Jim Keenan, Keith J (and others): slide velocity in a 1911?

Ruben Nasser · February 23, 2001, 09:43 PM

(I've also posted this at "the smithy", and nobody has answered yet). As I understand it, the force that causes the slide/barrel to accelerate backwards (this is, reaction -recoil- from the moving bullet) ceases after the bullet exits the barrel and the pressure is released (well before unlocking, in a properly timed pistol, the slide recoiling only about 0.1"). Does this means that slide velocity is highest at this point? From this point on the slide is working against the recoil spring, friction in the locking lugs, extracting and ejecting the case, resetting the disconector, lowering the hammer, etc. Do you have any idea of the slide velocity in a 1911? I contacted a tech representative from ParaOrd, and he says that Rob Leatham told him it is about 0.06 seconds in a stock 45 1911, which translates into 1000 rpm, sounds logical, I was thinking above 800 rpm, higher than a typical submachine gun).
Regarding the locking time, this is covered very confusingly (...at least for me) in Kuhnhausen's book. It seems he implies that the slide and barrel are somehow in a static position while the bullet is moving in the barrel, and that they only start moving backwards when the bullet exits. This is impossible, as equilibrium of momentum dictates that the slide/barrel must be moving backwards (albeit at a much slower velocity due to their larger mass) while the bullet is moving forward. See this previous thread:
http://thefiringline.com/forums/show...threadid=52358
Does the recoil spring contributes in this stage to the locking of the slide/barrel?
Do you have any idea of the cycle time in a 1911?
I did some ballpark calculations and the barrel time is in the order of 0.4 miliseconds or lower, wich seems about right.
I also did a mass x velocity equilibrium equation for the bullet and powder gasses on one side, and slide/barrel on the other, getting an initial (free recoil)slide/barrel velocity of about 28-30 fps, wich "feels" slow considering I even disregarded the influence (if any) of the recoil spring and other factors. On the other hand, this initial slide velocity is more that enough to get the pistol to cycle in .06 sec, since the average slide velocity required to complete the cycle (about 4.2") in this time is only about 5.8 fps, or 20% of this initial slide velocity.
Please help me understand this!!!

[Edited by Ruben Nasser on 02-23-2001 at 09:19 PM]

James K · February 24, 2001, 12:04 AM

Wow! I don't know the slide velocity and I don't think I have ever seen a figure, but that .06 seconds sounds reasonable. (Don't try to get a rounds per minute figure out of it, RPM is influenced by more than breechblock velocity.)

You are right in that Kuhnhausen, in spite of his excellent books on accurizing the .45, does not seem to really understand how the gun works. Your statement is correct, except that pressure has little to do with it except to get the bullet started. After that, it is all recoil and inertia.

This site is heck tonight. Either they have flow control on or the site is just slow, but I will look at your figures and see if I can figure things out a bit and get back to you.

Jim

Ruben Nasser · February 24, 2001, 08:56 AM

Posted by Cmc, at "the smithy":
"see the link below for photo of bullet exiting 1911 colt:
http://www.woodselec.com/photos/gun0011M.jpg
You have the same presure applied to the bullet and the cartridge, for the bullet to start moving the slide and barrel have to be static, the barrel lugs lock the slide/barrel together , the barrel wants to move forward and the slide wants to move back, they're locked, the bullet can start moving, the barrel lugs are taking the stress. Your calculations are too simplistic, to find the answers you are looking for you may have to set up a differential equation of motion of second order."

My reply (just to make this discussion easier to follow):
Thanks for the link to the photo of the departing bullet (I've seen many, some at the very moment the bullet exits the muzzle). It CONFIRMS what I'm saying, the slide/barrel recoil about .1" at this moment. It is not that they are magically motionless while the bullet departs (it is impossible), but rather their speed is so much slower that it appears so.
By the way, I've contacted the tech staff at ParaOrdnance, Springfield and Kimber. Only a Para rep. has answered yet, and he confirms that the barrel/slide start to move back at the very moment of firing and by the moment the bullet departs the muzzle the slide has recoiled a little but the locking lugs have not disengaged yet. I can forward you a copy if you want.
My calculation ARE simplistic, I'm just looking for a ballpark number of the initial slide/barrel velocity while locked. I would also like to get an idea of the slide velocity in a 1911 throughout the cycle, but don't know how to estimate/calculate it.

Jim K:
Thanks a lot for your answer!! maybe I'm biased, but I feel you are one of the few that really understands this system of operation. Two points:
- If equilibrium of momentum is to be mantained, and the bullet is accelerating while in the barrel (due to gas pressure - action vs reaction), then the slide is also accelerating backwards. I don't understand well your statement "pressure has little to do with it except to get the bullet started. After that, it is all recoil and inertia", please illuminate me.
- The theoretical 1000 rpm I've mentioned is correct (I made the comparison just to have a yardstick), because is based in cycle time, .06 seconds, not slide velocity.

Cmc · February 24, 2001, 12:03 PM

I have to agree with Jim ,Ruben The slide and barrel recoil together, As far as finding the velocity doing calculations you are only going to get a ballpark idea.
I have witnessed some test , work related and we used a high speed video with a calibrated grid behind in the background to find the exit velocity of the projectile.The test did not involved a firearm ,we used an airgun powered by a scuba bottle and I wont go into any details about
what we tested.
To calculate the speed we looked at the frames to find the distance the projectile move and the time between frames, simple calculations distance over time = velocity.
To find the velocity doing calculations you can
use energy conservation, that is the energy of the bullet
is 1/2x bullet mass x velocity squared minus the potential energy stored in the spring, neglect friction and solve for
the slide/barrel velocity.The kinetic energy of the slide and barrel together is 1/2 x their mass x squared velocity.
You will have to setup a control volume and all that,
I wont go into those details.

Ruben Nasser · February 24, 2001, 12:48 PM

Thanks Cmc, solving for energy gives a slide speed of about 120 mph; which seems high to me. The energy stored in the spring is very small (at this stage it is only compressed about 1.35"), and the energy of the gas mass going out even smaller. I would really like to see real life values of slide speed.
I hope you are convinced that the slide/barrel are never static, but rather moving backwards at a slow velocity.

Peter M. Eick · February 24, 2001, 06:32 PM

I am no expert, but I think it would beyond a second order differential eqn. I really could see it being more like a 5th order, but I would probably make it to complex (trouble with being a physicist!)

On page 185 of the book the m14 from Garand to m-21(I forgot the exact title walking to the computer) they have a measured graph of op rod velocity during firing. It peaks out at 28 feet per second. Since the M-14 cycles at about 750 rnds/minute, which seems in the ballpark for the 45, I would therefore conclude that 120 MPH or 176 feet per second is to high by a factor of 10.

If I really wanted to know I would go out with my digital video camera at high fixxed shutter speed and shoot the firing and count the frames from trigger pull to full lock up.

I will have to think about it a bit. I must admit that 28 feet per second seems a bit slow.

johnwill · February 24, 2001, 08:42 PM

One point about slide velocity. Obviously, when it first starts moving, it can't be moving at the maximum speed, unless the laws of physics have suddenly been repealed. The slide has to accelerate to whatever maximum speed it attains before the forces reach equilibrium, then it will deaccelerate until it reverses direction to chamber the new round. I won't even hazard a guess as to the speeds, or what the maximum speed might be, but it's clearly much slower than the bullet travel, and the maximum speed is reached well past .1" of slide movement.

Ruben Nasser · February 24, 2001, 08:59 PM

Johnwill, you are right about the acceleration curve, but why would the slide keep on accelerating after the bullet departs and pressure is released? To accelerate something you must apply a force, which is no longer present.
I don't know the energy content in the powder, or the heat losses, so the energy equations are a bit difficult for me. I would guess the friction losses are small.
I thought the mass x velocity equilibrium equation for the bullet and powder gasses on one side, and slide/barrel on the other would give an idea of the initial slide velocity
(after all, this is a "recoil" system), neglecting the influence of the recoil spring (small energy storage at this stage) and linkage (which does very little work in the initial .1"), but I forgot the influence of the mainspring.
Peter, that 28 fps value correlates well with the value obtained in my very rough calculation, but I'm not saying that's the max velocity, I would like somebody come up with a better calculation or a real life value.
On the other hand, even a fairly slow initial slide velocity is could be enough to get the pistol to cycle in .06 sec, since the average slide velocity required to complete the cycle (about 4.0") in this time is only about 5.6 fps. The difficult part is to estimate all the energy lost/stored (compressing the recoil spring, friction in the locking lugs and rails, extracting and ejecting the case, resetting the disconector, lowering the hammer, etc.) in the rearward travel to have an idea of the initial velocity. I guess the slide velocity should be highest right after the bullet departs, but I could be wrong, what do you think?

Cmc · February 24, 2001, 10:09 PM

RubenI agree with ya,
I was wrong, the barrel and slide recoil together and it is a kinematics problem,
therefore we have to deal with acceleration and forces.
I was trying to suggest a way do do a quick calculation.
If you want to get technical you will have
to use as you said F=MA,but it will have to analyzed for different intervals.
The first period of motion the barrel and all recoiling parts are moving together and
it starts when the bullet starts to move and ends when the breech unlocks from the barrel.The equation of motion for this period is
(mass barrel+mass slide+1/3mass return spring)Acceleration = Force on thelugs-force of recoil spring-Force of hammerspring-Friction force.
You know everything in the equation above so you can solve for the acceleration for this period of motion.you can also include any other mass that gets accelerated.
And By the way This is still mass times acceleration.If you want velocity you will have to integrate this equation.
You can do this In MathCad and change all the parameters really quick.
There are on this pistol two or more periods of motion to be analyzed and determine the forces to extract a cartrige and insert a new one.also the force and the mass of the disconnector The force of the spring and on and on.
There are many parameters to be evaluated in the design of weapons, I personally dont have any firearms design experience , I apologize to you all for giving you an answer of the top of my head, I did some research and this is what I came up with.
This is a nonlinear problem and does not have an easy solution, that is the reason prototypes are built , tested and fine tuned.

Peter M. Eick · February 25, 2001, 01:09 PM

Ruben,

My feeling is that the maximum slide velocity occurs right after the bullet exits because the gases from combustion are now acting like a rocket to combine with the momentum from the recoil. With no cork (bullet) to constrain the pressure, the gases will dissipate quickly and this will be the end of the "rocket assist". Now we are fighting friction, springs and disconnects to reduce the velocity.

Relooking at the graph of time verses displacement for the m-14 (earlier reference). The velocity peaks quickly then drops off to a relatively steady exponential decay till it impacts the back of the receiver. This would be expected if the main retarding force would be the spring.

It would be a very interesting problem to model in mathcad or matlab, but I think to get accurate information for a reasonable model we would need a lot of experimental results, and this would answer our question.

James K · February 25, 2001, 06:40 PM

What I get is pretty much in line with what you folks are saying. But in any case you will get only an average velocity, and it would take a lot of math to get anything else. Need I add that heavy math is not my bag?

I do know that there would be a lot of equations, though. The average slide velocity will only be a number.

But there is no question that pressure does not open the pistol; the whole purpose of the locked breech is to PREVENT pressure from opening the breech. The reason for slide/barrel unit movement backward is the forward movement of the bullet (and to a lesser extent the unburned powder and the powder gas). Anyone who does not believe this, check with Mr. Newton.

As I noted before, I have blocked the barrel of a 1911A1 pistol to prevent any bullet motion; the cartridge fires, but the bullet does not move and the slide does not move. The gas leaks out with a "pfft" sound.

This has also been done with a rifle, though I have not done it. A Model 1903 rifle with the barrel completely blocked ahead of the bullet was suspended from strings and fired remotely. It never moved. (Don't try this at home; I think they used a special steel case since the weakest part of pressure containment in a firearm is the cartridge case.)

Jim

Cmc · February 26, 2001, 03:39 AM

Jim
The task of the breech is to close the chamber safely during firing.It has to remain closed to support the cartridge case and prevent the gases from leaking and doing damage to the shooter and the weapon.
Now we know the 1911 is a locked breech weapon.
We also know is a short recoil tilting breech.
As I said above in short recoil the barrel and the Breech/slide start moving back together when the bullet starts moving, hence the term short recoil,the pressure in the barrel translates into a force at the breech face this force is in turn acts on the barrel lugs as they are locked to the slide.Pressure is not constant as it changes along the barrel lenght.Newton said for every action there is a reaction.also after the slide unlocks from the barrel if the pressure is still above the atmospheric pressure this energy will add to the slide velocity.
The test you said you did do not allow the pressure to rise and the puff you talked about maybe just the primer and gases leaking around the cartridge since the pressure did not rise to a level to even expand the case, I think for the pressure to rise the bullet has to reach the forcing cone.
Please dont take my comments offensive as I dont intend to offend you,I dont doubt the veracity of your test and your opinion is respected.
Can you elaborate on the test,purpose, procedure,results,etc?.

Ruben Nasser · February 26, 2001, 07:58 AM

Hola Cmc, what Jim K is explaining is that you MUST have some mass accelerating on one side (the bullet) for the slide/barrel to move in the other direction. If the bullet doesn't move then the chamber is only (hopefully) containing the pressure in all directions until some leak occur in the case and pressure is released; the slide/barrel won't move back and will remain locked.
If you don't have motion of the bullet then there is no reaction (recoil), and you could have the gun basically hanging motionless from strings. This is also true for other locked breech weapons, like his example of a 1903 rifle.
A blowback pistol with a blocked barrel will still move the spring and slide back (probably at a dangerous velocity because of the overpressure), and recoil to the front if hanging freely!!

James K · February 26, 2001, 11:18 PM

Hi, CMC and Ruben,

Ruben is correct. In a recoil operated pistol, the pressure itself does not open the breech, and I can assure you that in my experiments the powder did burn and the pressure expanded the case to the point where extraction was difficult. The bullet moving to the forcing cone does not cause the pressure; the pressure (normally) causes the bullet to move. But it is NOT the pressure that opens the gun, it is the barrel/slide mass reacting in the opposite direction to the motion of the bullet mass. Again, consult Mr. Newton.

A blowback pistol is a different story. In those pistols, the pressure pushes directly on the breech, which is free to move away from the barrel. The breech is held closed only by the mass of the parts. Note that it is not the spring that keeps the breech closed, except for that part of the recoiling mass that is the spring. A blowback pistol will work fine without a spring (and some have been made that way), except that the breech has to be closed manually. (The old S-M single shot .22 pistol had no recoil spring; pressure opened it and ejected the empty case. The shooter loaded another round and pushed the bolt closed.)

Jim

Ruben Nasser · February 28, 2001, 09:20 AM

I've contacted Sandy Strayer, of SVI fame, and other manufacturers, and this is what I've got so far with their kind help:
1- The most important factor for slide acceleration is reaction from the bullet moving forward, while it is still in the barrel. The slide and barrel move backwards, albeit at a much slower velocity due to their larger mass. Contrary to what many people believe, the slide/barrel does move backwards (and start to unlock just a little, probably depending on the load) before the bullet exits.
The friction in the locking lugs does not keep the slide and barrel "freezed" in a position, but rather the velocity and movement of the slide is so small compared to that of the bullet that the bullet exits well before the unlocking is of much importance. Remember that there is a .020 flat on the barrel lower lug link ramp area (connecting arcs) that allows for a reward movement of the slide before the barrel starts to unlock.
Aceleration is the most important component to study as the slide is unlocked. Aceleration is at maximum at the moment of unlocking and starting rearward movement. Velocity is very low at that point. It is aceleration that is retarded by the friction between the slide and frame and the springs. The greater the aceleration the greater the wear on the slide/frame fit. The down force of the slide against the front of the frame rails is the point of greatest friction and wear.
2- I'm still a bit confused regarding why acceleration is peaking at the moment of unlocking but velocity is not, since apparently the muzzle blast further accelerates the slide backwards, but its effect is minute compared to the big bang.
3- The recoil and main springs contribute to retard the acceleration of the slide, and diminish the peak velocity. Using a heavier mainspring or changing the hammer's leverage on the slide will have a one time effect on the slide startup.
4- I have tried to get an idea of the initial slide velocity in a 1911, and how it varies along its travel, but the velocity is constantly changing, and so is the acceleration or deceleration. In the words of S.Strayer: "You cannot get an idea of the "initial" velocity. It doesn't exist".
5- I still don't have a real life value for barrel time or peak slide velocity.
6- The cycle time is in the order of 0.06 seconds.

Cmc · February 28, 2001, 08:00 PM

Jim and Ruben
I agree with both of you and I think we are on the same wave
all I tried to explain was the dynamic analysis as I would analyse this problem.
As a design Engineer an analyst I am looking for forces,pressures and depending on the problem velocities and accelerations, lets not forget large deflections,trying to estimate friction and if possible reduce the problem or parts of it to a static analysis.
The load on the locking lugs on the barrel is calculated at least to 1.5 times the max pressure which is usually called proof test pressure, this is an example of a static calculation done for a dynamic load.
I do study firearms operation on my spare time, but literature on the design of firearms is not abundant.
Calculations I showed in the previous post above are from a chech college text book called "Military small arms design" , I think,I left it at the office.
The point is the slide is constantly changing velocity, accelerating and decelerating because of either a change in mass, forces acting on it and on an on.
So you see I would be interested mostly in forces to design components for a weapon.
Anyway this was a good exercise for the mind and I thankyou all for your you kind words of wisdon.
I love this country
I am the NRA

Ruben Nasser · March 3, 2001, 08:26 AM

For all still interested, check the calculations on: http://yarchive.net/gun/pistol/1911_dynamics.html
They make sense to me, he's going through the same reasoning and getting the same initial slide velocity I've got in my very simple calculations (...but I'm not saying these are definite numbers).
I'm still looking for real life values in slide velocity and barrel time, and the still-motion (or even x-ray) pictures that will bring an even better understanding of this issue.

February 28, 2001, 09:20 AM	#15
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	In summary.... I've contacted Sandy Strayer, of SVI fame, and other manufacturers, and this is what I've got so far with their kind help: 1- The most important factor for slide acceleration is reaction from the bullet moving forward, while it is still in the barrel. The slide and barrel move backwards, albeit at a much slower velocity due to their larger mass. Contrary to what many people believe, the slide/barrel does move backwards (and start to unlock just a little, probably depending on the load) before the bullet exits. The friction in the locking lugs does not keep the slide and barrel "freezed" in a position, but rather the velocity and movement of the slide is so small compared to that of the bullet that the bullet exits well before the unlocking is of much importance. Remember that there is a .020 flat on the barrel lower lug link ramp area (connecting arcs) that allows for a reward movement of the slide before the barrel starts to unlock. Aceleration is the most important component to study as the slide is unlocked. Aceleration is at maximum at the moment of unlocking and starting rearward movement. Velocity is very low at that point. It is aceleration that is retarded by the friction between the slide and frame and the springs. The greater the aceleration the greater the wear on the slide/frame fit. The down force of the slide against the front of the frame rails is the point of greatest friction and wear. 2- I'm still a bit confused regarding why acceleration is peaking at the moment of unlocking but velocity is not, since apparently the muzzle blast further accelerates the slide backwards, but its effect is minute compared to the big bang. 3- The recoil and main springs contribute to retard the acceleration of the slide, and diminish the peak velocity. Using a heavier mainspring or changing the hammer's leverage on the slide will have a one time effect on the slide startup. 4- I have tried to get an idea of the initial slide velocity in a 1911, and how it varies along its travel, but the velocity is constantly changing, and so is the acceleration or deceleration. In the words of S.Strayer: "You cannot get an idea of the "initial" velocity. It doesn't exist". 5- I still don't have a real life value for barrel time or peak slide velocity. 6- The cycle time is in the order of 0.06 seconds.

February 23, 2001, 09:43 PM	#1
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	(I've also posted this at "the smithy", and nobody has answered yet). As I understand it, the force that causes the slide/barrel to accelerate backwards (this is, reaction -recoil- from the moving bullet) ceases after the bullet exits the barrel and the pressure is released (well before unlocking, in a properly timed pistol, the slide recoiling only about 0.1"). Does this means that slide velocity is highest at this point? From this point on the slide is working against the recoil spring, friction in the locking lugs, extracting and ejecting the case, resetting the disconector, lowering the hammer, etc. Do you have any idea of the slide velocity in a 1911? I contacted a tech representative from ParaOrd, and he says that Rob Leatham told him it is about 0.06 seconds in a stock 45 1911, which translates into 1000 rpm, sounds logical, I was thinking above 800 rpm, higher than a typical submachine gun). Regarding the locking time, this is covered very confusingly (...at least for me) in Kuhnhausen's book. It seems he implies that the slide and barrel are somehow in a static position while the bullet is moving in the barrel, and that they only start moving backwards when the bullet exits. This is impossible, as equilibrium of momentum dictates that the slide/barrel must be moving backwards (albeit at a much slower velocity due to their larger mass) while the bullet is moving forward. See this previous thread: http://thefiringline.com/forums/show...threadid=52358 Does the recoil spring contributes in this stage to the locking of the slide/barrel? Do you have any idea of the cycle time in a 1911? I did some ballpark calculations and the barrel time is in the order of 0.4 miliseconds or lower, wich seems about right. I also did a mass x velocity equilibrium equation for the bullet and powder gasses on one side, and slide/barrel on the other, getting an initial (free recoil)slide/barrel velocity of about 28-30 fps, wich "feels" slow considering I even disregarded the influence (if any) of the recoil spring and other factors. On the other hand, this initial slide velocity is more that enough to get the pistol to cycle in .06 sec, since the average slide velocity required to complete the cycle (about 4.2") in this time is only about 5.8 fps, or 20% of this initial slide velocity. Please help me understand this!!! [Edited by Ruben Nasser on 02-23-2001 at 09:19 PM]

February 24, 2001, 12:04 AM	#2
James K Member In Memoriam Join Date: March 17, 1999 Posts: 24,383	Wow! I don't know the slide velocity and I don't think I have ever seen a figure, but that .06 seconds sounds reasonable. (Don't try to get a rounds per minute figure out of it, RPM is influenced by more than breechblock velocity.) You are right in that Kuhnhausen, in spite of his excellent books on accurizing the .45, does not seem to really understand how the gun works. Your statement is correct, except that pressure has little to do with it except to get the bullet started. After that, it is all recoil and inertia. This site is heck tonight. Either they have flow control on or the site is just slow, but I will look at your figures and see if I can figure things out a bit and get back to you. Jim

February 24, 2001, 08:56 AM	#3
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	Posted by Cmc, at "the smithy": "see the link below for photo of bullet exiting 1911 colt: http://www.woodselec.com/photos/gun0011M.jpg You have the same presure applied to the bullet and the cartridge, for the bullet to start moving the slide and barrel have to be static, the barrel lugs lock the slide/barrel together , the barrel wants to move forward and the slide wants to move back, they're locked, the bullet can start moving, the barrel lugs are taking the stress. Your calculations are too simplistic, to find the answers you are looking for you may have to set up a differential equation of motion of second order." My reply (just to make this discussion easier to follow): Thanks for the link to the photo of the departing bullet (I've seen many, some at the very moment the bullet exits the muzzle). It CONFIRMS what I'm saying, the slide/barrel recoil about .1" at this moment. It is not that they are magically motionless while the bullet departs (it is impossible), but rather their speed is so much slower that it appears so. By the way, I've contacted the tech staff at ParaOrdnance, Springfield and Kimber. Only a Para rep. has answered yet, and he confirms that the barrel/slide start to move back at the very moment of firing and by the moment the bullet departs the muzzle the slide has recoiled a little but the locking lugs have not disengaged yet. I can forward you a copy if you want. My calculation ARE simplistic, I'm just looking for a ballpark number of the initial slide/barrel velocity while locked. I would also like to get an idea of the slide velocity in a 1911 throughout the cycle, but don't know how to estimate/calculate it. Jim K: Thanks a lot for your answer!! maybe I'm biased, but I feel you are one of the few that really understands this system of operation. Two points: - If equilibrium of momentum is to be mantained, and the bullet is accelerating while in the barrel (due to gas pressure - action vs reaction), then the slide is also accelerating backwards. I don't understand well your statement "pressure has little to do with it except to get the bullet started. After that, it is all recoil and inertia", please illuminate me. - The theoretical 1000 rpm I've mentioned is correct (I made the comparison just to have a yardstick), because is based in cycle time, .06 seconds, not slide velocity.

February 24, 2001, 12:03 PM	#4
Cmc Member Join Date: December 25, 2000 Posts: 59	I have to agree with Jim ,Ruben The slide and barrel recoil together, As far as finding the velocity doing calculations you are only going to get a ballpark idea. I have witnessed some test , work related and we used a high speed video with a calibrated grid behind in the background to find the exit velocity of the projectile.The test did not involved a firearm ,we used an airgun powered by a scuba bottle and I wont go into any details about what we tested. To calculate the speed we looked at the frames to find the distance the projectile move and the time between frames, simple calculations distance over time = velocity. To find the velocity doing calculations you can use energy conservation, that is the energy of the bullet is 1/2x bullet mass x velocity squared minus the potential energy stored in the spring, neglect friction and solve for the slide/barrel velocity.The kinetic energy of the slide and barrel together is 1/2 x their mass x squared velocity. You will have to setup a control volume and all that, I wont go into those details.

February 24, 2001, 12:48 PM	#5
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	Thanks Cmc, solving for energy gives a slide speed of about 120 mph; which seems high to me. The energy stored in the spring is very small (at this stage it is only compressed about 1.35"), and the energy of the gas mass going out even smaller. I would really like to see real life values of slide speed. I hope you are convinced that the slide/barrel are never static, but rather moving backwards at a slow velocity.

February 24, 2001, 06:32 PM	#6
Peter M. Eick Senior Member Join Date: August 4, 1999 Location: Houston, Texas Posts: 2,991	I am no expert, but I think it would beyond a second order differential eqn. I really could see it being more like a 5th order, but I would probably make it to complex (trouble with being a physicist!) On page 185 of the book the m14 from Garand to m-21(I forgot the exact title walking to the computer) they have a measured graph of op rod velocity during firing. It peaks out at 28 feet per second. Since the M-14 cycles at about 750 rnds/minute, which seems in the ballpark for the 45, I would therefore conclude that 120 MPH or 176 feet per second is to high by a factor of 10. If I really wanted to know I would go out with my digital video camera at high fixxed shutter speed and shoot the firing and count the frames from trigger pull to full lock up. I will have to think about it a bit. I must admit that 28 feet per second seems a bit slow.

February 24, 2001, 08:42 PM	#7
johnwill Senior Member Join Date: April 6, 2000 Location: PA Posts: 3,451	One point about slide velocity. Obviously, when it first starts moving, it can't be moving at the maximum speed, unless the laws of physics have suddenly been repealed. The slide has to accelerate to whatever maximum speed it attains before the forces reach equilibrium, then it will deaccelerate until it reverses direction to chamber the new round. I won't even hazard a guess as to the speeds, or what the maximum speed might be, but it's clearly much slower than the bullet travel, and the maximum speed is reached well past .1" of slide movement.

February 24, 2001, 08:59 PM	#8
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	Johnwill, you are right about the acceleration curve, but why would the slide keep on accelerating after the bullet departs and pressure is released? To accelerate something you must apply a force, which is no longer present. I don't know the energy content in the powder, or the heat losses, so the energy equations are a bit difficult for me. I would guess the friction losses are small. I thought the mass x velocity equilibrium equation for the bullet and powder gasses on one side, and slide/barrel on the other would give an idea of the initial slide velocity (after all, this is a "recoil" system), neglecting the influence of the recoil spring (small energy storage at this stage) and linkage (which does very little work in the initial .1"), but I forgot the influence of the mainspring. Peter, that 28 fps value correlates well with the value obtained in my very rough calculation, but I'm not saying that's the max velocity, I would like somebody come up with a better calculation or a real life value. On the other hand, even a fairly slow initial slide velocity is could be enough to get the pistol to cycle in .06 sec, since the average slide velocity required to complete the cycle (about 4.0") in this time is only about 5.6 fps. The difficult part is to estimate all the energy lost/stored (compressing the recoil spring, friction in the locking lugs and rails, extracting and ejecting the case, resetting the disconector, lowering the hammer, etc.) in the rearward travel to have an idea of the initial velocity. I guess the slide velocity should be highest right after the bullet departs, but I could be wrong, what do you think?

February 24, 2001, 10:09 PM	#9
Cmc Member Join Date: December 25, 2000 Posts: 59	RubenI agree with ya, I was wrong, the barrel and slide recoil together and it is a kinematics problem, therefore we have to deal with acceleration and forces. I was trying to suggest a way do do a quick calculation. If you want to get technical you will have to use as you said F=MA,but it will have to analyzed for different intervals. The first period of motion the barrel and all recoiling parts are moving together and it starts when the bullet starts to move and ends when the breech unlocks from the barrel.The equation of motion for this period is (mass barrel+mass slide+1/3mass return spring)Acceleration = Force on thelugs-force of recoil spring-Force of hammerspring-Friction force. You know everything in the equation above so you can solve for the acceleration for this period of motion.you can also include any other mass that gets accelerated. And By the way This is still mass times acceleration.If you want velocity you will have to integrate this equation. You can do this In MathCad and change all the parameters really quick. There are on this pistol two or more periods of motion to be analyzed and determine the forces to extract a cartrige and insert a new one.also the force and the mass of the disconnector The force of the spring and on and on. There are many parameters to be evaluated in the design of weapons, I personally dont have any firearms design experience , I apologize to you all for giving you an answer of the top of my head, I did some research and this is what I came up with. This is a nonlinear problem and does not have an easy solution, that is the reason prototypes are built , tested and fine tuned.

February 25, 2001, 01:09 PM	#10
Peter M. Eick Senior Member Join Date: August 4, 1999 Location: Houston, Texas Posts: 2,991	Ruben, My feeling is that the maximum slide velocity occurs right after the bullet exits because the gases from combustion are now acting like a rocket to combine with the momentum from the recoil. With no cork (bullet) to constrain the pressure, the gases will dissipate quickly and this will be the end of the "rocket assist". Now we are fighting friction, springs and disconnects to reduce the velocity. Relooking at the graph of time verses displacement for the m-14 (earlier reference). The velocity peaks quickly then drops off to a relatively steady exponential decay till it impacts the back of the receiver. This would be expected if the main retarding force would be the spring. It would be a very interesting problem to model in mathcad or matlab, but I think to get accurate information for a reasonable model we would need a lot of experimental results, and this would answer our question.

February 25, 2001, 06:40 PM	#11
James K Member In Memoriam Join Date: March 17, 1999 Posts: 24,383	What I get is pretty much in line with what you folks are saying. But in any case you will get only an average velocity, and it would take a lot of math to get anything else. Need I add that heavy math is not my bag? I do know that there would be a lot of equations, though. The average slide velocity will only be a number. But there is no question that pressure does not open the pistol; the whole purpose of the locked breech is to PREVENT pressure from opening the breech. The reason for slide/barrel unit movement backward is the forward movement of the bullet (and to a lesser extent the unburned powder and the powder gas). Anyone who does not believe this, check with Mr. Newton. As I noted before, I have blocked the barrel of a 1911A1 pistol to prevent any bullet motion; the cartridge fires, but the bullet does not move and the slide does not move. The gas leaks out with a "pfft" sound. This has also been done with a rifle, though I have not done it. A Model 1903 rifle with the barrel completely blocked ahead of the bullet was suspended from strings and fired remotely. It never moved. (Don't try this at home; I think they used a special steel case since the weakest part of pressure containment in a firearm is the cartridge case.) Jim

February 26, 2001, 03:39 AM	#12
Cmc Member Join Date: December 25, 2000 Posts: 59	Jim The task of the breech is to close the chamber safely during firing.It has to remain closed to support the cartridge case and prevent the gases from leaking and doing damage to the shooter and the weapon. Now we know the 1911 is a locked breech weapon. We also know is a short recoil tilting breech. As I said above in short recoil the barrel and the Breech/slide start moving back together when the bullet starts moving, hence the term short recoil,the pressure in the barrel translates into a force at the breech face this force is in turn acts on the barrel lugs as they are locked to the slide.Pressure is not constant as it changes along the barrel lenght.Newton said for every action there is a reaction.also after the slide unlocks from the barrel if the pressure is still above the atmospheric pressure this energy will add to the slide velocity. The test you said you did do not allow the pressure to rise and the puff you talked about maybe just the primer and gases leaking around the cartridge since the pressure did not rise to a level to even expand the case, I think for the pressure to rise the bullet has to reach the forcing cone. Please dont take my comments offensive as I dont intend to offend you,I dont doubt the veracity of your test and your opinion is respected. Can you elaborate on the test,purpose, procedure,results,etc?.

February 26, 2001, 07:58 AM	#13
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	Hola Cmc, what Jim K is explaining is that you MUST have some mass accelerating on one side (the bullet) for the slide/barrel to move in the other direction. If the bullet doesn't move then the chamber is only (hopefully) containing the pressure in all directions until some leak occur in the case and pressure is released; the slide/barrel won't move back and will remain locked. If you don't have motion of the bullet then there is no reaction (recoil), and you could have the gun basically hanging motionless from strings. This is also true for other locked breech weapons, like his example of a 1903 rifle. A blowback pistol with a blocked barrel will still move the spring and slide back (probably at a dangerous velocity because of the overpressure), and recoil to the front if hanging freely!!

February 26, 2001, 11:18 PM	#14
James K Member In Memoriam Join Date: March 17, 1999 Posts: 24,383	Hi, CMC and Ruben, Ruben is correct. In a recoil operated pistol, the pressure itself does not open the breech, and I can assure you that in my experiments the powder did burn and the pressure expanded the case to the point where extraction was difficult. The bullet moving to the forcing cone does not cause the pressure; the pressure (normally) causes the bullet to move. But it is NOT the pressure that opens the gun, it is the barrel/slide mass reacting in the opposite direction to the motion of the bullet mass. Again, consult Mr. Newton. A blowback pistol is a different story. In those pistols, the pressure pushes directly on the breech, which is free to move away from the barrel. The breech is held closed only by the mass of the parts. Note that it is not the spring that keeps the breech closed, except for that part of the recoiling mass that is the spring. A blowback pistol will work fine without a spring (and some have been made that way), except that the breech has to be closed manually. (The old S-M single shot .22 pistol had no recoil spring; pressure opened it and ejected the empty case. The shooter loaded another round and pushed the bolt closed.) Jim

February 28, 2001, 08:00 PM	#16
Cmc Member Join Date: December 25, 2000 Posts: 59	Jim and Ruben I agree with both of you and I think we are on the same wave all I tried to explain was the dynamic analysis as I would analyse this problem. As a design Engineer an analyst I am looking for forces,pressures and depending on the problem velocities and accelerations, lets not forget large deflections,trying to estimate friction and if possible reduce the problem or parts of it to a static analysis. The load on the locking lugs on the barrel is calculated at least to 1.5 times the max pressure which is usually called proof test pressure, this is an example of a static calculation done for a dynamic load. I do study firearms operation on my spare time, but literature on the design of firearms is not abundant. Calculations I showed in the previous post above are from a chech college text book called "Military small arms design" , I think,I left it at the office. The point is the slide is constantly changing velocity, accelerating and decelerating because of either a change in mass, forces acting on it and on an on. So you see I would be interested mostly in forces to design components for a weapon. Anyway this was a good exercise for the mind and I thankyou all for your you kind words of wisdon. I love this country I am the NRA

March 3, 2001, 08:26 AM	#17
Ruben Nasser Senior Member Join Date: March 29, 2000 Location: Asuncion, Paraguay Posts: 280	For all still interested, check the calculations on: http://yarchive.net/gun/pistol/1911_dynamics.html They make sense to me, he's going through the same reasoning and getting the same initial slide velocity I've got in my very simple calculations (...but I'm not saying these are definite numbers). I'm still looking for real life values in slide velocity and barrel time, and the still-motion (or even x-ray) pictures that will bring an even better understanding of this issue.