[winfried]

Hi, two question

what would happen if the main spring of a pistol is removed and the pistol fired?

what will happen if you place your hand firmly on a pistol slide and fire?

Regards Winfried

[JDGray]

Locked breech or blowback design? Locked breech will fire normal and the slide will stay back, blowback may fill your face with hot gasses, as the recoil spring is the only thing holding the slide or bolt in battery. I heard you can hold the slide while firing, but never had a reason to try.

[1911Tuner]

Since the mainspring is assumed to drive the hammer...it won't fire. If you mean the action spring...or "recoil spring" if you prefer...the locked breech pistol will do as JD described. The spring doesn't have a lot to do with delaying the slide, as its primary function is returning the slide to battery...not keeping the breech locked or absorbing recoil. I've demonstrated that several times with various 1911 pistols.

I wouldn't be comfortable doing it with a straight blowback design, though...especially those with low-mass slides and uber-stout springs.
The .380 Beretta 81/84 Series comes to mind.

[winfried]

Actually, it makes almost no difference in a straight blowback system either. The mass of the slide prevents it from opening, spring or no spring, a slide will move no more than 3mm until the bullet leaves the barrel, but longer barrels require heavier slides.
Naturally, the spring will slow down the slide.

Yes and you can hold any pistol slide (have not tried a 44 Desert Eagle) and you would do that if for some reason, if you would not want the pistol to eject.

Regards

[1911Tuner]

Actually, it makes almost no difference in a straight blowback system either.

Depends on the slide's mass. Some aren't quite up to it. That's why most blowbacks have relatively stiff springs.

Back in the mid-80s, Llama marketed a scaled-down 1911 in .22 LR....32 ACP...and .380 ACP. The .380s came in two flavors. One was a blowback and the other one was a true to Browning locked breech/recoil operated pistol. Other than that, they were identical.

The LB/RO pistol had a very light spring in it. The blowback was almost unbelieveable, and if you didn't have pretty strong hands...it was a struggle to manipulate the slide.

spring or no spring, a slide will move no more than 3mm until the bullet leaves

A breech opening 5/64ths inch before the bullet exits is risky. If the headspace is toward the high end of allowance to start with, it can cause a case head to blow.

[Zerodefect]

No recoil spring?

Guns like the 1911 and Glock will fire normal enough but won't chamber the next round. Recoil may feel a bit "different" when the slide hits the back of the frame hard.

Holding your hand against the slide of an assembled autoloader? Shouldn't do much. It won't hurt you if you get a firm push on it and it deosn't have sharp edges. I've seen a shooter try this with a Glock 34. Wasn't a big deal.

Could you use your hand as a recoil spring? Sure, with a smooth backed Glock slide. I wouldn't want to try with a 1911.

(If you try any of this and lose a finger or something that's your fault not mine, so don't bother)

[JDGray]

So were your questions a quiz for us? As you seem to already have tried all you described, and knew the answers:cool:

[1911Tuner]

Firing an autoloader with your hand against it will result in your hand absorbing an equal level of momentum as the bullet has. The larger mass is accelerating at a slower rate...but momentums are equal, or nearly enough that they're assumed to be equal. Short answer...it wouldn't likely be pleasant, but you'll survive.

And before the jumpin' starts...

Momentums are theoretically equal. Momentums are only equal in the absence of outside force, or in the presence of equal outside force. Since a gun doesn't present equal outside force on both bullet and slide...breechbolt...the momentums can't be precisely equal. Close enough for gub'mint work and for the purposes of this discussion, though.

[Zerodefect]

Yes.

An arm weighs a whole lot more than a 230gn bullet. I'd imagine that if the back of your slide was shaped like a bullet and weighed only230gn.....this could be a really nasty test to try.

But the surface area of the back of a Glock slide is much larger,flat with dull edges. And the weight of the slide is also much more than the bullet. And the slide is only propelled a millimeter or two.

I've seen it done. At first my jaw hit the ground with what I had just witnessed some idiot do. But if you crunch the numbers......it's not much worse than the insane nun english teacher with the yardstick across the finges bit. Nasty bruise is worst case senerio. Of course if you don't firmly hold the slide forward and you let it build some momentum, it's going to sting a bit more.

I ain't trying it.

[1911Tuner]

Of course if you don't firmly hold the slide forward and you let it build some momentum, it's going to sting a bit more.

Even if you hold it firmly, it'll be like the recoil from a 14-ounce revolver firing whatever caliber you have...with all the recoil focused on a small area...straight line. That's gonna leave a mark.

[chris in va]

I'm not trying it. Interesting it can be done, but sounds like one of those "hey, hold my beer" moments.

[winfried]

Hi Tuner, I did not mean to hold the slide closed from behind, but place the hand over the slide like in catching a cartridge while unloading. It leaves no mark whatsoever. You effectively have a slide that now weighs say 5 000gram rather than 500 gram.

In a blow back system, the recoil spring has not much effect initially either.

The gas pressure exerts 1000kg of force initially, therefore it makes little difference whether there is a spring or not. Easy to calculate in the metric system.

Regards

Winfried

[1911Tuner]

Understood winfried, but it still depends on the slide's inertial mass and the accelerating forces imposed on it as to whether the breech will open while pressure is too high to be safe. If the slide is heavy enough to resist acceleration...delay it...for a long enough period of time, then firing a straight blowback without a spring can be done without a case failure. This is the reason that straight blowback pistols in calibers heavier than .380 ACP have massive slides and why even .32 and .380 caliber pistols wit small slides have heavy springs.

Not a requirement with locked breech/recoil operated pistols because the delay is accomplished in a different way...as with the comparison above between the two Llama .380 pistols. If the slide has low mass, the spring must be heavy. If the slide has high mass, the spring can be lighter. It's a matter of either/or.

[BluEyes]

blowback may fill your face with hot gasses, as the recoil spring is the only thing holding the slide or bolt in battery.

Not true. Compare the effort required to cycle the slide on a blowback pistol after dryfiring it to normal effort when the hammer isn't in the way.

Especially on some .22 rifles that have a very light cycling effort, I suspect that the force of recocking the hammer has more to do with initial opening speed than the recoil springs.

[rcmodel]

Inertia of the slide has more to do with it then anything else.
The hammer cocking & recoil spring force happens only after the slide inertia has been overcome by the recoil energy.

rc

[winfried]

Thanks RC and Tuner, you are both right as well. On a blowback system SMG.s included, any shell may move only a certain distance under pressure. On 9mm it is about 2mm before the shell blows, usually a half moon shaped hole appears where the feed ramp is. Then (maybe not always) the grips may blow off, the magazine bottom is blown out and the shooter has a stupid expression on his face, but nothing more serious happens.

On locked system, the slide and the barrel make up the mass.The slide while locked can move 5mm or so for the pressure to drop and the unlocking begins. Since the barrel mass has come to rest, the slide mass is now less, the spring to retard the moving slide can be weaker.

The slide velocity can be calculated very easy. If a slide weighs 400 grams and a bullet weighs 8 grams, the slide is 50 times as heavy and moves 1/50th in distance and 1/50th of the muzzle velocity.
So if a barrel is 100mm long the bullet will move 100mm and the slide will move 1/50th which is 2mm. BUT, the barrel length shall be the distance of the base of the bullet while still in the shell to the muzzle of the barrel.
A 100mm overall barrel will actually be shorter.
We, in the metric system determine the kick as impulse in grams/sec, in the imperial system it would be called slugfoot. Ever heard of that??
The recoil energy is of no consequence to operate a weapon, but the impulse is factor.

I like such technical discussions.

Regards

[1911Tuner]

On locked system, the slide and the barrel make up the mass.The slide while locked can move 5mm or so for the pressure to drop and the unlocking begins. Since the barrel mass has come to rest, the slide mass is now less, the spring to retard the moving slide can be weaker.

While the barrel's mass is part of the equation for a short time, that's only a small part of the mechanism that works to delay the slide acceleration and breech opening. The bullet itself offers the greatest delaying force.

I'll throw out a hint and let you study on it.

The bullet passes through the barrel under high frictional resistance. Whatever resistance that the barrel offers to the bullet, the bullet also offers to the barrel.

Because...

Force forward is force backward. This, regardless of whether the force is pushing or pulling...compelling or resisting. Force forward=force backward. That it all happens quickly doesn't mean that these frictional forces go away.

The slide is driven backward by recoil...grabs the barrel by the lugs...and takes it along for the ride, at the same time as the bullet is making its trip...in the opposite direction.

While the bullet is making the trip, the barrel is under a dragging force in the direction of the bullet's travel and in the opposite direction as the slide's travel.

Thus, the barrel is fighting the bullet's forward movement and the bullet is fighting the barrel's rearward movement. Because the slide is connected to the barrel during the event...whatever resists the barrel resists the slide...by default.

Clear as mud?

[JohnKSa]

Has anyone done experiments with an oversize smoothbore that doesn't create much friction between the bullet & barrel? I don't doubt that the friction plays a part in the overall system, I'm just curious to know how much of a contribution it makes in terms of retarding the unlocking process.

Some pictures like the one Winfried posted earlier in the thread would tell the story. One taken with a standard barrel, the second taken with a smoothbore that was the same size as the bullet so friction was essentially a non-issue.

[1911Tuner]

[quote] I don't doubt that the friction plays a part in the overall system, I'm just curious to know how much of a contribution it makes in terms of retarding the unlocking process.[quote]

It would have no effect on unlocking. Unlocking...barrel linkdown and disengagement...is a timed event, and occurs at its appointed place in the cycle, regardless of the cycle's speed.

Without the friction...or with less friction...the gun would start to operate like a straight blowback. In addition, there would be more gas blowby past the bullet due to inefficient seal. Of course, reduced resistance would mean reduced pressure and lower bullet velocity...which would mean less slide velocity...so it would probably balance out and the breech opening...separation of breechbolt and barrel...would still be delayed enough to prevent a case rupture.

Remember...Momentums would still be equal...or at least close enough that it wouldn't matter.

[khegglie]

Good thought out explanations guys. The orig. question sounds like a stunt I saw once... a karate guy broke a baseball bat with a kick.....WHY.... because he could.

[JohnKSa]

While the barrel's mass is part of the equation for a short time, that's only a small part of the mechanism that works to delay the slide acceleration and breech opening. The bullet itself offers the greatest delaying force.It would have no effect on unlocking. Unlocking...barrel linkdown and disengagement...is a timed event, and occurs at its appointed place in the cycle, regardless of the cycle's speed.
Clearly I'm missing something.

If the bullet creates a delaying force in the mechanism that works to delay breech opening then how can it have no effect in terms of retarding the locking process?

[1911Tuner]

If the bullet creates a delaying force in the mechanism that works to delay breech opening then how can it have no effect in terms of retarding the locking process?

I assume that you meant "unlocking" process.

Unlocking...barrel linkdown...is timed. Delay or no delay, that event will occur when it's designed to. In this case...it starts at about 1/10th inch of rearward slide travel and is complete at 1/4th inch. Doesn't matter how fast or slow the slide is moving and it doesn't matter if it happens while the breech is under pressure.

Example:

If you were to disable the automatic advance in your car's distributor, and set the static ignition timing at 10 degrees before top center...the plug will fire at 10 degrees BTC whether the engine is idling or at redline. That's timing.

The amount of time that it takes the piston to move from BDC to TDC is time, and it varies with engine rpm.

Timing is mechanically fixed, while time...being a function of speed and distance...is variable.

[JohnKSa]

Gotcha.

Ok, here's a clarified version of my comment.

"I don't doubt that the friction plays a part in the overall system, I'm just curious to know how much of a contribution it makes in terms of slowing slide velocity and therefore delaying the unlocking process."

To clarify further, the slide velocity is a function of conservation of momentum and can be calculated without knowing anything about the bullet/barrel friction. That is, the slide & barrel velocity can be calculated by knowing the mass and muzzle velocity of the ejecta (I'm going to call this "muzzle momentum") and the mass of the slide/barrel.

In other words, the presence of significant bullet/barrel friction probably* results in a lower muzzle momentum since it's fighting the force created by the expanding gases. Lower muzzle momentum would obviously mean slower slide velocity.

Stated yet another way, if one created two different loadings that fired the same bullet at the same muzzle velocity, one through a smoothbore, sized to present very little friction to the bullet, and one through a typical rifled bore, then the slide velocity would be identical as well--assuming the same slide/barrel mass.

On the other hand, if one fired the same loading through the two different barrels, the slide velocity would probably* be lower in the rifled barrel because the added friction would probably* result in a lower muzzle velocity.

So while the friction plays a part in the overall system, you can ignore it if all you're interested in is determining slide velocity and finding how long it will take the barrel to unlock. Not because it's negligible but because the effects are already wrapped into the muzzle momentum.

* I say "probably" because if you reduce friction too much you might create a situation in which there's not enough pressure to maintain a proper combustion process and that might actually reduce muzzle velocity.

[1911Tuner]

Sounds about right, John. Smoothbore=reduced friction=reduced force=reduced velocity and momentum...in both directions. Whatever resistance the bullet would get from the barrel would likewise be given to the barrel...and hence to the slide. There would almost have to be some change in the dynamics...such as maybe the bullet exiting at a little more or a little less slide travel...and my bet would be less because the other outside forces on the slide are unchanged. Everything means something.