This thread has become to full of ignorant ranting by others to be worth maintaining.



ASS_CLOWN

quote:

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Originally posted by ASS_CLOWN:
For example:

A 458 Lott (chamber pressure ~ 60 ksi)has a cross sectional area of 0.16475 in^2.

A 30-06 (chamber pressure of ~ 60 ksi) has a cross sectional area of 0.075 in^2.

So the 458 Lott "kicks" about 2.2 times harder than the '06.


ASS_CLOWN

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Having alternated between a .308 and a 458 Lott at the range one day, I can gurantee you haven't got a clue.

I don't know about you but I can look at the two case and reach the same conclusion.

The bigger the bullet and bigger the powder space is about as good as all that tech knowledge you can waste your time on.

A_C

Your post makes sense to me - the tradational recoil formulas assume that the gun is in free space and is an isolated system.

I've been toying with mounting 2 axis accleratometers (sp?) on a big bore.

To go off subject - I think that percieved recoil is a function of both the translational and rotational forces.

quote:

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Originally posted by ASS_CLOWN:

For example:

A 458 Lott (chamber pressure ~ 60 ksi)has a cross sectional area of 0.16475 in^2.

A 30-06 (chamber pressure of ~ 60 ksi) has a cross sectional area of 0.075 in^2.

So the 458 Lott "kicks" about 2.2 times harder than the '06.

ASS_CLOWN

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ASS we all know, you've never pulled the trigger to know what "2.2 times harder" IS...

you forgot rifle weight..

you forget ejecta

you forget bullet weight

HINT, ASS, a HEAVY 30-06 bullet is a 180... and a 458 lott shots a 500 grain...

a lott (at 8.5), with a 500 at 2300 has 84 ft/lb of recoil...

a 30-06, with 180s at 2750, 8.5#

has TWENTY ft/lbs of energy...

Since you have never shot either one, you can't be held accountable for having NO CLUE as to the difference.

quote:

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will fracture almost any steel alloy that measures is 0.5" X 0.375" X 2" long,

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Ah, dumb-ass... so you are telling us all that 40-100 ft/lbs will shatter any alloy?

How about one that's .375 in diameter?

and calls for SIXTY FIVE ft/lbs of TORQUE...

like a chevy main or wrist bolt?

Oh, yeah.. that doesn't apply in your world.

have you called your parole officer today?

The comparison of 06 and Lott saying that Lott
kicks 2.2 times harder than 06 is only valid, if
guns have weight set up as a match to the size of the bore.Like an 8 lb 3006 and a 14 lb Lott,Otherwise it is as Jeff says if guns weigh same, a 4 times difference.Ed.

In same wt gun the recoil energy of a Lott is 4 times greater than 3006.I use term kick as slang for recoil energy. Sorry for confusion.Now its affect in
actual action to the shooter is more like you say, in comparing the two, IE half of the
difference in energy.So if a gun has has 4 times recoil of another one, it affects the shooter about 2 times as much.A gun with
10 times thr recoil energy will affect the shooter about 5 times as much, in comparison.IE the gun has to get somewhere,recoil back to
expend its force on the shooter.In process of
getting there gun as to accelerate and while pushing shooter,slows down,and energy is lost, so that its affect isn't as much as the energy differential suggests, from low recoil gun to a high recoil gun. That is not to say like in last example above, where a gun has 10 times the recoil energy compared to 3006 that the 5 times greater affect it has on shooter, that
it is of no consequence to shooter..We will be in that area of recoil when 4bores are going.And more.Ed.

quote:

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Originally posted by geo45_70:
A_C

Your post makes sense to me - the tradational recoil formulas assume that the gun is in free space and is an isolated system.

I've been toying with mounting 2 axis accleratometers (sp?) on a big bore.

To go off subject - I think that percieved recoil is a function of both the translational and rotational forces.

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You have idnetified two of the factors - but I also believe perceived recoil is greater if there is more muzzle blast and louder noise, particularly to one who is not wearing proper hearing protection.

I definetly agree with you on the hearing protection. I now use both ear plugs and a pair of ear muffs when shooting from the bench.

Funny thing is - when I deer hunt (with Ruger No 1 in 375 HH) I don't remember any recoil - I don't even hear the shot.

AC,
I think you are right, comparing sectional density [bullet weight/4x bore area] and peak pressure WILL give something proportional to the guns peak reaction.

The units and being PEAK recoil make it not that useful, however I do believe the there is a place for peak recoil in subjective recoil felt with SOME guns, but not those with recoil pads. Revolvers and Trapdoors with black powder are better examples.

But I am still a big fan of yours. If someone wrote an end all reloading and gunsmithing book, your Von Mises calculations of max cartridge case head pressure would have to be in that book.

I think the end all book would use the regular recoil calculations, and not you musings for this thread.

Interesting discussion. However, I do think that other factors i.e. powder burn rate, have something to do with the actual recoil. I am not into physics at this point of my life, so I really can't tell you why. I can state that over my 35 years as a shooter and reloader, I've experienced many different calibers. I personally feel that recoil is grossly over rated!!!One experience that I've had, and have had others agree with, is the case of .300 Win Mag vs .338 Win Mag. A few years back I owned these 2 calibers in the exact same rifle, Browning A-Bolt Composite Stalkers with 26" barrels. Both guns were even scoped the same and both weighed 8 1/2 lbs scoped. I would load the .300 Winnie to 3050 fps with a 180 gr and the .338 to 3050 fps with a 210 gr Partition. I will have to assume similar pressures since chrono data is all I have. The odd thing is, I didn't care to shoot the .300 much at all. The recoil to me seemed more violent and much sharper. The .338 seemed to have a softer recoil, more of a push, similar to a black powder rifle. The powder loads in these was almost identical and It was with IMR 4350 in both. Logic would seem to tell me that the .338 was 30 grains heavier at the same speed, it should have kicked harder. It just didn't seem to. I even had some friends shoot both rifles and asked their opinions, and they seemed to concur with my findings. Long story short, I still have the .338 and have long since sold off the .300 Winnie. All I know is that even my wife shoots my .338 and doesn't mind the recoil. I wish there was some explanation for this phenomenom. Has anyone else experienced a similar situation to this??

Ass_ Clown I`ve believed the muzzel pressure caused the so called "jet effect" and was the cause of varying sharpness, for lack of a better term, of the FELT recoil. The recoil energy in other words stays the same, just the velocity of the recoil varies depending on the volume of gas and the pressure and bore diameter.
Are you saying you believe the pressure at the muzzel varies the accual recoil or just the percieved sharpness of it?

Recoil is aresult of momentum (Mass x Velocity).
Newton spoke tothe conservation of Momentum not Energy.
All the momentum forward is equaled by the momentum to the rear.
The forward momentum is composed of; The bullet mass time it's velocity. The powder mass times 1/2 the bullet velocity (average), and the effect of the Gas acceleratinr from 1/2 the bullet velocity to the velocity of the muzzle blast. It is this last term that can be modified by a muzzle brake. and I believe that
it is also responsible for the sharp sudden recoil ("fast kicker" felt with some firearms.
The pressure integrated over the bore length x the bore diameter will certainly give the force on the bullet,as well as the velocity of the bullet. But it is only part of the recoil source.
Good luck1

What ever happened to F = M*A? Nobody has ever discussed the accelleration of the bullet. Remember, F stands for FORCE!!!! ... felix

The free recoil of your firearm can be found using the formula below:

((weight of the bullet in pounds x the velocity of the bullet in feet per second)+(4,700 x the weight of the powder in pounds))squared / (64.348 x the weight of the gun in pounds)

"Free recoil"? What the hell is that? I've heard of free recoil energy and free recoil velocity, but this thread is about perceived recoil. It is specifically NOT about canned formulas.

In 1962 there was an article about recoil in Field and Stream magazine where they hung rifles and shotguns on strings with rubber bands pulling the triggers and string holding the triggers off. They set the trigger string on fire and noted how high the gun swung on the string. That heighth times the weight of the gun was the "free recoil". You can measure it, and you can calculate it.

Then there is the max force if the guns is constrained [not free]. That would be equal to what AC is talking about, minus bullet friction in the bore.

So in the former, there is a maximum of recoil energy, and in the latter, none.

So when someone says "free recoil", they mean "not the constrained kind of recoil".

Who really gives a shit about all of this anyway?

AC - sounds interetsing but sum of us may hava problem respecting tech talk from someone named ASS CLOWN(?) methinks perhaps??

Maybe u shood give us your tech/education/degree background?

Might hep.

AC, you can talk about pressure curves and bore diameters all day...

You can take the integral of the pressure curve on the cross sectional area of the bore to find force till your fingers bleed on the keyboard...

At the end of the day, all that force on the bullet does is accelerate it, and the end result is the same as the momentum imparted to the bullet...so just measure bullet mass and velocity and the time it takes to get there and be done with it...MUCH SIMPLER, and probably more accurate to your purposes. It it built in that it takes into account VERY difficult to measure things like friction with the bore...

For once, try the simple route first, you miight get to a respectable answer much faster!

CDH,


With regard to creditials, I am an ASS_CLOWN, perhaps they came from a Cracker Jack's Box. It was a very expensive Cracker Jack box though. I could also simply be a mechanic without ASE certification. Of course it may also be that "I am an IMBECILE".

Anyway I will let you all ponder this in peace.

ASS_CLOWN

quote:

The problem with your analysis method is that the bullet's momentum has NOTHING to do with recoil.

Impossible. Newtonian physics, equal and opposite reaction and all that. That big 'ol hunk of lead flying out the muzzle requires something to push against to make it fly. That is your shoulder via the propellant gas.

If what you say were true (and I say with VERY high confidence that it is not) then I could load my trusty 300WSM with 65 grains of 4350 behind a 180 grain bullet (my pet load) and 65 grains of 4350 behind a 150 grain bullet and have the same recoil. Guess what, IT ISN'T! Same shoulder, same rifle, same powder, same bullet type, different bullet mass only.

Should a .22 blank recoil as much as a .22 short in my revolver? Very similar powder charge...noticably different recoil.

Ignoring projectile mass makes the remainder of your analysis incorrect. The reason the bore ratios seem to give respectable numbers is that the larger bore shoots a heavier bullet...usually with more powder to boot!

But to keep things simple, compare a 30-06 to a 35 Whelen with like bullet weights and muzzle velocity. Reloading makes this simple to actually perform...you will find their recoil to be very similar in otherwise identical rifles...

Recoil does start before bullets leave bore.
The linear move of the gun in recoil is not
a direct relationship to the the recoil energy figures when restrained by the shoulder.Say
a gun with 20 ft lbs recoil energy held tightly
moves shoulder an inch and compared with gun having 200 ft lbs recoil energy(10 times
as much), the bigger
recoiling gun wouldn't move shoulder 10 in
but half of of that.That is what his formula is showing as to the felt recoil.Ed

quote:

Originally posted by ASS_CLOWN:
Recoil doesn't happen until the bullet leaves and the high pressure gases REACT against the atmosphere.
ASS_CLOWN

I am 100% sure you are 100% wrong about that. Recoil begins when the bullet starts accelerating, in the bore. Conservation of momentum applies ALL THE TIME, including while the bullet is in the bore.

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The force applied to the bullet is contained 100% by the structure of the firearm.

Wrong again. The firearm does not exist in a vacuum. The firearm contains all the propellant EXCEPT that which presses on the base of the bullet.

THE BULLET IS NOT FIXED TO THE FIREARM!

Last time I studied Kirchoff's laws they were applied to voltage and current in a circut...so I am not familiar with their application to a mechanical system, but hey, I think I see where you are going...

The force equation for the burning propellant shows that the force is contained by the breech of the firearm in all directions, except down the bore. Since the bullet moves forward, the firearm must move rearward. Period!

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4.) Barrel acceleration (just forward of chamber)

If you put an accelerameter on the barrel (or any other rigidly attached portion of the firearm) and did not measure movement until the bullet left the bore, your measurement is in error. Either you have a measurement delay somewhere or you are not accurately timing the bullet exit.

So YES, I AM arguing with measured data. Your data is wrong.

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Bullet mass speaks DIRECTLY to the pressure build up in the barrel. That is why your 22 blank doesn't kick as hard as you 22 short.

I'll give you this as true...in that the mass partially determines the length of time the pressure is contained...

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The force acting upon the bullet causes it to accelerate. An equal an opposite force is acting upon the LOCKED breech of the rifle's structure.

...which (the breech) is attached to the receiver, which is attached to the rest of the rifle, making the system a (relatively) rigid unit.

Parts moving freely within a unit can cause the system to move, even if the system is still a contained unit. Note the fact that revving my truck motor causes the truck to rock. Just because the system is intact (because you have defined the system as the rifle AND the bullet) does not mean that the system cannot move within its environment.

Look at it another way...the system (as you defined it) has not changed, but its center of mass will as the bullet travels down the bore. If you shift the center of mass within a system, either the system moves or an external force is required to keep it stationary. Simple mechanics.

Bottom line, what you are describing is not consistent with Newtonian physics.

Either you are right or Newton is right. You can't both be.

Read my signature very carefully.

I believe Newton.

Did you factor in Your Jack Daniels internalgraymattervariationellussionalvortexintoxication index for felt recoil comparison.

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I can argue your argument as well. A 460 Weatherby magnum. Using your erroneously applied Newtonian physics the recoil velocity is ~ 18 fps. How far does the bullet and powder move the gun. Lets see the bullet's momentum acts on the rifle (in your argument) for 0.0012 seconds.
So 18fps * 12in/ft * 0.0012 sec = 0.259"!!!!

So after the gun has moved .259" and it is moving 18 fps, what happens next? It instantanelously stops moving? Or does it continue moving and give you a good shove on the shoulder. And compress the recoil pad the rest of the way.

Imagine for a moment a 8 lb. gun with an electronic firing pin (no rearward pull on the trigger). Said gun is on a frictionless sliding block. And imagine it fires a 8 lb bullet. At the moment of detonation, the bullet will fly forward, and the gun rearward, at the same rate (in a perfect world). The gun does NOT wait to move backwards until the bullet leaves the barrel; it is moving as soon as the bullet is.

Ok, that is impossible, so now let's look at the same 8 lb gun driving a 180 gr bullet down range. Same frictionless support system. What is the difference in velocities now? Well, the gun weighs 56,000 gr, and the bullet 180. So the gun has 311.11 times the mass of the bullet. Guess which is going to move faster & farther? The bullet. And NOT when it leaves the barrel, but NOW.

Say the bullet leaves the barrel at 2500 fps, how fast is the gun travelling backwards on this frictionless support? Approximately 311.11 times slower than the bullet, or 8.04 fps. Don't believe it? Remember that for every action there is an equal and opposite reaction? Well, at 8.04 fps * 8 lbs = 64.24 ft*lbs/sec, and 180 gr/7000 gr/lb * 2500 fps= 64.29 ft*lbs/sec. Purty durn close to equal, no?

Let's weld the end of the barrel shut, then drill a 3/16" hole in the weld. Now chamber that 180 gr bullet, and push that electronic button to fire it off. Will the gun remain perfectly stationary? No. It cannot. The expanding gases would drive the gun backwards with the same energy as the bulet travelling forawrds, minus barrel friction. All of this would be in equilibrium until the point when the bullet hits the weld (and if pressure is greater than the weld strength, it all blows up in your face, thereby rendering all collected data invalid because the reporter cannot report it due to his sudden demse).

Ok, I am sure holes can be shot in all of this, and we all know that the things we learned in physics class (wherein all things were tested in a vacuum on frictionless surfaces) cannot exist in the real world, but I think my example illustrates what actually happens.

F6

quote:

Originally posted by sonofagun:
AC - sounds interetsing but sum of us may hava problem respecting tech talk from someone named ASS CLOWN(?) methinks perhaps??

Maybe u shood give us your tech/education/degree background?

Might hep.

There are supreme court justices with degrees in law that find the words "bussing" and "abortion" in the constitution, but can't find an individual right to a gun.

There is alot of bunk in load books.

But the more I learn about science and engineering, the more I see in AC's posts.

I don't care if he is Kindergarden drop out and taught himself by reading brail.

The gun starts recoiling before bullet leaves barrel, hence the taller front sights on pistols, as being lighter they are recoiling up before bullets leaves and so taller sights compensate so you can hit target. It is not so noticable on long guns, but they still start recoiling as soon as bullet starts moving.Ed

It is against my better judgement to get into this ,however here goes; The rifle moves back at the same time the bullet moves forward. This is newtons law of conservation of momentum. The muzzle blast does not "react against the atmosphere". It's force is the reaction of it's velocity increase when it is "uncorked". If a "reaction against atmosphere" were required a rocket would not work in space. a rocket does work in space, and a gun would recoil in space. As to the plugged barrel; When the bullet came to a stop in the barrel it's reaction (forward) would cancel the recoil (back) but in the meantime the gun would have moved (but it's center of gravity would not have).
This is not "rocket science" it is High school Physics.
Good luck!

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This is not "rocket science" it is High school Physics.
Good luck!

There is no gun movement and there is no recoil energy if the gun is restrained. Then the acceleration on the gun sees the peak force on the restraint.

If the gun is unrestrained, there is no recoil force, but the peak recoil energy is seen in gun velocity.

And everything in between is in between, like your shoulder, which in neither perfectly free or perfectly restraining. This means recoil operating guns must somethimes be tuned to the mass and compliance of the shooter.

Clark; Recoil is a reaction that involves momentum not energy. the gun will put a force to the rear even if it can't move.
Good Luck!

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So after the gun has moved .259" and it is moving 18 fps, what happens next? It instantanelously stops moving? Or does it continue moving and give you a good shove on the shoulder. And compress the recoil pad the rest of the way.

Yep. Someone's (AC) mind is made up and can't get out of the vicious circle of logic he has 'engineered' himself into.

Break out of it man!

Imagine this experiment, and follow closely. Do not attempt, this is a theoretical exercise only.

Weld a strong plug into the barrel of a rifle.

Port the barrel several inches behind the plug so that the propellant gases can escape straight up (+Y direction) before the bullet hits the plug.

Fire the gun.

What motion do you expect from the rifle system?

By your argument the only sensation felt would be a full recoil (the same as if fired normally) but down (-Y direction) from the propellant gases being directed straight up through the barrel porting.

By my argument you would feel a push back during the time the bulelt is accelerating down the bore, a downward push when the bullet passes the porting from the gases excaping, and a forward push when the bullet hits the end of the barrel. The down (-Y) force would be much less than the full recoil of the rifle being fired normally, because it is just the gases, not the gases and bullet action.

And one last try...just answer this question. Since you brought up equilibrium...if the bullet fired from a firearm hits a steel plate, the plate shudders/flips/falls, sometimes violently if it is light enough or the bullet fast and heavy enough. This positively shows that the bullet has momentum.

Where did this momentum come from? Keep in mind that Newton proved that for any action there is an equal and opposite reaction. Trace the origin of the bullet momentum back all the way until it is dissapated by the shooters shoulder.

AC-You ever run a chainsaw that wasn't balanced
enough.Actually none of them are balanced.I speak from experience. Ever feel a one lung stationary engine vibrate, said vibration, being the motor moving back and forth in reaction the piston movement, said movement
being imparted to piston by crank and rod
connected to the engine.You said you used a
DEERE for hydraulic system to do something,
and that alone should explain action,reaction.
Well in a gun the bullet is a piston of sorts
and even though it moves in the forth direction
so to speak it is connected to gun by exploding powder, pushing against gun, that makes it move forth.That connection """forces"" gun to move opposite when bullets starts moving. Runs like
DEERE as they say,HAhHAh..Ed.

How many ways do I try to illustrate things to you? Now we are into insults.

The imbecile (your word for yourself) says:

quote:

Summation of forces = 4773 (bullet force) - 4773 (breech reaction force) = 0

What is keeping the breech from moving? It has a force on the front from the pressurized gases and nothing on its rear. Or does it!?!?

You have basterdized the math. You have artifically isolated the system by ignoring the remainder of the material involved in the action and reaction.

quote:

In order for momentum to transfer there MUST be a collision. Where is the collision between rifle and bullet? There isn't one.

The rifle and bullet do not collide. The propellant gas 'collides' with both, pushing each in opposite directions. It does not take a violent all at once collision to transfer momentum (speaking of red herrings), just enough force to overcome resistance.

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If you plug the bore of a rifle the bullet's impact on the plug will drive the rifle forward. Until the bullet strikes the plug there will be NO force external to system. The external force happens when the bullet transfer's it's momentum to the muzzle plug.

Basic thermodynamics. Energy cannot be created, only changed in form. If there is a force external to the system (as you admit above) when the bullet hits the plug, there MUST be a force external to the system before. The force CANNOT be created solely within the system. You have completely closed yourself into the trap of your own flawed logic.

It is that simple.

You can make it as hard as you like, but you are wrong. It does not take a long math equation to prove this. I'm an engineer because I like to work with facts and results, not vodoo theories, or I'd be a scientist.

I'm through here. Either you are wrong or the rest of the world is wrong. You are wrong.

quote:

Originally posted by ASS_CLOWN:

The expansion of gases does push back on the action's locked breech with an equal force to that applied to bullet's base. That is about the only thing you have stated correctly.

So the gases push back on back of the cartridge case, which pushes back on the bolt face, which is pushes back on locking lugs into the reciever, which pushes through a recoil lug into the stock, which is connected to a recoil pad. Hereafter we will call this collection of parts a "gun".

So if you have a force rearward on the "gun" and a force forward on the bullet, where is the balancing force that keeps the gun from moving? Where is this force applied to the "gun". The unbalanced force on the bullet shoots it out the barrel. If you want anyone to believe you, you will have to show where the balancing force acts on the gun, to counteract the "force on the breech" that you admitted to above.