Long stroke SureFire Carrier

[lysander]

Lysander,

Thank you for posting the pressure graph and the two integrals. Did you use an Excel integration routine?

I have dreams of high-speed cameras, to get confirmation of these results.

In #203 you wrote that an time difference of .5ms was necessary to match carbine and rifle. On your graph, it looks like the unlocking time difference between the H1 carbine and the rifle is about .35ms, if that is linearly extrapolated back to the standard carbine buffer we get a difference of about .38ms.

The 1.6 oz difference between the H1 and H3 buffer gives a delay of .05ms (your number), linearly extrapolating this to the 3.8oz difference between the standard and the A5H4 buffer we get a delay of 1.2ms. The 5.5oz difference between the standard and XH buffer from Heavy Buffers gives a delay of .17ms.

Switching to these two buffers from the standard gets us 31% and 45% of the way along the standard carbine/rifle gap at unlocking.

At the beginning of extraction (which should take place at the piston max travel, or thereabouts, since the cam is at the end of the track) the H1/rifle gap looks like .38ms, which extrapolates to maybe .44ms with the standard buffer. The H1/H3 gap here looks like .13ms. An analysis similar to the above gives .31ms delay for the A5H4 buffer (70% of the carbine/rifle gap), and a .45ms delay for the XH buffer (102% of the carbine/rifle delay).

According to this the time benefit of a buffer change can give a substantial proportion of the carbine/rifle benefit at unlocking, and a bigger proportion at extraction.

The hash marks on your graph are a bit small for these eyes, you could do the above analysis with the precise numbers. Also, the time gaps from the H1-H3 buffers at extraction look a bit nonlinear, and I am using linear extrapolation.

Speaking of error analysis, I don't see any in my cursory reading of the papers you referenced.

I have added a curve for a theoretical 10 ounce buffer.



The time intervals are taken from the calculated data points.

Unlocking start (sec):
H1 - 0.00160
H2 - 0.00160
H3 - 0.00165
10 oz - 0.00185
Rifle - 0.00200

Max Piston Stroke (sec):
H1 - 0.00295
H2 - 0.00300
H3 - 0.00305
10 oz - 0.00355
Rifle - 0.00335

Carrier Velocity Max (m/s):
H1 - 6.049
H2 - 5.777
H3 - 5.597
10 oz - 4.295
Rifle - 5.929

Doubling the H3 buffer weight gets you just over half way to the the rifle unlock time, but at the expense of 30% of your carrier velocity. You are not going to have reliable cycling with that much velocity reduction. In order to bring the velocity up, you are going to have to increase the piston pressure, which will move the line back to the left. A 50% increase in cavity pressure is required to bring the carrier velocity back up to around the 5.8 to 5.9 m/s at this is what you get for time

Unlocking start (sec): 0.00160
Max Piston Stroke (sec): 0.00290
Max Velocity: 6.000 m/s

Right back where you started. Changing the mass, while trying to maintain both energy and momentum constant, does not influence the time to unlock to any useful degree, period.

Time to the start of extraction is of little value at this point, the maximum twisting of the bolt occurs at the start of unlocking. Go back to the chamber pressure curve in one of my first posts and look at the chamber pressure at around .003 sec (3.0E-03). It is about a tenth of the value at .001 sec (1.0E-03). Guess on the high side with a chamber pressure at .003 sec of 10,000,000 Pa and figure out just how inflated a brass case will be at this pressure. You will see that time at the start of extraction, in all cases, is not very important.

[lysander]

Sullivan says the design was quite hurried, I would not make the assumption above. Recall the Edgewater buffer.

Thank you, Bruin for the fun video.

Well, hurried or not the resulting system worked out damn near spot on.

With the average P/T curve for M198 with a peak chamber pressure of around 55,000 psi and a port pressure of around 12,000 psi, the bolt is almost perfectly balanced between the residual chamber pressure force and the piston pressure force.

And, remember the Edgewater buffer is the exact same concept (conical rings and washers) as the buffer in the M2 HB used at the time (they have since been replaced by hydraulic buffers), and the M2 HB was well known as a reliable weapon....

[Shootin' Bruin]

I'm happy you enjoyed it Stainless. Although I understand that it isn't a perfect demonstration, the setup in the video at least serves to help one visualize the concepts.

Changing the mass, while trying to maintain both energy and momentum constant, does not influence the time to unlock to any useful degree, period.

Time to the start of extraction is of little value at this point, the maximum twisting of the bolt occurs at the start of unlocking. Go back to the chamber pressure curve in one of my first posts and look at the chamber pressure at around .003 sec (3.0E-03). It is about a tenth of the value at .001 sec (1.0E-03). Guess on the high side with a chamber pressure at .003 sec of 10,000,000 Pa and figure out just how inflated a brass case will be at this pressure. You will see that time at the start of extraction, in all cases, is not very important.

Lysander, while I agree that the time to the start of unlocking of the bolt is important specifically for bolt longetivity, don't you think emipircal data suggests that time to extraction is important as well? Otherwise, why do carbine gassed ARs exhibit less extraction problems with heavier buffers?

And, remember the Edgewater buffer is the exact same concept (conical rings and washers) as the buffer in the M2 HB used at the time (they have since been replaced by hydraulic buffers), and the M2 HB was well known as a reliable weapon....

Functionality in one application doesn't necessarily equate to functionality in another, don't you think? Otherwise, why would the Edgewater buffer need to be replaced?

For those who are wondering how this all relates to the OBC, I'd like to restate that Surefire's claim is that the OBC delays both time to bolt unlock and time to extraction via a combination of increased reciprocating weight and a revised cam path. I hope someone out there is willing to put their claims to the test.

[flenna]

You know, my 20 year old Colt BCG with untold number of rounds through it still works like a charm. I am not sure of unlocking speed, reverse velocity or gas expenditure, though. But I am enjoying this thread.

[lysander]

Lysander, while I agree that the time to the start of unlocking of the bolt is important specifically for bolt longetivity, don't you think emipircal data suggests that time to extraction is important as well? Otherwise, why do carbine gassed ARs exhibit less extraction problems with heavier buffers?

Carbine length gas system, coupled with light buffers and weak extractor springs, suffer from extraction difficultly due to high bolt velocity and the extractor bouncing over the case rim, not bad timing. The stiffer gold extractor spring or an o-ring provide sufficient tension to prevent this problem, as numerous carbine length system with std buffers and o-rings shows. Anything that will reduce bolt velocity will aid in extraction until the point where you have robbed the system of sufficient energy to complete the cycle.

Functionality in one application doesn't necessarily equate to functionality in another, don't you think? Otherwise, why would the Edgewater buffer need to be replaced?

It doesn't. But, at the time the AR-15 was designed with the Edgewater type buffer there was no evidence that it would not be a good idea. With hindsight, everything is "obvious".

[Shootin' Bruin]

Anything that will reduce bolt velocity will aid in extraction until the point where you have robbed the system of sufficient energy to complete the cycle.

So increased time to extraction as provided by the increase is mass in addition to increased extractor tension is important yes? You state that the latter alone is sufficient, but it is also readily demonstrable that the former by itself helps with reliability in extraction as well (up to the point you mentioned in which the added mass prevents the system from achieving the necessary cycle distance.) Of course, there is a reason the M4 eventually got both.

It doesn't. But, at the time the AR-15 was designed with the Edgewater type buffer there was no evidence that it would not be a good idea. With hindsight, everything is "obvious".

This does not change the fact that the Edgewater buffer was clearly suboptimal which was the example Stainless used in trying to make the point that the M16's original design parameters should not be taken as gospel. The gun was far from being "spot on" in its early iterations.

Sullivan says the design was quite hurried, I would not make the assumption above. Recall the Edgewater buffer.

Well, hurried or not the resulting system worked out damn near spot on.

With the average P/T curve for M198 with a peak chamber pressure of around 55,000 psi and a port pressure of around 12,000 psi, the bolt is almost perfectly balanced between the residual chamber pressure force and the piston pressure force.

And, remember the Edgewater buffer is the exact same concept (conical rings and washers) as the buffer in the M2 HB used at the time (they have since been replaced by hydraulic buffers), and the M2 HB was well known as a reliable weapon....

Functionality in one application doesn't necessarily equate to functionality in another, don't you think? Otherwise, why would the Edgewater buffer need to be replaced?

[StainlessSlide]

Doubling the H3 buffer weight gets you just over half way to the the rifle unlock time, but at the expense of 30% of your carrier velocity. You are not going to have reliable cycling with that much velocity reduction. In order to bring the velocity up, you are going to have to increase the piston pressure, which will move the line back to the left. A 50% increase in cavity pressure is required to bring the carrier velocity back up to around the 5.8 to 5.9 m/s at this is what you get for time

Unlocking start (sec): 0.00160
Max Piston Stroke (sec): 0.00290
Max Velocity: 6.000 m/s

Right back where you started. Changing the mass, while trying to maintain both energy and momentum constant, does not influence the time to unlock to any useful degree, period.

Well of course if you bring the carrier velocity back up to that of the milspec carbine system, the unlock time will go back to milspec! The whole point is to reduce carrier velocity to increase unlock and extraction time.

To repeat, it's not carrier velocity that compresses the recoil spring, it's carrier energy. With that seriously heavy buffer, the velocity doesn't need to be anywhere near 6 to have milspec energy. And the milspec energy is usually quite a bit higher than necessary anyway, since most carbine systems are overgassed with 5.56. That's why carbines usually function fine with heavier-than-milspec buffers (when shooting 5.56). The milspec carbine buffer may have been designed to fit a cost constraint, and tungsten supply and fabrication would have been much more expensive than steel.

[lysander]

Momentum matters:



Assume no spring for the moment, we'll add that in later.

M1 is the buffer mass
M2 is the bolt carrier mass
M3 is the bolt mass

Before the cam pin bottoms out in the cam track (bolt velocity is zero) the momentum of the system is:

momentum = (M1 + M2) x v1

-or-

= (M1 x v1) + (M2 x v1)

What is going to happen when the bolt carrier (M2) picks up the bolt (M3)?

M1 will continue back at v1, but M2 will have to slow down because of the conservation of momentum.[1]

M2 x v1 = (M2 + M3) x v2

with v2 being the new velocity of the bolt carrier, something less than v1.

Since the energy = 1/2 x mass x velocity squared, the energy of the bolt carrier/bolt combination will be less that the energy of the bolt carrier before picking up the bolt.

E1 = 1/2 x M2 x (v1)^2 - before picking up the bolt

E2 = (1/2 x M3 + M2) x (v2)^2 - after picking up the bolt

E1 > E2

Note that E2 is completely independent of M1, the buffer mass.

Since the energy available to operate the system (E2) is independent of the buffer mass, the energy tied up in the buffer is lost to the system. The bolt carrier/bolt combination alone, must have sufficient momentum (which is proportional to velocity and thus energy) to operate the system. If we are dealing with a 'standard" AR system the mass of the carrier (M2), bolt (M3) and spring are fixed. So, no matter what mass buffer you put it, the momentum of the bolt carrier/bolt combination (which is proportional to velocity and thus energy) required to operate the system is fixed, therefore , no matter what buffer you chose, the minimum bolt carrier/bolt velocity required to operate the system is the same. And since increasing the buffer mass can only decrease the initial velocity (v1), it can only decrease the energy of the system (E2). This is why when we put in the super heavy buffer, we needed to increase the gas pressure to restore the velocity.

And, the minimum velocity required is a lot closer to what the rifle runs than what you might think, note the number of "my rifle won't lock back..." threads here and elsewhere on the web. Also, I'll wager about 20% -25% of the perfect running, 3:00 ejecting, smooth shooting, ARs out there will fail the lock back on an empty magazine when fired straight down. Gravity retards the system just enough to reduce the velocity below minimum.

Now the spring.

The spring only acts on the back of the buffer (M1), but before the bolt carrier picks up the bolt the bolt carrier and buffer travel as one and the spring provides a retarding force to both, but when the bolt carrier picks up the bolt the spring will now only act on the buffer alone. If the energy in the spring is equal to or greater than the energy of the buffer alone the buffer will maintain contact with the bolt carrier, if not the buffer will separate from the bolt carrier.

Spring energy = 1/2 x k x (d)^2

Buffer energy= 1/2 x m1 x (v1)^2

with

k = spring constant
d = spring compression

In either case, the spring/buffer combination cannot add energy to the bolt carrier/bolt combination.

EDIT: Obviously, if you change the weight of the bolt carrier you will get a different minimum velocity requirement, but it will still be independent of the buffer used. Since the system being discussed in this thread does use a non-standard carrier weight, that might make a "kinder, gentler" rifle.....

____________________________
NOTES:
1. Let's keep it simple and assume a completely inelastic collision.

[lysander]

This does not change the fact that the Edgewater buffer was clearly suboptimal which was the example Stainless used in trying to make the point that the M16's original design parameters should not be taken as gospel. The gun was far from being "spot on" in its early iterations.

The kinematics of the system were well done, aka, they did the math right, which is the one of the most important things.

The other important thing is comprehensive testing of the complete system BEFORE fielding, which was not done and the root of all the early problems with the rifle.

[pyrotechnic]

Momentum matters:



Assume no spring for the moment, we'll add that in later.

M1 is the buffer mass
M2 is the bolt carrier mass
M3 is the bolt mass

Before the cam pin bottoms out in the cam track (bolt velocity is zero) the momentum of the system is:

momentum = (M1 + M2) x v1

-or-

= (M1 x v1) + (M2 x v1)

What is going to happen when the bolt carrier (M2) picks up the bolt (M3)?

M1 will continue back at v1, but M2 will have to slow down because of the conservation of momentum.[1]

M2 x v1 = (M2 + M1) x v2

with v2 being the new velocity of the bolt carrier, something less than v1.

Since the energy = 1/2 x mass x velocity squared, the energy of the bolt carrier/bolt combination will be less that the energy of the bolt carrier before picking up the bolt.

E1 = 1/2 x M2 x (v1)^2 - before picking up the bolt

E2 = (1/2 x M1 + M2) x (v2)^2 - after picking up the bolt

E1 > E2

Note that E2 is completely independent of M1, the buffer mass.

Since the energy available to operate the system (E2) is independent of the buffer mass, the energy tied up in the buffer is lost to the system. The bolt carrier/bolt combination alone, must have sufficient momentum (which is proportional to velocity and thus energy) to operate the system. If we are dealing with a 'standard" AR system the mass of the carrier (M2), bolt (M3) and spring are fixed. So, no matter what mass buffer you put it, the momentum of the bolt carrier/bolt combination (which is proportional to velocity and thus energy) required to operate the system is fixed, therefore , no matter what buffer you chose, the minimum bolt carrier/bolt velocity required to operate the system is the same. And since increasing the buffer mass can only decrease the initial velocity (v1), it can only decrease the energy of the system (E2). This is why when we put in the super heavy buffer, we needed to increase the gas pressure to restore the velocity.

And, the minimum velocity required is a lot closer to what the rifle runs than what you might think, note the number of "my rifle won't lock back..." threads here and elsewhere on the web. Also, I'll wager about 20% -25% of the perfect running, 3:00 ejecting, smooth shooting, ARs out there will fail the lock back on an empty magazine when fired straight down. Gravity retards the system just enough to reduce the velocity below minimum.

Now the spring.

The spring only acts on the back of the buffer (M1), but before the bolt carrier picks up the bolt the bolt carrier and buffer travel as one and the spring provides a retarding force to both, but when the bolt carrier picks up the bolt the spring will now only act on the buffer alone. If the energy in the spring is equal to or greater than the energy of the buffer alone the buffer will maintain contact with the bolt carrier, if not the buffer will separate from the bolt carrier.

Spring energy = 1/2 x k x (d)^2

Buffer energy= 1/2 x m1 x (v1)^2

with

k = spring constant
d = spring compression

In either case, the spring/buffer combination cannot add energy to the bolt carrier/bolt combination.

EDIT: Obviously, if you change the weight of the bolt carrier you will get a different minimum velocity requirement, but it will still be independent of the buffer used. Since the system being discussed in this thread does use a non-standard carrier weight, that might make a "kinder, gentler" rifle.....

____________________________
NOTES:
1. Let's keep it simple and assume a completely inelastic collision.

Lysander, thanks for all the info you've put into this thread. I'm having hard time with understanding the statement you bolded.

Once M2 picks up M3, M1 is going to continue on compressing the spring until all of its kinetic energy is added into the spring's potential energy.

M1*v1^2=k(x3^2-x2^2)

x2 being the point where M2 picks up M3
x3 being the point where rearward motion of M1 would stop.

M2+M2 catch back up before it stops or they don't. Either way there would be another impulse/mometum solution to find out what the velocity of the combined masses would be after said collision.

I think it would be fair to say that changing the mass of M2 will have a greater overall impact on the minimum required carrier velocity, but I don't see how changing M1 won't also have an effect on that number.

Thanks.

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