Question about recoil
With the same parent case, say 260 rem and 308 or 270 WSM and 325 WSM. Why is it that the smaller caliber seems to have less recoil. Yes I'm assuming same weight rifle, size and fit of stock(but that's more felt recoil). Even though you gain velocity do you lose efficiency?
Also, when do you feel the recoil, when the powder is burnt and gases expanding as bullet travels down barrel? Or when the bullet leaves the barrel and the blast comes out pushing the rifle back?
Also, when do you feel the recoil, when the powder is burnt and gases expanding as bullet travels down barrel? Or when the bullet leaves the barrel and the blast comes out pushing the rifle back?
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1. Bigger bullet = more weight = more energy to move it = you feel it more.
2. Bullet is out of the barrel and on it's way before you feel it.0 -
Getting to Newtonian physics, recoil is primarily a function of MOMENTUM, which is itself the product of mass and velocity.
So given two otherwise identical guns with similar energy loads, the one firing the HEAVIER BULLET will have more recoil. Bigger calibers generally mean heavier bullets, and there is your answer.
Note that its possible for a lighter bullet to offer more recoil *IF* the velocity multiplied by its mass is higher, but in the real world, cartridges are not generally loaded that way.
To answer the second question, the same hot gas that pushes the bullet out of the barrel simultaneously pushes the gun against your shoulder. The two things HAVE to happen at the same time since they are the same process.
Again, its simple Newtonian physics: for every action there is an equal (and opposite) reaction.
So when the bullet starts moving, the gun starts moving. In other words the recoil necessarily STARTS before the bullet has left the gun! You can even see this happen if you watch high-speed/ stroboscopic photography of a gun firing.
Now, as to when you FEEL the recoil, that's a little bit of a different thing. The human perception of touch is dependent on the nervous system. The recoiling gun presses against your skin, activating mechanoreceptors, those chemically synapse to nerves, the nerve signal goes to the brain, where it is processed in the cortex.
All this happens extremely fast, but its still going to take maybe 1/10 of a second after being touched before you become consciously aware of it.
So by the time you PERCEIVE the recoil, the bullet should be out of the gun.0 -
Skyfish:
It is an interesting question, and I am glad that you asked.
I am no phycisist, nor am I a scientist. But what little I know suggests that the recoil is the result of the energy created, and more energy means more recoil, other things being equal.
An example to illustrate is garnered from the Weatherby.com website, where I compared the energy of two Weatherby cartridges that use the same case: the .257 and the 7mm. The .257 generates 2826 ft. lbs. of energy at the muzzle with the 87 grain factory load, where the 7mm generates 3662 with the 175 grain factory load. This is almost a 30 percent difference in energy! The 7mm will recoil more.0 -
quote:Originally posted by stevecrea
Skyfish:
It is an interesting question, and I am glad that you asked.
I am no phycisist, nor am I a scientist. But what little I know suggests that the recoil is the result of the energy created, and more energy means more recoil, other things being equal.
Actually its not true that more kinetic energy necessarily means more recoil (see below).
Again, recoil is mostly dependent on momentum, which is a product of mass and velocity.
Kinetic energy is mass x velocity squared. . .its far more velocity sensitive than momentum. That means that lightweight bullets moving very fast can have high kinetic energies, even though we know in the real world that fast light-moving projectiles tend to offer low recoil.
It is true that given identical guns, and identical weight bullets, the one moving faster will have more kinetic energy, and generate more recoil. . .but it will also have more momentum, a better explanation for the increased recoil.
If the mathematics/physics here isn't clear, I have a good real-world example to think about.
Which pistol offers more recoil, 9mm luger or .45 ACP?
I think virtually everyone who has shot both rounds from similar pistols (eg 1911 pistols) would agree that the .45 has more recoil.
Yet if you look at the ballistic tables, the two actually have essentially IDENTICAL kinetic energies!
Both a standard 115 grain ball 9mm luger at 1200fps and a 230 grain .45 ACP at 850 fps each generate about 350 ft-lbs muzzle energy. The exact figure will depend on the load and gun in question, but they are in fact quite close.
So why does the .45ACP have more recoil? Because it fires a HEAVIER bullet, and therefore even though the energies are the same, the .45 has more MOMENTUM. (This also explains, by the way, why the .45 is generally the better performing round).
The guys who shoot IPSC competition are well aware of the value of momentum. They call that measurement "power factor", and they use it to qualify their weapons for competition. Why do they use that instead of kinetic energy? Simply because power-factor (eg momentum) helps normalize recoil between calibers better.
If they didn't use momentum, some wise-acre would show up with a gun firing 40 grain .22 bullets at super high velocity. That gun would generate negligible recoil, but equivalent energy to .45s, giving them an unfair competitive advantage.
As another real-world example that most shooters will be familiar with, in general recoil goes up as you go up to heavier bullets in a given caliber. Same thing. . .similar energies, but more momentum with the heavier bullets.
If you're interested, you can read more about the differences here:
http://128.110.226.21/guns/energy.pdf0 -
There's a rather complex formula for determining free recoil in the Lyman reloading manuals. It takes into account bullet weight, velocity, gun weight, and powder charge. 0 -
Here is how free recoil energy is calculated:
Free Recoil Energy = 0.5*(Wg/32.2)*((Wb*Vb+Vg*Wc)/(7000*Wg))^2
Free Recoil Energy [foot-pounds]
Wg = Weight of gun [lbs]
Wb = Weight of bullet [grains]
Vb = Velocity of bullet [feet/second]
Vg = Velocity of gun [feet/second]
Wc = Weight of powder charge [grains]
Here is how bullet energy is calculated:
Bullet Energy (BE) = 0.5*((Wb/7000)/32.2)*(Vb^2)
Bullet Energy [foot-pounds]
Essentially, both equations are Energy = 0.5MV^2
M = Mass
V = Velocity
There will be different variations on the above equations. I tried to use the same terms to illustrate the similarity of the equations.
Regards,
Heavyiron0 -
quote:Originally posted by Ambrose
There's a rather complex formula for determining free recoil in the Lyman reloading manuals. It takes into account bullet weight, velocity, gun weight, and powder charge.
Well, one fairly simple way to measure recoil in terms of how much "ouch" its going to put on your shoulder is to measure the kinetic energy of the recoiling rifle in foot-pounds.
Since (as above) momentum is conserved, all else being equal heavier guns lead to lower kinetic energy in the recoiling gun and less perceived recoil.
Put differently, if the mass of the gun is higher, its recoil velocity will be lower, creating less perceived recoil.
If you want to get all mathematical about it the momentum of the shot will equal the weight of the bullet x its velocity; the momentum of the gun will be equal to the momentum of the bullet, though in the opposite direction (since again, its conserved); the velocity of the recoiling gun will equal its momentum divided by its mass; and the kinetic energy of the recoiling gun will equal half its mass x the square of this velocity.
So, deriving the formula, from the above paragraph I think we get:
Recoil energy = (bullet weight x muzzle velocity)^2/ (2 x gun mass)
Of course this is now getting away from the original question, about why smaller bullets from the same case give lower recoil.0 -
skyfish,
Continuing on with this as sort of an Axiom to this. Two bullets of the same weight from different calibers and the same case, will recoil differently. Meaning a .257 Rob with a 120 gr. will give less recoil than a 7mmx57 shooting a 120 gr. bullet at equal velocities. The difference is the amount of square area the powder plasma has to push on. It has to push harder(have a higher pressure created for it) than does a 7mm bullet to get to the same speed.0 -
Heavyiron:
If I am interpreting you correctly then, recoil is directly correlated to the kinetic energy generated by the particular round, because they are largely the same equation?0 -
Stevecrea,
Correct.
Free recoil energy and bullet energy have the same governing equation which is:
E=0.5MV^2
One equation calculates the energy of the gun in motion the other calculates the energy of a bullet in motion. The same equation could be used to calculate the energy of a freight train in motion.
The energy for the gun is a slightly more complicated system and must therefore include the additional variables of weight powder charge (Wc) and the velocity of the gas leaving the barrel (Vg) plus the weight of the gun (Wg).
The equations will not factor correctly unless the weights are converted to masses with conversion factors which is what the other numbers are all about.
Mass is not synonymous with weight.
Mass = weight/acceleration due to gravity.
Regards,
Heavyiron0
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