Interesting: Recoil due to powder weight.

[jstein650]

Don't want to beat this to death, but I think it's important, and interesting to a lot of folks. I posted above (#18), but was a little vague. Here is the formula for calculating free recoil energy. The k here is a constant applied to the velocity of the powder gasses. Different constants are used because of the different dynamics of firearms. Shotguns with a long barrel use a low of 1.25 due to the relatively large bore, small powder charge, and low velocity. HP rifles use a high of 1.75, again relatively small bore, large charge and high vel. The reason it's not an 'exact' is that the powder gas column is expanding and the front of that expanding column will be different than the rear.

R.E. = [0.5 / M(gun)] * [M(bullet)*V(bullet) + M(powder)*k*V(bullet)]^2

Of course, you need to keep all the units in check; lbs, ft., sec.

 

[Clark]

The book "Elements of Ordinance" by Hayes 1938 to get the gas portion of the recoil. Chapter VII page 241 to 278 "The theory of recoil and recoil systems". On page 241 it says that:
Mvf = mv + uVc
That is the [mass of the gun] times [the free recoil gun maximum velocity] is equal to the [mass of the bullet]times [ the velocity of the bullet] plus the [mass of the powder] time the [velocity of the powder].
On page 242:
"experiments with the Sebert Velocimeter indicate..... the center of mass of the powder leaves the muzzle at 4700 fps"

My father used that book in the design of the M55, M107, and M110 for Detroit Arsenal, and the XM-70 for the marines.

 

[jstein650]

I'm not very familiar with those weapon systems, but I believe they are pretty high velocity numbers, which puts my figure in the ball park with regard to powder gas speed. I doubt though, that any part of the ejecta from a shotgun gets anywhere close to 4700 fps, hence the lower constants for those guns.

 

[1911Tuner]

The bullet exerts no force on the breechface.

In any action/reaction system, there must be a reactive mass or resistance. The bullet is the reactive mass. It provides a resistance for the force to push off of.

Recoil is more about the rate of acceleration and the force requirement to achieve that rate. The higher the rate of acceleration, the higher the force requirement. Force forward is force backward.

Let's do a hypothetical.

Two .44 Magnum revolvers, identical in every way...including weight...except for the barrel. One has a 10-inch barrel, and the other, a 2-incher.

Let's use a 240-grain bullet for both, and let's assume equal muzzle velocities of 1200 fps.

For 240/1200 in the 10-inch barrel, we'll load up with 2400

For 240/1200 in the two inch barrel...we'll use Bullseye.

Let's assume that the Bullseye charge doesn't blow up the gun, and...for the sake of simplicity...let's assume equal pressures.

Let's also assume that acceleration is linear, for the same reasons.

Let's ignore the powder/gas/ejecta exit mass and its effect on total recoil, for simplicity's sake, and just focus on a 240-grain projectile.

In the 10-inch barrel, the rate of acceleration is 120 fps per inch.

In the 2-inch barrel, the rate of acceleration is 600 fps per inch.

Which revolver will kick harder?

 

[jstein650]

OK. Same gun weight, even though barrel lengths are different, fine. You say ignore powder weight/ejecta. (I understand the 2400 & B'Eye hypothetical, but again we're ignoring those weights)
"Kick" is a subjective thing, that's why real recoil energy is used for a consistent definable quantity. Time is a factor in acceleration. Long barrel less force over more time, short barrel, more force over less time. The free recoil energy will be the same. One probably will perceive the short barrel to be 'sharper', and the long barrel to be 'more as a shove' as they say, but the energy will be the same.

 

[LiveLife]

1911Tuner did a good job of illustrating.

Hence they call Newton's laws of motion "laws" and not "theories" (like the "theory" of evolution which has yet to be proven ).

The law of physics is something that's been proven and although we could talk all we want and "theorize" but in the end, it will be the law of F=ma.

Yay, I remembered something from my physics class!

 

[Lost Sheep]

To clear it up

When I stated the bullet exerts no force on the breechface it was to emphasize the fact that


The bullet does not actually touch the breechface, thus cannot apply force to it.

I regret any misunderstanding

Lost Sheep

 

[jstein650]

bds: F=ma is true. Again, as I've stated above, recoil energy involves the factor of time, which is being ignored in 1911's example. Energy is not the same as force.

 

[LiveLife]

bds: F=ma is true. Again, as I've stated above, recoil energy involves the factor of time, which is being ignored in 1911's example. Energy is not the same as force.

The factor of time is part of acceleration as it is defined as distance/time squared - http://en.wikipedia.org/wiki/Acceleration

In physics, acceleration is the rate at which the velocity of a body changes with time

The bullet does not actually touch the breechface, thus cannot apply force to it.

A fan may not actually "touch" the curtain but will "exert" force on the curtain to move.

In the same manner, a bullet may not "touch" the breechwall but will "exert" force on the breechwall and push it back which we feel as recoil.

 

[jstein650]

I think where we're getting our signals crossed with regard to how one looks at recoil. Yes, we're talking about a recoil FORCE, but how that force is applied over time is, what I would argue, is more relevant. Recoil ENERGY accounts for force over time. This is why in 1911's example, the recoil energy is identical; although the force on the bullet, and its opposite counterpart recoil force, will differ, if the same velocity is reached, the total recoil energy will be identical. More specifically, in the short barrel the force would be 5 times higher, but applied for a fifth of the time. I the long barrel would require a fifth of the force applied for 5 times longer.

 

[LiveLife]

OK. Same gun weight, even though barrel lengths are different, fine ... The free recoil energy will be the same. One probably will perceive the short barrel to be 'sharper', and the long barrel to be 'more as a shove' as they say, but the energy will be the same.

I think I am beginning to understand at least one aspect of your posts.

Let's say two different bullet weights/powder charge loads produce the same amount of muzzle energy but different muzzle velocities. This would mean they will produce the same amount of "total" recoil but not the same "felt" recoil.

The lighter bullet with more powder charge will need greater acceleration or higher chamber pressure which will produce more "felt" recoil than heavier bullet with less powder charge that needs less acceleration or lower chamber pressure which will produce less "felt" recoil. Even though muzzle velocities will vary for the two bullet weights, acceleration/chamber pressures will differ inversely to produce the same amount of muzzle energy or total recoil.


I am not sure if this will help but here's one example using 230 gr 45ACP bullet vs 155 gr 40S&W bullet to produce the same 165 power factor (PF) USPSA major loads:

PF = (bullet weight in grains x velocity in fps) / 1000 or Velocity = (PF x 1000) / bullet weight

So for the heavier 230 gr bullet, only 717 fps and about 10,000 PSI is needed but for the lighter 155 gr bullet, over 1060 fps and about 28,000 PSI to make 165 PF.

 

[jstein650]

Kind of... and this sort of brings us back to the beginning. Muzzle energy normally only refers to the energy of the moving bullet after it's been propelled out of the muzzle. If recoil, however it is perceived, only depended on its velocity and mass, the charge required to get it there wouldn't matter. Since the the cartridge does require the expulsion of nearly all the powder in the case at higher than the bullet velocity, it does matter. In the same rifle, if I can get a 180 gr bullet to 3000 fps with a 45 gr charge it will have less recoil than a load that required 60 gr, by a calculable amount.
1911's example sort of took that out of the equation, and brought up the possible dynamics of how a short quick acceleration would 'feel' compared to a necessarily longer gradual acceleration. The free recoil concept is an attempt to quantify ejecta mass, gun mass, and the energy that imparts to the shooters shoulder or hand as 'recoil'. And felt recoil is a rather nebulous thing to really measure. The effect of chamber pressure (and we usually only see 'peak' pressure), combustion time, barrel friction, firearm design and its center of gravity, among other factors have all been debated and certainly contribute to 'felt' recoil though.

 

[Clark]

jstein650
I'm not very familiar with those weapon systems, but I believe they are pretty high velocity numbers, which puts my figure in the ball park with regard to powder gas speed. I doubt though, that any part of the ejecta from a shotgun gets anywhere close to 4700 fps, hence the lower constants for those guns.

I have been building some rifles and one shotgun that have gas escapement below 1 atmosphere above ambient. That is the threshold of super sonic gas.


50CB short rifle


12 ga shotgun with long choke tubes that have choke tubes

But when Quickload calculated the muzzle pressure of my loud guns, 257 Roberts Ackley and 7mmRemMag, they are often 10kpsi at the muzzle.

And gas is going to come out of a big bore faster than a small bore, given the same pressure.

 

[jstein650]

bds: PF = (bullet weight in grains x velocity in fps) / 1000 or Velocity = (PF x 1000) / bullet weight

So for the heavier 230 gr bullet, only 717 fps and about 10,000 PSI is needed but for the lighter 155 gr bullet, over 1060 fps and about 28,000 PSI to make 165 PF.

One thing though, Power Factor is really momentum - mass x vel., while energy figures, whether muzzle or recoil use a square of the velocity x mass.

 

[LiveLife]

Well, I am still trying to grasp the OP questions and thought PF example might shed some light on the discussion.

I am not sure if this will help but here's one example using ... 165 power factor (PF) USPSA major loads:

 

[Clark]

We have a book that says for gun design, the center of mass of the gas leaves at 4,700 fps.
And I have guns that no gas, not even the leading edge, goes faster than 1,125 fps.

The variables to deal with are multiple;
1) the gas pressure at the muzzle [not max chamber pressure]
2) the impedance of the bore [small hole, slow flow]
3) the total volume of the bore and chamber

 

[jstein650]

Clark:
Exactly. By the way, those are some pretty wild looking guns! I imagine that shotgun is pretty quiet too, no?
I think the 4700 fps figure could only apply to a relatively short barrel, high muzzle velocity gun, on the order of 3000+ fps. Which is why as I was trying to explain different constants are used for different types of guns when figuring free recoil energy. The factors you mention, bore diameter, chamber and bore volume, etc. do affect this then as your experiments have shown.

 

[1911Tuner]

More specifically, in the short barrel the force would be 5 times higher, but applied for a fifth of the time.

Let's revisit this for a minute, and define what recoil actually is.

What we perceive as recoil is mostly momentum after recoil has ended. Recoil itself is only present while the bullet is being accelerated...because recoil is nothing more than backward acceleration.

In my hypothetical, time is much more of a factor than it is in an actual recoil event.

With even slow burning pistol powders, pressure and force rise rapidly, often reaching peak within a half-inch of bullet travel. With really quick numbers, it can happen before the bullet clears the case mouth. Bullseye is such a powder.

Since peak pressure equates to peak force on the system...maximum recoil acceleration occurs before the bullet even gets close to the muzzle, and any acceleration that occurs after the peak only adds to the total. We perceive this as a longer
recoil event...a "push" after the initial impulse...for lack of a better description, and by the time our brains can even process the fact that something has happened...it's over, and momentum is all that's left.

Free recoil energy is one factor and momentum is another...but the sudden acceleration...the "punch"provided by acceleration... is what it's about. The boxer's hard right cross stays in contact with his opponent's jaw for a brief instant after the initial shock and pushes him backward...but it was the violent acceleration...the punch...that put his lights out. The shove after the punch is just a bonus.

In the hypothetical, the 10-inch revolver would recoil pretty hard, but it's nothing that we can't handle. The 2-inch gun would probably fracture a few bones.

 

[jstein650]

How about your example like this: Same gun in every way. Same bullet and MV. Same 10" barrel. One uses a heavy charge of 4227 or H110, the other a light charge of, say B'Eye. How do you think the recoil would compare?

 

[ATLDave]

I don't think you can count it that way, the powder is a potential energy. Upon ignition it become kinetic energy, if there was another force acting on the powder and the powder was not acting as a source of work on the bullet then you might be able to include it into your recoil calculation

We're not doing nuclear fission in the chamber. The powder is not "converted to" energy. Chemical bonds are broken and created during combustion, and that releases energy, but the mass remains the same. That mass is ejected from the gun. It contributes to recoil. Whether the difference in the weight of one powder charge versus another is detectable by a human shooter is another matter.

 

[1911Tuner]

How about your example like this: Same gun in every way. Same bullet and MV. Same 10" barrel. One uses a heavy charge of 4227 or H110, the other a light charge of, say B'Eye. How do you think the recoil would compare?

The recoil impulse would naturally be less due to ifferent force requirements to achieve the rate of acceleration.

You can lift a 50-pound weight over your head at a rate of one foot per second pretty easily. If you try to lift the same weight...accelerate it off the floor...at a rate of 100 feet per second, it becomes a much different problem due to the force requirement needed to achieve that rate.

Another look at it, just to clarify.

An action/reaction event requires an accelerating force on to interacting objects. Once that accelerating force has been removed...as in at the point of bullet and residual ejecta exiting the muzzle...the system is no longer closed, and action/reaction ends.

Another way:

The only time that recoil is occurring is while the reaction side of the system is being accelerated by a force. No force, no acceleration. No acceleration...no recoil.

If you stand close to a wall and shove yourself away from the wall, you are being accelerated by the force provided by your arms. Once you've pushed yourself far enough for your hands to lose contact with the wall, you are no longer being accelerated, and an action/reaction system no longer exists. You have momentum and energy...but there is no recoil.

Back to the gun.

When the bullet/ejecta leave the system, the force that drove the system leaves behind it. The action side of the equation no longer exists, and without action, there is no reaction. Neither the bullet nor the slide/breechbolt can accelerate to a higher velocity. All they can do is decelerate from outside forces.

 

[Macchina]

Well, I am still trying to grasp the OP questions and thought PF example might shed some light on the discussion.

It was not a question, it was a statement. Rephrased: we often forget to add the weight of the powder when THINKING (not calculating) recoil as we plan new loads or decide on new calibers.

The point of this thread is that when we think of recoil, we tend to think of bullet weight but forget the total load weight which includes powder. My favorite .38 Special load (158g LSWC over 4g W231) has only 2.5% of the load made up by powder. Let's compare the powder/load ratios of a 300 RUM to a 300 WSM:

300 RUM: 130g bullet over 107g of H1000 gives us 45% of the load as powder. Velocity: 3615 fps.

300 WSM: 130g bullet over 77g of Superformance gives us 37% of the load as powder. Velocity: 3645 FPS

That's an 8% decrease in total load weight with a slight increase in velocity! You may not think 8% is that big a deal, but think about the recoil difference between a 130g bullet and a 160g bullet traveling at the same speed: it's very noticeable.
P.S. I used the first offered load data for 130g bullets from Hodgdon's Website for the above figures.

 

[jstein650]

I'm afraid, that after all of this discussion, and it has been enlightening, that a lot of casual readers do not equate, that, differences in the 'LOAD' (powder mass in your example) really contribute to recoil energy (however that may 'perceived'). And, I agree, these differences are realized in real life much more in the realm of higher powered rifles, than in, say a .45 Colt Cowboy load or a 12 ga. skeet load.

 

[jstein650]

"The recoil impulse would naturally be less due to different force requirements to achieve the rate of acceleration."

"With even slow burning pistol powders, pressure and force rise rapidly, often reaching peak within a half-inch of bullet travel. With really quick numbers, it can happen before the bullet clears the case mouth. "

I'm not sure in the 10" vs. 10" barrel example, which you think would have been different due to 'different force requirements to achieve the rate of acceleration'. Or which you think would have been greater.

"recoil is mostly momentum after recoil has ended".

The momentum imparted to the firearm; M=vm , which is why my (Lyman's) formula neatly takes into account the momentum of the mass of the firearm, as well the ejecta, based upon the VELOCITY of the ejecta, in the real world, and the effect the atmosphere has upon that moving mass. 'Clark' above, has experience with this, since he's been kind of basing his subsonic loadings based on atmospheric pressures.

Your information thus far, seem to concur with mine that the nature of the INTERNAL dynamics that occur during the cycle from ignition to muzzle exit; the DIFFERENCES in peak chamber pressure, the linearity of the acceleration, etc., all occur so very quickly, that the effect on felt recoil is negligible or intangible at best. The OP's question was always whether or not, and by how much, the powder mass contributed to recoil.

 

[Captcurt]

The online recoil calculators work well to calculate recoil.

And if you check you will see that powder weight is in the equation used to calculate recoil.