Who needs cocking serrations?

[TrailerparkTrash]

Again… 5lb spring, 3lb, 4lb, whatever…. Shooting his gun basically without “shocks,” is beating his gun senseless. It’s just dumb, but I love watching people destroy their guns for the sake of my FREE entertainment value on YouTube.

[subscriber]

Trailer,

You sound like someone that believes the stiffer a recoil spring the better:

Yes, a stronger recoil spring reduces the frame impact when the slide opens, but in increases impact when the slide closes. On a 1911 type pistol an overly stiff recoil spring tends to break off the two bits of steel on the barrel "foot" that the slide stop pin impacts when the slide closes:

If you look at the amount of steel involved in a 1911 (or 2011) and how it is loaded in slide opening VS slide closing, a "soft" recoil spring is less likely to break anything than an "extra power" one. These forums are a good source of 1911 type info:

https://forum.m1911.org/showthread.php?102434-Damage-due-to-incorrect-recoil-spring-strength

https://www.1911forum.com/threads/16-1-2-recoil-spring-to-weak.401817/

https://forum.m1911.org/showthread.php?1964-Recoil-Spring-Function

The guy doing the air racking appears to be shooting a 9 mm rather than a .45. So his frame is taking about 6 foot pounds of energy on every shot, if he left the recoil spring out (and there clearly is one capable of feeding the next round and closing the slide). So, how much damage can a 5 lb recoil spring do; compared to the standard 10 lb spring for 9 mm 1911 style mechanism. The standard recoil spring in 9 mm is that light because slide mass was designed for the .45 ACP.

The .45 1911 started with a 14 lb recoil spring when it was introduced, but the small radius firing pin stop made the slide hard to rack. So the FPS was given a larger radius and the recoil spring "weight" was increased to 16 lb. If you are shooting very light loads you can round over the firing pin stop to ease hammer coking and still use a recoil spring capable of reliable operation:
https://forums.brianenos.com/topic/2...ring-pin-stop/ \

https://forum.m1911.org/showthread.php?67515-EGW-Pin-Stop-Modification&highlight=reduced+recoil

[Dan_Eastvale]

Trailer,

You sound like someone that believes the stiffer a recoil spring the better:

Yes, a stronger recoil spring reduces the frame impact when the slide opens, but in increases impact when the slide closes. On a 1911 type pistol an overly stiff recoil spring tends to break off the two bits of steel on the barrel "foot" that the slide stop pin impacts when the slide closes:

If you look at the amount of steel involved in a 1911 (or 2011) and how it is loaded in slide opening VS slide closing, a "soft" recoil spring is less likely to break anything than an "extra power" one. These forums are a good source of 1911 type info:

https://forum.m1911.org/showthread.php?102434-Damage-due-to-incorrect-recoil-spring-strength

https://www.1911forum.com/threads/16-1-2-recoil-spring-to-weak.401817/

https://forum.m1911.org/showthread.php?1964-Recoil-Spring-Function

The guy doing the air racking appears to be shooting a 9 mm rather than a .45. So his frame is taking about 6 foot pounds of energy on every shot, if he left the recoil spring out (and there clearly is one capable of feeding the next round and closing the slide). So, how much damage can a 5 lb recoil spring do; compared to the standard 10 lb spring for 9 mm 1911 style mechanism. The standard recoil spring in 9 mm is that light because slide mass was designed for the .45 ACP.

The .45 1911 started with a 14 lb recoil spring when it was introduced, but the small radius firing pin stop made the slide hard to rack. So the FPS was given a larger radius and the recoil spring "weight" was increased to 16 lb. If you are shooting very light loads you can round over the firing pin stop to ease hammer coking and still use a recoil spring capable of reliable operation:
https://forums.brianenos.com/topic/2...ring-pin-stop/ \

https://forum.m1911.org/showthread.php?67515-EGW-Pin-Stop-Modification&highlight=reduced+recoil

So Browning and all other firearms engineers had and continue have it all wrong?

[subscriber]

Dan,

If you want to argue, be specific. Browning started with a recoil spring spec for the 1911, and then changed it after a year. That is because he started with a firing pin stop radius (the surface that cocks the hammer), then he changed it. This was because of complaints from the field that the slide was too hard to rack with the hammer down. Was Browning's first set of specs wrong; or his second? Or is the 1911 design such genius that it will operate with a wide range of spring specs?

Browning designed the 1911 as a .45. Much later, Colt added versions in 9 mm; and later in .40 S&W and 10 mm. Does that mean Colt (and other manufacturers, subsequently) know better than Browning? Argue with them, but be specific.

If you can't argue specifics then you really don't have a case. I am not arguing with Browning. I am arguing with you. And you don't have a case. Because you don't understand 1911s. At least your instincts are good: Don't change things you do not understand.

[Dan_Eastvale]

Specifically….
Will also increase failure to go into battery

[subscriber]

Definitely a risk - especially when the gun is dirty. Reliability is most important for a self defense gun. The guy air racking in the video is a trick shooter. Air racking is one of many stunts he does with guns - check his other videos.

I have made specific statements about force and no one has asked me where I got that from. So, here is a sample calculation:

I said that the average force accelerating the .45 bullet down a 1911 government model barrel was 1000 lb. Where did I get that from?

The effective barrel length is 4.5" for a Govt model 1911. That is the nominal 5" barrel minus the distance the bullet base is from the breech face inside the cartridge case. The bullet is accelerated up the barrel until the bullet base reaches the muzzle, pretty close to 4.5" later.

4.5" / 12" = .375 foot

.375 x 1000 lb = 375 foot.pound

(850 FPS x 850 FPS x 230 grain)/450240 = 369 foot.lb

So, the average force of 1000 lb acting on the bullet and the slide (locked to the barrel) is 6 ft.lb off from the exact value for hardball. Close enough for govt work. Yes, the total force acting on the bullet is higher by the average friction of the bullet down the barrel. But it is the net force accelerating the bullet that causes recoil.

As a trick shooter, the air racker in the video does a lot of shooting (watch his other videos). If his pistol runs only 50 k rounds instead of 100 k rounds before a major part breaks, he replaces the part or the pistol. I am not concerned with him wearing out or fracturing his pistol. Only that he does not endanger anyone else.

If you were actually concerned about premature wear on the guns of others, you should tell all the 3-gun guys to strop dropping their mags on the ground. Don't they realize how much grit their guns ingest from dirty mags, and how that increases wear? Not a problem (for them), until their guns jam from too much dirt (or weak recoil springs on top of the dirt). Their guns are consumable, like their ammo.

[Scotty]

It has a STI Recoil Master guide rod which uses 2 springs. The initial travel spring is like 6 lbs. There's a much heavier second spring for the remainder of the travel.

A 9mm 1911 usually has around 9-10 lbs recoil spring which is actually slightly over sprung. I shoot a 11 lb spring for 40's.

[mk2dave]

I said that the average force accelerating the .45 bullet down a 1911 government model barrel was 1000 lb. Where did I get that from?

The effective barrel length is 4.5" for a Govt model 1911. That is the nominal 5" barrel minus the distance the bullet base is from the breech face inside the cartridge case. The bullet is accelerated up the barrel until the bullet base reaches the muzzle, pretty close to 4.5" later.

4.5" / 12" = .375 foot

.375 x 1000 lb = 375 foot.pound

(850 FPS x 850 FPS x 230 grain)/450240 = 369 foot.lb

Can you walk me through you math? It's been a minute since I've done physics, and I have questions.

If we assume a 12" barrel and a velocity of 1000fps (http://www.ballisticsbytheinch.com/45auto2.html), we get 511ftlbs.

12"/12"=1 foot

1 ft * 511 lb = 511 ft pounds

So the force acting on the bullet is halved just because of the longer barrel?

[subscriber]

You are correct, Dave,

The reason the dropping average accelerating force seems wrong is because it is average force multiplied by the total distance that the bullet is being accelerated that correlates with measured velocity and energy. The average pressure drops as the barrel gets longer. This is because a pistol cartridge case has only a small amount of fast burning propellent, so the actual pressure profile is very spikey. The peak pressure is normally multiples of the average pressure. The shorter the barrel, the closer the peak and average pressures are, with more waste energy in the muzzle blast. The longer the barrel, the further the peak and average forces are from each other.

This means that an inch of bullet travel ahead of the chamber adds a lot more kinetic energy to the bullet than the inch before the muzzle. With a long enough barrel, no energy is added in the last inch. For a longer barrel still the bullet will slow down from the "ideal" length.

The highest velocity for the .45 ACP is reached from about 10" of barrel length. It depends on the exact load being shot, and the tightness and surface finish of that particular chamber and bore; so absolute statements cannot be made.

In some instances 12" may provide more velocity. In others it may be less that 10". Certainly, a 16" long barrel shooting a pistol caliber is extracting all that caliber has got, and in some instances, the bullet may be slowing down very slightly from it peak velocity reached earlier in the bullet travel. Which pistol caliber makes a difference, so what is true for a 9 mm is not true for a .44 Mag. Mostly because of the mass of powder used and the internal volume of the case when the bullet is seated.

The quantity that matters for energy extraction is expansion ratio. A 9 mm case has such a tiny volume that the space behind the bullet reaches optimum in a shorter barrel length than a .44 Mag. A .45 Colt revolver case has a large volume, but it is a carry over from black powder days. The 45 Colt is a low pressure caliber with a small mass of smokeless propellant; so using a 20" barrel does not add that much energy over a 5" barrel. And a large part of the gain is because a lever action rifle does not have the cylinder flash gap of a revolver. Ballistics by the inch has data on the effect of the flash gap, if you are interested.

Yes, the .45 Colt can be loaded to .44 Mag levels in guns that are strong enough to handle it. In which case, longer barrels would add more energy, than with black powder equivalent commercial smokeless loads.

[subscriber]

So the force acting on the bullet is halved just because of the longer barrel?

In a nutshell, the average force is reduced for a longer barrel, but that does not reduce the energy transferred to the bullet, because the reduced force is applied over a longer distance.

A longer than 5" barrel length obviously has no effect on the gas pressure profile, or average force applied to the bullet within the first 5" of travel.

The longer the barrel gets, the less velocity you lose per inch, if you cut the barrel shorter.

[mk2dave]

Thanks for the explanation. Guess I have to dust off the ol' textbooks....