At what point in the firing process does the slide start to cycle ?

[BritinUSA]

I have always thought that when a 1911 fires, the bullet leaves the barrel, followed by the hot gas that propelled it, then the slide begins to cycle. The bullet will have left the barrel BEFORE the cycle of the slide, otherwise the shots would always go high as the barrel tilts during the process.

In the following photograph, the bullet can be seen a few inches beyond the barrel (just ahead of the flash) and the slide has not yet moved.

But in this photo below, there appears to be a bullet, just before the crown of the barrel and the slide is already half-way back So is this normal ? Is this actually a bullet that I can see inside the barrel, because it certainly looks like one? And if its not a bullet, what the heck is it ?

[BritinUSA]

Closeups of the above images:

[6Liter240Z]

The bullet in the second pic is probably almost to the target already.

[BritinUSA]

The bullet in the second pic is probably almost to the target already.

Thats what I would have thought, but if thats the case what is that copper colored thing in the barrel ?

[Bkreutz]

I think I'd call the second picture a squib.

[BritinUSA]

I think I'd call the second picture a squib.

I just checked the video, all the shots sounded the same to me.

[GrumpyOne]

Muzzle flash...

[BritinUSA]

Muzzle flash...

I think I see what you mean, the flash has lit up part of the groove in the rifling.

[busdriver02]

The slide starts traveling back as soon at the charge ignites. As the bullet accelerates one way the slide/barrel starts in the opposite direction, equal and opposite reaction and all.

Given bullet travel down the barrel time is around .5ms (quickload estimate) the slide will have traveled something less than .15 inches (.15 is 25fps for .5ms).

Edited May 12, 2016 by busdriver02

[bikerburgess]

Playing with heavy bullets in a Tanfoglio 9mm (170gr) I ran into a load that hit very high, compared to the lighter bullets. best I could figure is the bullet was still in the barrel when it started to unlock from the barrel.

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[Steve RA]

It will vary from gun to gun depending on a number of variables. Spring weight, contour of the firing pin stop, etc. As long as it stays constant it won't affect the accuracy. Mechanical accuracy, not yours necessarily!

[StraightSh00ter]

I like this question - it's pretty neat how the whole thing works.

Maybe it's a hot ember in the second pic?

The ignited charge initiates the motion of the slide/barrel and the bullet, so they happen at the same time. The bullet is much lighter than the slide and barrel so it moves much faster. Initially the slide and barrel travel rearward together (perpendicular to the link) before the barrel starts moving down.

The motion of the barrel could still affect the path of the bullet, but with a tight barrel lugs and bushing it should be repeatable enough to compensate with the sights as others have mentioned.

The 1911 animations on YouTube are fun to watch.

[peterthefish]

The slide starts traveling back as soon at the charge ignites. As the bullet accelerates one way the slide/barrel starts in the opposite direction, equal and opposite reaction and all.

Given bullet travel down the barrel time is around .5ms (quickload estimate) the slide will have traveled something less than .15 inches (.15 is 25fps for .5ms).

No, it doesn't. There is a rearward force exerted by the case on the breach face, and a forwards force exerted on the barrel by the bullet. Those forces are equal until the bullet leaves, at which point recoil begins.

In practice, a little gas blows by the bullet and adds net rearwards force but the practical impact is minimal.

The second pic is most likely embers / flash as others have noted.

[Flexmoney]

Don't believe your lying eyes digital camera?

[zzt]

There should be a slight rearward bias to the recoil forces when the pistol is fired. The lugs are still engaged at this point, so there is no rearward movement of the barrel or slide. Once the bullet leaves the barrel, the balancing forces are gone. A large part of the recoil is actually the jet of gasses exiting the front of the barrel. Without them, action of the slide would be anemic.

Consider this. Take a 185gr JHP @ 775fps 50 yard bullseye load. Fire it in a 1911 with a 13 or 14lb recoil spring and everything works normally. There is recoil. The slide operates normally, ejects the empty casing and loads a new round.

Now keep everything the same except use a comp'd barrel. When you fire, there is a little bit of recoil straight back, but no upward rotation of the gun. The slide does not budge. Why? There is very little gas jetting out of the front of the comp. You would have to go down from a 14 to a 7lb recoil spring to get the action to cycle. Even then, the weak spring has problems stripping the next round.

[Nightops]

Has to be an ember from the powder. No mechanical way it could still be the bullet, Too much smoke out the muzzle. The barrel is already unlocked, tipped up in the slide

[PatJones]

Playing with heavy bullets in a Tanfoglio 9mm (170gr) I ran into a load that hit very high, compared to the lighter bullets. best I could figure is the bullet was still in the barrel when it started to unlock from the barrel.

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Unlikely, heavy bullets hit higher in revolvers as well and Revo barrels are fixed to the frame.

[PatJones]

The slide starts traveling back as soon at the charge ignites. As the bullet accelerates one way the slide/barrel starts in the opposite direction, equal and opposite reaction and all.

Given bullet travel down the barrel time is around .5ms (quickload estimate) the slide will have traveled something less than .15 inches (.15 is 25fps for .5ms).

No, it doesn't. There is a rearward force exerted by the case on the breach face, and a forwards force exerted on the barrel by the bullet. Those forces are equal until the bullet leaves, at which point recoil begins.

How does your theory explain the function of blowback pistols? In a blowback design there are no locking lugs and the barrel is rigidly attached to the frame.

All semi auto firearms exhibit very little slide movement while the projectile is present because the mass of the slide takes time to accelerate, and the barrel dwell time of the projectile is very short. In a blowback design, the mass of the slide and the force of the recoil spring are the only things locking the breech during firing.

In a short recoil pistol the barrel and slide are locked together by the lugs and recoil together as a single unit until there has been enough movement to unlock the barrel.

[bikerburgess]

Playing with heavy bullets in a Tanfoglio 9mm (170gr) I ran into a load that hit very high, compared to the lighter bullets. best I could figure is the bullet was still in the barrel when it started to unlock from the barrel.

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Unlikely, heavy bullets hit higher in revolvers as well and Revo barrels are fixed to the frame.

That was not the case in this instance, I am used to and have observed POI changes with different bullet weights and velocities in this and other guns, this was WAY outside that range, if I recall correctly it was about 8" higher at 10yd than a similar PF 147gr load

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[MikeRush]

Not my video, but may contain the answer to the OP's question.

[peterthefish]

The slide starts traveling back as soon at the charge ignites. As the bullet accelerates one way the slide/barrel starts in the opposite direction, equal and opposite reaction and all.

Given bullet travel down the barrel time is around .5ms (quickload estimate) the slide will have traveled something less than .15 inches (.15 is 25fps for .5ms).

No, it doesn't. There is a rearward force exerted by the case on the breach face, and a forwards force exerted on the barrel by the bullet. Those forces are equal until the bullet leaves, at which point recoil begins.

How does your theory explain the function of blowback pistols? In a blowback design there are no locking lugs and the barrel is rigidly attached to the frame.

It's not a theory, it's physics. In a blowback gun there is a rearwards force against the slide. Because the slide is not locked to the barrel, and the barrel, attached to the frame, is being held by the shooter, the slide begins to move immediately, with movement damped by slide mass and spring weight.

With a locked breach firearm the rearward force on the slide is cancelled out by the force on the barrel because they are locked together.

[CZinZA]

There should be a slight rearward bias to the recoil forces when the pistol is fired. The lugs are still engaged at this point, so there is no rearward movement of the barrel or slide. Once the bullet leaves the barrel, the balancing forces are gone. A large part of the recoil is actually the jet of gasses exiting the front of the barrel. Without them, action of the slide would be anemic.

Consider this. Take a 185gr JHP @ 775fps 50 yard bullseye load. Fire it in a 1911 with a 13 or 14lb recoil spring and everything works normally. There is recoil. The slide operates normally, ejects the empty casing and loads a new round.

Now keep everything the same except use a comp'd barrel. When you fire, there is a little bit of recoil straight back, but no upward rotation of the gun. The slide does not budge. Why? There is very little gas jetting out of the front of the comp. You would have to go down from a 14 to a 7lb recoil spring to get the action to cycle. Even then, the weak spring has problems stripping the next round.

Unfortunately the science is not quite right. The bullet weighs more than 100 grains and the ejected gas only about 4 grains (since its equal to the powder plus some oxygen). Since we are talking about conservation of momentum, which is directly proportional to mass, its easy to see that the gases probably contribute 5% to the recoil at most.

The comp doesn't work by stopping the gas going forward. It works by jetting the gas upwards, thus producing a force that counteracts the upwards component of recoil

Sent by Jedi mind control

[bikerburgess]

The slide starts traveling back as soon at the charge ignites. As the bullet accelerates one way the slide/barrel starts in the opposite direction, equal and opposite reaction and all.

Given bullet travel down the barrel time is around .5ms (quickload estimate) the slide will have traveled something less than .15 inches (.15 is 25fps for .5ms).

No, it doesn't. There is a rearward force exerted by the case on the breach face, and a forwards force exerted on the barrel by the bullet. Those forces are equal until the bullet leaves, at which point recoil begins.

How does your theory explain the function of blowback pistols? In a blowback design there are no locking lugs and the barrel is rigidly attached to the frame.

It's not a theory, it's physics. In a blowback gun there is a rearwards force against the slide. Because the slide is not locked to the barrel, and the barrel, attached to the frame, is being held by the shooter, the slide begins to move immediately, with movement damped by slide mass and spring weight.

With a locked breach firearm the rearward force on the slide is cancelled out by the force on the barrel because they are locked together.

You may want to watch the above video before you dig into your position that the slide doesn't move until the bullet leaves

[peterthefish]

The slide starts traveling back as soon at the charge ignites. As the bullet accelerates one way the slide/barrel starts in the opposite direction, equal and opposite reaction and all.

Given bullet travel down the barrel time is around .5ms (quickload estimate) the slide will have traveled something less than .15 inches (.15 is 25fps for .5ms).

No, it doesn't. There is a rearward force exerted by the case on the breach face, and a forwards force exerted on the barrel by the bullet. Those forces are equal until the bullet leaves, at which point recoil begins.

How does your theory explain the function of blowback pistols? In a blowback design there are no locking lugs and the barrel is rigidly attached to the frame.

It's not a theory, it's physics. In a blowback gun there is a rearwards force against the slide. Because the slide is not locked to the barrel, and the barrel, attached to the frame, is being held by the shooter, the slide begins to move immediately, with movement damped by slide mass and spring weight.

With a locked breach firearm the rearward force on the slide is cancelled out by the force on the barrel because they are locked together.

You may want to watch the above video before you dig into your position that the slide doesn't move until the bullet leaves

You should read my whole post first.

[zzt]

CZ, you are quite wrong. 5% is not nearly enough to account for it being necessary to drop 7lb in recoil spring strength to get the slide to operate in the example I gave above.

If you look at the formula for recoil, you will find that the powder weight is multiplied by 4700 before being squared. So it isn't 4 gr, or 0.0006 lb that is added to muzzle velocity times bullet weight squared. It is 0.0006 * 4700 =2.6857 squared, or 7.2131 lbs that is added in that section of the formula.

So what you are considering as the effect of powder weight (0.0006 lb) is actually 1202.1769 times too small.