24" Disco Barrel. Fired without pellet. 2' / 0.0018s = 1,111.1fps
Looks like some awesome tech. But you lack any detail (or maybe I'm blind). What is the gun? If PCP, how much regulator pressure? Caliber? Maybe - 24" Disco barrel - says enough and I'm simply too ignorant to know...
OK now tell us how this info can be used in the real world
Looks like some awesome tech. But you lack any detail (or maybe I'm blind). What is the gun? If PCP, how much regulator pressure? Caliber? Maybe - 24" Disco barrel - says enough and I'm simply too ignorant to know...
Sorry. .22 Benjamin Discovery, stock except for two-stage (Super)sear, hammer debouncer and hammer spring preload adjuster, unreg'd, reservoir pressure ~1500psi.
The data gathering gadgets in play were described on this forum in recent threads...
https://airgunwarriors.com/community/airgun-talk/so-were-the-cardews-wrong/paged/2/#post-50333
Etc.
OK now tell us how this info can be used in the real world
Well, for starters, because it seems pretty unlikely that an airflow could ever push a pellet faster than it's moving itself, (e.g., Mach-1 as here), anyone interested in achieving supersonic MVs probably needs to begin with a very different flow path design than what's found in the Disco, which itself is fairly typical for the generic knock-open valve, PCP.
What said differences might be, and the theory behind them, might be a topic for a very interesting discussion.
Just curious, and WAY out of my league, but I assume the measurement you took was an AVERAGE of the velocity of unobstructed air over time in the barrel. Could it be possible that the initial release of air had a higher velocity/energy, and could therefore impart more energy to the obstructing pellet than the average velocity of unobstructed air in the bore? If I'm FOS, please disregard.
Just curious, and WAY out of my league, but I assume the measurement you took was an AVERAGE of the velocity of unobstructed air over time in the barrel. Could it be possible that the initial release of air had a higher velocity/energy, and could therefore impart more energy to the obstructing pellet than the average velocity of unobstructed air in the bore? If I'm FOS, please disregard.
Hi, Jim. Very reasonable question.
You're correct. The time interval measured is the delay from impact of hammer with valve stem until the arrival of the air blast at an acoustic sensor an inch or so away from the muzzle. Therefore the velocity calculated is the average flow velocity down 24" of bore.
However, any conjecture that the pellet could be made to accelerate ahead of and actually outrun the column of air pushing it (then leaving a hard vacuum in its wake?), seems kinda' implausible. After the pellet was moving as fast as the leading edge of the air column, what agency could push it any faster?
Again, I'm way out of my league on this, but intuitively (at least to me) an unobstructed column of air expanding against only atmospheric pressure seems like it would expend all of its energy much faster than a "capped" column of air, even one with a sliding cap. If initial velocity of expelled air were somehow higher than your documented average of 1,111, and the pellet slowed expansion due to imparted energy, there would be no wake vacuum. Perpetual motion machine? 🙂
I know that spring guns must never be fired without the necessary resistance of a pellet in the bore, lest damage occur. I'm just wondering what the pellet resistance/capped air column has to do with the functioning of a pcp/msp, and frankly, what the relevance is of measuring an unobstructed column of air. Once again, if I'm FOS, please so indicate, and I will respectfully fold my tent.
Again, I'm way out of my league on this, but intuitively (at least to me) an unobstructed column of air expanding against only atmospheric pressure seems like it would expend all of its energy much faster than a "capped" column of air, even one with a sliding cap. If initial velocity of expelled air were somehow higher than your documented average of 1,111, and the pellet slowed expansion due to imparted energy, there would be no wake vacuum. Perpetual motion machine? 🙂
I know that spring guns must never be fired without the necessary resistance of a pellet in the bore, lest damage occur. I'm just wondering what the pellet resistance/capped air column has to do with the functioning of a pcp/msp, and frankly, what the relevance is of measuring an unobstructed column of air. Once again, if I'm FOS, please so indicate, and I will respectfully fold my tent.
I agree. Intuition suggests an unobstructed column would expand faster than one burdened by the inertia of a pellet, which would accept a share of available energy, thus depriving the air of that share.
But doesn't that logic make the unobstructed case suggest to your intuition a higher average velocity and shorter breech-muzzle flight time, rather than slower and longer? It does to mine.
As for the relevance of the no-pellet case, it's pretty well documented that shooting lighter pellets reliably results in higher MVs than shooting heavier ones. No-pellet is the embodiment of the limiting case of the zero-weight pellet, and therefore should exhibit the fastest acceleration and highest possible MV.
Shouldn't it?
OK now tell us how this info can be used in the real world
Well, for starters, because it seems pretty unlikely that an airflow could ever push a pellet faster than it's moving itself, (e.g., Mach-1 as here), anyone interested in achieving supersonic MVs probably needs to begin with a very different flow path design than what's found in the Disco, which itself is fairly typical for the generic knock-open valve, PCP.
What said differences might be, and the theory behind them, might be a topic for a very interesting discussion.
As for what a possible air flow path suitable for supersonic MVs might look like, here's a (totally speculative and untested) possibility.
If initial velocity of expelled air were somehow higher than your documented average of 1,111...
Actually, Jim, I just grasped the true significance your excellent point. 1,111.1fps is indeed only the average velocity of the air column in my experiment -- not its final exit velocity -- which may indeed have been different -- maybe higher.
Maybe much higher.
For example, if the rate of acceleration from breech to muzzle was constant, it would be exactly twice as high -- 2,222fps = Mach 2!
However, I doubt that was the case. I think a supersonic report would have been much louder than the one I heard.
However, I doubt that was the case. I think a supersonic report would have been much louder than the one I heard.
Fascinating experiment, Steve.
Would the mass of the air molecules be enough to create an audible sonic crack?
However, I doubt that was the case. I think a supersonic report would have been much louder than the one I heard.
Fascinating experiment, Steve.
Would the mass of the air molecules be enough to create an audible sonic crack?
I don't recall offhand the air utilization efficiency of this gun, but I'd guess the volume of the air expended in the shot to be ~150bar-cc = a mass of ~230mg = ~3.6gr. That's much heavier than the flea-weight pellets (plastic or aluminum) often used in max-MV experiments, and those scholars always report loud sonic cracks.
If initial velocity of expelled air were somehow higher than your documented average of 1,111...
Actually, Jim, I just grasped the true significance your excellent point. 1,111.1fps is indeed only the average velocity of the air column in my experiment -- not its final exit velocity -- which may indeed have been different -- maybe higher.
Maybe much higher.
For example, if the rate of acceleration from breech to muzzle was constant, it would be exactly twice as high -- 2,222fps = Mach 2!
However, I doubt that was the case. I think a supersonic report would have been much louder than the one I heard.
Actually Steve, I was thinking about a reverse case, wherein the initial blast of air would be much higher than your 1,111, and slow down, rather than continue to accelerate, due to the resistance of the pellet. In that case, (in my uninformed mind,) the column of air would continue to expand and continue oi impart energy to the pellet.
I apologize for the relevance comment. Your empty barrel test seems to me to have asked more questions than it has answered. At this point, I will VERY respectfully fold my tent and watch the rest of the show.
Actually Steve, I was thinking about a reverse case, wherein the initial blast of air would be much higher than your 1,111, and slow down, rather than continue to accelerate, due to the resistance of the pellet. In that case, (in my uninformed mind,) the column of air would continue to expand and continue oi impart energy to the pellet.
I apologize for the relevance comment. Your empty barrel test seems to me to have asked more questions than it has answered. At this point, I will VERY respectfully fold my tent and watch the rest of the show.
Well, if you so choose, Jim. But I am still curious about why an "initial blast" would at first move faster than sound, but then slow juuuust enough below Mach-1, that the average valve-to-muzzle speed would mysteriously work out exactly (or at least "exactly" within the limits of accuracy of my measurement -- ~+/-1%) to Mach-1.
By pure coincidence?
OK now tell us how this info can be used in the real world
Here's another thought about potential practical real-world usefulness of this seemingly academic result, in the form of a question.
We see here that the effect of opening the valve propagates through the bore at Mach-1. Therefore isn't it reasonable to conclude that propagation of the effect of closing the valve will also be limited to travelling at the speed of sound?
And for extra credit, what might that say about valve timing for optimum air utilization efficiency?
Still thinking? Here's a hint from a decade and a half ago.
the bore is not empty, it contains a static column of air. once the valve closes the pluse of air most also fill or replace said static column. just a couple of strange thoughts that popped up in my demented mind.
the bore is not empty, it contains a static column of air. once the valve closes the pluse of air most also fill or replace said static column. just a couple of strange thoughts that popped up in my demented mind.
Good point. But since the mass of the charge is about 10 times greater than the mass of the air column initially filling the bore (150bar-cc vs 15 = 190mg vs 19mg), the effects of said column on the velocity of the charge shouldn't really be very significant.
And in any case, said air will be there whether or not a pellet is fired, and since a typical .22 pellet will weigh several times more than the air charge, the mass of the initial ambient air fill of the bore will be even more trivial by comparison whenever a real shot is taken.
@steve-in-nc
Alright, I tried to calculate the time based on an adiabatic expansion of air. That is an awful expression! Let us simplify and assume an approximately constant acceleration of the pellet. If the final speed of the pellet is v and the length of the barrel is L, then with approximately constant acceleration the time taken is 2L/v. I am quite sure you're right that the effect of the valve closing will propagate at most at speed of sound through air. Then the effect of valve closing takes L/ v_s (v_s= speed of sound) to get to the end of the barrel. The optimum time for valve closure should be 2L/v-L/v_s.
Caveat: 1. The acceleration is not constant. But if I did my calculation correctly, it is largely decided by the chamber volume if that's the right term assuming the barrel volume is going to be significantly larger than the chamber volume.
2. The speed of sound depends on the square root of temperature. The adiabatic assumption would indicate that temperature should fall quite a bit as volume expands which should slow the speed of sound. That would significantly reduce the optimal valve opening time.
@steve-in-nc
Alright, I tried to calculate the time based on an adiabatic expansion of air. That is an awful expression! Let us simplify and assume an approximately constant acceleration of the pellet. If the final speed of the pellet is v and the length of the barrel is L, then with approximately constant acceleration the time taken is 2L/v. I am quite sure you're right that the effect of the valve closing will propagate at most at speed of sound through air. Then the effect of valve closing takes L/ v_s (v_s= speed of sound) to get to the end of the barrel. The optimum time for valve closure should be 2L/v-L/v_s.
Caveat: 1. The acceleration is not constant. But if I did my calculation correctly, it is largely decided by the chamber volume if that's the right term assuming the barrel volume is going to be significantly larger than the chamber volume.
2. The speed of sound depends on the square root of temperature. The adiabatic assumption would indicate that temperature should fall quite a bit as volume expands which should slow the speed of sound. That would significantly reduce the optimal valve opening time.
Terrific! Continuing the argument to its exciting climax by factoring out first L and then 1/v-s and recalling that MV (Muzzle Velocity) is the popular term for your v-L the optimum time for closure becomes =
L(2/MV - 1/v-s) =
L/v-s(2v-s/MV - 1)
I wonder if it should be interesting (or just comical) that this expression goes to zero (i.e. 1 - 1) as MV approaches Mach-2? Is this saying that Mach-2 MVs are possible with zero valve duration = no air expenditure at all?
Or that Mach-2 MVs are not possible at all?
Or that if we can just design a valve that closes before it opens then there'd be no limit to MV?
At all?
@steve-in-nc
I think for any speed greater than or equal to Mach 2 the equation says the pellet will leave the barrel before the effect of valve closing catches up no matter how short the interval of valve closing.
However, how is it possible to have supersonic speed when the air flows into the barrel through an orifice. Shouldn't the air flow have a choke speed of 1 mach? Since the pellet can not move faster than the air column pushing it, and I don't see how the air can accelerate inside the barrel, is it even possible to have supersonic pellet speeds? Shouldn't the pellet speed be capped at whatever happens to be Mach 1 for the temperature of the air inside the chamber? And since the valve closing effect also propagates at whatever happens to be Mach 1 for the air, the minimum valve open time should be L/ Mach 1 for MV= Mach 1
I should hasten to add I am in no way an expert on either fluid dynamics or internal dynamics of an airgun and I could very well be missing something obvious and basic.
@steve-in-nc
I think for any speed greater than or equal to Mach 2 the equation says the pellet will leave the barrel before the effect of valve closing catches up no matter how short the interval of valve closing.
However, how is it possible to have supersonic speed when the air flows into the barrel through an orifice. Shouldn't the air flow have a choke speed of 1 mach? Since the pellet can not move faster than the air column pushing it, and I don't see how the air can accelerate inside the barrel, is it even possible to have supersonic pellet speeds? Shouldn't the pellet speed be capped at whatever happens to be Mach 1 for the temperature of the air inside the chamber? And since the valve closing effect also propagates at whatever happens to be Mach 1 for the air, the minimum valve open time should be L/ Mach 1 for MV= Mach 1
I should hasten to add I am in no way an expert on either fluid dynamics or internal dynamics of an airgun and I could very well be missing something obvious and basic.
Actually what can (theoretically) happen if MV > Mach-1 is, as the pellet accelerates but isn't yet up to v-s, the pressure wave from valve closure can catch up and pass it, but later change places as the pellet continues to accelerate through sonic speed and ultimately overtakes the wave. So we'll need to solve a quadratic to calculate how long the valve must remain open to prevent the wave from ever catching the accelerating pellet before it leaves the muzzle.
I've made a start at that (it's kind'a messy) and hope to have the solution later today.
Just in case it means anything.
Oh! What a good point and an excellent little problem. So here's my stab at the solution:
x_pellet = 1/2 at^2. I am going to assume a constant acceleration with the caveat listed in my first post. That gives a= MV^2/ 2L where MV= final muzzle velocity and L= length of barrel.
x_air (location of wave front generated by valve closing)= v_s( t- t_0) where v_s= speed of sound and t_0 = time the valve remained open.
When they meet up x_pellet=x_air i.e the quadratic you were referring to is,
1/2 at^2-v_s t+v_s t_0=0
The roots are,
t= [v_s +/- Square root of ( v_s^2- 2a v_s t_0)]/ a
If we don't want the pressure wave to ever catch up with the accelerating pellet we need,
2a v_s t_0 greater than/ equal to v_s^2 which gives,
t_0 > (or equal to) v_s/ 2a= v_s L/MV^2
For MV=2v_s this gives t_0 should be equal to/ greater than L/ 4 v_s.
(I wish I could tex my solution for better readability. Don't know how to in this interface)
Hope the algebra was correct! I am notoriously bad at keeping track of 1/2 s...
Oh! What a good point and an excellent little problem. So here's my stab at the solution:
x_pellet = 1/2 at^2. I am going to assume a constant acceleration with the caveat listed in my first post. That gives a= MV^2/ 2L where MV= final muzzle velocity and L= length of barrel.
x_air (location of wave front generated by valve closing)= v_s( t- t_0) where v_s= speed of sound and t_0 = time the valve remained open.
When they meet up x_pellet=x_air i.e the quadratic you were referring to is,
1/2 at^2-v_s t+v_s t_0=0
The roots are,
t= [v_s +/- Square root of ( v_s^2- 2a v_s t_0)]/ a
If we don't want the pressure wave to ever catch up with the accelerating pellet we need,
2a v_s t_0 greater than/ equal to v_s^2 which gives,
t_0 > (or equal to) v_s/ 2a= v_s L/MV^2
For MV=2v_s this gives t_0 should be equal to/ greater than L/ 4 v_s.
(I wish I could tex my solution for better readability. Don't know how to in this interface)
Hope the algebra was correct! I am notoriously bad at keeping track of 1/2 s...
I wish you (and I) could improve readability too!
However, I seem to have taken a slightly different route (based on exactly the same assumptions) to exactly the same result.
1. Pellet's muzzle arrival time: T = 2L/MV
2. Pellet's velocity at time t where 0 < t < T: v = MV*t/T
3. Time when pellet goes sonic: t_s = T*v_s/MV
4. Pellet's distance from breech at t_s: l_s = L(t_s/T)^2 = L(T*v_s/MV/T)^2 = L(v_s/MV)^2
5: Minimum valve dwell time greater than or equal to: t_0 = l_s/v_s =
L*v_s/MV^2
If it looks a bit damp, it's because I worked on it while swimming today! At least by taking this path, I avoided those damned 1/2s of which you speak. I hate them too -- especially when I have nothing to write on!
@steve-in-nc May be the swimming unlocked the "cuter" solution, as the expression goes. I like your solution better than mine-- I rather brute forced it. In any case our solutions agree which means I did not muck up any 1/2s along the way!
Now the interesting question is whether these equations give numbers that are more or less in line with measurements. In the old Yellow Forum you wrote quite a few chains of posts on optimal efficiency of pcps. Did you calculate/ measure the optimum valve dwell time when you were playing with tuning the pcps?
@steve-in-nc May be the swimming unlocked the "cuter" solution, as the expression goes. I like your solution better than mine-- I rather brute forced it. In any case our solutions agree which means I did not muck up any 1/2s along the way!
Now the interesting question is whether these equations give numbers that are more or less in line with measurements. In the old Yellow Forum you wrote quite a few chains of posts on optimal efficiency of pcps. Did you calculate/ measure the optimum valve dwell time when you were playing with tuning the pcps?
Actually, Funky (please pardon the familiarity if unwelcome), back then I would have killed (well, gotten pretty pushy) to have the kind of ability to directly and accurately measure interior ballistic minutia like valve dwell times (e.g., the 2.29ms in the Disco dry-fire example above) that simply falls out of Jim-in-UK's VM invention.
A candle in the darkness.
Cheap, too!
Another air-only experiment: Air-conserving 1377, 10" barrel, 10 charge pumps, 4 recharge, 5fpe with CPLs (if you give it one). 761us = 1095fps.
@steve-in-nc That's pretty close to the theoretical value of optimal valve open time. Assuming the MV is 900 f.p.s the theoretical optimal valve dwell time for a 24" barrel by our expression is v_s L/ v^2= 2.77 ms which in the same ball park as the 2.29 ms for the dry fire case that you measured. For 1000 fps MV, the minimum optimal dwell time per our expression is 2.25 ms which is very close to your measured value. That's pretty encouraging for a quick calculation.
I don't understand pcp mechanism well enough to understand how the pellet in the barrel versus dry fire change the dwell time of a self regulating mechanism such as the Discovery's.
Funky
p.s: I got into airgunning again and joined this forum-- rather the old Yellow-- when I was in grad school. Let's just say my imagination in choosing my profile name was limited by my narrow research interests.
I don't understand pcp mechanism well enough to understand how the pellet in the barrel versus dry fire change the dwell time of a self regulating mechanism such as the Discovery's.
Funky
p.s: I got into airgunning again and joined this forum-- rather the old Yellow-- when I was in grad school. Let's just say my imagination in choosing my profile name was limited by my narrow research interests.
Sans pellet, the expected pressures downstream of the valve stem should be lower, making the pressure differential across the valve, and therefore the closing force acting to return the stem to its seat, higher. So dwell during a dry fire should be shorter, helping to make these numbers plausible and a nice result -- so long as we don't take them too seriously for MV >> v_s!
Thanks for the help!
Graph of To (Valve dwell time) vs MV
Drawn for a 24" barrel and M1 = 1120fps. Yellow of course denotes the highly speculative wonderland of MV > M1, the how of which is touched upon - however whimsically - below.
Looking at the enclosed formula...
Note that if T = 300K and Pe/P = 1/3 (e.g., 100bar breech pressure vs 300bar reservoir pressure), then Ve = 1321fps.
If Pe/P = 1/300 (300bar dry fire), Ve = 2546fps.
