Long stroke SureFire Carrier

[StainlessSlide]

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.

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

Ok, we are moving to a different level of approximation.

In completely inelastic collision, there will definitely be some energy loss, that's what inelastic means. This assumption fixes an energy loss for a given collision. I concur that this is a good assumption.

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)...

You mean E2 = (1/2 x (M3 + M2)) x (v2)^2 , since you don't want M1 in your equation, and you need the parentheses.

E2 is not the energy available to operate the system, though, since the buffer energy is going into the spring also (we are using "energy" as a shorthand for "kinetic energy"). You need to add on 1/2 x M1 x v1^2 to get the total energy, which of course does depend on the buffer mass.

Since v2 = v1 * M2 / ( M2 + M3 ), the energy loss is E2/E1 = M2/(M2 + M3) (The derivation is a bit lengthy for this venue). E2/E1 should be very roughly .75. This factor is independent of buffer mass, and the buffer does not lose energy in this collision. If the buffer mass is, say, 70% of the carrier mass, it will carry about 40% of the initial energy. The energy loss to the total system buffer/carrier/bolt will then be a factor of about .85.

... the energy tied up in the buffer is lost to the system.

Where did it go? The buffer is moving backwards and compressing the spring, if it's in contact with the carrier or not. Its energy is still going into the spring. The buffer will eventually stop, and it is stopped by the spring (ignoring a possible low velocity collision with the extension). Since the out-of-contact buffer will decelerate rapidly, it will recontact the carrier, and there will be more slightly inelastic collisions that will be difficult to model. If this happens over a significant period of time, we cannot assume that total carrier/bolt/buffer momentum is conserved, because the spring is exerting a force. But we could assume that there is one recontact collision and that it is completely inelastic. The relative velocity will be small though, so I don't think the energy loss will be significant.

the momentum of the bolt carrier/bolt combination (which is proportional to velocity and thus energy) required to operate the system is fixed

No, as we have said, the energy of the buffer/carrier/bolt is what operates the system (to our current level of approximation). And no, energy is proportional to momentum squared at fixed mass, E = p^2/(2m). This came up in an earlier post. And we are changing the mass by changing the buffer.


A good time series of carrier position with different buffers from a high speed camera would be much more conclusive than this freshman-level analysis. This should be the next step, especially since we can't seem to agree on our application of elementary classsical mechanics.

[lysander]

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+M3 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.

The minimum velocity is the lowest possible velocity under any circumstances that will allow the the rifle to cycle.

What is required for an AR to cycle?

The bolt carrier must move sufficiently rearward so as to place the face of the bolt behind the bolt catch. [1] So, the energy in the bolt carrier/bolt combination alone must be sufficient to get the bolt face where it needs to be. The buffer cannot transfer positive energy to the bolt carrier/bolt combination, it cannot make the bolt carrier/bolt combination go faster, therefore, adding a heavier buffer will not allow the system to work at speed slower than minimum. If, as you speculate, the buffer looses contact with the carrier and the two eventually come back into contact resulting in a reshuffling of momentum and velocity, the carrier/bolt combination velocity will be below its original, and this new velocity must still be above the minimum required to cycle the action.

In short adding a heavy buffer cannot allow the system to work with a lower initial velocity, but may, it the case to separation, require an even higher initial velocity (something above minimum) to properly function. If the two remain in contact there is no momentum transfer and the minimum is the minimum. In either case, if the initial velocity is below the minimum, nothing can be done to make it work.

And, the minimum seems to be it the 5.4 to 5.5 m/s range.

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1. Plus a little bit to give the catch time to rise, plus a little bit to account for friction etc, but basically, if the face of the bolt doesn't make it that far back the rifle cannot function properly, ie, lock back on an empty magazine.

[lysander]

Ok, we are moving to a different level of approximation.

In completely inelastic collision, there will definitely be some energy loss, that's what inelastic means. This assumption fixes an energy loss for a given collision. I concur that this is a good assumption.



You mean E2 = (1/2 x (M3 + M2)) x (v2)^2 , since you don't want M1 in your equation, and you need the parentheses.

E2 is not the energy available to operate the system, though, since the buffer energy is going into the spring also (we are using "energy" as a shorthand for "kinetic energy"). You need to add on 1/2 x M1 x v1^2 to get the total energy, which of course does depend on the buffer mass.

Since v2 = v1 * M2 / ( M2 + M3 ), the energy loss is E2/E1 = M2/(M2 + M3) (The derivation is a bit lengthy for this venue). E2/E1 should be very roughly .75. This factor is independent of buffer mass, and the buffer does not lose energy in this collision. If the buffer mass is, say, 70% of the carrier mass, it will carry about 40% of the initial energy. The energy loss to the total system buffer/carrier/bolt will then be a factor of about .85.



Where did it go? The buffer is moving backwards and compressing the spring, if it's in contact with the carrier or not. Its energy is still going into the spring. The buffer will eventually stop, and it is stopped by the spring (ignoring a possible low velocity collision with the extension). Since the out-of-contact buffer will decelerate rapidly, it will recontact the carrier, and there will be more slightly inelastic collisions that will be difficult to model. If this happens over a significant period of time, we cannot assume that total carrier/bolt/buffer momentum is conserved, because the spring is exerting a force. But we could assume that there is one recontact collision and that it is completely inelastic. The relative velocity will be small though, so I don't think the energy loss will be significant.



No, as we have said, the energy of the buffer/carrier/bolt is what operates the system (to our current level of approximation). And no, energy is proportional to momentum squared at fixed mass, E = p^2/(2m). This came up in an earlier post. And we are changing the mass by changing the buffer.


A good time series of carrier position with different buffers from a high speed camera would be much more conclusive than this freshman-level analysis. This should be the next step, especially since we can't seem to agree on our application of elementary classsical mechanics.

Please explain how positive energy in the buffer can be transferred to the bolt carrier/bolt combination to get them to increase velocity.

The energy in the buffer is absorbed by the spring, and is lost to the system, in as far as establishing the minimum required velocity to function: getting the bolt as far back as it needs to be.

The collision of the cam pin in the cam track would not be completely inelastic, however, at the same time the cam pin hits the end of the cam track the extractor starts pulling on the spent case in the chamber which will keep the cam pin, and bolt from bouncing against the carrier. The assumption that the collision is inelastic is not that bad and the model still matches published experimental data.

The theoretical separation of the buffer and carrier actually has never been mentioned in any modeling exercise I have read and I don't think it really happens, and since all the models leave it out yet still match experimental data, even if it does occur its effects are negligible.

[StainlessSlide]

As an illustration, consider the following thought experiment.

The carrier picks up the bolt, loses a bit of energy, and the buffer separates in the aft direction. The carrier .and the bolt then disappear, while the buffer is still moving aft with energy 1/2 x M1 x v1^2. Does the spring stop its compression? No, it will eventually stop the buffer, which means the buffer energy has gone into compressing the spring, into spring potential. The buffer energy is not lost to the system in this case, nor is it lost if the carrier and bolt do not disappear.

The buffer does not need to give energy to the carrier, it give it to the spring, operating the action.

[lysander]

As an illustration, consider the following thought experiment.

The carrier picks up the bolt, loses a bit of energy, and the buffer separates in the aft direction. The carrier .and the bolt then disappear, while the buffer is still moving aft with energy 1/2 x M1 x v1^2. Does the spring stop its compression? No, it will eventually stop the buffer, which means the buffer energy has gone into compressing the spring, into spring potential. The buffer energy is not lost to the system in this case, nor is it lost if the carrier and bolt do not disappear.

The buffer does not need to give energy to the carrier, it give it to the spring, operating the action.

Just how does that help get the bolt carrier bolt combination get where it needs to be?

What you just wrote is the perfect example of why the buffer weight cannot effect the minimum velocity required to get the bolt and carrier to their required position for proper cycling....

[StainlessSlide]

The energy in the buffer is absorbed by the spring, and is lost to the system, in as far as establishing the minimum required velocity to function: getting the bolt as far back as it needs to be.

Compressing the spring is operating the action. The carrier will follow in the wake of the buffer as long as the buffer is out of contact, the carrier then being unimpeded by the spring. Thus the carrier/bolt will be still moving aft during this phase, with the energy it had after separation (less any dissipation).
When there is recontact, or if we assume that the buffer never left contact (as you suggest), then the buffer mass is integrated into the total mass, and it is significant as in my previous analysis.

Unfortunately time does permit my further conversation on freshman classical mechanics. Thanks again, it has been food for thought.

[Clint]

I would tend to agree with this perspective.

The buffer mass should contribute its energy to the spring.

The buffer mass contributes nothing to bolt pickup or case extraction, as that is up to the carrier mass and residual chamber pressure alone.

As an illustration, consider the following thought experiment.

The carrier picks up the bolt, loses a bit of energy, and the buffer separates in the aft direction. The carrier .and the bolt then disappear, while the buffer is still moving aft with energy 1/2 x M1 x v1^2. Does the spring stop its compression? No, it will eventually stop the buffer, which means the buffer energy has gone into compressing the spring, into spring potential. The buffer energy is not lost to the system in this case, nor is it lost if the carrier and bolt do not disappear.

The buffer does not need to give energy to the carrier, it give it to the spring, operating the action.

[Clint]

A couple further thoughts and questions.

What is the action spring efficiency and how much energy is lost to friction during the full stroke?

Does the mass of the action spring itself contribute in any meaningful way to the calculations?

Does the gas piston deliver constant impulse/momentum, constant energy or something in between?

[StainlessSlide]

A couple further thoughts and questions.

What is the action spring efficiency and how much energy is lost to friction during the full stroke?

Does the mass of the action spring itself contribute in any meaningful way to the calculations?

Does the gas piston deliver constant impulse/momentum, constant energy or something in between?

Cogent questions, all. Data is what we need to answer them.

The papers lysander linked have data for the very first part of the cycle, but only for milspec buffers of carbine and rifle.

With few more displacement time series for the whole cycle, with other buffers, we could look for answers.

[JediGuy]

Question probably best answered by moderators...

Could the extremely interesting technical information that deviated from the title of the original post be separated into a new thread?