kuhnhausen kuhnfusion re: 1911 linkdown

[olyeller]

Hi,
This here is for all y'alls input, regarding barrel bedding at link down.
I understand that the barrel is properly linked down when the rear face of the barrel foot/lug is solidly and evenly contacting the vertical surface of the frame, and the bottom of the barrel is, as 1911tuner has said, a onion skin's distance away from the horizontal concave barrel bed. -and, of course, 1/32" or so in front of the frames' feed ramp.

However; in Vol.1, pages 66-67, Kuhnhausen says to remove material from the bottom lug rear face until, "the barrel just links down into full contact with the frame bed. The rear face should not be in contact with the frame at this point."

?huh?

Then, in Vol.2, pg. 191, Figure 178, the vertical frame surface is marked, "Stop Surface" , and there is text that reads "the barrel must clear the top of the frame at linkdown, and stop against this vertical surface."

Vol.2's explanation is jibing what Im gathering from these posts here and other forums, mostly by 1911tuner.

Now I know that the barrel could hit the vertical surface prematurely, and leave a big gap between the concave frame surface and screw up feeding, and oppositely, the barrel at linkdown could smash into the bed without any vertical contact of the barrel lug rear face, which could put severe stress on the link, pin, and entire bottom lug.

So whats the key here?

what is complete barrel linkdown;? -when the vertical frame surface and rear lug face contact, the top of the barrel has a nice .010" clearance from the top of the slide, and a round is easily chambered?

During "proper" linkdown, is there some deceleration of the barrel before vertical frame contact that allows this contact to not be damaging?

Or is it that the "damage" that can occur during premature vertical lug contact, not damage to the bottom striking surfaces, but the non-clearanced lugs and slots that are getting battered at the top, by the slide?

Gosh I hope Im clear with this post; if not, Ive at least helped myself by thinking this through a little more.

thanks!

 

[1911Tuner]

Linkdown

Ideally...The barrel should stop against the vertical impact surface, with the lugs well clear of the slide...with the bottom just a few thousandths off the bed, and drop the rest of the way to the bed. Perfectly, it won't quite hit the bed...but most do, and it doesn't seem to do any harm as long as the barrel doesn't hit the bed before it stops against the VIS. I've seen lugs pull clear through chamber wall from hitting the bed first.

Kuhnhausen can be a little confusing in some of his texts, and as many of us have noted...he can also make a mistake here and there. Volume 2, Page 124 is one that stands out. Top illustration identifies the rear barrel face as a "Recoil Face" when that simply can't be. The barrel doesn't recoil and drive the slide rearward. The slide recoils and pulls the barrel rearward with it.

 

[Jim K]

"The barrel doesn't recoil and drive the slide rearward. The slide recoils and pulls the barrel rearward with it."

It would be more accurate to say that the slide and barrel recoil together, since they are, in effect, one unit until unlocking is complete. The above is a bit like saying a Mauser 98 bolt recoils and pulls the barrel and stock with it.

Jim

 

[1911Tuner]

Mauser

>The above is a bit like saying a Mauser 98 bolt recoils and pulls the barrel and stock with it.<
****************
Well...That's actually kinda what happens. The bolt gets thrust back and pushes the receiver backward...(because the bolt is locked to the receiver via the lugs)....which the barrel is screwed into...and attached to the stock.
When the bolt is thrust backward, it takes the barrel and stock with it.

Think of this, Jim...

Tight-fitting cork on a rope in a plastic pipe. Rope in your left hand, pipe in your right. Pull in opposite directions, gradually increasing the force until the cork slips forward...the pipe will slip in the opposite direction via the force applied. When the cork and pipe move, the arms pulling on both also move.
So...while the bullet is being pulled out of the pipe, the pipe is being pulled off the cork...by your arm. Not because the cork moved forward...but because when the cork moved forward, it allowed the pipe to move backward.

Equate to the 1911...or the Mauser...as you wish.

The cork is the bullet. The pipe is the barrel. Your right hand is the locking lug. Your arm is the slide. Is the pipe/barrel recoiling...or is it being pulled by your arm? Does the pipe/barrel move because the cork/bullet moves...or because the movement of the cork/bullet frees the pipe/barrel to move backward? It doesn't matter where the force comes from...either pushing from within, or pulling from without. Two objects and two forces are involved.
As long as there are two interacting objects with two forces acting equally on each...recoil happens. You have the criteria for action/reaction.

"For every action, there is an equal and opposite reaction."
The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

 

[olyeller]

very interesting; the cork, or bullet, has to be bleeding off some rearward force from the locked slide/barrel while its applying friction to the barrel, right?

I wonder if this friction is noticable for different bullet weights.

 

[1911Tuner]

re:

olyeller pondered:

> the cork, or bullet, has to be bleeding off some rearward force from the locked slide/barrel while its applying friction to the barrel, right?

Right. The bullet being forced through the barrel is under a lot of friction, and pulls the barrel forward with it. As the bullet moves, the frictional resistance drops. Harder to get something moving that it is to keep it moving.
As soon as the bullet moves, so can the barrel....just like in the cork/pipe analogy. When the barrel can move, so can the slide. Many believe that the bullet movement is what causes recoil...when it actually allows recoil because it allows the barrel to move backward.

Consider this hypothetical situation:

If you, a genius inventor/designer came up with a miracle alloy that was 10,000 times as dense as lead...so that a bullet with the identical dimensions of the 230-grain round nose bullet weighed a thousand pounds instead of a half-ounce. This miracle alloy also had a near-zero coefficient of friction.
If loaded and fired from a .45 ACP case with a standard powder charge, the bullet wouldn't move because of its sheer mass...but the slide would...because the barrel could. No friction to keep the barrel from slipping off the bullet. The force against the breechface would drive the slide...the slide would pull the barrel with it...and the gun would literally be pushed off the stationary bullet. So, the gun would cycle, even though the bullet didn't move. Prevent the barrel from moving, and the slide won't move either.

Another demonstration that's easy to do, to show that the force pushes the breechblock and drives the rifle backward:

Insert a steel rod through the bore and against the bolt face. The rod extends past the muzzle a foot or so. Hit the end of the rod with a sledge hammer. Will the rifle "recoil"? Of course it will. Why? Because a force was exerted on the bolt...and because the rifle was free to move. Simple, what?
*********************

And:
>I wonder if this friction is noticable for different bullet weights.<

Sure. As a rule, as weight goes up, so does surface area in the barrel. The inertial resistance is also a player, as is the lead and jacket alloy. Harder lead=more frictional resistance. Harder jacket, ditto. So...it's all subjective to
what the coefficient of friction is for a given size/weight/material. Oil the bore, and it goes down. Thoroughly degrease it, and it goes up.