ammunition being fired. The slide was clearly in recoil before the projectile had ever left the barrel.

    Bullseye shooters knew this decades ago when they stressed 'followthrough' and 'not giving up on the shot'. 'Course this was with softball loads moving along at a leisurely pace.

    salty
     
  24. 1911Tuner

    1911Tuner Moderator Emeritus

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    Salty...I'll go find the link to Tripp's stop-action videos that'll show it. Might take a while, so sit tight.

    You'd be surprised at the number of people who really believe that recoil starts after the bullet exits...even after ol' Isaac Newton explains in detail that it can't happen like that in this universe.
     
  25. Gun Slinger

    Gun Slinger member

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    Tuner,

    That's the point that I was trying to convey when I said that your statement,

    was contrary to Newton's Third Law. I don't dispute at all that the bullet makes an impressive initial gain in velocity during its first few inches of travel as internal pressure "spikes" or that as the maximum expansion volume of the burning proprellant is reached that pressure falls off rapidly towards equlibrium at atmospheric pressure or that while such a pressure decrease is occuring that the bullet continues to make velocity gains (accelerate) albeit not as great as that made during the first few inches of barrel dwell time and that recoil begins at the moment of bullet movement and ends when it leaves the muzzle (ΣW). If you'll take a look at the posts that I have made above, never have I made such assertions contrary to what I have just stated.

    Indeed, the only disagreement that I had with what you posted was the statement quoted above because when we calculate free recoil energy the only velocity that matters is muzzle velocity since we cannot choose to stop the being effected by the recoil impulse when the bullet is only a third of the way down the barrel and ignore the smaller, but not necessarily insignificant, incremental gains made in velocity after that point. Calculating free recoil energy is a function of the cummulative (total) acceleration of the mass down the entire length of the barrel or essentially:

    ∫ ma δa = m ∫ a δa = ½ma² = ΣW = the sum of all the work done to the bullet during its travel down the entire length of the barrel

    So after all of this, we find ourselves in agreement...:eek:

    Who'd a thunk? :scrutiny:

    :)

    Regards,
     
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