felt recoil??

This is a discussion on felt recoil?? within the Defensive Ammunition & Ballistics forums, part of the Defensive Carry Discussions category; I am a little surprised with some results that I got while testing a new pistol. I shot 5 different 45 SD rounds and (2) ...

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Thread: felt recoil??

  1. #1
    Distinguished Member Array Dragman's Avatar
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    felt recoil??

    I am a little surprised with some results that I got while testing a new pistol. I shot 5 different 45 SD rounds and (2) 230 grain regular pressure (1) 165 grain regular pressure (1) 185 grain +P and (1) 165 grain +P.

    I thought the 230's would give me the most felt recoil and I exspected the lighter +P's to give me more muzzle flip. but the most recoil felt from any of them was the 165 grain +P and it wasn't a snappy flip if was a straight back driving recoil. Was I just wrong in my assumption that light fast is snappy and big slow hits hard????
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    I am not a physicist but the way I understand it is:

    1. The vector of recoil (push back or flip up) is based on the geometry of the weapon. The closer your hand is to the centerline of the bore the more the recoil is a push.

    2. The energy "in" the bullet is mass x velocity squared all divided by 2. In this way a smaller but faster bullet may have more energy than a larger slower bullet - if the smaller bullet is going fast enough.

    This is my layman's understanding but if I am wrong, please enlighten me.
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    Ex Member Array Ram Rod's Avatar
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    It also makes quite a difference in what you shoot them through as far as semi-autos are concerned. An all steel pistol will absorb more of the energy than an aluminum or polymer frame. Then there's the recoil spring and length of barrel to take into account. Muzzle flip and recoil......muzzle flip will actually lessen the felt recoil since it's direction of motion deviates from the straight line force. The grip of the modern semi-auto pistol was never really designed to be a balance point even though some come close to being just that. Its just a way for us to hold on to and control the pistol, and often times a good place for the magazine.

    I am not a physicist but the way I understand it is:

    1. The vector of recoil (push back or flip up) is based on the geometry of the weapon. The closer your hand is to the centerline of the bore the more the recoil is a push.

    2. The energy "in" the bullet is mass x velocity squared all divided by 2. In this way a smaller but faster bullet may have more energy than a larger slower bullet - if the smaller bullet is going fast enough.

    This is my layman's understanding but if I am wrong, please enlighten me.
    The E=mc² and the "vector" (actually a radian) may be a good way to explain part of the theories involved, but with many variables, then calculating the origin of the axis. 2√MxC² works well on paper and the x/y axis on a single plane. In three dimensions, there will be another axis 'z' (perpendicular to x and y) which may determine that the x and y values don't actually originate at the intersection of the x/y axis and they may indeed be negative values according to 'z' which would be the center of where the other forces are acting upon it and also a possible "vector" of the x/z relationship according to 'y'.
    I'm no physician either. It's just fun to think about and talk about. Shooting a pistol and choosing ammo isn't rocket science. A sore thumb beats a headache any day. LOL!

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    Distinguished Member Array Dragman's Avatar
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    So with the fact that I fired all the rounds out of my Colt New Agent and used the same type of grip why would the lighter one recoil the hardest??? Is this just a case of that brand of ammo is hoped up to its limits and the others aren't as hot??
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    Quote Originally Posted by Ram Rod View Post
    The E=mc² and the "vector" (actually a radian) may be a good way to explain part of the theories involved
    Actually RR, "E=mc²" is the relationship between energy and mass postulated by Einstein, where c is the speed of light. For the current discussion, the kinetic energy is one half the mass times velocity squared - E = 1/2 mv². Just my geek side poking through.
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    To the best of my recollection....I think I stayed home from school with a runny nose the day the teacher covered that.

    Quote Originally Posted by gasmitty View Post
    Actually RR, "E=mc²" is the relationship between energy and mass postulated by Einstein, where c is the speed of light. For the current discussion, the kinetic energy is one half the mass times velocity squared - E = 1/2 mv². Just my geek side poking through.

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    Distinguished Member Array Dragman's Avatar
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    Quote Originally Posted by QKShooter View Post
    To the best of my recollection....I think I stayed home from school with a runny nose the day the teacher covered that.
    Me too!!!
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    Ex Member Array Ram Rod's Avatar
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    Quote Originally Posted by gasmitty View Post
    Actually RR, "E=mc²" is the relationship between energy and mass postulated by Einstein, where c is the speed of light. For the current discussion, the kinetic energy is one half the mass times velocity squared - E = 1/2 mv². Just my geek side poking through.
    Told you I was no physician. Substitute the intended equation.

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    Or... 'If it won't go in, use a bigger hammer'. Actually, I studied ballistics in the Army as part of the Master Gunner Course at Ft. Knox. Use to be quite good at drawing all those fancy 'Surface Danger Area Diagrams' based on the various 'firing tables' produced by Aberdeen Proving Grounds. Dang if I retained much of any of it, but it was fun to play that gig at the time.

    Currently, I’m using Buffalo Bore 158+P and it recoils like crazy no matter what size revolver frame I use. No, I would not care to shoot that stuff out of some vapor-weight. My medium stuff includes 135+P and it kicks milder though in similar backward fashion. For me, the lighter 110 or 110+P CD tends to flip/recoil upward.
    Last edited by Saber; January 1st, 2011 at 02:47 PM.
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    Quote Originally Posted by Dragman View Post
    So with the fact that I fired all the rounds out of my Colt New Agent and used the same type of grip why would the lighter one recoil the hardest??? Is this just a case of that brand of ammo is hoped up to its limits and the others aren't as hot??
    Force equals mass times acceleration. The actual transferred force depends on the amount of time it takes for the object to absorb the applied force. this is typically thought of in terms of bullet impact, but actually hods true with felt recoil as well. There is a point of momentum at which the smaller mass reaches the same amount of energy and points of momentum beyond that in which it is exceeded.

    Another way to think of it. Take 2 balls both made of the same material (baseball material). A person can throw a 5 lb ball at you and hit you with it. They won't be able to get a lot of speed out of it, but it will hurt. The impact point will be spread across a larger surface area and will displace the impact object (more time to absord the impact and you will move backwards). A MLB pitcher can throw a standard 5oz baseball at you and hit you with it travelling at 100 mph and it may hurt more than the 5lb shot. Less impact area and less displacement and time to absorb.
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    VIP Member Array Janq's Avatar
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    As correctly noted by Smitty.

    Energy = One half of Mass x Velocity as Squared

    Normally one would need to take into account the weight of the gun too when figuring recoil energy, but as in this case the gun did not change in any way...You can then wholly drop/eliminate that value altogether from your math.

    A decent article explaining the math, which is physics, can be found here.

    Calculate Recoil Energy
    Legendary shooting editor Jim Carmichel gives you the formula to calculate your gun's recoil energy
    Uploaded on September 18, 2007
    http://www.outdoorlife.com/articles/...-recoil-energy

    For shooters this stuff is important to know to _properly_ evaluate ammunition be it commercial or handloads, as related to specific application and use.

    The reason why the 165 gr. recoiled differently in felt characteristic has to do with mass differential of the projectile.
    It takes less energy AND time (!) to move a lighter in mass object (less dense) than to do same toward a heavier object.
    This in turn on it's own with all other variables (i.e. powder charge qty. and type) being equal will result in lower barrel timing as the projectile leaves the barrel quicker upon a theoretical constant of X psi pressure within the chamber and barrel. This is why 2 3/4" #4 buck feels like it hits harder and _different_ than #4 birdshot.

    Mass...It's what hits you, in the hands.

    Such a light projectile would very likely cycle same by felt characteristic and reduction in visual barrel jump (actual recoil...Which is different than felt recoil) as a 1911 chambered in 9MM.
    I own and run both; Fullsized full steel construction.

    Shooting involves nothing but math and physics.
    And as I tell my kids at home; "Math is fun"...And so is shooting. : )

    - Janq

    P.S.
    If you had instead reduced or increased the weight of the gun, rather than that of the projectile, very similar neat effects in noticed recoil characteristics would have occurred.
    Mass both imparts felt recoil (projectile weight) and enhances/reduces it (firearm weight).
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    VIP Member Array Tubby45's Avatar
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    A heavy bullet with less powder charge will recoil less than a light bullet and heavier powder charge.

    This is why gun gamers shoot heavy bullets with fast, light powder charges.
    07/02 FFL/SOT since 2006

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    Quote Originally Posted by gasmitty View Post
    Actually RR, "E=mc²" is the relationship between energy and mass postulated by Einstein, where c is the speed of light. For the current discussion, the kinetic energy is one half the mass times velocity squared - E = 1/2 mv². Just my geek side poking through.
    Gasmitty - what are you proposing is multiplied by 1/2 - m or mv2. In other words, which of the following formulae would you deem accurate?

    E= 0.5*(mv²) or
    E= (0.5*m)*v²?

    I think it is the first and we are on the same page.
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    Quote Originally Posted by Tubby45 View Post
    A heavy bullet with less powder charge will recoil less than a light bullet and heavier powder charge.

    This is why gun gamers shoot heavy bullets with fast, light powder charges.
    Agtee 100% ; )
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    VIP Member Array Janq's Avatar
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    KS it is the first.
    See my post secondary where I wrote the equation out in English (rather than just state the formula).

    Tubby the reason for that is because less/lower powder charge results in lower pressure and lower pressure nets lower energy, regardless of projectile weight. All other variables being equal.

    Gun gamers shoot heavy (i.e. 230 gr.) to make major as related to power factor minimums.
    You can run lower projectile weight but then that means making up for that reduction in energy by increasing powder charge cpability...Which is not as easy to do as simply backing down on powder charge to just above major minimum cutoff, running a heavy projectile at .230 gr., with the start point being high (maximal) rather than low (minimal).

    Math, it's about everything! : )

    - Janq
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