Hello, recoil question

[brow_tines]

Hello everyone, I have been trolling this forum for awhile now. I actually think I have read the last 25 pages or so just in the reloading section. My question is, recoil, the bigger the bullet the less recoil? Thanks for the info and help.

[chris in va]

No, usually the opposite. With the same powder charge, a heavier bullet will recoil more than a lighter one.

Think of it like moving a rock. A small one doesn't give much resistance, but a large one will 'push back' as you apply force.

[brow_tines]

Thank you, I guess what kinda confused me was bigger bullet=less powder, smaller bullet=more powder

[Brian Pfleuger]

As with so many things, there is no single answer that is always correct.

The discussion is also complicated by whether or not you're referring to "felt recoil" or "mathematical recoil".

Felt recoil is HIGHLY subjective and varies heavily between individuals. So heavily that it's not even a reliable indicator.

Also, some people perceive the "quicker" recoil that is likely to be produced by a light, fast bullet to be worse than the "slower" recoil that is produced by a heavy, slow bullet.... and some people don't.


On the question of powder weight versus bullet weight, the powder can indeed be a significant portion of recoil, up to 50% IIRC. There are many variables though. How completely the powder burns before the bullet exits the barrel and peak pressure are two of the important ones.

[Casimer]

The recoil is proportional to the inertial mass of the bullet, so the heavier the bullet the more recoil is imparted to the gun. But you're right that the charge will play a role. There will be charges for lighter bullets that create forces equivalent to heavier ones, by accelerating the bullet more. Recoil is basically the 'equal and opposite' force in Newton's third law of motion.

[Brian Pfleuger]

Originally Posted by Casimer

There will be charges for lighter bullets that create forces equivalent to heavier ones, by accelerating the bullet more.

Actually, what's being accelerated is the combined mass of the bullet and the powder. It's not quite as simple as powder plus bullet because some of the powder essentially stays in the chamber (as burn gases) and some of it accelerates with the bullet.

It's not as simple as "a heavier charge makes the bullet go faster".... the heavier charge has a DIRECT influence on recoil.

[FlyFish]

The combustion products of the powder are expelled out of the barrel at the same velocity as the bullet, and so form part of what's referred to as the "ejecta" - along with the bullet, of course. From the standpoint of calculating free recoil, it's the mass of the total ejecta that counts, so a grain of mass is a grain, whether it comes from the bullet or the powder (or, in some cases, from a sabot). Although it's a bit of an oversimplification, we can pretty much assume that the powder's contribution to the ejecta is approximately equal to the mass of the original powder charge. Peetzakilla is right that some of it stays behind, but most of it is expelled.

For pistol cartridges, the powder mass is usually small compared to the bullet (for example, I load my .45 ACP target rounds using 4.0 gr of Bullseye under a 200 gr LSWC), so doesn't contribute much to recoil. For rifle cartridges that may not be the case - in my .22-250 loads I may have 35 or more grains of powder under a 52-gr bullet, so the mass of the powder is contributing a significant fraction of the recoil energy.

[Brian Pfleuger]

Originally Posted by FlyFish

For rifle cartridges that may not be the case - in my .22-250 loads I may have 35 or more grains of powder under a 52-gr bullet, so the mass of the powder is contributing a significant fraction of the recoil energy.

Indeed. In fact, the powder charge can be MORE than the bullet... I load 22-250 with a 35gr Nosler ballistic tip and 40.7gr Win748 and my 204 with Win748 uses just 1gr less than the 32gr bullet.

[Loader9]

Up until I read how Hornady made their Super Performance ammo that is not only faster but has less recoil did I understand that recoil as we would like to identify by saying it is the reaction of the bullet accelerating down the barrel. A Reaction of the rate of acceleration versus mass of the bullet. Well, guess what, that ain't so. Not according to Hornady. In their findings, they state that 70-80% of recoil is AFTER the bullet leaves the barrel. So much for our physic classes in high school. While they didn't exactly state why that is, they did state that having a quick building pressure curve that has pretty much peaked at about the time the bullet is mid way down the barrel to very little pressure left when the bullet leaves the barrel. After I've thought about this for a few days, I'm going to assume (note, my assumption doesn't mean it's right) that the jetting action of the pressure as the gases leave the barrel account for a significant amount of recoil. It stands to reason that there will be a "recoil" from the jetting or thrust of the gases when they exit the barrel. I would not have thought that it would be near as much as the 70-80% quoted by Hornady. YMMV

[TXGunNut]

Welcome to the asylum!
I'm a little confused by Hornady's rocket science but the theory may have merit. I'm on board with the "equal and opposite" school of thought and I agree that some rifles, cartridges and even shooters handle recoil better than others. As I understand it the Hornady bullet continues to gain speed after it leaves the barrel, pushed along by the still-expanding gases. I'm pretty sure this final push is a minor effect and very inefficient but for all I know some of my reloads work the same way.
Kinda funny, some ammo companies are marketing ammo with less flash, now Hornady wants to sell a bigger one.

[Shoney]

The light fast recoil vs. the heavy slower recoil question makes more sense if you realize there is a difference in the perceived recoil.

The fast light bullet delivers a slap recoil, while the heavy slow bullet delivers a long slow push recoil. The heavy bullet in fact, delivers more recoil, but our perception is that the quick slap has more recoil.

[Unclenick]

Hatcher's chapter on recoil explains the three sources of it:

One is equal and opposite reaction force to the bullet acclerating down the tube.

Two is the equal and opposite reaction force to a portion of the powder going down the tube with the bullet.

Third is the rocket effect at the muzzle from the propellant gas escaping and accelerating briefly to a velocity greater than that of the bullet.

For two and three, above, a perception pratfall that can occur is we are accustomed to think of gas as that thin, light fluid that is all around us. But inside the gun barrel under very high pressure, the gas is much more dense. Indeed, absent a nuclear reaction, matter is neither created nor destroyed in burning the powder, so the gas has the exact same mass as the powder that burned to make it. Generally, one assumes about half the mass of the powder, on average, is accelerated with the bullet. This is based on the gas at the breech remaining still, gas at the bullet base going the speed of the bullet, plus a bit of unburned powder going forward.

Reality is more complicated, of course. Some chambers on overbore cartridges are so big that a lot more than half the gas stays near the chamber end before the bullet exits. And we still haven't counted whatever unburned powder is ejected. The high pressure gases, being lighter than the bullet, scoot right past it. Schlieren photographs show the muzzle blast sphere expanding out ahead of the bullet until air resistance slows it toward the speed of sound and the bullet then blows past its leading edge. Hatcher found about 4700 fps was a good number to use with the .30-06 and 24" barrel. I've seen up to 5,500 used with some slow powders or short barrels, where the pressure at the muzzle is higher.

The bottom line for rocket effect is that it can contribute sometimes over 60% of the recoil impulse in the worst case. If you've ever looked at one of those muzzle brakes that vents the gas evenly in all directions perpendicular to the bore, you may have wondered how it could work when it isn't directing gas up to push the muzzle down, or rearward to pull the gun forward. The answer is it works simply by preventing (or at least mitigating) the rocket effect recoil you would otherwise have had without relieving the pressure before the bullet uncorks the muzzle.

[mehavey]

Read here:

http://www.chuckhawks.com/recoil_table.htm

Unless you step things way down (read lighter bullet and slower bullet for the same cartridge -- generally counter to the whole cartridge rationale/use), a cartridge class generally stays in the same recoil range up & down its bullet weight spectrum.

[FlyFish]

Unclenick: As always, thanks for weighing in on this. I learn something every time I read one of your posts.

[brickeyee]

The combustion products of the powder are expelled out of the barrel at the same velocity as the bullet

No, they move at the speed of the bullet in the barrel until the bullet exits, then they accelerate greatly.

You can see the muzzle blast passing the bullet in just about any high speed photography of a gun being fired.

To correct for this (AKA 'jet effect') the projectile velocity is used to estimate a gas velocity (a simple multiple) and the larger number used to calculate the momentum of the gases.
The fraction of mass from the gases that does NOT get ejected is painfully small and normally ignored.

Handguns normally have much lighter powder mass than bullet mass.

This is not true for some rifles were the powder mass can exceed the bullet mass (especially in smaller calibers at high velocity).

The free recoil calculation is not very complicated.

Weight of the bullet times speed + weight of the powder times bullet speed times multiplier (1.5 is often used).

Dived the weight of the gun by this sum to get recoil velocity of the gun.

Use 0.5 * mass * velocty^2 to find recoil energy.
The mass of the gun needs to be used here, not the weight.

A common mass unit is 'slugs' calculated by dividing the weight of the gun in pounds by the acceleration of gravity (32.174 ft/s^2).

How this force feels has a lot to do with jerk (how quickly it is applied) and how the gun 'fits' the shooter (including recoil pads, stock shape, etc.).

[Unclenick]

To be more complete, you need to include item 2 on my list, above. Hatcher's approach to covering that was simply adding a little less than half the powder weight to the weight of the bullet before making the bullet recoil calculation. He then ran the gas at 4700 fps, or about 1.7 times the bullet velocity. Since the calculation precedes more modern measuring methods, I assume that was deduced by firing a rifle mounted to a ballistic pendulum or some other means of measuring free recoil, then working backward to account for it.

One practical thing about free recoil that should be mentioned is that how hard recoil feels to the shooter depends a lot on how the butt of the stock is pulled into the shoulder. Many people instinctively hold the gun away from the shoulder on the bench thinking to avoid the recoil, but this actually lets the gun build up some speed so it smacks into your shoulder, adding to the perceived sting. If you pull the gun firmly back into the shoulder, you avoid this.

[zippy13]

As usual, Unclenick is spot on. Even back in Gen. Hatcher's day, with limited technology, it could be demonstrated that free recoil was more than just an evaluation of the bullet's energy and the mass of the gun. Yet, some still hold to the belief that the bullet's muzzle energy tells all.

Something frequently not mentioned, when discussing perceived recoil (or kick) it the event time duration. For a given free recoil, as the event time increases, the kick will lessen. By expanding the event the maximum unit stress is reduced, and we judge kick based on the maximum unit stress. Increasing the event duration, or buffering, is a classic method of reducing kick.

[brow_tines]

Thanks for all the replies, everything was sum good reading.

[brickeyee]

"Even back in Gen. Hatcher's day, with limited technology, it could be demonstrated that free recoil was more than just an evaluation of the bullet's energy and the mass of the gun. Yet, some still hold to the belief that the bullet's muzzle energy tells all."

You are correct.

Even way back then they new it was the bullets momentum, not its energy.

[TXGunNut]

Laws of physics haven't changed. Technology has given us better ways to measure, evaluate and understand them.

[mapsjanhere]

Use 0.5 * mass * velocity^2 to find recoil energy.

No, the energy is not equal, the impulse is (otherwise a 10,000J 50 BMG would go straight through your shoulder, the actual recoil energy is only 100 J).
Impulse is mv, so while in light bullets the faster speed gives you lot more energy (due to the velocity squared term) the recoil goes down for the same energy delivered.

[Unclenick]

What is equal and opposite is imparted momentum, as Brickyee said. You find the free recoiling velocity of the gun by dividing the bullet and gas momentum by the mass of the gun. Then you perform the energy calculation for recoil based on that velocity and the gun's mass.

The reason for the energy calculation, after already having worked out the momentum, is perceived pain in the target. A typical passenger car rolling forward on a flat at less than half a mile per hour has as much momentum as a 5.6 ounce baseball thrown at 80 mph. Yet, putting your hand out to stop the car rolling doesn't hurt like putting it out to stop the baseball does. Because of the square term in velocity, even though their momentum is the same, the baseball has lots more kinetic energy than the car. The pain felt comports well with the energy number, but not with the momentum number in that case.

[brickeyee]

"No, the energy is not equal, the impulse is (otherwise a 10,000J 50 BMG would go straight through your shoulder, the actual recoil energy is only 100 J).
Impulse is mv, so while in light bullets the faster speed gives you lot more energy (due to the velocity squared term) the recoil goes down for the same energy delivered."

The velocity in this case is from the momentum of the gun recoiling, and the mass is the gun's mass.

This is how free recoil energy is calculated.

Conservation of momentum gives a velocity for the recoiling gun, 0.5*M*V^2 gives the recoil energy the gun now has.

[Doodlebugger45]

I can't argue with all of the scientific evidence presented here. It is all sound. But from a very practical standpoint, no matter how many formulae I see, my own loads in rifles tells me that the heavier the bullet, the heavier the recoil will be. It's no big deal in my .243 loads. They don't seem to have any recoil at all regardless of the combination. But when I get into my .325 WSM, I can tell the difference immediately when firing a 170 gr bullet compared to a 220 gr bullet. I think the answer is that even when we go up in bullet weight, we try to attain velocities that are as close as we can get to even lighter weight bullets. The laws of physics don't change to suit our whims. But our whims do change. All I know is that my heavier bullets do recoil and hurt more than the lighter bullets.

[Unclenick]

That's not contrary to the science, it turns out to be exactly what you'd expect. The heavier bullets are normally propelled with greater momentum than the lighter ones. Again, it's because energy is proportional to the square of velocity.

Example, take a 100 grain bullet with 1000 ft-lbs of energy and a 75 grain bullet with 1000 ft-lbs of energy. Divide 1000 ft-lbs by the bullet weight, take the square root, then multiply the result by the bullet weight. You now have a number proportional to momentum. (If you are in the right unit system it actually will be momentum, but proportionality is all that matters for the comparison.)

75×√(1000/75)=274
100×√(1000/100)=316

So, even though the muzzle energy is the same, the momentum is 318/274= 1.16 times bigger for the 100 grain bullet. There are the other recoil factors involved, but that gives you the general principle. So it's not your imagination.