Since recoil is all about the conservation of momentum, what effets do the weight of individual shooters, and even shooting position play in ACTUAL muzzle velocity?

For example: Conservation of momentum would seem to dictate that if a 100 pound shooter and a 300 pound shooter both shoot the same rifle/load from the prone position that the heavier shooter will achieve a higher muzzle velocity.

You know M1V1 = M2V2 and all that.

Also would a direct blow back action have lower muzzle velocity than a locked breech rifle, all other things being equal?

Thanks,
ASS_CLOWN

Ac-- Don't know for sure. But the fact initial
recoil of gun while bullet is in barrel, say in an 3006, 10 lb gun, occurs in about 1.4
m-sec,and gun moved about .06 inches, if you couldn't stop that movement before the recoil pad compressed it might not make a difference.
Now if barrel was anchored solid to 200 lb weight who knows, it may gain a couple fps..Same for blowback action.IE,
does blowback occur before or after that initial
recoil event of 1.4 s-sec and .06 in gun
recoil, during which the bullet has went the length of the barrel.??

If you assume that recoil doesn't occur until after the bullet exits the muzzle, the weight of the shooter, and even the gun, should have no effect on muzzle velocity.

Someone with a better physicsbackground can tell me if this is right. The bullet will always have the same relative velocity to the gun. Ignore the shooter for now, and vary the weight of the gun. For a very very light gun, the gun will be blasted backwards and the bullet will be nearly at rest. For a very heavy gun, the opposite happens. But, if you measure the velocity difference between the gun and the bullet it will always be the same. So the total velocity of the bullet, plus the velocity of the gun will be constant. You can put a gun with no recoil pad against a building that won' move, and you get no increase in MV. a better way to think of it would be:

m1v1-m2v2=0 (a constant)

From the calculations I have seen, the gun moves just a little over a quarter inch before the bullet leaves the muzzle. A quarter inch is just taking up the slack in a shirt and beginning to push into the shoulder tissue. No meaningful resistance in either.

Thus, the weight of the shooter should be imperceptible. An immovable rest would have an impact, though the math is beyond me to calculate how much. FWIW, Dutch.

If you find a way to test I can be your 300 pound guy.

Ed,

In a direct blow back the bolting system is free to move the instant the bullet begins to move. Actually, since the springs holding the bolting in place are typically in the 10 lb/in to 50 lb/in range they will undoubtedly move before the bullet since it takes considerably more force than that to engrave riflings.

Since the impulse of the system is exponential, I suspect that the shooters weight will have a much more profound influence on muzzle velocity than many seem to comprehend.

For example, the bullet is accelerating the rifle rearward at different rates during the firing sequence. In the blow back system it takes 25 lbs to push back the bolting, but it takes 250 pounds to engrave the rifling and overcome friction with the bore. The rifle would be receiving 10 times the acceleration as the bullet. This would/could continue for practically the entire time the bullet is in the barrel having a signficant effect on muzzle velocity. In the blow back model when the bolt has moved back so far the case is pulled and the gas pressure bled off, this too would slow the bullet.

ASS_CLOWN

Need a test on a blowback gun, fired regular
and fire with bolt locked tight. Chrono results.
If the bolt weighs a lb and bullet is 100
gr(aren't most blowback shoulder guns smaller calibers),with the pressure of the spring, and
the bolt is 70 times as heavy, and it shouldn't start back out of battery as quick and as fast as bullet moves.Should be some delay with the weight differential.It and gun as a whole will still start back as bullet starts.

an interesting test....
(don't try this at home kids)

rest a rifle on a tire, do NOT tie it down.

put a string on the trigger...

chrono the shot of the rifle "flying"


perform a substaintially same test, this time hold the rifle firmly and over the chrono.

there ya go...

little thank of equal and opposite reactions, don't ya know

jeffe

To answer the question, the effect of the weight of the shooter is so small that you will not be able to measure it because it will be too small compared to regualar shot to shot velocity variation.

Recoil comes in two parts:

While the bullet is in the barrel, you have a closed system, and the center of mass of that system stays in one place. Since, within the barrel, the bullet and gas are moving forward, the rifle moves backward, to keep the center of mass of the whole system in the same place.

Once the bullet exits, the barrel becomes a rocket, exhausting supersonic gas, pushing back on your shoulder. You no longer have a closed system.

The force of the rifle butt against your shoulder is probably well modeled by an inelastic collision. Energy is conserved, but momentum is not.

I believe that hugh has the right idea...the relative velocity between the rifle and bullet will be the same. If the rifle is accelerated backwards a few fps, that will show as a corresponding loss in the bullet velocity with respect to a staionary object (chrono sitting on a tripod, for example).

Yes, all else being equal, the blowback would have less MV than a locked breech (bolt action or similar)...

The few fps will probably be hard to detect, though, as the standard deviation from shot to shot will likely be much higher.

A direct blow back, by design (mass of the bolt/slide), has to maintain the bolt almost closed until the pressure has dropped off in the firing process...hence the delayed 'locked breech' action of 'floating barrel' type pistols (Glock, Browning, et. al.)...usually higher power calibers (roughly 9mm and up) than true blowback pistols (.380 ACP and below). I was just looking at this the other day on a web site illustrating the difference in the .380 Glocks vs. a 9mm Glock of the same frame size...the .380 had to be made a true blowback because of underpowered cartridge...

CDH,

You are mixing up true blow back actions with 'delayed blow back actions' there is a difference.

I can think of several blow back actions which shoot either +P 9mm Parabellum or +P 45 ACP. You may not be familiar with them as they are mostly class III NFA firearms. The can handle pressures that will literally destroy semi-auto pistols in short order (pistols with the 'delayed blow back' design).

How much muzzle velocity is lost with a gas operated action (self loading or full auto doesn't matter) with a locked breech? Some of these rifles (AR15 for instance) have the gas tap a considerable distance back from the muzzle, while other are very close to the muzzle. So are they adversely affected, with regard to there achieved muzzle velocity?

Also in my original question which shooter would get kicked harder? The little guy/gal who would accelerated to a higher recoil velocity and moved further, or the big guy/gal who aborbs the kicking and is not accelerated as much or moved as far (as the little shooter)?

Also, if the bullet momentum is doing most of the kicking, and only accounts for a rifle movement of 0.06" to 0.250" why do we see shooters moving a 1" to 4"? What is keeping them moving? You must agree that a shooter will not move unless a force is actively pushing them.

ASS_CLOWN

I "think" the velocity wouldn`t be effected much by the gas bled off. The difference in a few inches of barrel is usually minimal depending on the cartridge, and the gas bled by the system is minor in comparison. The few fps loss as the others have stated will be almost undetectable. The bullet will still be under pressure when it exits the barrel as will the bolt when it`s fully open.
The recoil of the rifle by the two types of people you describe would depend on their position, sitting prone, standing. The heavier person will likely absorb (resist) more of the recoil while the lighter one will move with it and allow it to dissapate through their body. A fixed body (heavier) will "feel" the recoil more. Both will recieve the same amount of energy.

AC- the movement that the shooter makes in
stopping the recoil is just because the gun has
recoil speed(momentum), like energy in motion as they say.In 3006 example the 10 lb gun has attained about
9 fps in .06 inch of acceleration distance while
bullet is in barrel, and jet effect adds a
couple more fps.Say you take my 510HE and accelerate a bullet 3 times as heavy(180gr
times 3 to 540gr) in same wt gun(10 lbs).
It would attain 3 times the speed of recoil
in 3 times the distance while bullet in barrel.And jet effect adds a little more.In both instances the gun has momentum and the shooter will be shoved back in reaction to it.
Said shooter reaction is how the moving gun, a projectile of sorts, is stopped.Shooter will be shoved harder by the big one of course..Ed.

Ed,

quote:

In 3006 example the 10 lb gun has attained about
9 fps in .06 inch of acceleration distance while
bullet is in barrel,

If that is true, why do I see guys shooting 30-06s getting moved ~ 1" from recoil? That is 16 times more than the movement the bullet imparts to the gun. These are guys that are benchshooting and laying into their rifles big time, so their body mass is definitely resisting the recoil velocity, recoil momentum, recoil energy, recoil etc.

Ol' Joe,

quote:

Both will recieve the same amount of energy.

I don't understand that comment. How can they both receive the same energy? One shooter has three time the mass of the other shooter ( 100 lbs vs 300 lbs). Since conservation of momentum is M1V1 = M2V2 and energy of motion is 0.5MV^2 I cannot understand how both shooters would receive the same energy.

Lets consider the 30-06 and 180 grain bullet that Ed uses. Let assume the muzzle velocity is 2700 fps.

Momentum of bullet at time it leaves the muzzle is = 69.4286 lb-ft/sec

Velocity of 100 lb shooter due to bullet momentum =69.4286/100 = 0.6943 fps

Energy absorbed by 100lb shooter = 0.5*(100/32.174)*(0.6943)^2 = 0.749 lb-ft

Velocity of 300 lb shooter due ot bullet momentum = 69.4286 / 300 = 0.2314 fps

Energy absorbed by 300 lb shooter = 0.5*(300/32.174)*(0.2314)^2 = 0.2497 lb-ft

So the heavier shooter absorbs ~ 1/3 the energy of the lighter shooter.

Is this right or am I wrong again?

ASS_CLOWN

Several antecedal reports (here and published)of before/after porting reports from handguns generally show some velocity loss, but much less than actually shortening the barrel to the length of the ports. Porting generally allows much more gas loss than an autoloading gas system...and think about it...how big is the gas port vs. how big is the bore...who gets more gas? Some gas operated firearms have cutoffs, allowing for 'normal' operation...go get one and a chrono and tell us what you find.

The rifle is ACCELERATED during the very short recoil generation time and moves the short distance as stated previously, but once momentum is imparted it has to be opposed, and since we humans cannot withstand the same level of instantenous force as a solid metal gun, we have to compensate...by spreading out the decelleration time. This leads to several inches of recoil, rubber buttpads, bruises, etc.

The light person and heavy person absorb the same amout of energy from firing the same gun...they just do it with differing amounts of 'grace'. The big person has more body mass to distribute the load into...more M, less V, less pain...all else being equal.

You seem quite capable of figuring all this out yourself...why do you persist in asking questions for which you already know the answers. Ref. blow back vs. delayed blow back...yes you are correct, an UZI will handle ammo that my Glock would not like a steady diet of...why ask in the first place!?!?

quote:

quote:
Both will recieve the same amount of energy.


I don't understand that comment. How can they both receive the same energy? One shooter has three time the mass of the other shooter ( 100 lbs vs 300 lbs). Since conservation of momentum is M1V1 = M2V2 and energy of motion is 0.5MV^2 I cannot understand how both shooters would receive the same energy.

Ass_clown, The mass of the shooter has no bearing on the amount of energy they recieve from a recoiling rifle. They both get the same recoil, how it affects them or the effects of their mass on the recoil is what varys.
How they react is what will change. Add the persons mass to the wgt of the gun (tightly shouldered rifle) and it becomes part of the equation. the 10 lb rifle now has some of the persons body mass added to it-resistance. The energy is the same, just the effect of it on their body has changed.

quote:

Originally posted by ASS_CLOWN:

quote:

Both will recieve the same amount of energy.

I don't understand that comment. How can they both receive the same energy? One shooter has three time the mass of the other shooter ( 100 lbs vs 300 lbs). Is this right or am I wrong again?

ASS_CLOWN

yes, you don't understand this


The rifle does the SAME TIME every time , assuming indentical velocity/pressure/bullet/etc. What happens to the USER is variable... just as it hurts more to let a big kicker FLY into you, than a firm hold.


Have you ever actually fired a gun?


or are you actually understand what RECOIL is?

jeffe

AC askes "if the gun only moves .075 to .25 inches while the bullet is in the barrel, what makes the shooter move 2 to 4 inches".
This shows his depth of knowledge on the subject. He should also ask " if the powder pushes on the bullet for only 24 inches, what keeps it moving to the target"?.
As an exercise; Since the C.G of the system remains fixed a 8 pound gun shooting a 150 gr bullet out of a 24 inch barrel using 50 gr of powder will have moved back 0.075 inches when the bullet reaches the muzzle.

Hawkins-You hit the crux of the process, once bullet and gun are in motion, and have momentum,it takes resistance to stop them.Ed.

Weight of the shooter has no impact on the velocity of the bullet. M1V1=M2V2 doesn't work in reverse. A higher M2 results in a lower V2. In other words a heavier gun gains less velocity from recoil than a light one, and therefore less recoil energy. M1 and V1 are inputs they are constant and not changed by the other factors.

Hawkins & Ed,

How is this for resistance. 300lb shooter literally laying to the rifle. The shooter moves 3" (460 WM ) while the bullet only moved the rifle ~ 0.25". How did the 10 lb rifle moving a 0.25" push the 300 pound shooter over ten times as far?

It seems to me that there is FAR more to this than simply bullet momentum. Heck the shooter is soft to boot, the rifle's 0.25" rearward movement probably did little more than compress the recoil pad and some soft shoulder tissue.

Now if the muzzle blast is moving the shooter the 3" then it seems to me that the VAST majority of recoil would be due to the rocket effect of the gas and not the bullet's momentum. However, we have all discussed that on another thread and you clearly illustrated how that was not the case! So please forgive me while I attempt to get an understanding of what is going on.

Why do you all say that the 'center of gravity' of the rifle is constant? I don't get that at all.

ASS_CLOWN

ASS_CLOWN

CDH,

I ask these questions due to the responses I received on the other recoil threads. You and several others clearly illustrated how ignorant I was. I am merely trying to understand the recoil, bullet momentum, shooter mass, muzzle velocity relationship better.

quote:

You seem quite capable of figuring all this out yourself...why do you persist in asking questions for which you already know the answers. Ref. blow back vs. delayed blow back...yes you are correct, an UZI will handle ammo that my Glock would not like a steady diet of...why ask in the first place!?!

This statement of yours really has me confused! First you tell me I am an ignorant imbecile, and now you seem to say that I am very knowledgeable and 'know all the answers'. So which is it?

ASS_CLOWN

scotty,
have you been seeing dr j. Daniels again?

you aren't making sense, and are parsing the problem whereever you feel fit.

have you called your parole officer today?
jeffe

Ac-In 3006 example the gun accelerated to
about 8-9 fps in .06 inches of movement while
bullet was in barrel,and a couple more fps from jet effect.For a total of 11 fps.And the bullet accelerated
to 2700 fps in 24 inches of movement up the barrel.So something has to stop them, like air resistance and target for bullet, and shooter for the gun. I mean your the one who was talking about guns hitting the shooter like sledgehammer.Which would only happen if you were holding gun away from you, and even
the little 9fps variety, might hurt.. But by gun being tight. it is in a deceleration mode
by the time the pad is half compressed, after being up to speed, with the shooter the means of deceleration.THe acceleration total time is about 2 milliseconds, 1.4 m-sec for bullet in barrel plus .06 m-sec for jet effect.Now it takes longer to bring it to a stop, hence
the 1-4 inches shooter moves, and more if
a hairy caliber.Ed.

Ed,

Please explain how it takes longer for it to stop? By conservation of momentum the shooters velocity is a 1/10 to a 130th that of the rifle.

Also the only way the rifle could have a recoil velocity of 9 fps in your example, is if it WASN"T held tightly to the shooter. Otherwise the shooters mass would be added to that of the rifle and slow everything down.

So lets assume a shooter of 100 pounds has a shoulder quandrant mass weight of 20 pounds. The shooter holds the rifle in tightly against their shoulder just as you say they must.

The rifle weighs 10 pounds, in your example, and the bullet is 180gr @ 2700 fps.

So since the rifle is held TIGHTLY against the shooter's shoulder the rifles weight and the shoulder quandrant weight of the shooter must added together to acquire the total weight resisting the bullet's momemtum.

So that would be 20lbs (shooters shoulder) + 10lbs (rifle) = 30 lbs.

Now to conserve the bullets momemtum.

Bullet momentum = [180/(7000*32.174)] * 2700 = 2.158 slug ft/sec

Conserving the momentum them:

(30/32.174)*Velocity of shooter = 2.158 slug ft/sec

Velocity of shooter = 2.314 fps

Rearward movement of shooter due to bullet's momentum = 2.314fps * 0.0014 sec = 0.0032' or 0.0384".

The observed shooters, who typically weight ~200 pounds move an 1" or better.

The 0.0014 seconds is the time the bullet is in the barrel.

Decelerating that velocity is the remainder of the shooters mass times the acceleration of earth's gravity. That would be 80 pounds.

If you do the math, and I can do it for you if you need it done, the shooters inertia (big word for their mass at rest in this case) will for all practical purposes instanteously decelerate the shooter's shoulder (it simplifies to 30/80 times 0.0014 secs for the time it takes to stop the shooter's shoulder or 0.0005 sec).

Explain what I am doing wrong.

ASS_CLOWN

Turd Boy,

Screw your "theoretical" math shit.

Put the butt of your rifle against a solid wall.
Put chronograph in front of rifle.
Fire rifle, note MV.

Repeat procedure, only use your shoulder.

Record your results and post them here, so we can all marvel at your brilliance.

This assumes that you have a rifle, a chronograph, and that you know how to use both. I'm not worried about the brilliance part, 'cause despite what you think, you ain't got none.

I seriously doubt the above assumption, and I really wonder why I even bothered to reply to your post. But I know you just love all the attention you get here.

Remember, freedom of speech and posting is a great thing, it makes it easier to ID the morons.

Your assuming that he shooter can hold the
gun tight enough to negate the .06 inches the
gun in example moves by adding his weight.
Impossible. The gun will move that 16th of an inch without the shooter affecting it 1% because rubber pad, clothing, skin and fat under skin,
all have many times the give as that 16th of
an inch the gun moves,in getting up to recoil speed, even with a tight hold.Once up to speed it starts to slow down, due to the mass of the shooter stopping it.. just like bullet does after leaving muzzle, due to air resistance and target...

For crying out loud! You guys are rising to the bait like trout to a fly! Want this guy to be frustrated? DO NOT reply to any of his posts at all! You're playing right into his pathetic mental game.

Nuffsaid

Latigo your right, but think of the entertainment value.

Ed,

quote:

The gun will move that 16th of an inch without the shooter affecting it 1% because rubber pad, clothing, skin and fat under skin,
all have many times the give as that 16th of
an inch the gun moves

Now if the gun moved without affecting the shooter how the heck does the gun move the shooter? I think all this recoil & momentum stuff is just made up high school physics examples and is way way too simplified to be anywhere near reality. Sorry, but it just doesn't add up.

In one post you say the rifle moves the shooter, in the next you say it doesn't. The reason for the reversal, because if the rifle actually moved the shooter, then the shooters mass would be involved and the recoil velocity and energy would be reduced significantly.

Seems to me that if the recoil pad and fat are compressing to accomondate the rifle's movement in you latest example, then after the bullet left the muzzle the recoil pad and fat would push the rifle forward and the shooter back at the ratio of their relative masses to get back to where they all started.

So if the portion of the shooter's mass involved in the fat was 5 pounds and the rifle and recoil pad weighed 10 pounds and the whole affair was compressed 0.06" then the shooter would move 0.045" backward (away from the rifle) and the rifle would move 0.015" forward (away from the shooter). Still a long long way from an inch.

I am not trolling either. I want these guys to explain the mechanics of recoil to me.

ASS_CLOWN

No where did we say that rifle doesn't move shooter,But the first .06 in of gun movement,
that bullet accelerating in the barrel has imparted to the gun, plus a tenth or two from jet effect,
isn't stopped immediately by shooter, because of the give in pad,clothing, skin,etc, allowing
gun to get up to speed, where it then has momentum to move shooter.Gun moves shooter after it gets going, and regular pads, hold,
etc can't stop it from getting up to speed.
It happens in less than 2 milliseconds, a 50th
the time iyt takes to blink.Then gun moves shooter, till shooters mass stops it.

quote:

I am not trolling either. I want these guys to explain the mechanics of recoil to me.

ASS_CLOWN

Easy...

rifle fires... rifle kicks.

Little bullet... little kick.

Big bullet... big kick.

There you go! Now that wasn't so hard...

An object in motion tends to remain in motion until acted upon by an opposing force. The rifle pushes the shooter back until his wieght and the msucles in the body resist it. After the force is dissapated, and the recoil pad decompresses there is not enough force in the in the pad to push the rifle forward any measurable amount.

quote:

Seems to me that if the recoil pad and fat are compressing to accomondate the rifle's movement in you latest example, then after the bullet left the muzzle the recoil pad and fat would push the rifle forward and the shooter back at the ratio of their relative masses to get back to where they all started.

elkhunter,

The recoil pad and the shooter's soft tissues are springs. Therefore, if they are compressed by the moving 0.06" due to the bullet's momentum, they will spring back to their original position (decompress as you say) and the shooter WILL NOT be moved rearward any more.

Try it. Push a spring down 0.06" and immediately release it. Does the spring continue to compress or does it immediately decompress back to it's original position?

Heck use an actual recoil pad, push it in 0.06" and let go, does the recoil pad continue to compress deeper (even though you are no longer pushing on it) or does it spring (decompress) back to it's original shape/position?

Hint the answer is that is springs back to it's original position.

ASS_CLOWN