Limp Wristing?

What do you believe provides the resistance to hold the frame in place while the slide is compressing the recoil spring to come to a full stop?
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A:The same thing that does what you describe while the bullet is travelling forward.
Try to understand that there are two equal forces at work and the frame simply
holds the device that produces those forces.
 
45_Auto said:
What do you believe provides the resistance to hold the frame in place while the slide is compressing the recoil spring to come to a full stop?
[...]
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That was a very good explanation. One way to help visualize that is to think about the two extremes: (1) the grip is solidly held in a vice, or (2) the gun is floating in outer space, and discharges.

I think my relaxed grip makes my 10mm 1911 run better because with full-spec 10mm ammo, the slide tends to move back TOO fast ... it throws the brass completely across width of the range, where they hit the wall and fall to the floor. A relaxed grip slows that down at least a little.
 
I'm about to adopt a Lightweight Commander. So mass of energy, to bullet weight and energy absorbed. To size of barrel to hold on gun. Equals lowest energy absorbed though hand? I do need the basics you guys are sick with this science. I believe JMB design the 1911 for a typical man with moderate strength. Don't let gun fly out of hand and hold on target. Gives the appropriate base for gun to function. It does need a hold like any gun ever designed.
 
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I use the sidewalk VS water-bed to teach this lesson;

If you held a coil spring vertically and compressed it against the sidewalk and then took the same spring and tried to compress it against a water bed mattress you would find that the same amount of pressure would not compress the spring against the bag of water as it did when it was against cement.

Such is the dynamic with autos.

If you don't hold the gun in a way that allows the spring to fully compress you can get short cycles.

I had a very small and thin woman (84 pounds) come to my shop just 2 weeks ago with a sub-compact 9mm and said the gun just didn't work right.
She and I went out to my firing range and she shot it in front of me. A 7 shot mag malfunctioned 5 shots.
I took the gun, reloaded and shot it 7 for 7.
Then I did it 3 more times. So I shot 21 rounds just fine and she could not get the gun to work 3 shot in a row ever.

I explained the problem to her just as I did above using the spring against a water-bed as a lesson.

When she would squeeze the gun firmly and use the left hand to squeeze her right hand, and lean forward a bit at the hips the gun started to work for her perfectly.

I took out my CZ75 and let her shoot it. She was amazed at how easy it was. I explained that the CZ was very heavy compared to her little sub compact, and the standing momentum of the larger gun was helping act as "the sidewalk" to compress the spring.

I have seen many 1911s jam from limp writing when I was teaching combative tactics in the USMC back in the 70s and 80s Most times when the man or woman started to grip the gun tighter and keep their wrists straight the jamming stopped. I have seen Browning A5 Shotguns do it too, when their owners try to shoot them from the hip. The arms often will not hold the gun forcefully enough to allow full compression of the spring.
So it is not uncommon.
The "fix" for it is often as simple as holding on firmly and solidly.
 
I'll go along with Brit. I tried thumb and forefinger and can't get a malfunction !
In a similar test I took my M1 90 Benelli and fired it like a pistol -one handed, butt not touching my shoulder.Worked perfectly ! Must not be doing it right .:confused:
 
Yep. The idea is that the hand on the frame holds said frame in place while the slide recoils over it. If that isn't done, the physics get messed up and the slide doesn't work right.
"...lacking in upper body, arm and hand strength..." It's not upper body, arm and hand strength. It's upper body, arm and hand tone.
The slide does not recoil against the bullet. It recoils from the forces created by the gasses against the return/recoil spring. The bullet going forward is just one result of said gasses being created.
 
polyphemus said:
Try to understand that there are two equal forces at work and the frame simply
holds the device that produces those forces.
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Interesting theory.

What part of the gun do you believe restrains the rear of the recoil spring while it is being compressed?

Do you believe that any force is exerted on the frame by that compression force?

If you're interested, it's relatively easy to compute the effects of those equal forces on the components of the gun. I did that 4 or 5 years ago during a discussion on slide forces, it could easily be extended to the frame if you want to expend the effort:

If anyone is really interested, you can use Newton’s stuff from 400 years ago to understand how your gun works. We used it to get us to the moon back in the 60’s, we use it on the rockets and guns we’re designing today. If you don’t care to know, don’t worry about it, keep believing whatever makes you happy! No big deal.

For example, how long does your bullet remain in your gun barrel, what kind of acceleration does the bullet experience, how much force is being applied to the slide or frame?

This kind of stuff is REAL important if you’re designing things like rocket mechanisms or artillery fuses or proximity detonators.

Not real easy to effectively transmit mathematical concepts in a forum format, but we can try. I’m going to ignore the contribution of the unburned powder and gases exiting the barrel with the bullet.

We’ll call the starting position of the bullet the zero of our coordinate system, so X(initial) = 0. We’ll abbreviate it as Xi cause I’m lazy. That means Xi = 0.

V(initial) is the starting velocity. We’ll call that Vi. For a bullet in a gun, it’s initial velocity is 0. So Vi = 0.

We’ll call the last position we consider, the end of the barrel, X(final). I’ll abbreviate that as Xf.

T = time in seconds.
T^2 means “T raised to the second power” or "T squared" or “T x T”.
X = position in inches.
V = velocity in feet per second.
A = acceleration in feet per second squared (fps^2).

The basic motion equation in terms of time is:

Xf = Xi + V x T + (A x T^2) / 2

If you want more on this basic motion equation, you can go here:

http://en.wikipedia.org/wiki/Equations_of_motion

Introductory calculus teaches us that the first derivative of position is velocity, and the first derivative of velocity and the second derivative of position is acceleration (I’m not going into calculus for you).

So Velocity = Vi + Acceleration x Time, or

Vf = Vi + A x T

We’ll look at a typical .45 caliber 1911. First thing we have to do is get everything into common units.

Bullet weight = 230 grains = 230/7000 pounds = .033 pounds = .001025 slugs (slug is mass unit in english system).
Barrel length = 5 inches = 5/12 feet = .417 feet
Bullet velocity when it exits the barrel at 5 inches = 830 feet per second.

Now we just use the basic motion equation:

Xf = Xi + Vi x T + (A x T**2) / 2

Put in our 1911 values that we know:

.417 = 0 + 0 x T + (A x T**2) / 2 or

.417 = (A x T**2) / 2 (NOTE – This will be EQ1)

Notice that we have 2 unknowns (A and T) but so far only one equation. From basic algebra we know that we need as many equations as unknowns if we hope to solve them.

This is where we use the rest of the stuff we know and the velocity equation:

Vf = Vi + (A x T)

We know Vf (Velocity final) is 830 fps. We know Vi (Velocity initial) is 0, so

830 = 0 + (A x T)

Since we loved algebra, we solve for A in the above equation and find

A = 830 / T (NOTE – This will be EQ2)

Now we can cleverly plug the above relationship back into our basic motion equation EQ1, replacing A with 830 / T.

.417 = (830/T) x (T**2) / 2

Simplifying this (basic algebra again) gives

.417 = 415 x T

Now we solve for T:

T = .417/415 = .0010048 seconds

In other words, T is right at 1/1000 of a second. This is the total time that the bullet is accelerating, which is the time that the bullet is in the barrel.

Since we now know T, we can easily solve for A from EQ2:

A = 830/ T = 830 / .001 = 830,000 feet per second squared.

Acceleration due to gravity = 32.2 fps**2, so our bullet is experiencing:

830,000 / 32.2 = 25776.4 G’s as it goes down the barrel.

Kind of important to know if you want to hang some electronics on your bullet. They better be able to handle more than 25,000 G’s.

How much force is the bullet and therefore the locked together barrel/slide seeing?

We know that the force from the bullet equals its mass times acceleration, and we know that for every force there is an equal and opposite reaction (that Newton stuff again).

So the force delivered to the locked together barrel/slide from the bullet is:

F = M x A

F = mass of bullet x acceleration of bullet

F = .001025 x 830,000 = 851 pounds
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I'll leave it to others to explain mathematically, but the observed results are that some recoil operated guns don't seem to care about how they are held, and other do.

What is the math to explain how a gun that is jamming repeatedly in the hands of one shooter, functions flawlessly in the hands of a different shooter??

Same gun, same ammo, same time on the range, same everything, except the person holding the gun (and how they hold it?)

I've seen it happen with others, I've been personally involved myself, it IS a real thing. What differs seems to be the way people explain it.

When the only difference is the person holding and shooting the gun, logic suggests that if the gun acts differently, it must be because of something the shooter is doing, or not doing, differently.

If it not the way they hold the gun during recoil, then, what is it??
 
44 Amp said:
What is the math to explain how a gun that is jamming repeatedly in the hands of one shooter, functions flawlessly in the hands of a different shooter??
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Very simple math, using a 16 pound recoil spring and ignoring moments due to bore axis geometry: If you have a 16 pound recoil spring in your recoil operated weapon, the inertia of your weapon plus the resistance of your grip must provide at least 16 pounds of force to the frame to allow full compression of the spring thus full travel of the slide.

The shooter experiencing the jams is not providing enough force to allow the velocity of the slide relative to the frame to be high enough to properly eject the brass and transfer enough energy into the firing mechanism and recoil spring to load the next round.

To operate properly a recoil operated weapon requires a force on the frame in the opposite direction of the recoil. This force must slow the frame enough to allow the relative velocity of the slide to eject the fired casing while at the same time cocking the firing mechanism and compressing the recoil spring far enough to store sufficient energy to load the next round.

The recoil force on the gun is only exerted during the 1/1000 of a second that the projectile is being accelerated (F = MA). Everything after that is due to the inertia of the barrel/slide. If one shooter's grip does not provide sufficient force in the opposite direction of that recoil force, then the frame is moving rearward with the slide, thus the velocity of the slide relative to the frame is low (poor ejection) and the force delivered to cock the firing mechanism and compress the recoil spring is low (feed problems).
 
Amazingly it seems,the recoil spring has no bearing on the recoil part of the cycle it's on the return part when it functions as intended.Assuming we are talking about a 1911 it has been shown that the pistol will function without a reaction spring and that includes ejection.If no recoil spring is present no recoil is felt.A 13# reaction spring should reliably strip a round from the magazine and possibly a lighter one functions just as well so no great need of wrist support is needed for normal operation.
 
Polyphemous said:
Amazingly it seems,the recoil spring has no bearing on the recoil part of the cycle
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It IS true that ejection will occur no matter how weak the recoil spring is, but that's not true for too strong a spring. Imagine that the spring is extremely strong ... much stronger than would ever be used on a gun. What do you think would happen when the round fired?
 
Imagine that the spring is extremely strong ... much stronger than would ever be used on a gun. What do you think would happen when the round fired?
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A: I think that should the slide manage to complete the recoil cycle it would return to battery so fast that it would cause a stoppage of some sort including failure to eject.
OP would like to know whether a weak wrist could cause a failure to eject.
And my point is that shell ejection takes place even without a recoil spring
so if you fire a pistol that is within specs all you are doing is holding it and aiming steady.It will eject fine whether you death grip or not.
 
Polyphemous said:
I think that should the slide manage to complete the recoil cycle [...]
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With a strong enough spring, it won't. Strong enough, and the slide won't move at all with respect to the frame. (The slide can't move wrt the frame without compressing the spring, and if the spring gets strong enough, it won't compress).
 
I've done the thumb-and-one-finger deal, trying to induce a malf, and it's never worked.
I handed the gun to my then-roommate's sister, and she had two or three malfs per mag.
She was clearly holding the gun with a firmer grip than I was.
I've heard the argument that an all-steel gun can't be limp-wristed, as the weight of the gun alone is enough resistance to ensure function, and while that would appear to be true while I'm holding the gun and trying to induce a malfunction, it's not true when someone else held the gun and tried not to.
 
.If no recoil spring is present no recoil is felt.
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You might want to think about rephrasing that statement into something actually correct and accurate. ;)

To operate properly a recoil operated weapon requires a force on the frame in the opposite direction of the recoil
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Higher math/physics defines terms in ways that allow formulas to work. This is not always the way people define the same terms in regular conversation.

Resistance, is a word we generally recognize. It is "force in the opposite direction" in math, but our common perception of force is some kind of "push", and we don't really see resistance as a push.

I was once shown a higher math problem where a swimmer swims the length of the pool, and back, climbing out at the same spot where he began. There is a math formula(s) that "prove" the swimmer went nowhere.

Most of the folks I know, would disagree. It's a matter of perception and definition of terms.

With a recoil operated firearm, mass matters. Mass is the "force in the opposite direction" via inertia. Mass of the entire weapon and mass of the moving parts.

30lb recoil operated machineguns don't care about how you hold them in order to operate. They have a large excess of mass, above what is needed to operate.

A 3lb (or less?) handgun does not have as much of an excess of mass. Some have enough, others seem to be right on the line, and need a certain degree of resistance (through the shooter's grip). If they don't get it, malfunctions result.
 
I have found that people who limp wrist are lacking in upper body, arm and hand strength.
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I don't have your extensive experience as a teacher, but my more limited observations indicate that it is more a problem of technique than raw strength. My daughter didn't get stronger - she understood how her technique was bad and corrected it. I have seen, heard, and read numerous similar experiences.

With all due respect to 44 AMP and Tailgator's opinions I have to respectfully disagree with the notion that the preference for tools which resist operator error is entirely predicated on misunderstanding of how tools function. I think the idea is that such tools can save your skin if you make a mistake during a stressful situation.

. . . Of course with the proper technique any properly made gun should function properly. However, there is always a chance of a failure and if you get that stovepipe in a self-defense situation, it could be very, very, bad. It's why many people still carry revolvers despite the other obvious disadvantages. It mitigates the danger of failure.
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I don't think I said that bold part, and if I implied it I did not intend to insult anyone's knowledge. I said that the problem was real, that it could be managed by technique and that some pistols very well may be more susceptible to that technique problem than others.

A revolver is certainly a solution, but it comes at a trade-off, usually, of capacity. A heavier gun is, all else being equal, very likely to be less susceptible to the type of failure being discussed, but there is a trade-off of carrying around the extra weight. A less flexible frame (metal versus polymer) may also reduce the risk of such failures, but has the same trade-off of carry weight and, in some cases, a difference in perceived recoil.

By all means, use what you are comfortable with; just be aware that you are, in almost every case, trading one shortcoming for another. That is, in my opinion, the essence of handgun selection.
 
polyphemus said:
If no recoil spring is present no recoil is felt
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I've always wondered why I can't feel the recoil of my revolvers, Contenders, and double barrel shotguns! :rolleyes:

Thanks for explaining that, it's obviously because they have no recoil spring!
 
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