The Great Thrust Vector Debate

[1911Tuner]

Another Experiment

Duplicate post on the other thread.
_____________________

Okay....Let's recap Jim Keenan's experiment in which he threaded the muzzle end of a barrel and inserted a rod to block bullet movement. A
set screw was threaded into the muzzle, positively blocking any chance of
bullet movement.

We learned two things from this.

One is that the 1911 pistol is far stronger than many of us imagined.

Two is that if the bullet doesn't move, the slide won't move. This
SEEMS to fly in Kuhnhausen's face on his "Balanced Thrust Vector"
theory of recoil, in which he appears to state that the bullet exits and
THEN the slide recoils...which is wrong. If the bullet is gone BEFORE the slide moves...the slide won't move. There is no resistance(the bullet) to the pressure to act against the slide and cause it to move. No launching pad, if you will.

I submit that because the barrel was pushed forward...and because the
slide and barrel were mechanically connected...the slide COULDN'T
move in the opposite direction. The barrel was keeping it static via the locking lugs. it would be the same as two cars linked together by a chain
and pulling in opposite directions. Assuming equal vehicle weight, power,
gear ratios/torque multiplication, and equal traction...neither one CAN move...until something breaks to make the balanced vector UN-equal.

I propose another experiment. I'll remove the locking lugs from a USGI
Colt barrel and mail it to Jim. Hopefully. he still has the rod and screw,
so that all he needs to do is thread the barrel and go. I realize that the gun will then be blowback operated, but the point is this:

If the mechanical connection between the barrel and slide is broken,
the pressure will be free to pursue the path of least resistance, and the slide will move...probably violently. I suggest a strong recoil spring and a shock buffer. I'll include a buffer with the barrel.

 

[misANTHrope]

I propose another experiment. I'll remove the locking lugs from a USGI
Colt barrel and mail it to Jim. Hopefully. he still has the rod and screw,
so that all he needs to do is thread the barrel and go. I realize that the gun will then be blowback operated, but the point is this:

If the mechanical connection between the barrel and slide is broken,
the pressure will be free to pursue the path of least resistance, and the slide will move...probably violently. I suggest a strong recoil spring and a shock buffer. I'll include a buffer with the barrel.

Indeed, this should be the case. If we examine the pistol with the rod in place, we will see that once the powder is ignited and pressurizes the case, the result can be easily split into two force systems. First, the case walls exert a force out radially against the chamber, but since this force is equal in all directions, there will be no tendency for motion unless there is a mechanical failure. In practice, and for the purposes of our discussion, this radial force can be ignored- I bring it up merely to cover all the bases, so to speak. Second, the force resultant from the case pressure will act upon the base of the bullet and the inside of the case head. Whatever gap may exist either between the case and breech face or the bullet nose and the rod will be closed as the bullet and case are force in opposite directions. For our purposes, we'll assume this gap to be negligible.

So the barrel is being pushed forward via the bullet pushing on the rod, and the slide is being pushed rearward via the case. However, the barrel and slide will remain locked together by the lugs, and there is no unbalanced force to the rear to make the slide and barrel cycle rearward together and thus unlock. But without the lugs to hold the two parts together, the slide should indeed recoil rearward, very violently indeed. I definitely wouldn't put anything of value astern of the guinea pig pistol.

The fundamental difference between the pistol with and without the rod is the amounf of force exerted on the barrel in a forward direction. Without the rod in place, the only force is the result of kinetic friction between the bullet and barrel, which is going to be small compared to the force exerted on the bullet and case by the burning powder. But with the rod in place, the force that previously accelerated the bullet to high velocity before it exited the barrel will all be giving the barrel a violent shove forward.

 

[Tamara]

JohnKSa,

Force is measured in mass times acceleration, not pounds--the units are mass x distance divided by seconds squared.

Happily, this is the first question answered on the Measurements page of the Physics Tutoring Website.

Pound, as explained by the bright young engineer type posting before me, most certainly is a measurement of force.

 

[NMshooter]

So, now that we have discussed Browning's short recoil system, anyone want to discuss his long recoil system?

 

[Andrew Wyatt]

SO the short recoil system works like this:

1. primer ignites.

2. pressure rises in case, expanding case

3. projectile starts to move, barrel and case recoil together, pushing on the slide.

4. pressure drops, freeing case from chamber

5. barrel unlocks, slide continues rearward, removes case from chamber and ejects it.

Is that it?

 

[JohnKSa]

pound is indeed a unit of force

Pound, as explained by the bright young engineer type posting before me, most certainly is a measurement of force.

You are both correct. That was a really stupid mistake on my part... That's what I get for posting without thinking through the units carefully--not used to working in the English system.

My point about psi not being force is still correct--when you divide by the area, the resulting units (which are correct in the metric units I posted) no longer imply motion.

I can't exactly recreate my thoughts since I put this in the same post:

Pressure is Newtons per area. That is Force divided by Area.

kind of like pounds per square inch is force divided by area... oh well.

 

[sm]

Tamara ,

Thank you for that Physics Tutoring link

Good discussion folks - I thank you all.