Strength of slide return spring in a 1911

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facts and logic do not apply.

I beg to differ, sir. I strive to stick to the facts.

Apparently you didn't read the part where we discussed that "about" 90% of the bullet's velocity is obtained within a half inch of its travel...which means that the slide's velocity isn't wholly determined by muzzle velocity. It's a factor...but it's just one of several.

But, do carry on.
 
You guys are about to discover why discussing basic physics with 1911tuner is like discussing concealed carry with Rosie O'Donnell ..... facts and logic do not apply. :)


From what I've read so far, I would suggest a change to what you wrote:

You guys are about to discover why discussing basic physics with 1911tuner is like discussing physics with Albert Einstein ..... what you previously knew to be true, wasn't.​

I'm not saying '1911tuner' is or isn't another Einstein, only that some things he's posted here seem completely wrong, based on what we (at least I) thought we knew before the discussion, as he has a far better understanding of what is going on then we do.

It's like the poor fellow way back when, trying to convince people the Earth wasn't flat, which they absolutely knew for a fact that all you need to do is look around you, to know it's flat!
 
From what I've read so far, I would suggest a change to what you wrote:

In spite of that little hijack attempt...

Recoil...which is what the slide does...is nothing more than backward acceleration, and its velocity is determined by the rate of acceleration and the force required to achieve that rate. By the time the bullet reaches the muzzle, the slide has gotten all the speed it's ever going to with that round.

One of the characteristics of smokeless powder is that it peaks rapidly, and...depending on its burn speed...drops rapidly, so that the pressure and force initially delivered has fallen off to maybe 15 or 20% of its peak and rather than continuing to deliver a meaningful accelerating force against the slide, about all it's doing is maintaining velocities.

Think about JohnK's post. If the bullet's top velocity is realized within 2 inches or so of barrel, the slide's top velocity is also realized at that point and the only way to accelerate it further is to introduce a larger force than the one that's working on it...and a larger force isn't available during the event.

We're also factoring in the outside forces that are fighting the slide and trying to bring it to a stop. i.e. The hammer's mass...the mainspring...and the recoil/return spring, so that by the time the slide impacts the frame, its lost a lot of its peak velocity.

Which brings us back to the original topic of discussion...

Don't overthink the recoil spring. The slide just doesn't hit the frame all that hard.
 
Some related points:

1. On one occasion when he was introduced by a Colt reprehensive to a VIP as an "engineer," Browning replied: Sir, I am not an engineer, I am a MEACANIC! And a very fine one. When it came to firearms design, he could eat engineers for breakfast.

2. The ammunition was developed before the pistol. When Browning knew the specifications for the .45 ACP cartridge he designed/developed the pistol specifically for it. He was not concerned about endless kinds of different ammunition. The platform is very versatile, but if you change the weight of the slide/barrel. and hammer (main) spring as well as the bevel on the firing pin stop, you may enter a whole new ballgame that the inventor never foresaw.

3. Sometimes current day manufacturers or individuals try to compensate for substantial changes made to either the platform or ammunition by tinkering with various spring tensions, the recoil spring in particular. This again is something Browning didn't foresee. Sometimes this may seem to work, but far from always.

4. Browning designed and developed what became the model 1911 pistol to be a military service sidearm. As such it's likely he wouldn't have made its reliability hostage to recoil spring tension.
 
Browning designed and developed what became the model 1911 pistol to be a military service sidearm. As such it's likely he wouldn't have made its reliability hostage to recoil spring tension.

Just so.

A point of interest is that...in the patents...Browning referred to it as the "re-action" spring and not the recoil spring. Since it's accepted that recoil is the reaction side of Newton 3, there may be a misunderstanding as to just what he meant by "re-action" spring, and it led many people to believe that its job is controlling recoil instead of returning the slide to battery.

If he had dropped the re- and just called it an action spring...or if had simply called it a "return" spring...about 90% of these arguments would never have started.

I said "about" 90%. Don't take that as exact. Just a WAG.
 
Maybe I can ask a related question, now that the above has been settled.

Given that the action of the slide is to eject the previous case, and the action of the spring is to bring the slide back to battery, as this is happening the ejected cases will be tossed some distance, with that distance presumably determined by the speed at which the slide is traveling at the moment of ejection.

In my case, all the cases (both with my 18# spring and 15# spring) end up centered around an area about 18 feet away from where I'm shooting, behind me and to my right). I have zero FTL and zero FTE now that the gun is broken in. (This is with both Winchester 230gr White Box ammo, and now also with 230gr Hornady FMJ with 5.2gr of Unique powder, the lightest load in the Hornady reloading book.)

Since the only thing "bad" about this is that my spent cases feel like they're on the way to Cuba, it's not a "problem", and there's also that saying "If it ain't broke, don't fix it"..... but from reading various forums, I think 6 to 10 feet might be the expected distance.
 
Given that the action of the slide is to eject the previous case, and the action of the spring is to bring the slide back to battery, as this is happening the ejected cases will be tossed some distance, with that distance presumably determined by the speed at which the slide is traveling at the moment of ejection.

There's more to the ejection distance than slide speed in recoil.

I set up a friend's Springfield exactly like my range beaters. 14-pound recoil spring...23-pound mainspring...1/16th radius on the bottom of the firing pin stop. Our ammunition is pretty much identical.

My brass lands about 5 feet from the gun, slightly behind my right foot. His tries to go into the next county. So forceful that we had to switch lanes and put him in the far left lane in the falling plate bay because his brass was being damaged from hitting the cinder block divider 12 feet from him when he was in the right side lane.
 
After all this....

Someone's still gonna have to explain to me why spring strength
is largely irrelevant to loading differences in "good guns"...
...and how the actual chart/data in Post 45 is all wrong.
 
After all this....

Someone's still gonna have to explain to me why spring strength
is largely irrelevant to loading differences in "good guns"...
...and how the actual chart/data in Post 45 is all wrong.

Because the Recoil Spring's primary function is to return the slide from the rear position to battery, loading the next round as it goes. This action is independent of what size or power the rounds being used are.

The recoil spring has a relatively small impact (but not none) on the slide's deceleration while it's moving to the rear. You could use a strong enough spring to retard the slide enough to stop it from reaching full rearward and loading the next round, especially if you are using lighter loads, but as long as it has enough force to strip and load the next round you can't really get too weak on the recoil spring.

It looks like your chart is showing the results of trying to get pretty severely underpowerd rounds to function by using recoil spring tension to adjust slide velocity. It should be noted that the MOST powerful round on that chart is 75% of the energy of GI Ball. So someone was trying to get a gun to work pretty far outside the envelope. they probably would have had better luck leaving a strong enough spring in there to cycle (forward) reliably, and then easing the slides decelleration with something else. (Lighter slide, FPS, maybe mainspring. So the chart is not "All Wrong" as I'm sure it's their data. Whoever did it was just tweaking a variable to make a gun work outside designs. Not really relevant to other guns and loads.
 
Actually not.

In partial reponse to the OP's original question, what the chart shows is that a 75% load will
still operate a HB spring sufficiently to not only get behind/load the next round, but also
rearward enough to lock the slide on an empty magazine

That is, fully function.

That the HB spring will also fully function w/o harm up to.including full GI hardball was neither
the question nor the objective.

If people remember, here was the OP's question:
"...use either my 230gr FMJ reloads with a light powder charge,
or 185gr bullets, also with a light charge. ... how to know what
spring to use? Bullet weight may vary, as well as the amount of
powder for custom loads. Is there a chart posted somewhere,
that gives some guidance...
That chart answers that question with actual data.
 
Someone's still gonna have to explain to me why spring strength
is largely irrelevant to loading differences in "good guns"...
...and how the actual chart/data in Post 45 is all wrong.

The chart isn't all wrong. It's just a little too complicated. There's no sense in trying to precisely match a spring rate to the ammunition...because, the spring's function is returning the slide to battery, and whatever else it does is incidental.

You can get by nicely with two springs. One for hardball and one for "softball" reduced velocity ammunition. 16 and 12 would do nicely.

But, because the spring must be compressed in order to do its job, and it has to allow the slide to make full travel rearward...which underpowered ammunition may not have sufficient recoil impulse to do...a light spring becomes necessary with such ammunition.

As far as having enough strength to return the slide to battery...if the gun is functioning correctly, the slide will strip a round and send it home with a 10-pound spring. Again, my litmus test is to assemble the gun without the spring, and hand feed rounds from the magazine by pushing on the rear of the slide with the tip of one finger.

Then, why might you ask, did Browning use a 14-pound spring?

Because the gun was designed with a battlefield in mind...and the battlefields would likely be in Europe or in the jungles, and trench warfare would likely be the order of the day...which as we've seen was a mucky affair.

The US Army wasn't concerned with frame battering or any such nonsense. The frames were designed to take it and the slides were considered expendable...to be replaced as needed...but they were greatly concerned with reliability under adverse conditions.

That's why the Army Ordnance Department ordered a dozen or more slides and barrels for every complete pistol delivered. As long as they had a viable frame, they could assemble a serviceable pistol over and over.

Because the slide and barrel assembly is the gun. The frame is little more than the gun mount. Stop worrying about the frame. It's the slide and barrel that catch all the hell.
 
The chart isn't all wrong. It's just a little too complicated.
It's not complicated, 1911, just tell me the energy (i.e, bullet weight/velocity)
of the custom-loaded low-power cartridge, and the chart gives you a starting
point for full-function slide op'n.

...which was the OP's request.
 
The point was that slide velocity rearward isn't dependent on muzzle velocity. Muzzle velocity is a factor, but it's only part of the whole picture.
Well, sorta...

At the precise point that the bullet exits the muzzle, the slide velocity rearward is wholly dependent on muzzle momentum (ejecta velocity x ejecta mass). The ejecta is everything that leaves the muzzle, gases, bullet & unburned powder/fouling. In practice the bullet's momentum is, by far, the dominant factor and for most purposes the rest of the ejecta can be ignored.

After the bullet exits, the slide will slow down based on friction, recoil spring force, cocking effort, etc.
Apparently you didn't read the part where we discussed that "about" 90% of the bullet's velocity is obtained within a half inch of its travel...which means that the slide's velocity isn't wholly determined by muzzle velocity.
The fact that most of the bullet's velocity is attained very early in its travel is absolutely not proof that the slide's velocity isn't wholly determined by muzzle velocity. The slide velocity at the point of the bullet's exit is a function of recoil which is purely based on conservation of momentum.

Therefore, AT THE INSTANT THE BULLET EXITS the momentum of the ejecta is absolutely the only factor that needs to be considered. For practical purposes, the bullet's muzzle momentum (and the mass of the slide & barrel) is all you need to know.

BEFORE the bullet exits, it is necessary to calculate the momentum of the bullet and gases in the barrel to determine what the slide velocity is. However as pointed out, at least in the type of firearms under discussion here, the majority of the bullet's velocity (& momentum) is imparted very early in travel and therefore it's sufficient for practical purposes (you'll get an answer correct to within 10-20%) to use the muzzle momentum for the purpose of determining slide velocity before the bullet exits.

AFTER the bullet exits, then it is necessary to take friction, spring force, etc. into account to determine slide velocity precisely.
Recoil...which is what the slide does...is nothing more than backward acceleration, and its velocity is determined by the rate of acceleration and the force required to achieve that rate.
You can calculate recoil by working with acceleration and forces, but it is much simpler to calculate it by working with momentum. Trying to work with acceleration and forces needlessly complicates the issue in this case and is likely to, as it has here, result in confusion.
By the time the bullet reaches the muzzle, the slide has gotten all the speed it's ever going to with that round.
Absolutely correct, and that's why, at the point the bullet leaves the barrel, there's no need to have spent any time trying to figure out acceleration rates, etc. while the bullet was still in the barrel. Everything you need to know about slide velocity at the point of bullet exit can be calculated from ejecta momentum and the mass of the slide+barrel.

That's the simple version. For a detailed analysis, including a comment on the effects of the recoil spring on slide velocity, the link below should suffice.

http://yarchive.net/gun/pistol/1911_dynamics.html
 
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It's not complicated, 1911, just tell me the energy (i.e, bullet weight/velocity)
of the custom-loaded low-power cartridge, and the chart gives you a starting
point for full-function slide op'n.
...which was the OP's request.

It's way too complicated because it operates on the assumption that the spring is there to save the frame. It's not. It's there to return the slide.

All the slide has to do is make full travel rearward and reliably return to battery. All else is irrelevant.

A 16 pound spring for full power loads and a 12 pound spring for the powderpuff stuff is all you need...and 14 will probably do for all but the most powderpuffy...like 200 grains at 600 fps.
 
.......If people remember, here was the OP's question.......


If the OP had know then, what he knows now, he would never have asked that question.

Of course, had he not asked it, he wouldn't have learned all this new (to him) information and understanding.

The OP fell asleep last night, trying to justify to himself that this made sense, and got into bed thinking about the weight of the bullet vs. the weight of the gun, until it became hard to tell thought from dream, but it wasn't until reading the posts from this morning that it all "clicked into place" for him. (me).
 
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Well, sorta...

If you're determining the slide's maximum velocity, then all that applies.

But the slide's velocity /energy/momentum at the terminal point of its travel...when it hits the impact abutment...is the point of my post.

The slide's maximum velocity is only part of the story...and in this case, only a small part...because the slide never reaches the velocity that the equations suggest...because the instant that it starts to move, outside forces start fighting it.

Remove the effect of those outside forces...recoil spring...hammer mass and mainspring...and any friction between the rails, and the slide will reach the theoretical maximum.

But not as long as they're in play.

Equal momentums are only possible in the absence of outside force, or in the presence of equal outside force...and the outside forces are never equal in a real gun.

Once the bullet and all the other junk has exited, the slide can only decelerate, and any outside force that can cause it to decelerate, will cause it to decelerate...and by the time the slide hits the impact abutment, it's lost a good bit of its initial velocity.

And that's the point.
 
But the slide's velocity /energy/momentum at the terminal point of its travel...when it hits the impact abutment...is the point of my post.
At the point when the barrel hits the frame and stops, the slide velocity will still be pretty close to maximum as determined by the muzzle momentum and the mass of the slide plus barrel. The significant effects on the slide's velocity are the muzzle momentum and perhaps the force required to begin cocking the hammer. Friction and recoil spring force at this point in slide travel aren't going to have much of an impact.

At the point where the slide hits the frame the slide velocity will have slowed somewhat due to spring force and other factors. But probably not too much. The link in my previous post has a good discussion of why, as you properly state, recoil spring force isn't a major factor.
The slide's maximum velocity is only part of the story...and in this case, only a small part...because the slide never reaches the velocity that the equations suggest...because the instant that it starts to move, outside forces start fighting it.
At the point of the bullet's exit, the slide IS at maximum velocity as demanded by conservation of momentum. All of the forces (outside and/or inside) are wrapped up into the ejecta momentum figure. It will, at the moment of bullet exit achieve precisely the velocity that the equations suggest--but only for that instant.

It will slow down from that point on due to the forces acting on it, and we can determine how much it slows down (not much) by looking at those forces. But we don't need to analyze those forces to determine the maximum slide velocity or to determine the slide velocity while the bullet is in the bore. Momentum tells you all that you need to know.

This is what I meant when I said that focusing on forces and accelerations can be confusing. It's tempting to assume that we're missing something because we're ignoring everything but muzzle momentum but that is simply not true. In reality all of the forces acting on the bullet and the firearm are automatically taken into account through the principle of conservation of momentum, a fundamental principle of physics. At least up until the point of bullet exit. After that, it's a little more complicated.
Equal momentums are only possible in the absence of outside force, or in the presence of equal outside force...and the outside forces are never equal in a real gun.
The momentums are only equal until the bullet exits. After that, the outside forces will cause the two momentums to diverge. But at any point while the bullet is in the barrel and up to the exact point that the bullet exits the two momentums ARE precisely equal and all of the forces acting on the problem are wrapped into the momentum figures. Established science demands it.
Once the bullet and all the other junk has exited, the slide can only decelerate, and any outside force that can cause it to decelerate, will cause it to decelerate...and by the time the slide hits the impact abutment, it's lost a good bit of its initial velocity.
That is correct. Maximum slide velocity occurs at the moment of bullet exit and only decreases from that point onward.

You're getting to the right answer in the end, but you're taking the long way around and some of your intermediate steps are problematic from a scientific standpoint. Fortunately the errors are cancelling out.
 
mikemeyers,

the distance and direction your cases land from the gun depends on, and is unique to, the gun. like 1911tuners example, the same ammo ejects differently from two supposedly identical guns. you have to use the same gun (and same recoil spring) if you want to equate case ejection distance with slide velocity. no fair having two variables here!

murf
 
mikemeyers,

the distance and direction your cases land from the gun depends on, and is unique to, the gun. like 1911tuners example, the same ammo ejects differently from two supposedly identical guns. you have to use the same gun (and same recoil spring) if you want to equate case ejection distance with slide velocity. no fair having two variables here!

murf


Yeppers, one more mis-understanding on my part. Prior to this discussion, I had other ideas, which I guess don't apply. Apparently Les Baer built my gun such that the shells land about 20 feet from where I'm standing, and that hasn't changed with different ammunition and different springs - which is the way it should behave, based on what you and 1911tuner have posted.

I'm curious about this, and may call Les Baer anyway - it's a Premiere II, and I guess it doesn't matter either way - I just get more exercise searching for my shells. :)
 
But at any point while the bullet is in the barrel and up to the exact point that the bullet exits the two momentums ARE precisely equal

Only in the absence of outside force or in the presence of equal outside force...and there are outside forces that are acting on the slide the instant that it starts to move. Outside forces that aren't acting on the bullet. Thus, the slide can never reach its theoretical maximum velocity and momentum...and momentums can't be equal. If the slide weren't affected by any outside forces, then yes...the momentums would be equal. They'd have to be.

The estimations that I made earlier concerning the 32:1 slide to bullet mass didn't allow for any thing except the acceleration and velocities of the slide assembly and the bullet. I didn't factor in the other forces acting on the slide.

We're addressing only the slide's velocity and momentum.



The same way that a bullet fired in a barrel that's full of water will never reach the same velocity that it would without the outside force imposed by the water.

Think of it this way:

Momentum is the product of mass X velocity.

If you mounted a rifle solidly into a 2-ton steel block and fired it, the rifle would never develop any momentum because it couldn't move. No movement...no momentum.
 
At the point when the barrel hits the frame and stops, the slide velocity will still be pretty close to maximum as determined by the muzzle momentum and the mass of the slide plus barrel. .......


There's a lot going on in this discussion about slide velocity, but after watching 1911tuner's video, all you need to do is wrap your hand around the slide, and it has 0 velocity:

https://www.youtube.com/watch?v=Xyis5h9MvUU

"something" is reacting to all that energy from firing the bullet, but as I watch the video, it can't all be in the slide. Even if your hand wasn't resisting the slide movement, there's friction and two springs......
 
I referred to this earlier Mikemyers:

. What weight recoil spring should I use with a particular load?
This is a very common but hard question to answer in exact terms and in most cases an exact answer is not possible. There are many factors which influence the correct weight recoil spring to use. These factors include the particular ammunition brand and load, individual pistol characteristics, individual shooting styles and your individual, subjective feeling of how the gun shoots and should feel.

The factory spring weight is designed to operate the pistol with what would be considered average loads, plus or minus a little. It is not uncommon for manufacturers to specify what they consider a factory ammunition load.
In general terms, the heaviest recoil spring that will allow the pistol to function reliably is the best choice - tempered by the above factors. As a rule of thumb, if your spent casings are first hitting the ground in the 3 to 6 foot range, then the recoil spring is approximately correct. If you are ejecting beyond the 6-8 foot range, then a heavier recoil spring is generally required. If your casings are ejecting less than 3 feet, a lighter recoil spring may be needed to assure reliable functioning.

Taking these factors into consideration, it then comes down to how the gun feels and performs when shooting - in your judgment. However, using too light a recoil spring can result in damage to the pistol and possible injury to you.

https://www.gunsprings.com/faq#Faq3

Simpson's book and charts can be found here:

http://www.amazon.com/Custom-Govern...&qid=1441669237&sr=8-3&keywords=layne+simpson

I quoted from them earlier, they are straight forward. They are suggestions. Your gun and the role you want it to play and how you want it to feel when shooting it will tell you more.

tipoc
 
It's way too complicated because it operates on the assumption that the
spring is there to save the frame. It's not. It's there to return the slide.
No, it's not. Otherwise we'd just have 10# springs and everything would be hunky-dory.

Again, look at the Chart. It says nothing about saving the frame.
It says (literally) "Max Spring Weight to Cycle & Lock"
It it doesn't do that, the OP has a single-shot pistol.
 
I referred to this earlier Mikemyers:.....


"As a rule of thumb, if your spent casings are first hitting the ground in the 3 to 6 foot range, then the recoil spring is approximately correct. If you are ejecting beyond the 6-8 foot range, then a heavier recoil spring is generally required. If your casings are ejecting less than 3 feet, a lighter recoil spring may be needed to assure reliable functioning."​

Tipoc, my gun seems to be listening to 1911tuner instead....

  • With the 18# spring and Winchester White Box 230gr FMJ, the shells are tossed around 20 feet.
  • With the 18# spring, and someone else's "very light target loads, with 185gr bullets, it was the same.
  • With my 18# spring and 230gr FMJ reloads with a the minimum allowable loading, again it was the same.
  • With my 15# spring, and the same 230gr FMJ reloads, again it was the same.
My Les Baer seems to eject all the cases to mostly land from 15 to well over 20 feet from where I'm standing. They are mostly clustered close to each other, with some a bit closer, and some a bit further.

I also have a 13# spring from Les Baer - I suppose I should try that as well, but I don't see any way that an even lighter spring will reduce that distance... maybe it won't increase the distance either.

I suspect, that as was posted earlier, this particular gun just does it this way, and maybe other Premier II guns eject the shells differently.
 
Only in the absence of outside force or in the presence of equal outside force...and there are outside forces that are acting on the slide the instant that it starts to move. Outside forces that aren't acting on the bullet. Thus, the slide can never reach its theoretical maximum velocity and momentum...and momentums can't be equal. If the slide weren't affected by any outside forces, then yes...the momentums would be equal. They'd have to be.
I understand this is a sticking point for you.

There is nothing theoretical about the principle of conservation of momentum and all of the forces acting on the slide and bullet are taken into account by conservation of momentum.

I'm not saying that the forces don't exist or don't affect anything. They do. But those effects are part of what makes the momentum what it is. Because the force of friction slows the bullet down in the barrel, the muzzle momentum will be less. That doesn't mean we ignored it by focusing only on the muzzle momentum, we didn't. We didn't because it is already reflected by the muzzle momentum figure.

The momentums WILL be equal and the slide will reach the velocity that the conservation of momentum indicates. That is something you can take to the bank. Right up until the bullet and the firearm are no longer in contact, the momentum of the bullet and the slide+barrel are precisely equal. Once they part company then the momentums will diverge, but not until then.

The slide will only attain that maximum velocity for the instant of bullet exit and then it will begin to slow.
The same way that a bullet fired in a barrel that's full of water will never reach the same velocity that it would without the outside force imposed by the water.
That is correct. And since it won't reach the same velocity the muzzle momentum will be less and therefore the slide momentum will also be less. They will still be equal and they will still take into account all of the forces involved but both will be reduced by the introduction of the additional resistance.

Think of it like this. I put a number of things in a bag and weigh it and tell you the weight. You tell me that the weight can't be correct because right before I weighed it you sneaked in an item I didn't know about. You tell me that because I don't know what's in the bag I can't possibly know what it weighs. But the weight IS correct. Even though I don't know what's in the bag due to your sleight of hand, I DO know what the bag weighs because I weighed it after you put the item in.

The same principle applies here. Momentum is the weight of the bag--it contains all the results of all the forces. Even if a force we don't know about sneaks in somehow, it will still be reflected in the final momentum figure.
If you mounted a rifle solidly into a 2-ton steel block and fired it, the rifle would never develop any momentum because it couldn't move. No movement...no momentum.
What happens then is the momentum (minus any momentum/energy absorbed by the flexing of the gun/mount/block/etc.) is imparted to the earth and will infinitesimally slow or speed up the rotation of the earth. But it doesn't just go away--the momentums will still be equal. It's just that when you divide the momentum of a .45 bullet at muzzle velocity by the rotational inertia of the earth, the resulting change in rotational velocity is completely undetectable.
There's a lot going on in this discussion about slide velocity, but after watching 1911tuner's video, all you need to do is wrap your hand around the slide, and it has 0 velocity:
Yes, I've done that experiment. If you restrain the slide with your hand, then the momentum that would normally be imparted to the slide/barrel is imparted to your hand instead. It is an interesting experiment but doesn't provide a lot of insight into what happens when you fire a gun with the slide operating normally--i.e. unrestrained.

Just to be clear, the science does seem to support Tuner's assertion that the recoil spring's function in a 1911 is primarily to return the slide and certainly supports the idea it has very little effect on the slide's velocity early in the rearward travel of the slide. The problem isn't his basic conclusion, it's that some of the assertions he's making on the way to the basic conclusion don't jibe with established principles of science. It's sort of like getting to your destination by driving the wrong way down a one way street. You end up in the same place, but you broke some rules getting there and it would be a lot less confusing if you had followed directions that didn't end up putting you head-on to one-way traffic.
 
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