Light mag springs/minimum reset speed, and the affect the bolt weight has...

[QUINCYMASSGUY]

OK folks, an all-AGD topic of discussion this time. This is all in relation to the mag and the bolt/spring relationship. For this discussion, suspend the concept of the bolt needing to have air go around it and out the front of the barrel in order to fire, this relates specifically to the minimum spring/hammer weight/air the mag would need in order to cycle at, let's say, 25cps.

The possible variables:
X=weight of bolt
Y=spring resistance
Z=minimum force required to cycle forward
And friction on the Lvl 10 oring is involved in it but lets avoid that one for now.

Now, the first thing to note is that Z could be increased to make it faster but mainly the return motion is what I am keeping in mind. If too light a spring is used, the bolt would take long enough to reset that the user could potentially pull the trigger again without the bolt having reset fully. Basically a shortstroke that is not really end user error, but system error. This would be a problem and the spring needs to be strong enough to allow the full cycle rate to be approx. 25-30cps so no user can outshoot it in any situation, even lower than normal performance (lack of oil for example). The other thing to keep in mind is that once the air has pushed the bolt forward and it hits the front, no extra momentum besides the spring is being used to push it back. So I don't see a heavier bolt being useful, in fact a lighter one would most likely accelerate more. So would a lighter spring be counteracted by a much lighter bolt? Feasible? maybe, maybe not (superbolt 1 sucked, but looking at it in a different way could help), but this is mostly theoretical. What if the Lvl 10 bolt was half the weight it is now? And a delrin coated piston (not whole bolt, just the piston) to prevent friction on the Lvl10 oring, could a lighter spring be utilized in this fashion?

The other thing I'm keeping in mind that is an advantage of Mags is that the air has to flow from the absolute back to the absolute front of the bolt, unlike other bolts where the air comes from under and enters the bolt halfway, the force of the air pushing against the back of the bolt can counteract the force the air in the barrel pushes back on the bolt with, allowing a lighter bolt to be used without it popping out the minute the barrel pressurizes.

I guess this is all about looking at ways to need less air to actually cycle the mag, taking the air needed to fire out of the equation could be a useful way for looking at it. Thoughts? Creative and Positive Criticisms (this is very welcome, key words are creative and polite)?

[QUINCYMASSGUY]

quick followup

Quick followup question: Would any of these things have an effect on the weight of the pull?

1.) A lighter bolt being held back by the sear

2.) A lighter spring so less pressure needed in the chamber to fire

3.) A thinner passage between the reg and the chamber for air to flow through? (Ignore shootdown, though I know it could be an issue)

Just brainstorming....

[bjjb99]

I'll take a shot at your first post. :)

If you ignore friction and the like, you've got three parameters to work with. I'll take a lead from your original post as far as the actual parameters, but I'm going to change the symbols used to something more in line with what folks would see in Physics.

m = mass of the bolt (your "weight")
F = force pushing the bolt forward due to gas pressure behind the bolt
k = spring constant (your "spring resistance")

I am going to assume the following firing cycle:

1. Sear is released, allowing force F to push the bolt forward, compressing the spring.
2. As the bolt moves forward, the spring force pushing back against the bolt increases.
3. At the point of maximum forward movement, the spring force and the force behind the bolt (force F) are equal for an instant.
4. At the point of maximum forward movement, the gas is released from behind the bolt, reducing F to zero.
5. The spring force pushes the bolt back to its rest position.
6. As the bolt moves rearward, the spring force decreases.
7. At the rearmost position of the bolt, the spring force is zero.
8. Sear is engaged, locking the bolt.

I've uploaded a picture to show the net force on the bolt as a function of the bolt's position during it's travel... hope the link works:

http://www.geocities.com/jeremiahswinson/images/bolt_position_force.gif

I will further assume that F is constant during the entire forward motion of the bolt. In reality the pressure behind the bolt drops slightly as the bolt moves forward, but we're going to ignore that. :)

When you run through the equations governing the motion of the bolt, what you end up with is basically two simple harmonic motion quarter-cycles stuck together. Here's a link to a position -vs- time plot showing the bolt's motion:

http://www.geocities.com/jeremiahswinson/images/bolt_position_time.gif

The shape of the curve is sinusoidal for each quarter-cycle. The time required for the bolt to move from one extreme of position to another is equal to the following equation:

t = pi/2 * sqrt(m / k)

Interestingly, since we've assumed that the force due to gas pressure is precisely equal to the maximum force delivered by the spring, there's no F term in this equation... it just vanishes into the ether. :)

So, if we want a cyclic rate of S cycles per second, we need to adhere to the following relationship:

m = k / (pi * S)^2

Any mass lower than the value determined above will yield a cyclic rate greater than S. Here's my third and final diagram showing this relationship for S=25 and S=30 cycles per second.

http://www.geocities.com/jeremiahswinson/images/cyclic_rate.gif

This graph is linear, so if the mass of your bolt is off the edge of the plot, you can easily extrapolate what spring constant you will need or vice-versa.

Hope this helps a bit...

BJJB

EDIT: I have put the three plots into an html file. Here's the link:

http://www.geocities.com/jeremiahswinson/index.htm

Hopefully this link will be viewable...

[bjjb99]

As for your second post, I think that (2) and (3) both contribute to the trigger pull weight. In an Automag, the effect of (3) is most likely far greater than that of (2).

BJJB

[QUINCYMASSGUY]

bjjb99, great to hear from you and thank you for actually converting my rough ideas into more formal physics principles. And here's the accelerations/decelerations if everything remained constant (Lvl 10 throws a wrench in the works):

As the bolt moves forward:
1.) Bolt speed increases but acceleration decreases, so more force but the amount it's gaining by drops until the end when it hits the front and acceleration goes to nothing.
2.) Spring resistance increases and most likely accelerates, till the point it goes equal with the air force and halts the momentum.
3.) Air pressure is metered, so as it expands down the power tube, the psi decreases, meaning the force actually decreases. If air continued to flow, the pressure would probably be the same.

So basically, when the increasing force created by the decreasing air pressure slows enough that the accelerating spring force overrides it, or when it hits the front and just gets a serious shot of inertia, the spring takes over as the dominant force. A different way to phrase the bolt motion. A cross-graph illustrating the two forces and where they meet would be cool. Your links are not working, so if you already graphed that, excellent. I'm very interested in seeing what you did. So if the force used to push it and the force used by the spring are both decreased in a proper ratio, less force would be needed to push it forward but ROF would be sacrificed, however, if it already allowed a 40bps ROF then 10bps could easily be sacrificed for other benefits, see where I'm going with this? If anything a thicker or thinner (probably thinner) piston might be used to decrease the friction of the oring in case that might prevent proper bolt return as long as the decrease is consistent with the rest of the system, if the airhole could allow 40bps before it could be decreased to the point it'll still allow 30bps since that's what the bolt/spring allows. It's the theory of bottlenecks, used in manufacturing and in a sense is exactly what we're doing here, manufacturing an act of a paintball being shot. If anywhere in the system is a hindrance, the whole system suffers.

The rest of your first post I will decipher tomorrow, I'm an imaginative guy but weak on the physics. I am definitely going to toss the 2 cents around tomorrow but I'm ready for bed. Great convo man, keep it up and I'd love to see those graphs!

[bjjb99]

I have put the three plots into an html file. Here's the link:

http://www.geocities.com/jeremiahswinson/index.htm

Hopefully this link will be viewable...

BJJB

[QUINCYMASSGUY]

bjjb99, great resources, thank you!

Here is my next question: at what point in the Mag's movement is the bolt's velocity the greatest? With the spring eventually causing it to stop (or at least slowing down before impact) at what point in a normal as well as Lvl 10 bolt does it achieve maximum velocity? My guess would be around the point where it's 80-90% forward but I may be wrong. Thoughts?

[bjjb99]

Under the conditions and assumptions I described earlier in this thread, the bolt will be moving fastest just before the gas is released. This hypothetical bolt is moving fairly close to its maximum velocity after about 40-50% of the its forward motion.

Now, in a real Mag valve, there's a lot more going on than the hypothetical case we've used so far in this thread. The most significant difference is that the gas behind the bolt is not released instantaneously. Instead, the bolt "floats" at the forward end of its travel as the gas pressure behind it is released. The pattern and duration of this "floating" depends on the shape of the valve tube tip and the bolt stem, the pressure behind the bolt, and the spring constant. Level 10 probably complicates matters even further.

BJJB