How much power does the action consume

[TheRoland]

I said it imparted force to counteract the backward slide and only momentarily.

But where does this force come from? This is the key to our whole argument. I argue there is no force which could possibly stall the barrel's movement.

 

[TheRoland]

Yes, the forces are equal. But mass x velocity=acceleration. While the mass of the bullet is tiny compared to the mass of the slide, the velocity is WAY WAY WAY more. Thus, the bullet accelerates much faster than the slide, which is how the bullet can go anywhere in the first place.

Well, sure, but "much faster" is not the same as "prevents the breech-face from moving backward". Which is what I was responding to.

But not as much as it would without it. Do you have xray pictures of the 1911 further into the action?

I'm not sure, I'll look for them. I think I disagree with "not as much as it would without", but my position has been that the slide and barrel accelerate backwards the instant the bullet begins to move. Do we agree now?

 

[Admin]

 

[TheRoland]

Some people play and have fun on merry-go-rounds, though.

 

[HooVooLoo]

But where does this force come from? This is the key to our whole argument. I argue there is no force which could possibly stall the barrel's movement.

The force comes from the friction of the bullet attempting to accelerate through the barrel. That frictional force is imparted on the barrel and attempts to "pull" the barrel/slide forward.

 

[TheRoland]

The force comes from the friction of the bullet attempting to accelerate through the barrel. That frictional force is imparted on the barrel and attempts to "pull" the barrel/slide forward.

But, as I just said, and think you just agreed, the bullet moving is an unimpeachable indication that this force is tiny compared to the force due to gas pressure, and couldn't hold anything anywhere.

EDIT: I think I've got a bad case of "PEOPLE ON THE INTERNET ARE WRONG!" and should step away. Thanks everyone for a fun argument! I'll check back much later.

 

[HooVooLoo]

Well, as I indicated a couple of posts up there...Attempt to push a bullet through a rifled barrel with a stick really really quickly, and see how the barrel feels in your hand.

 

[terraformer]

That looks fun if you add in some cotton candy and a big loli-pop...

 

[HooVooLoo]

I invoke Occam's Razor. Therefore, the thread is over. Yes? LOL

 

[terraformer]

I invoke Occam's Razor. Therefore, the thread is over. Yes? LOL

Well, lets not lock it. Roland is out looking for more pics and maybe I can rent a high speed camera and get a beater 1911 to experiment with. The experiment would be easy to do (though I am betting skirting the safe side) with two identical barrels in a single gun. With lugs, without. Camera rolling and the gun on a rest with a and pull cord. The results are pretty easy to verify. If the slide with no lug barrel moves earlier, the side I am advocating for wins. If no change in slide speed occurs then the lock simply holds the two together and everything moves. The high speed would confirm.

 

[lupis42]

Hey, at least we're not calling each other nazi's

 

[TheRoland]

Hey, at least we're not calling each other nazi's

Oh, I knew there was an argument I forgot to make!

 

[strangenh]

Well, that was fun. I learned a little about how modern short-recoil operated firearms work. . . and I learned which NESers to not ask for help on mechanical engineering projects I might have.

 

[HooVooLoo]

Well, lets not lock it. Roland is out looking for more pics and maybe I can rent a high speed camera and get a beater 1911 to experiment with. The experiment would be easy to do (though I am betting skirting the safe side) with two identical barrels in a single gun. With lugs, without. Camera rolling and the gun on a rest with a and pull cord. The results are pretty easy to verify. If the slide with no lug barrel moves earlier, the side I am advocating for wins. If no change in slide speed occurs then the lock simply holds the two together and everything moves. The high speed would confirm.

No no, I do not mean locking the thread. I'm just calling it over like the answer 42, lol!

In all seriousness, that would be great if you could get a high-speed camera and do some films of various actions so we can see it all in super-duper slo-mo.

P.S. I am voting for result #3. Once the muzzle is clear, the slide/barrel will both move to the rear as a unit until the barrel hits a stop if the locking mechanism works as intended. If the barrel separates from the slide/bolt/breechface before the muzzle is clear, the gun either did not have a "locking mechanism", or my understanding of the first few milliseconds of physics of inertial-recoil is skewed.

P.P.S. Terra... If, during your filming experiments something blows up and you die, can I have ur stuff?

 

[smokey-seven]

And a little more

explosion: a violent expansion or bursting with noise, as of gunpowder

deflagrate: to burn, esp. suddenly and violently

HooVooLoo said:

Small-arms ballistics occur strictly through deflagration, not an "explosion over time".

I will grant you the terminology win credit if you will agree that all explosions happen over a period of time rather than instantaneously.

HooVooLoo said:

P.S. I am voting for result #3. Once the muzzle is clear, the slide/barrel will both move to the rear as a unit until the barrel hits a stop if the locking mechanism works as intended.

The slide and barrel move opposite to the bullet as soon as the mass of the slide and components has been over ridden. There is 20,000 PSI of force that has acted on the breech face of the slide while the bullet is inside the barrel for .0007 seconds. The resultant mechanical action is to the rear and the toggle is in the process of pivoting while the bullet is in the barrel. Frictional resistance of the barrel to bullet pulling it forward has been over ridden by the slide in motion, locked to the barrel and pulling it to the rear.

Terraformer said:

The pivot on the 1911 is a circular arc. As movement occurs in the barrel there MUST be a gap at the breach face that opens up. That would allow pressure to escape.

No.

The barrel's movement is in the vertical direction at the breech end (as well as towards the rear). There is no gap at the breech until the lugs (male and female) have separated and by that time, the pressure has been dropping off because the barrel is empty. This is the reason for the lugs. The toggle needed to be of the length at its final design so that the timing of the action coming to the slide release pin and thus stopping the barrel from rearward movement was of sufficient length so that the bullet was gone from the barrel and pressure dropping.

That 20K PSI force has already acted on the moving components and is dropping off when the toggle reaches its end of travel. Stick a dowel in the barrel of a closed and empty 1911 so that it is resting on the breech face. Hit that dowel with a hammer with 20K PSI force. Do I have to remove the dowel before there is any motion of the slide? Are you guys still trying to tell me that there is no motion of the slide until the bullet leaves the barrel?

Terraformer said:

Your idea also doesn't explain why the locking lugs need to be ground down to have blanks reliably cycle a 1911 sans modification.....

Yet, you need to restrict the barrel in order to get the gas pressure high enough to push the slide back.

No. I have previously stated I can load up a blank and have the slide move all the way to the rear and lock open. A second blank won't chamber because it jams on the barrel ramp / chamber because there is no bullet in it to guide it home. If there was a live fully loaded round in the magazine, then the 1911 will fully function and lock up on the live round. All it takes is enough pressure in the barrel to operate. You need to do nothing with the lugs if you had a functional weapon prior. There is no restriction in the barrel.

 

[HooVooLoo]

I will grant you the terminology win credit if you will agree that all explosions happen over a period of time rather than instantaneously.

Well, the term "instantaneous" opens up a whole other can of worms that would need it's own subforum to get into, much less a separate thread, heh.

From my understanding, an "explosion", or high-explosive (detonation) is a release of energy with a shock-wave front of extreme pressure. Once the material had "exploded (detonated)", the shock-wave has been generated "instantly". That is the extent to which I mean that an explosion does not happen over time, but "detonates instantly" or whatever.

Hell, even a fusion bomb explosion happens "over time", since the initiating charge must cause the fission process to "start to happen", which then leads to the fusion process to "happen".

As for this whole deal about how blank cartridges work the action of a SA, I have no idea. I've never tried it. Only in a pump-shotgun, with blackpowder-based blanks. Loud, tons of smoke and a decent amount of recoil.

 

[lupis42]

Well, the term "instantaneous" opens up a whole other can of worms that would need it's own subforum to get into, much less a separate thread, heh.

From my understanding, an "explosion", or high-explosive (detonation) is a release of energy with a shock-wave front of extreme pressure. Once the material had "exploded (detonated)", the shock-wave has been generated "instantly". That is the extent to which I mean that an explosion does not happen over time, but "detonates instantly" or whatever.

Even in high-explosive (detonation), the explosion propagates through the explosive over time. The difference between "high" explosives like C4, and "low" explosives like gunpowder is that in high explosives, the reaction is propagated through the explosive by the shock wave (detonation) while in low explosives, the reaction is propagated by thermal conductivity (deflagration).

 

[HooVooLoo]

Even in high-explosive (detonation), the explosion propagates through the explosive over time. The difference between "high" explosives like C4, and "low" explosives like gunpowder is that in high explosives, the reaction is propagated through the explosive by the shock wave (detonation) while in low explosives, the reaction is propagated by thermal conductivity (deflagration).

Indeed. That's why I won't argue the semantics of this bit any longer. It's like Occam's Razor inverted.

 

[smokey-seven]

HooVooLoo said:

Indeed. That's why I won't argue the semantics of this bit any longer. It's like Occam's Razor inverted.

I smile and agree. There was no reference in your reply about the 20K PSI hammer and dowel and not driving the slide to the rear tho. That's not semantics.

 

[smokey-seven]

HooVooLoo said:

As for this whole deal about how blank cartridges work the action of a SA, I have no idea. I've never tried it. Only in a pump-shotgun, with blackpowder-based blanks. Loud, tons of smoke and a decent amount of recoil.

I have a Winchester 10 GA black powder, blank only please cannon that will roll the 10 -15+ (pound in weight) cannon back a foot or two on the BLANK cartridge going off. Nobody can tell me blanks have no recoil and not get a verbal fight from me. BTW, there is absolutely no restriction in that barrel nor any locking lugs.

I have observed a blank 45 shell driving the slide of my 45, all the way back. Please people, this is real and not mythical.

 

[TheRoland]

The whole "what does a blank do" discussion, while interesting, is somewhat irrelevant to the main point, I think. You can show via basic laws of physics that the slide/barrel/cartridge begin to move backwards as soon as the bullet begins to accelerate. This is mathematically provable.

I found that there's actually an X-Ray high speed film out there of a 1911 firing, but have yet to find the actual file.

 

[smokey-seven]

Moving on....

HooVooLoo said:

Well, the term "instantaneous" opens up a whole other can of worms that would need it's own subforum to get into, much less a separate thread, heh.

From my understanding, an "explosion", or high-explosive (detonation) is a release of energy with a shock-wave front of extreme pressure. Once the material had "exploded (detonated)", the shock-wave has been generated "instantly". That is the extent to which I mean that an explosion does not happen over time, but "detonates instantly" or whatever.

Feel free to start a new thread on this subject, but as I have understood the difference between, "high" and, "low" explosives, is the TIME of the duration of the energy. High being short time and low being longer time. Sure you are talking thousandths of a second, but it is the duration that makes the difference.

A contained explosion in a barrel, (deflagration I admit) is happening over a longer time, and that is what it is, a more controlled explosion, and a burn rate over time that has not burst the metal restriction by virtue of an outlet for the force less than the constricting environment strengths, and allows the force imparted out of one end instead of being contained in 360 degrees. The EQUAL and OPPOSITE reaction acts on the slide as soon as the mass of the slide and components has been overcome. The barrel and slide moves rearward before the bullet leaves the barrel.

 

[smokey-seven]

TheRoland said:

The whole "what does a blank do" discussion, while interesting, is somewhat irrelevant to the main point, I think. You can show via basic laws of physics that the slide/barrel/cartridge begin to move backwards as soon as the bullet begins to accelerate. This is mathematically provable.

Thanks for having my back in this thread, it is nice to know there are people out there that still understand old man Issac.

Did you smile at my Winchester cannon comment?

 

[terraformer]

Thanks for having my back in this thread, it is nice to know there are people out there that still understand old man Issac.

See, this is the BS that has my blood boiling at this thread. The amount of times words have been put into my mouth incorrectly is beyond f'n belief. Point once where I said the slide NEVER moved. Once. Please. I said stalled and before that I said delayed. I never said it was locked to the slide. See, there are only 500 or so ft lbs of energy in a 230 grain bullet. Do you think that a 1911 only sees 500 ft lbs of energy in recoil? I finally got you to admit that with no bullet and just gas pressure can push a slide in recoil (though I still assert to do it safely and reliably you need to restrict the barrel instead of amping up the powder). Why can't you admit that after the bullet leaves the barrel the 20K PSI of gas pressure has an effect on the slide/barrel as it leaves the barrel? You have said twice that, ceteris paribus, a longer barrel does not increase the speed of the bullet. I call BS on that. The longer the barrel (within bounds and subject to diminishing returns), the longer that expanding gas has to push on the bullet to accelerate it. So long as the force of the expanding is greater than the increased friction on the bullet, the bullet will accelerate.

Now, why is this important? If the barrel and the slide are locked to each other and both ends are plugged, the chamber pressure simply increases until something ruptures or the gas ceases to expand. The bullet is simply a controlled means of relieving the pressure and harnessing the force of the gas for work. The bullet needs energy imparted to break it's inertia and energy imparted to break the force of friction to the barrel. Any energy imparted (by way of the expanding gases pushing behind it) beyond that, is energy that imparted into the bullet for work. Since at the point of the bullet leaving the barrel the chamber pressure is still over 15K PSI, that is 15K PSI of gas pressure that is moving in ONE DIRECTION. Pop a hole into a CO2 cartridge and let go. What happens? Well, that HAS TO HAPPEN here too. For all of the pressure inside the chamber to be used SOLEY to move the bullet, the chamber pressure has to be 1 ATM @ 0' at time of bullet exit. It isn't.

So, by fixing the barrel to the slide, you are ensuring that the full amount of the energy of the expanding gas is directed in one direction, out of the muzzle. We agree on that because someone told you once how smart Browning was and that the locking lugs where to make this happen. But I don't think you ever thought through why. I asked the question and someone answered but I knew the answer. Unfortunately no one questioned the answer. Anyhow, to do this there were any number of other ways Browning could have done it that were obvious from making the slide heavier to the spring stronger to gas actuation. Brownings first patent in firearms was a gas actuator. He created the system that Garand perfected for the rifle. He easily could have created something for a handgun if he wanted to.

Yet, he did something different.

Let me preface this next section by saying I am not going to go back over this and check calculations. They are merely to illustrate a point. If you have a problem with one, raise it and I will alter it if I agree. They are definitely overly simplistic.

So, the energy (e) or the gas is in the chamber. (f) for the forward friction of the bullet. (s) for the energy imparted to the bullet. So the amount of energy heading in recoil (re) for the now infamous .0007 seconds (which you are off on that as you took the OAL and not the case length when you calculated it) and it is re=e-f=s (this is true up until muzzle exit). That means that the slide recoils solely from the energy imparted to the bullet AND NOTHING else. That is purely mechanical recoil at play. This is the concept that you all have been swearing up and down is the end all be all of recoil. I have never once said it wasn't recoil, but the felt recoil of a 1911 firearm is greater than 500 ft lbs. That is because 500 ft lbs is all that makes it into the bullet. There is a lot left on the table in the form of gas exiting the muzzle. No one has accounted for it. It is a number far greater than the 500 ft lbs because if it wasn't, there would not have been enough energy in the .45 ACP to operate a Thompson with a blowback recoil mechanism and a 4x barrel length all with an increase -- not a decrease-- in muzzle velocity of the round. This is an increase after the friction of the barrel is accounted for (which yes, it is different rifling but are you going to really say a 20" barrel has less friction than a 5" barrel?). Additionally, compensators would be useless, etc. This is energy that NEEDS to be accounted for and is PART OF recoil.

The only graph/data I could find to illustrate this is here. You will see the units are unreadable and it's for a 30-06 but the point remains. The bottom one is muzzle pressure (ie; the gas pressure acting on the fluid of air in the atmosphere) and the top one is chamber pressure. You will see it occurs over time, which is why I made a point about this is something that evolves over time.

So, with a pure blowback mechanism the amount of energy imparted on the slide (re) prior to muzzle exit would not be s, but it would be s + f + x where x equals the amount of energy escaping into the atmosphere by the slide backing away from the barrel. With a strong spring and a heavy slide/block x may be marginal. x is zero in a 1911 or any other locked chamber. X also robs the overall system of energy and was one part of what browning was trying to do. But in order to do it, he needed to ensure the barrel didn't flip out of place while the bullet was still in the chamber. He could have either fixed the barrel to the frame and locking the slide like he did in his first rifle design or devise a way of putting to use forces that were already there to delay the full effect of the recoil on the one part (the slide) that controlled the barrel lock until after the bullet left the chamber.

Once the bullet exits the muzzle, the recoil energy is s (+f in a blowback mechanism) + the force of the rest of the gas pressure exiting the muzzle and pushing on the atmosphere. I don't know where to begin calculating minus knowing what the volume of gas is that x grains of powder produces. As I don't have that, I can't figure the actual ft lbs. You could probably back calculate it from the pressure and a few other variables, but this what I have been saying.

Anyhow, you will note in this video (which is unfortunately to short to see the full recoil) that the barrel pushes forward before the slide moves and takes the barrel back with it. (can you say sloppy tolerances in the lugs)
http://www.youtube.com/watch?v=ltyEyNfdp8I&feature=related

Here is a blowback operated pocket gun
http://www.youtube.com/watch?v=POE6NT4xKCs&feature=related
Note how it jumps forward (more like muzzle down) a tad before it kicks back.



PS: I said above the Arc of the pivot is circular, not the movement of the barrel. I am right. If you doubt me you have to prove how moving a straight line drawn between two points when one point is fixed makes anything other than a circle. As a result, the bottom of the chamber releases from the slide face BEFORE the top does because the front of the barrel is fixed by the bushing. Hence the cant that the barrel makes because the front of the barrel is hung up on the barrel bushing.

 

[terraformer]

OK, this was still bugging me. I figured somewhere out there in the ether the data existed that we needed. Luckily enough the data is for a .45 acp and a 1911.
The relevant figures for 230 grain ball are:
peak pressure = 2430 ft lbs.
friction = 260 ft lbs

http://yarchive.net/gun/pistol/1911_barrel_length.html

For 230 grain ball, friction is half of the muzzle energy of the bullet. So a slide and barrel locked together will exhibit no movement RELATIVE TO EACH OTHER and it's movement RELATIVE TO SPACE will be a function of 500 ft lbs of energy. Without the barrel lockup the slide would have received 760 ft lbs of force. How much does the slide/barrel combination weigh and what is the spring weight of the recoil spring? I doubt it requires 500 ft lbs worth of force to move, but you can see that when the bullet is in the barrel, the force acting on the sprung weight of the slide/barrel is significantly smaller compared to the overall potential energy in the system without the barrel lockup to the slide.

A choice quote:

From the tests it appears that the pressure in the 45 auto cartridge peaks after the bullet has moved 0.3 inches. This causes an insignificant movement of the slide and hence no degradation of the state of the barrel/slide lock up.

 

[TheRoland]

OK, this was still bugging me. I figured somewhere out there in the ether the data existed that we needed. Luckily enough the data is for a .45 acp and a 1911.
The relevant figures for 230 grain ball are:
peak pressure = 2430 ft lbs.
friction = 260 ft lbs

http://yarchive.net/gun/pistol/1911_barrel_length.html

For 230 grain ball, friction is half of the muzzle energy of the bullet. So a slide and barrel locked together will exhibit no movement RELATIVE TO EACH OTHER and it's movement RELATIVE TO SPACE will be a function of 500 ft lbs of energy. Without the barrel lockup the slide would have received 760 ft lbs of force. How much does the slide/barrel combination weigh and what is the spring weight of the recoil spring? I doubt it requires 500 ft lbs worth of force to move, but you can see that when the bullet is in the barrel, the force acting on the sprung weight of the slide/barrel is significantly smaller compared to the overall potential energy in the system without the barrel lockup to the slide.

A choice quote:

I haven't read your previous argument here yet (I promise I wlll in the morning) but that's a great link! However, you've made a big mistake in your interpretation of it.

Specifically, you're mixing up force (lbs) and energy (ft*lbs). You can't just take muzzle energy (ft*lbs) and subtract a force (lbs) as you've done.
The friction number you're using there is a force, not an energy amount. You need to compare it to the force not the energy, on the bullet. This force is around 2803 lbs, according to that link. That sounds about right to me.

So the numbers you need to be subtracting are 2803 (lbs) and 260 (lbs), not 500 (ft*lbs) and 260 (lbs). This leads to 2543 (lbs) of force on the slide, which, I might add, is the same net force on the bullet.

EDIT: So, in conclusion, once the bullet leaves the barrel, force on the slide will increase by about 1/10th, then fall off rapidly. I don't think that a very brief 9% reduction in force constitutes any real delay in the cycle of the action. That's probably about the force difference between two different factory loads.

 

[terraformer]

I haven't read your previous argument here yet (I promise I wlll in the morning) but that's a great link! However, you've made a big mistake in your interpretation of it.

Specifically, you're mixing up force (lbs) and energy (ft*lbs). You can't just take muzzle energy (ft*lbs) and subtract a force (lbs) as you've done.
The friction number you're using there is a force, not an energy amount. You need to compare it to the force not the energy, on the bullet. This force is around 2803 lbs, according to that link. That sounds about right to me.

So the numbers you need to be subtracting are 2803 (lbs) and 260 (lbs), not 500 (ft*lbs) and 260 (lbs). This leads to 2543 (lbs) of force on the slide, which, I might add, is the same net force on the bullet.

hmmmm... That's a big mistake. I will look it over again and adjust. I need to work on occasion so I was clearly not paying close attention to this one. I work in energy efficiency and bitch when people mix up power and energy. You would think I would have caught that one... The 2K+ ft lbs of gas pressure force seemed high to me too as the resistance of air is not that high. That would explain it.

 

[TheRoland]

hmmmm... That's a big mistake. I will look it over again and adjust. I need to work on occasion so I was clearly not paying close attention to this one. I work in energy efficiency and bitch when people mix up power and energy. You would think I would have caught that one... The 2K+ ft lbs of gas pressure force seemed high to me too as the resistance of air is not that high. That would explain it.

It happens, especially with English units.

 

[TheRoland]

Hey, I just realized, that with Terraformer's numbers (which I think prove me right, by the way ) , it's possible to actually numerically calculate the answer to the OP's original question, at least with regards to the experimental 1911. If we know all the forces on the bullet and breech-face, and the time T it takes for the bullet to exit the barrel, we can calculate the acceleration of the barrel and thus both velocities at time T! From velocities, we can get energy, and boom, there's the OP's answer. Of course, we'd have to figure out the recoil speed of a fixed action, too, as, strangenh pointed out long ago, it's not going to be 0.

I'll try to do this when I get a chance... tomorrow .

 

[terraformer]

Hey, I just realized, that with Terraformer's numbers (which I think prove me right, by the way ) ,

There is not enough info there to prove you right in part because we are both right depending on what points you are looking at. But what does exist, shows there is significant force in the forward direction on the barrel, but 10% of peak (it has greater impact the lower the gas pressure) is not significant enough if true that the pressure peaks in the barrel that early on. I wonder how true that is. That would put it's relative impact at it's weakest at the very point in the cycle where it could do the most good. ie; Acceleration v. velocity. You cut down the acceleration early and it has a longer lasting effect on velocity. Ultimately it's velocity that determines how long it will take to move the barrel off the correct plane.

So that ultimately makes me wrong based on contributing impact.

Given that early a pressure peak, there is also no way there is high enough pressure at muzzle exit to accelerate the recoil significantly. It's there, but the acceleration has to be happening at a lower rate than earlier in the cycle. In the end, I was assuming that the gas expansion was more inefficient a system than it may be.

There is one other force that this lbl researcher dances around that is of interest. The force holding the lock together but he makes a point of stating there is 770 lbs of force holding the two pieces together. Given how hard metal is, it shouldn't matter from a deformation perspective if the whole purpose of the lock is simply to keep the breech closed. There is a some shear force that one needs to overcome to get those two pieces of metal to move while touching. I don't know what the friction coefficient of steel is, but there is an interesting question if that is high enough to have any impact. Given there is no specific care and feeding instructions for 1911s or any other handgun wrt to oiling the locking lugs, I doubt it is very high but it is there. He probably just mentions it as matter of fact rather than to focus any attention on it.