Understanding Momentum Conservation in Isolated Systems

[Win_94]

I need the concept of momentum conservation in isolated systems described (in layman's terms,) as it pertains to the recoil of firearms, as described on this page.

Bsharp.org : The Physics of Everyday Stuff : Gun Recoil

I assume I am understanding it wrong, thinking he is saying that "while the bullet is still in the barrel" recoil doesn't happen but is "conserved" until it exits the barrel.

If he isn't saying that, what is the point of Newton's first law as far as "unless acted on by an external, unbalanced force?" I don't get the "external, unbalanced" significance.

I bring this up because a reoccurring claim on gun boards is that; heaver bullets impact higher than lighter bullets using the same firearm with the same point of aim.
The reasoning is that heavier bullets are slower to exit the barrel so the recoil forces the barrel to rise, which results in the higher point of impact.

I need a physicist to write a compelling argument as to why this is false or true. I believe it to be false but no amount of evidence I provide is considered since I'm a lowly marksman.​

I have some of my so called "evidence" that may be of interest?

I reload ammo, and when I load develop, I use the Audette Ladder Test. It is to find where barrel vibrations are least at the muzzle as well as finding when powder charge weights reach plateaus of velocity.
Since every consecutive cartridge gets a bit more powder, it hits higher than the previous one. Here is an example of one of my latest tests.

There was a major vibration at the muzzle on shot #4 and the barrel whip happened to throw it higher than the next two. All other shots are true to the notion that the faster bullets impact higher; due to reaching the target sooner, therefore dropping less.

Every Audette Ladder Test I have done, with snub-nose 357magnums to 26 inch barreled 30-06 rifles, at ranges from 35 yards to 400 yards, all faster bullets hit higher than slower ones.

I would also add I've noticed a exception to the rule with bullets shot at just under the speed of sound hitting higher than ones shot at just above the speed of sound. A inch or so difference that I attribute to the "bullet bow shock-wave" that supersonic bullets experience.

Just in case that term is unfamiliar.
YouTube: Ballistic Shell Fired In Slow Motion​

Thank you for your time; I hope this wasn't too scatterbrained of a post.

 

[Rap]

It might be true. If you clamp the gun, the barrel will not rise. Are these tests with a clamped gun? If not, a line drawn thru the barrel will generally be above the point where the recoil is absorbed, and the barrel will rise. If the gun is clamped, the same weight bullets with higher charge will always go higher. When you talk about heavy and light bullets, do they both have the same charge? If they do, then they will have the same energy (I think). The heavier bullet will have more momentum causing more recoil, and the barrel will rise more causing the bullet to tend to go high. However, the heavier bullet will be slower, and gravity will have more time to act on it causing it to tend to go low. Figuring out which effect will dominate would need more information, like barrel length, where recoil is absorbed, amount of charge, etc.

About them having the same energy, I will have to think about that.

 

[Win_94]

It might be true. If you clamp the gun, the barrel will not rise. Are these tests with a clamped gun?

No. When using a rifle during the ladder test, I rest the forearm on a sandbag and the butt end of the stock on a bean bag. The butt is against my shoulder as normally done, my hand is on the trigger as normally done, but my other hand is squeezing the beanbag as to level the rifle to acquire the proper point of aim.
You can see how it is set-up in this video.
Eddystone Ladder Test

With a handgun I'll hold with both hands, with the barrel resting on a sandbag. Sometimes I'll forgo the sandbag if I forget it like this time.
357 Magnum, 180 gr HP-XTP, 100 Yards
As you can see, there was a lot of erratic impacts during that ladder test.

When you talk about heavy and light bullets, do they both have the same charge?

No. When dealing with standard charge weights, the heavier bullets always get a lesser charge as to not cause overpressure. Although they can use the same charge weight; the lighter bullet can usually handle a lesser charge without issue.

Figuring out which effect will dominate would need more information, like barrel length, where recoil is absorbed, amount of charge, etc.

I can give you an example.

Using a Winchester 94 30-30 with a 20" barrel, shooting a 150 grain full metal jacket with a ballistic coefficient of .398, using 33grains of Varget with the speed of about 2160fps, my estimated pressure of 33,150CUP; will impact the 350 yard target at the point of aim.

Whereas the same rifle, using a 130 grain flat nose with a ballistic coefficient of .249, using 35.9 grains of Varget with the speed of 2360fps, a book pressure of about 35,700CUP; will impact the 350 yard target 2 feet higher than the point of aim.

I got the pressure figures here.
Hodgdon: Reloading Data Center
you'll need to "agree" then press "Cartridge Loads", then select "rifle" then under Cartridge select "30-30 Winchester."
Then reference the "Varget" powder beside the bullet weights listed.

The rifle looks like this;

With both shots, the forearm is resting on a log, the butt is against my shoulder squarely, my left hand working the trigger, while my right hand is holding the butt to the height to get the proper point of aim.

I would like to note that heavier bullets almost always are more aerodynamic.​

Thanks for your help!

 

[Win_94]

Now that I'm thinking of it.

Here is a tool that I use to calculate drop.
Honnady Ballistics Calculator
Notice if the only variable changed is the bullet weight, that the only change in data is energy.

I just thought it might be of interest.

 

[Rap]

Now that I'm thinking of it.

Here is a tool that I use to calculate drop.
Honnady Ballistics Calculator
Notice if the only variable changed is the bullet weight, that the only change in data is energy.

I just thought it might be of interest.

That's pretty interesting. The reason the energy goes down for lower weight bullets is because the muzzle velocity is fixed and energy is mv^2/2 where m is bullet mass, v is muzzle velocity. You said the charge changes for different weight bullets. Does the muzzle velocity always stay the same? In order to figure out if heavier bullets hit higher, we have to keep something the same, so we are comparing apples to apples.

 

[Michael C]

I need the concept of momentum conservation in isolated systems described (in layman's terms,) as it pertains to the recoil of firearms, as described on this page.

Bsharp.org : The Physics of Everyday Stuff : Gun Recoil

I assume I am understanding it wrong, thinking he is saying that "while the bullet is still in the barrel" recoil doesn't happen but is "conserved" until it exits the barrel.

Yes, I think you're understanding it wrong. The primary recoil does indeed take place while the bullet is still in the barrel. The momentum of the whole system of gun + bullet is conserved, so when part of this system (the bullet) changes its momentum, another part (the gun) must change in the opposite direction in order to conserve the total momentum.

If he isn't saying that, what is the point of Newton's first law as far as "unless acted on by an external, unbalanced force?" I don't get the "external, unbalanced" significance.

If there was an external force acting on the system of gun + bullet in a particular direction, the momentum of the system gun + bullet could be changed. Just throw the gun away from you and you have changed its momentum by exerting an external force on it.

 

[Win_94]

You said the charge changes for different weight bullets. Does the muzzle velocity always stay the same?

No.
It is complicated, but on average; the same charge weight results in a heaver bullet traveling slower than its lighter counterpart.

In order to figure out if heavier bullets hit higher, we have to keep something the same, so we are comparing apples to apples.

I do have two loads that have the same point of impact, with the same muzzle velocity, which are different bullet weights.
The 357 magnum load I've displayed above, and this load.
357 Mag, Screwed-Up Video

The gun is a 357 magnum, 8" barrel. Both average muzzle velocities are around 1060fps. The differences are bullet weights, 180gr for the first and 158 for the recent video; and ballistic coefficients, .230 for the heavier 180gr bullet, and .206 for the 158gr.

Does the muzzle velocity always stay the same? In order to figure out if heavier bullets hit higher...

But that isn't their argument, their argument is: Anything that results in less muzzle velocity, results in higher impact; simply because recoil had more time to take effect.

The bullet weight is just an aside, a means to an end. Just as less powder equals less muzzle velocity, resulting in more time for the barrel to rise before the bullet exits.​

I know their notion to be false; but since I'm not a "scientist" and my data isn't derived from "controlled environments," it isn't taken into consideration. I thought that page written by the physicist was going to help me argue the point; and I needed conformation that it would.

My contention is; less muzzle velocity results in the projectile dropping much more than the miniscule amount of barrel rise that may take place.​

I think I'm going to have to find a different approach.
Thanks for your time, I really do appreciate it!

 

[Win_94]

Yes, I think you're understanding it wrong.

As time went on, I knew it meant something different than what I initially thought; whereas, it was not going to help me in my argument.

Thanks for your input!

 

[Rap]

Don't give up yet - we can't solve a problem that is not clearly defined, and I think you may have just clearly defined it.

You say their argument is: Anything that results in less muzzle velocity, results in higher impact.

You can show that that is obviously wrong, because if the muzzle velocity is one foot per second, the bullet will drop at your feet and there will be practically no recoil to raise the barrel.

Your contention is: less muzzle velocity results in the projectile dropping much more than the miniscule amount of barrel rise that may take place.

That is certainly true when the muzzle velocity is one foot per second, two feet per second, etc., but whether it is always true, I don't know, but it might be easy to figure out. I will play around with it.

 

[Michael C]

My contention is; less muzzle velocity results in the projectile dropping much more than the miniscule amount of barrel rise that may take place.​

I don't think we have enough information. How much is "minuscule"? What about the better aerodynamic properties of the heavier bullets? If the heavier bullets present less air drag than the lighter ones, they will be slowed down less by the air and have a higher impact point when other factors are equal (if you fired a heavy bullet and a light bullet at the same speed and angle in a vacuum, they would follow the same trajectory).

You need, as Rap says, to find a way of comparing apples with apples. Here are variables that could affect the height of impact:

1. Muzzle velocity
2. Barrel angle
3. Aerodynamic properties of the bullet

For the moment I can't think of anything else that would have a noticeable effect, but I may have missed something. Assuming that these three variables are the only important ones, you might start like this:

- Clamp the gun down so that the direction of recoil will be in line with the muzzle. Now calculate charges so that the heavier bullets will have the same muzzle velocity as the lighter ones. Fire series of heavier and lighter bullets. If there is a consistent difference in the impact height between the heavier and the lighter bullets, then in this case it must be due to the aerodynamic differences between them.

That first experiment would give you an idea of how much difference in impact height is caused by the different air drag on the two sorts of bullet.

Then you could use the same setup to see what difference is made by the muzzle velocity alone: fire series of bullets of the same sort with different charges.

Then, with the same setup, you could see how the differences caused by muzzle velocity and aerodynamics add up by firing both heavier and lighter bullets with different velocities.

Now it should be possible to sort out the different influences when you hold the gun. For instance, you fire the gun, holding it as usual, for a series of lighter and heavier bullets all with the same muzzle velocity. Then you could compare the results of this test with those of the similar test where the gun was clamped down.

 

[Win_94]

What about the better aerodynamic properties of the heavier bullets?

I knew my scatterbrained posts would make this confusing. I'm starting over, forget everything above this line.

People have claimed that:

Slow bullets remain in the bore longer than fast bullets.
Therefore; with slower bullets recoil has climbed more while the bullet in still in the bore, resulting in a higher point of impact.​

My contention is:

Less muzzle velocity results in the projectile dropping much more than the miniscule amount of barrel rise that may take place. Slower bullets hit lower.​

I goggled the argument; I don't know these people.
http://rugerforum.net/ruger-single-action/45793-heavy-bullets-hit-high.html
http://glocktalk.com/forums/archive/index.php/t-1364952.html

The common theme is velocity.

I would like to stress that slower bullets shot at 1050fps, (subsonic,) will hit higher that those shot at 1250fps, (supersonic.) Many of the handgun people are seeing that effect.

Therefore, we must deal with both bullets traveling at either subsonic, or supersonic speeds.​

 

[Win_94]

Don't give up yet - we can't solve a problem that is not clearly defined, and I think you may have just clearly defined it.

You say their argument is: Anything that results in less muzzle velocity, results in higher impact.

You can show that that is obviously wrong, because if the muzzle velocity is one foot per second, the bullet will drop at your feet and there will be practically no recoil to raise the barrel.

Your contention is: less muzzle velocity results in the projectile dropping much more than the miniscule amount of barrel rise that may take place.

That is certainly true when the muzzle velocity is one foot per second, two feet per second, etc., but whether it is always true, I don't know, but it might be easy to figure out. I will play around with it.

Thank you! I really do appreciate this!

 

[Rap]

Well, I thought about it, and it seems that if you start out with a level barrel, the recoil will make the barrel tilt up by some angle when the bullet leaves the barrel. That angle is a constant times the mass of the bullet. The constant depends on geometry and mass distribution of the gun, but not on muzzle velocity, and not how long the bullet is in the gun.

So there are different cases we can talk about. Let's ignore the effects of air resistance on different weight bullets. If you have a heavy bullet and a light bullet, and they both have the same muzzle velocity, then the heavy bullet will hit higher, because the gun jumps up more. If you have a fast bullet and a slow bullet and both have the same mass, the fast bullet will hit higher, because the gun jumps up the same amount in both cases, but the fast bullet drops less.

In one case the muzzle velocity was the same, in the other case, the mass of the bullet was the same. If you have a case where the bullet is going slower because its heavier, then neither the mass or the muzzle velocity is held constant. Then we have to hold something else constant so we can compare the two cases, and I'm not sure what you want to keep the same.

We could hold the charge constant. We could hold the muzzle energy constant. Whatever we do hold constant, I think it will turn out that there will be a certain range where the heavier bullet will hit higher because the angle is higher, but then beyond that "critical range", the heavier bullet will hit lower, because its slower. I guess you are saying that that critical range is rather small. I could try to estimate that critical range for constant muzzle energy if you want to keep going with this.

 

[Michael C]

Well, I thought about it, and it seems that if you start out with a level barrel, the recoil will make the barrel tilt up by some angle when the bullet leaves the barrel. That angle is a constant times the mass of the bullet. The constant depends on geometry and mass distribution of the gun, but not on muzzle velocity, and not how long the bullet is in the gun.

I'm interested in how you arrive at this result. For a given bullet mass, won't a higher muzzle velocity result in a greater recoil force, and therefore a greater torque causing more barrel tilt?

 

[Rap]

I'm interested in how you arrive at this result. For a given bullet mass, won't a higher muzzle velocity result in a greater recoil force, and therefore a greater torque causing more barrel tilt?

As the bullet goes down the barrel, let's say its position along the barrel is $x(t)$. The bullet is subject to a force $F=m\ddot{x}$ where m is the mass of the bullet and $\ddot{x}=d^2x/dt^2$ is the acceleration, and an equal and opposite force of the bullet is exerted on the gun, on a line that runs down the middle of the barrel. The point at which the gun rotates around is behind and below the axis of the barrel. Let's say the perpendicular distance of the pivot point from the axis of the barrel is h. Then the torque $\tau$ about that point is $\tau=Fh=I\ddot{\theta}$ where $\theta$ is the angle of the gun above level and I is the moment of inertia of the gun about the pivot point. The moment of inertia has nothing to do with the bullet, its just a matter of the shape of the gun and how its mass is distributed inside that shape. (That assumes the bullet mass is much smaller than the gun mass). So now we have: $$\tau=mh\ddot{x}=I\ddot{\theta}$$ The general solution to this equation is $$\theta(t)=\frac{mh}{I}\left(x(t)+a+bt\right)$$ where a and b are constants. At time t=0, we have $x(0)=0$, $\dot{x}(0)=0$, $\theta(0)=0$ and $\dot{\theta}(0)=0$, which means that $$\theta(t)=\frac{mh}{I}x(t)$$ Let's say the barrel length is L. In other words, at the time $t_f$ that the bullet exists the barrel, $x(t_f)=L$. Then when the bullet exits the barrel, the angle is $$\theta(t_f)=\frac{mhL}{I}$$ which is just a function of the bullet mass, the gun geometry and mass distribution, and where the pivot point is. This ignores friction of the bullet and the barrel. I think that will change things because usually the friction force is proportional to the bullet velocity in the barrel and the area of contact, and I don't know what that constant of proportionality is. I might be able to estimate it, to see how much of an effect it is.

 

[Win_94]

Whatever we do hold constant, I think it will turn out that there will be a certain range where the heavier bullet will hit higher because the angle is higher, but then beyond that "critical range", the heavier bullet will hit lower, because its slower. I guess you are saying that that critical range is rather small. I could try to estimate that critical range for constant muzzle energy if you want to keep going with this.

I was thinking the distance was the major factor. Many hand-gunners shoot at 7 yards because they are self-defense oriented. I rarely shoot my handgun at less than 35 yards since I am hunting oriented.

Even though my bullets travel 5 times the distance of theirs, I would think at 7 yards the height difference would be too hard to discern. Much, much less than most shooters can group their shots. Maybe 1/4 an inch?

Take this an example from the post I linked to earlier.

I bought a box of 240gr Winchester bullets and at 25 yards they hit 6-7 inches higher than the 180's I set the sights in with last week!

Is there some way that you could discount or confirm that amount recoil difference at that range? That gun weighs 48oz. with a 7.5 inch barrel.

I'm thinking, at the most it would climb 1/2 an inch at 25 yards; and they would not notice it because they probably shoot 3 or 4 inch groups.
And at that rate, the "critical range" would be about 300 yards.

Here is my target using my 357, I didn't have enough bullets to load enough cartridgess to be thorough.

But I think that my experiment was flawed since my faster bullets were traveling supersonic and my slow bullets were subsonic. There is a big difference in velocity reduction between the two velocity ranges due to the shock-wave generated by the supersonic bullet.
But my load using 158gr bullets and the load using 180s, having near the same subsonic velocity, hit the same height from 50 yards to 125 yards.
125 Yard 357 Zero
I don't have a 50 yard target that I can find.

I think most people that see these great height differences are experiencing barrel harmonics; especially the ones not hand-loading in an effort to eliminate the harmonic effect. I would just like to point to reasoning as to why the POI difference is not that great due to recoil.​

I thank you for all your time and effort! You've gone above and beyond what I expected! Thank you!

 

[Rap]

I thank you for all your time and effort! You've gone above and beyond what I expected! Thank you!

Thats ok, I do this sort of thing for a living (computer modelling scientific instruments and experiments).

 

[Win_94]

I have a better understanding of what your saying than I had when I first read the explanation. The first paragraph in comment 13 completely dispels the notion; while the math in comment 15 proves it.


Just in case you might be interested, I have another few videos I made yesterday and today on the subject.

A rifle at 100 yards; POI Difference Again.
A pistol at 25 yards; POI Difference 38 Specials

Again, thanks for your time and effort; this really means a lot!

 

[Win_94]

Sorry to bother you again, but what you wrote, does it jibe with this comment I recieved?

Ok, the physics are correct but there's nothing like empirical data so, I called an old friend of mine today. He's a retired engineer from the Rock Island Arsenal. He's done work on the m198 Howitzer and the new 105mm Hawkeye.

His work involves the mechanics and stresses associated with recoil so I asked him about this subject.

He tells me that both sides of the argument have merit but only if we are willing to split hairs. At initial ignition of the charge, there is a small and barely perceptible motion of the barrel/breach. (only detectable by sensitive instrumentation) However, this motion is of so little significance that it has no practical effect on POI.

However, the motion widely accepted as recoil, were it to take place while the projectile were still in the barrel, would make POI so unpredictable as to make any sighting system useless; nothing more than decoration on the weapon.

In short, recoil occurs only after the projectile has exited the barrel and has no effect on POI.

Everyone satisfied?

...

I got home from a long drive and didn't refer to my notes. There are 2 points I didn't include in my post:

1. the initial and small movement is so brief that it ceases before the projectile moves.

2. The major barrel movement we know as recoil does not occur until after the projectile has exited the barrel.

I received those comments before I posted parts of your last explanations.

Also... Is it okay if I link this post from the other? Or how about I link you to there?
http://www.thefirearmsforum.com/showthread.php?t=100867&page=2
The person that posted the above information is "Hedge."

Thanks again!

 

[Rap]

Sorry to bother you again, but what you wrote, does it jibe with this comment I recieved?

It does not. I cannot agree that recoil only occurs after the bullet leaves the gun. The bullet is pushed down the barrel by the pressure of a gas in the barrel. This pressure force on the bullet is equal and opposite to the pressure force on the gun - Newton's law. As the bullet leaves the barrel, there will be some high pressure behind it and this pressure will continue to push the bullet a little, while pushing back on the gun a little, but almost all of the recoil will happen while the bullet is accelerating in the barrel.

I received those comments before I posted parts of your last explanations.

Sure, quote me anywhere you want. Also realize that I am not an expert on gun operation, I'm just good at figuring this stuff out, and I'm willing to learn from a good argument. If I realize I was wrong, I won't waste time beating a dead horse. I'm just not willing to accept mythology or "rules of thumb" as truth, without understanding the physics behind them.

I'll take a look at that other post.