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All Springer/NP/PCP Air Gun Discussion General => Machine Shop Talk & AG Parts Machining => Engineering- Research & Development => Topic started by: NCmountainShooter on August 10, 2014, 11:33:04 AM
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I just had a brainstorm idea but no time to think through all of the configurations. I've always liked the simplicity of a dump valve, but thanks to rsterne I understand all that is wrong with them with regard to wasted air. My question is whether we can use a regulated air chamber with a variable volume, ie a piston, to improve the inefficiencies of a dump valve. Fill a chamber to X psi that is reasonable and efficient in PCP mechanics then cut it off and compress it to 2X, 3X, or more. Release that with a dump valve. My understanding of pressure dynamics is that the smaller the space the easier it is to handle high pressures. While having a tank at 6000psi is costly in size and weight, having a small diameter chamber engineered to handle that should be much easier. I would propose the piston be either mechanical or hydraulic, linked to a side lever of appropriate size to make compression easy. Link the bolt to the same lever for economy of motion if possible or not if that's too complicated. I almost see this like the hybrid PCP FX guns only it's a hybrid PCP with a piston gun and dump valve minus the violent movement in a piston gun. I would have the piston release it's pressure with a gentle bleed/release after firing.
Back to the original question, would this improve the air inefficiency of a dump valve? It seems that it should be able to use the same sip of air as a regulated PCP at much higher force. This goes against the rule that it takes 1cc of air per FPE, but I'm not sure if that only applies to more traditional setups.
Thanks in advance for any input.
Chris
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I think what you are proposing may be possible.... My rough "rule of thumb" of 1cc per FPE will only work well over a limited range of pressures, at very high pressures, you can get away with a smaller volume, and at very low pressures you will need more.... It is based on a starting point of 1 FPE/CI where that cubic inch of air is, of course, at 1 bar (ie after expansion).... For typical regulated guns, the 1 cc per FPE is a good value, but perhaps some examples would help.... Let us use 50 FPE for a desired result, which at 1 FPE/CI means that we need to release 50 CI of air (at 1 bar), which is 820 cc of air that has to be released by the valve.... Here is what is happens if the reservoir is 50cc at various pressures....
1000 psi = 69 bar x 50 cc = 3450 cc.... 3450 - 820 = 2630 cc.... 1000 x 2630 / 3450 = 762 psi at the end of the shot.... average pressure = 881 psi (88%)
2000 psi = 138 bar x 50 cc = 6900 cc.... 6900 - 820 = 6080 cc.... 2000 x 6080 / 6900 = 1762 psi at end.... average = 1881 psi (94%)
3000 psi = 207 bar x 50 cc = 10350 cc.... 10350 - 820 = 9530 cc.... 3000 x 9530 / 10350 = 2762 psi at end.... average = 2881 psi (96%)
So you can see that using the same CI of air (at 1 bar) from a higher pressure plenum/valve causes less percentage drop in the pressure.... Just to compare what would happen in an unregulated PCP at 2000 psi, with a 250 cc reservoir, here is what happens....
2000 psi = 138 bar x 250 cc = 34500 cc.... 34500 - 820 = 33680cc.... 2000 x 33680 / 34500 = 1952 psi at end.... average = 1976 psi (99%)
To end up with that average of 1976 psi with our 50 cc plenum, we would need to increase the pressure to about 2100 psi....
2100 psi = 145 bar x 50 cc = 7241 cc.... 7241 - 820 = 6421 cc.... 2100 x 6421 / 7241 = 1862 psi at end.... average = 1981 psi (5 psi more than the unregulated PCP)
So in a 2000 psi regulated PCP with 1 cc per FPE of plenum, we only need to bump up the setpoint pressure by less than 100 psi to get back to the power we had in the unregulated version....
Now to your question.... what about taking that 50 cc at 2000 psi and compressing it to 6000?.... Well just based on the above calculations, here is what we get....
6000 psi = 414 bar x (50/3)cc = 6900 cc.... from there it's the same math.... average pressure ends up at 1881 psi (94%) compared to the unregulated gun.... So on the face of it, it seems like there is no advantage.... BUT, the gun is a lot more efficient at 6000 psi than at 2000.... so now I'll switch over to Lloyd's Internal Ballistics Spreadsheet and run some comparisons....
Unregulated, 2000 psi, 250 cc reservoir, 24" barrel, .25 cal, 25 gr, valve closes at 7".... 50 FPE @ 1.02 FPE/CI
Regulated, 2000 psi, 50 cc plenum.... 48 FPE @ 1.07 FPE/CI
Regulated, 2100 psi, 50 cc plenum.... 50 FPE @ 1.03 FPE/CI
Regulated, compressed 3:1 to 6000 psi, 16.7 cc plenum.... 95 FPE @ 0.54 FPE/CI (because of the higher velocity, the valve is closing much later even with the same dwell)
Regulated, compressed to 6000 psi, 16.7 cc plenum, reduced dwell.... 82 FPE @ 1.04 FPE/CI (at roughly the same efficiency)
Regulated, compressed to 6000 psi, 16.7 cc plenum, further reduced dwell.... 50 FPE @ 1.82 FPE/CI (much greater efficiency at original power)
OK, so before you start jumping off the walls, that is FAR from a dump valve, in fact the dwell is so short the valve is closing when the pellet has only moved less than an inch.... To get back to about where we were for power and efficiency, and with a dump valve, we need to reduce the plenum (in this case valve) volume.... I had to do some trial and error, but here is what I came up with.... It turns out you only need about 1/10th the volume, 1.67 cc of air at 6000 psi to get back to 50 FPE with a dump valve....
Regulated, compressed to 6000 psi, 1.67 cc valve with dump shot.... 50 FPE @ 1.29 FPE/CI.... So the answer is yes, what you are proposing should work.... IF you can build it....
Unfortunately, that's a big IF.... You need to start with 5 cc of air (0.305 CI) at 2000 psi and then compress it 3:1 to 6000 psi.... Let's say that you use a piston 1/2" in diameter, which is 0.196 sq.in.... you need a cylinder 1.56" long, and a piston with a 1.04" stroke to increase the pressure 3:1.... At the end of the compression stroke, the force on the piston would be 1178 lbs. so you would need quite a leverage system to achieve that.... If the lever was a foot long, you would still have a 100 lb. load.... You could trade off smaller diameter for longer stroke, or larger diameter for less stroke on the piston, but you still end up with a 100 lb. load on the end of a 1 foot lever to compress that 5 cc of 2000 psi air to 6000 psi....
So the answer to your question is, yes, by using a much higher pressure you can make a dump valve as efficient as a conventional PCP.... The question then becomes, can you build and operate it?....
Bob
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I can't recall the name of the British SSP that never was. It featured a a charge chamber that was of variable volume. As the chamber was charged it would drive a piston back against a large spring pressure increasing the volume. When the valve is fired, as the pressure dropped the spring/piston would squeeze the remaining air volume in an attempt to keep the average pressure high and hopefully increasing efficiency. At least that is my understanding of the concept. There may be merit to the variable chamber, for both PCP and pumpers. A PCP would need some type of check valve or a very small orifice to keep the chamber from refilling too early.
Tom
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I believe that was a relatively recent hybrid, I remember seeing it in concept form.... I wonder if the piston would produce any strange recoil tendencies?.... As an SSP, it would obviously be a dump valve configuration, so any benefit would have to be from a higher average pressure during the first portion of the shot cycle, as ultimately all the air you put into the gun ends up out the barrel....
A PCP version could also use a valve that is cycled by the action to refill the chamber.... but I wonder if the advantage compared to the simplicity of a larger plenum/reservoir volume in place of the (space taken by the) moving piston (and spring) would result in any real benefit?.... After all, the unregulated PCP in my above example has 99% of the pressure available over the shot while the valve is open, and a 1 cc / FPE plenum still has 94%....
Bob
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I can't recall the name of the British SSP that never was
Webley Paradigm was the Brit SSP ...
I've been working on an SSP of sorts but prefer to call it a High Pressure Springer - HPS. Belleville washers compress when you pump it and then release the charge with a dump shot. Spring washers are more efficient at storing energy than air and you get the added benefit of adiabatic heating when finally released.
Gamo had an overly complex version that kinda worked. My test platform was initially a 953 and with all the tweaks I got it from a humble 460FPS to around 670FPS. Definite promise so now comes the real deal based on a B50 platform - work, test, curse ... lather, rinse, repeat 8)
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Good name.... HPS.... and yes, it was the Webley Paradigm I'm thinking of.... I would dispute the adiabatic heating, however, because the pressure is not increasing during the shot, it's just not decreasing during the spring expansion phase.... However it won't have as much adiabatic COOLING like we get in a conventional PCP or pumper....
Bob
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Bob,
Thanks for the detailed reply. It's been a long work day and that's going to take some rereading and research to replicate in my own figures. You've given me some great info to start with. At least I have an idea that's good in theory. I'll do the math and drawing to figure how to make it happen.
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Chris.......the other day I offered up one of my old creations but, based on member response, it appears to have been
the wrong place to do it:
www.gatewaytoairguns.org/GTA/index.php?topic=71647.msg682905#msg682905 (http://www.gatewaytoairguns.org/GTA/index.php?topic=71647.msg682905#msg682905)
Have a look at the brief writeup (YF link) as it may well contain specific information you would find useful. That project
was much larger and more technical than my description would indicate but unfortunately very few people were able
to take advantage and ask questions. The gun continues to be a joy to shoot and I use it frequently.
The Webley gun mentioned above had, in my opinion, a couple of serious shortcomings but then, in all fairness, it
was designed for economical production (?) and not maximum performance.
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Chris.......the other day I offered up one of my old creations but, based on member response, it appears to have been
the wrong place to do it:
www.gatewaytoairguns.org/GTA/index.php?topic=71647.msg682905#msg682905 (http://www.gatewaytoairguns.org/GTA/index.php?topic=71647.msg682905#msg682905)
Have a look at the brief writeup (YF link) as it may well contain specific information you would find useful. That project
was much larger and more technical than my description would indicate but unfortunately very few people were able
to take advantage and ask questions. The gun continues to be a joy to shoot and I use it frequently.
The Webley gun mentioned above had, in my opinion, a couple of serious shortcomings but then, in all fairness, it
was designed for economical production (?) and not maximum performance.
Do you have a link to more details on the gun?
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Chris.......the other day I offered up one of my old creations but, based on member response, it appears to have been
the wrong place to do it:
www.gatewaytoairguns.org/GTA/index.php?topic=71647.msg682905#msg682905 (http://www.gatewaytoairguns.org/GTA/index.php?topic=71647.msg682905#msg682905)
Have a look at the brief writeup (YF link) as it may well contain specific information you would find useful. That project
was much larger and more technical than my description would indicate but unfortunately very few people were able
to take advantage and ask questions. The gun continues to be a joy to shoot and I use it frequently.
The Webley gun mentioned above had, in my opinion, a couple of serious shortcomings but then, in all fairness, it
was designed for economical production (?) and not maximum performance.
Here is the link.... http://www.network54.com/Forum/79537/thread/1257727426/SSP+rifle----------discovering+what's+possible (http://www.network54.com/Forum/79537/thread/1257727426/SSP+rifle----------discovering+what's+possible)
Just starting to read now. Very interesting so far.
Tom
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Thanks Tom. I couldn't get link to work on my phone from other page.
RGB1, wow. That's amazing. I can't tell is it just a single pump or is there a precharged component as well? I'm not sure what is meant by floating piston. The lever you have is more what I imagined over Bob's assumed 1' arm. I'm also toying with the idea of a lever to power linked hydraulic cylinders as to improve the power dynamic. Use a long stroke on a narrow chamber translated into a higher power short stroke in a wider cylinder. There have to be other ways to power this than just a direct lever on a piston. I would love to see more of your design if it's open to public scrutiny, though I understand if your tight with details. I've decided I won't get rich designing a custom airgun mechanism and don't worry about the idea getting stolen. Thanks for sharing the pics and narrative!
Chris
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One thing you might consider is that regulators are never perfect in their consistency, that is why there is some remaining shot to shot deviation in velocity.... Since you will be taking a relatively small volume and reducing it and raising the pressure, any inconsistency will be tripled.... This pressure variation MAY lead to significant shot to shot velocity swings in your "boosted" design.... It may or may not be an issue, just something to consider.... Your biggest challenge will be to engineer the mechanism to boost the pressure to the 6000 psi range, contain that safely, reliably and consistently, and the get the valve to release it consistently and completely.... Something else to consider when you decide on your working pressure is that Boyle's Law does not apply perfectly above about 3000 psi because air starts acting as a non-ideal gas.... At 4500 psi the amount of air molecules is about 9% less than expected, and as you increase the pressure the space between the molecules gets so small that the pressure continues to build faster than Boyle's Law would predict.... At 6000 psi, about 21% less air can be crammed in a given space than what you would expect.... Here is the relationship for some "Real" gases, compared to an "Ideal" gas.... using van der Waal's equations....
(http://i378.photobucket.com/albums/oo221/rsterne/Important/IdealGases_zps796cc652.jpg) (http://s378.photobucket.com/user/rsterne/media/Important/IdealGases_zps796cc652.jpg.html)
I couldn't find any data for air, so the green line is a geometric average of the data for Nitrogen and Oxygen in a 78:21 ratio, and ignoring the other 1% components of air.... As you can see, at 1500-2000 psi you actually have about 8% more air in a given space than Boyle's Law would predict, at 3000 psi it basically agrees with Ideal gas theory, at 4500 psi you have 9% less and at 6000 psi 21% less.... That is why a 60 minute SCBA tank only holds 88 CF of air at 4500 psi instead of 97 CF as predicted by Boyle's Law.... The air in a chamber at 6000 psi would be about 21% less dense than expected, so as it expands during the shot, I would expect the pressure to drop more rapidly than Lloyd's model would predict (it's based on Boyle's Law).... However, it might also mean that the "compression ratio" you need to get to 6000 psi from 2000 would be less than 3:1, as you would actually be going from 108% to 79% by density, so you may only need a compression ratio of about 2.2:1 instead of 3:1.... if I understand the process properly....
Bob
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rgb1.... I missed the Bullpup thread, but you have definitely created a very interesting, unique design.... If I understand it correctly, it uses a single stroke of the 26" long lever to pressurize the valve chamber, against which a piston exerts a counter force.... and the piston is in a precharged chamber, so in fact working like a gas ram?.... You can remove the precharge chamber (2.2 CI @ 925 psi) to change the pressure inside to alter the pressure at which the piston moves back, and hence the pressure in the active part of the valve (and the cocking force)?.... As configured in that thread, your maximum working pressure (in the valve) was about 1017 psi?.... I'm not sure what the active volume of the valve is, but when fired it dumps all the air (but not that in the precharge chamber), with the pressure staying above 925 psi until the piston completes its travel, at which point the pressure then drops as the pellet moves down the bore?.... Where that occurs depends on the ratio of valve volume to barrel volume.... If there is something I missed, a gentle smack up the side of the head is in order....
NMountainShooter.... It matters not how you achieve your force multiplication, through levers or hydraulics, you still need to input a given amount of energy to compress the 2000 psi air to 6000, ending with whatever size piston you use working against that 6000 psi.... Double the length of the lever (24"), half the force at the end (50 lbs.).... Use twice as large a piston (1") and use 4:1 hydraulics in between, and you're back to 100 lbs. on that 12" lever (or 50 lbs. on a 24").... There is no free lunch....
Bob
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For purposes of illustration and to ensure that we're all on the same page, the Olafsson patent
drawings # 6 and 7 work well (shown partway down in the old original post).
-- the chamber to the right of the piston is precharged to 925 psi
-- air from the pump is introduced at the corner thru passage 40
-- once the pump pressure reaches 925+ the piston is forced to the right........
for a 10 inch stroke the final pressure is ~ 1007 (both sides of the piston)
-- the valve is opened, air flows directly into the barrel
Bob.........." the piston is in a precharged chamber, so in fact working like a gas ram?"
Not really. It merely separates the precharge from the pump charge.
..........." I'm not sure what the active volume of the valve is"
Could you rephrase please?
............" with the pressure staying above 925 psi until the piston completes its travel"
Due to the speed of the process (therefore at least "partly" adiabatic) I'd say
it drops below that value.
Hope this helps.
Ron
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I think whether the piston separates the precharge from the pump charge.... or you consider the pump charge as compressing an air spring (gas ram).... is really just a matter of semantics.... Either way, the gun acts partly like a pneumatic, and partly like a springer, IMO.... You have replaced the Belleville washer stack in the Paradigm with an air spring (or vice versa), which I like MUCH better because your spring force is MUCH more constant than any mechanical spring, other than an extremely long one with huge preload.... It's a very compact and elegant solution, and extremely adjustable, my sincere congratulations....
By "active volume" I mean what is the volume of air at 1007 psi in the firing chamber (valve) before you pull the trigger?.... I can make a guess, since the precharged chamber is 2.2 CI @ 925 psi and you are raising that pressure to 1007 psi, that it must be decreasing to (925/1007) x 2.2 = 2.02 CI, which means you have added 0.18 CI (~ 3 cc) of air at 1007 psi plus whatever the headspace (volume after firing) is on the barrel side of the piston.... However, what that volume is, and hence the total volume available to drive the pellet, at 1007 psi, is what I was trying to get a feel for.... Your disclosure optional, of course....
We do know the barrel volume, .177 cal x 20" long, so that is 8.07 cc, so your valve is at least 37% of the barrel volume and likely larger by the time you add the headspace.... Therefore, we know that the pellet is seeing at least 925 psi of pressure for the first 37% of it's travel, which is similar dynamics to a conventional PCP or ACP where the valve is closing at 37%.... After that point, the expansion would be at least partly adiabatic as you state.... In terms of the firing cycle, your gun is acting like a PCP with a valve of roughly 2.2 CI (36 cc) at 1007 psi, but only dumping ~3 cc of that total volume.... leaving 36 cc at 925 psi remaining.... It's not a precise analogy, but close.... As I said, unique and well thought out....
Bob
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For purposes of illustration and to ensure that we're all on the same page, the Olafsson patent
drawings # 6 and 7 work well (shown partway down in the old original post).
-- the chamber to the right of the piston is precharged to 925 psi
-- air from the pump is introduced at the corner thru passage 40
-- once the pump pressure reaches 925+ the piston is forced to the right........
for a 10 inch stroke the final pressure is ~ 1007 (both sides of the piston)
-- the valve is opened, air flows directly into the barrel
Bob.........." the piston is in a precharged chamber, so in fact working like a gas ram?"
Not really. It merely separates the precharge from the pump charge.
..........." I'm not sure what the active volume of the valve is"
Could you rephrase please?
............" with the pressure staying above 925 psi until the piston completes its travel"
Due to the speed of the process (therefore at least "partly" adiabatic) I'd say
it drops below that value.
Hope this helps.
Ron
Wow, very slick design. I'm most impressed. Since this is an old project, have you done anything else with the idea?
Chris
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This is some high tech stuff! Would be very interested to see if this can go anywhere! :D
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Bob........
Measures taken to reduce clearance volume:
-- rectangular grooves for o-rings were not used. Instead, a round bottom groove placed
close to the edge was used throughout
-- passageway between pump tube and chamber made deliberately small... .38 long x .040
-- pump linkage is preloaded to eliminate all head space at end of stroke
The pressure available (~1007) was determined with clearance volumes included in the
calculation. They amounted to ~.007 CI.
initial volume of air (precharge @ 925)
(2.2)(925+14.7) = V(14.7)
V = 140.6 CI
air added from 10.5 inch pump stroke is 12.9 CI
volume of air is now 153.5 but clearance volume must be considered so the final
pressure is
(2.2+.007)(P) = (153.5)(14.7)
P = 1022
= 1007 gauge
the air added from the pump occupies this volume
(12.9)(14.7) = V(1022)
V = .186 CI
Air in the clearance spaces can certainly take part in producing power (that is, until
the piston reaches the end of it's travel...at which point they're held captive) but
since they only amount to < 4% (.007/.186) they aren't of much significance. Nevertheless,
the above figure could be modified to read .193 CI for an "active volume".
Additionally, when the valve opens a very short direct port is exposed.....~.0065 CI
Hope this helps.
Chris.....a pcp version (no pump) was halfway designed several years ago but other
airgun stuff got in the way. Way too many irons in the fire.
Ron
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So 0.193 CI = 3.16 cc / 8.07 cc = 39% for valve to barrel volume.... That gives us a pretty good model of what is going on, thanks.... Any way you want to look at it, your efficiency is stellar.... 10.5 gr. @ 903 fps = 19 FPE / 12.9 CI = 1.47 FPE/CI (11.2 barcc/FPE) and that is based on the air input by the pump.... My hat's off to you, Sir!....
Bob
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Given the amount of work it will take to compress to 6000psi, I make the assumption that it should be for a use that is worth the extra effort and not require quick follow-up shots. The best application that fits that description to me is a big bore single shot rifle. If we shoot for Bob's ideal max FPE for a given caliber from the "What makes sense" thread, can we work backwords to see what it would take to approach that max? For example with the goal of 950fps, the max weight and thus FPE for .308 is 115gr and 230 FPE. What volume of air at 6000psi would it take with a in-line dump valve? Then add some for dead space behind bullet and other predictable inefficiencies.
For this application, I think a very long lever is not reasonable. What about a screw driven piston with a geared wheel and maybe a flip out handle to up the leverage as the psi get higher? It's less unwieldy than a 2 foot lever and it can spread the work over numerous revolutions rather than in one hard pump. It obviously won't be quick to reload, but in single shot situations, like deer hunting, that might be acceptable.
Thoughts?
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First we need to look at what can be done with 3000 psi and a couple of barrel lengths.... Remember I said that 112 gr. @ 950 fps on 3000 psi with a 24" barrel is a "maximum power goal", but I am not aware of anyone getting there yet.... However it HAS been done with a 28"-30" barrel.... Just to give you an idea of what is required to reach that 224 FPE with a 500 cc reservoir, consider these numbers, compliments of Lloyd's spreadsheet....
24" barrel requires 80% efficiency and gives 0.67 FPE/CI (valve closing at muzzle)
30" barrel requires "only" 69% efficiency at 0.55 FPE/CI (valve closing at muzzle)
30" barrel at 75% efficiency would be 1.07 FPE/CI with the valve closing at 50%
At present, 70% has been done with a .308 cal, but 80% seems out of reach at the moment.... In order to get "acceptable" efficiency at 224 FPE, you need a 30" barrel if you only have 3000 psi to use.... Now, lets take that same gun, running at 75% efficiency, and run it at 6000 psi with a dump shot.... According to Lloyd's spreadsheet, you would need 8.5 cc to reach 224 FPE with a 30" barrel, and the efficiency should be 1.38 FPE/CI.... Shorten the barrel to 24", you need 11.3 cc @ 6000 psi, and the efficiency drops to 1.16 FPE/CI.... If the best you can do for efficiency is 70%, you need 14.3 cc with a 24" barrel (0.98FPE/CI) or 10.3 cc with a 30" (1.18 FPE/CI)....
So as you can see, the volume you need for the dump valve depends on 2 things, barrel length and efficiency.... If we pick 70% and a 30" barrel, which we should be able to achieve, you need a 10.3 cc valve.... Assuming you start with 2000 psi air, you need to start with 31 cc.... If you use a 1" ID chamber, that would be 2.4" long, with a 1.6" stroke.... The force required to compress to 6000 psi would be 4712 lbs.... You aren't going to do that with a simple lever, so your screw driven piston is looking good about now.... Let's use 10 TPI and a 6" long crank handle.... That would give a mechanical advantage of 60 x 2PI : 1, or 377:1 resulting on a force on the crank handle of about 12.5 lbs for the last turn (although it starts at 1/3 that).... and 16 turns required to achieve the 1.6" of compression.... If the friction losses are 50%, the force would double to about 25 lbs for the last turn (~8 at the start).... or use a longer crank to overcome the friction losses....
Bob
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One thing that is interesting is that the amount of energy YOU have to exert to compress the air is ~419 FPE.... (average force 3141 lbs. x 1.6/12 ft.) or (average force 8.3 lbs. x 6/12 x 2PI x 16 ft.).... It might be less than that because of the effect I showed from van der Waal's equations (only ~ 2.2:1 compression required instead of 3:1) to reach 6000 psi.... which would result in less stroke (fewer turns).... but I'm guessing that would be at least eaten up by friction losses, so lets call it a wash.... So where is the extra energy going?.... Well, compressing the air creates heat, which is then lost into the mass of the gun, so that's part of it.... Two thirds of the pressure used for the shot (224 x 2/3) = 150 FPE is energy you added to the shot, so that is the direct benefit of your hard work.... Then there is the energy lost inside the gun itself.... and the energy lost to accelerate that 10.3 cc of 6000 psi air out the barrel, which is, surprisingly, 80 FPE.... However, the net result is that nearly 2/3rds of the energy you put into cranking doesn't show up in the bullet.... something to consider for sure....
Bob
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So here's a similar idea, but no external input required, the gun does all the work.... How about a pressure booster?.... First we use a regulator to drop the pressure in a set volume chamber to 2000 psi, so that every shot is the same.... Then we use a piston with 3 times the area on one side to compress the air on the other side to 6000 psi.... That air is then dumped by the valve to produce the shot....
There are lots of things to work out, how to cycle the piston without wasting the air used to do so is the major one.... If that can't be done, then any gain in efficiency from the higher pressure would go out the window.... Anyway, for anyone with lots of time on their hands and an inventive side, it might be worth pursuing....
Bob
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So here's a similar idea, but no external input required, the gun does all the work.... How about a pressure booster?.... First we use a regulator to drop the pressure in a set volume chamber to 2000 psi, so that every shot is the same.... Then we use a piston with 3 times the area on one side to compress the air on the other side to 6000 psi.... That air is then dumped by the valve to produce the shot....
There are lots of things to work out, how to cycle the piston without wasting the air used to do so is the major one.... If that can't be done, then any gain in efficiency from the higher pressure would go out the window.... Anyway, for anyone with lots of time on their hands and an inventive side, it might be worth pursuing....
Bob
I picture this like Ron's gun, only the piston has a small side and a large side.
Starting from ready to fire, the front chamber is compressed to 6000psi and low volume. Open the valve and dump the air to fire. Front chamber is now at atmospheric pressure, rear of the booster is still at 2000psi. Close the valve. Open a connection from rear chamber to front chamber to equalize the pressure. This won't quite be 2000psi but that could be compensated for with a higher rear pressure. The pressure in the rear drops as it fills the front. The booster should move backwards until the two side are at equilibrium. I'm not sure if this gets all the way back to neutral or not. Close the connection from rear to front. Open the regulator into the rear chamber to get back up to 2000psi. This then pushes the booster forward, compressing the front chamber back up to 6000psi.
Does that sound close? The devil will be in the details of the pressures and the volumes to get actual numbers that work with that system. The good thing is that no air gets wasted moving the booster. I'm just not sure if it will generate 3:1 boost. We may have to force the booster back further rather than depending on equilibrium to do it. If you manually move the booster, the pressures still equalize and there shouldn't be any resistance. Once the volume at front is adequate, close the connection and open the regulator. Maybe connect the rearward movement of the booster to the bolt mechanism. Still no air gets wasted manipulated the booster.
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The booster should move backwards until the two side are at equilibrium.
That is where your idea falls flat.... Because the HP side is 1/3rd the area of the LP side, the piston won't move back, it will stay at the forward (6000 psi) position.... You need an intermediate chamber between the two (pushing on the HP side of the LP piston diameter) to move it back, then the air in THAT space needs to go somewhere to allow the piston to move forward again during the compression phase.... In a conventional pressure booster, that air is vented (and lost)....
An additional problem is that when the piston is in the forward (6000 psi) position, the LP chamber is full of 2000 psi air.... For the piston to be able to move back, that air has to go somewhere.... The easiest way I can see it to use a check valve in the piston to allow it to move into the front HP side, but you don't want it to move during the shot cycle.... and you sure don't want to vent it and lose it.... Your suggestion of opening a passage to the 6K chamber would help, but as I describe above on its own it won't force the piston back, you also need to connect it to the middle portion of the booster....
I'm sure somebody can figure it out, but that is the problem I've run into on my sketch....
Bob
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Because the pressure in the large chamber drops as it fills the small chamber, the piston should move some, just not enough to triple the small chamber's volume in prep to repressurize the large chamber. If the two chambers are open to one another there should be more play in the piston, I'm just not sure how easy it would be to move. If you can force it back to get the small chamber to max volume, ideally at 2000psi, then you close the connection and repressurize the large chamber. I'll look up 3 chamber systems as you describe. That may be the only way to make enough force to push the piston back all the way.
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Ok, I have an idea for a 4 chamber system where air moves from the rear large chamber to one of the same diameter on the other side of that end of the piston, then into a small diameter chamber behind the front of the piston, then into the front small chamber. The air used to equalize and thus power the piston vents into the firing chamber so none gets wasted. The key to this system would be high flow 1 way air valves. I need to look up one way air valves.
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Now it's complicated enough you need to do a drawing, you totally lost me....
Bob
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Now it's complicated enough you need to do a drawing, you totally lost me....
Bob
Here is a rough drawing. The 2000psi air fills the rear chamber compressing the piston. The gun fires, valve dumps then closes. One way check valves connect the rear chamber to the other side of the big piston surface. As air enters this chamber the piston now had equal forces on both sides and can slide to the rear. A spring in this chamber or a connection to a handle on the outside would work. As the piston goes back air fills the firing chamber from the other side of the small end of the piston. Close the check valves in rear and front. Fill rear chamber from regulator and it will press the piston forward again, maybe with some outside assistance. The air in the big end of the piston gets forced into the small end of the piston. The firing chamber is ready to fire again and air is primed to refill the firing chamber.
I hope this makes more sense. The piston won't move all by itself, but it can be made to have low resistance for a handle or spring mechanisms.
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The air in the big end of the piston gets forced into the small end of the piston.
I don't see how, once compression starts the pressure in the small chamber is higher.... Normally in a pressure booster that air is vented to the atmosphere.... The force available to compress the 2K air in the small chamber to 6K comes from the pressure difference across the large part of the piston.... If the ratio of areas of the pistons is 3:1, you need a full 2K across the big piston to reach 6K in the small chamber....
In addition, as drawn that skinny rod joining the two pistons would never stand 4700 lbs. of force.... Normally the piston would just be stepped like the one in a regulator, with 3 times the area on one end as the other.... I have a very similar drawing, but what to do with the air in the middle portion remains a problem.... It is necessary to reset the piston for the next stroke, but once reset that air has to go somewhere to allow the pressure across the large piston to reach 2K....
Bob
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The air in the big end of the piston gets forced into the small end of the piston.
I don't see how, once compression starts the pressure in the small chamber is higher.... Normally in a pressure booster that air is vented to the atmosphere.... The force available to compress the 2K air in the small chamber to 6K comes from the pressure difference across the large part of the piston.... If the ratio of areas of the pistons is 3:1, you need a full 2K across the big piston to reach 6K in the small chamber....
In addition, as drawn that skinny rod joining the two pistons would never stand 4700 lbs. of force.... Normally the piston would just be stepped like the one in a regulator, with 3 times the area on one end as the other.... I have a very similar drawing, but what to do with the air in the middle portion remains a problem.... It is necessary to reset the piston for the next stroke, but once reset that air has to go somewhere to allow the pressure across the large piston to reach 2K....
Bob
Inside the piston, if the small diameter half and the large diameter half have the same maximum volume at opposite ends of the piston cycle then pressure should be the same 2000psi. It's only in the firing chamber that you build up 6000psi. As the piston compresses, the small diameter half actually is expanding not compressing. I guess if the piston moves as one, the max length of the two chambers will equal, so the volumes can't be equal. It helps getting it on paper. I'll have to keep thinking it through.
Edit to add: The gist of my idea is to vent the middle area to the empty firing chamber after firing and closing the valve. It will be at atmospheric pressure and needs to be filled to 2000psi again anyway before we start compressing it again.
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Inside the piston, if the small diameter half and the large diameter half have the same maximum volume at opposite ends of the piston cycle then pressure should be the same 2000psi.
How can that be, they have different diameters and the same stroke?....
Redraw your piston with the part where you show a shaft the same diameter as the small end of the piston.... then it will be strong enough.... You then end up with a donut shaped space around the small part with the OD of the large part, which has essentially zero volume when at full compression and is the length of the stroke when at the start of the stroke.... That elongated donut, plus both the small chamber and large chamber, all need to be at 2K at the start of the stroke.... At the end of the stroke, the small chamber is at 6K, the larger chamber is at 2K, and the donut (which has shrunk to virtually zero length) has to be at atmospheric (0 psi gauge)....
Because of the very high forces involved you can't use levers, although you can use a lever or a spring to move the piston back to the beginning of the stroke, providing there is 2K both sides of it....
Bob
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Inside the piston, if the small diameter half and the large diameter half have the same maximum volume at opposite ends of the piston cycle then pressure should be the same 2000psi.
How can that be, they have different diameters and the same stroke?....
Redraw your piston with the part where you show a shaft the same diameter as the small end of the piston.... then it will be strong enough.... You then end up with a donut shaped space around the small part with the OD of the large part, which has essentially zero volume when at full compression and is the length of the stroke when at the start of the stroke.... That elongated donut, plus both the small chamber and large chamber, all need to be at 2K at the start of the stroke.... At the end of the stroke, the small chamber is at 6K, the larger chamber is at 2K, and the donut (which has shrunk to virtually zero length) has to be at atmospheric (0 psi gauge)....
Because of the very high forces involved you can't use levers, although you can use a lever or a spring to move the piston back to the beginning of the stroke, providing there is 2K both sides of it....
Bob
Yeah, I realized the error with the different diameters and the same stroke as I was writing that last text. Lets take the structural integrity question out of my drawing with 4 chambers for a second and just consider air volumes and pressures. The problem with my logic was that the two middle chambers can't hold the same amount of air. With your 3 chamber design we vent and lose all of the air in that chamber. In mine, if we move the air from the big end of the piston into the little end of the piston then dump the difference, that solves the problem with volume differences. If the small end of the piston's max volume, ie piston fully compressed, is the same as the compressed chamber's max volume, ie uncompressed, then we can vent all of that air from the smaller chamber into the firing chamber as the piston decompresses. The only wasted air is the difference between the two middle chambers. That makes the air lost at least 33% more efficient. Since the smaller chamber has 1/3 volume as the larger chamber, we can save 1/3 of the air and dump just 2/3.
Now for the issue of structural integrity for the piston. I drew mine that way to make the picture as simple as possible. I didn't assume that would be the shape of the piston. One option would be a hollow cylinder where the chambers are inside the piston, but that makes it hard to separate the two chambers in the piston. My drawing assumed a divider that was rigid. Another option might be to have two halves of the piston that move independent of each other for different parts of the stroke. I need to draw and think hard on how that would work. Another way might be to use two unconnected floating pistons where the pressure in the rear chamber presses on a piston that presses on a middle chamber that then has a smaller diameter floating piston. This idea has many more variables and its much more of a dynamic or squishy pressure booster. I will have to ponder these options and other ways to save air.
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The only wasted air is the difference between the two middle chambers.
Agreed.... and my design has no volume for the smaller of the middle chambers, and since the piston rod is the same size as the small end of the piston, the volume of the larger of the middle chambers is the same as yours, minus the volume of the piston rod.... which equals the volume of the smaller chamber.... In other words, we have exactly the same difference in volume.... That MUST be the case, because the stroke is the same for both ends, and the pistons are the same diameter....
You mention "venting the air from the middle chamber into the firing chamber as the piston decompresses"....
However, the firing chamber already has to be at 2K in order to provide 1/3 of the force needed to move the piston back to the "decompressed" state...To fully decompress the piston, both the firing chamber (1/3rd of the area) and the middle chamber (2/3rds of the area) have to be at 2K to match the driving side of the large piston (full area).... Slimming the connecting piston rod makes no difference, as whatever area is added to one end is added to the other, and the forces cancel out....
Keep trying, you may come up with something I haven't thought of....
Bob
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The only wasted air is the difference between the two middle chambers.
Agreed.... and my design has no volume for the smaller of the middle chambers, and since the piston rod is the same size as the small end of the piston, the volume of the larger of the middle chambers is the same as yours, minus the volume of the piston rod.... which equals the volume of the smaller chamber.... In other words, we have exactly the same difference in volume.... That MUST be the case, because the stroke is the same for both ends, and the pistons are the same diameter....
You mention "venting the air from the middle chamber into the firing chamber as the piston decompresses"....
However, the firing chamber already has to be at 2K in order to provide 1/3 of the force needed to move the piston back to the "decompressed" state...To fully decompress the piston, both the firing chamber (1/3rd of the area) and the middle chamber (2/3rds of the area) have to be at 2K to match the driving side of the large piston (full area).... Slimming the connecting piston rod makes no difference, as whatever area is added to one end is added to the other, and the forces cancel out....
Keep trying, you may come up with something I haven't thought of....
Bob
I hadn't considered that both chambers need to be 2000psi to move back. Having the same psi isn't enough. You need the same psi over the same surface area to equalize the forces. Alright, if you'll keep pointing out what I'm missing, I'll keep plodding through. Thanks!!!
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(EDIT TO ADD: Read the next set of posts in sequence before replying. It gets simpler and more efficient.)
Ok, here is a system that conserves 1/3 of the mid chamber air and also adds some more oomph as the valve dumps. The piston has 2 sections I'll call the cylinder and the ring. The ring can move up and down the rear 1/3 of the cylinder. At the front limit of the ring's movement it presses the cylinder forward, and at the rear it pulls the cylinder backward.
1) This is ready to fire, with 6k in the valve chamber (VC), 2k in the cylinder, 0psi in front of the ring, and 2k behind the ring. The valve then opens.
2) The VC is evacuated and the cylinder follows through with 2kpsi of force to complete the compression stroke. The valve closes on its chamber with 0psi, 2k in cylinder, 0psi in front of the ring and 2k behind the ring. Open check valve connections from the cylinder to the VC and the back of the ring to the front. An external force now needs to push the piston to the rear. This force would be proportional to the resistance of air moving between the chambers. As they equalize, the force should be reduced to just the friction between the parts. There will need to a plunger mechanism in the cylinder to force air out of the cylinder into the VC.
3) Piston is fully decompressed. VC is max volume and 2k psi, cylinder is 0psi, and 2k on both sides of the ring. Open the space in front of the ring into the cylinder.
4) The ring moves forward to its limit as pressure equalizes into the cylinder. Close off the cylinder and dump in front of the ring to get back to 0psi. That should compress the piston back to ready to fire.
Thoughts? The devil will be in the check valve connections and how to evacuate the cylinder into the VC.
I'd be curious to see what kind of ci/FPE this setup would have. I wonder if the dumped air could be used to push the piston back open somehow to improve efficiency and minimize external forces.
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Simpler. Just by allowing the ring to come back only 2/3 of the total stroke and the cylinder the full stroke, we get the same efficiency. Make the cylinder solid. It's your 3 chambers. Have a line between the VC and behind the piston, easiest is through the cylinder, and let that open and close with a simple valve mechanism. No need to connect the mid chamber to anything else. It just fills and dumps. Does that make sense? Now to see if we can do anything else with the mid chamber air.
Edit to add: with this mechanism we dump twice what we send down the barrel. The question is whether we gain enough FPE by increasing to 6000psi and the added force of the piston following through at 2000psi. We should also consider whether there is one ratio of boosting that gives the best efficiency. That sounds like a calculus problem, and I'm over 10 years past my physics major, maybe 12 past my last calculus class.
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For the mid chamber air, what about dumping it into a second valve chamber. If we have 1 chamber with X volume of air boosted to 6000 psi and a second chamber with 2X volume at 2000psi, they can both be dumped down the same barrel. Dump the high pressure chamber first and then the low pressure chamber when the pressure behind the pellet drops to 2000psi.
I suspect this design will work best for very long barrels where the added air can continue to make a difference.
Edit: Because the second VC acts more like a traditional dump valve with all of the inefficiency that entails, it won't triple the force down the barrel. My guess is that it may boost the efficiency of this system by 50-100% though.
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For the second VC, I need to know how much force it will take to dump the mid chamber into the second VC. Without a check valve, I think it will be the same as taking a 1000psi chamber and compressing it into half the size back up to 2000psi. If it's got a check valve between the two, it might not take much extra force to shift over, just enough to overcome the resistance of flow through the check valve and friction on the piston. As the piston compresses, the pressure in the mid chamber stays at 2000psi while the 2nd VC goes from 0psi up to 2000psi. When the volume of the mid chamber hits 0 we now have 0psi opposing the piston.
Another option is to use the same type of reservoir as above only use it through another check valve to fill the VC after closing the dump valve and as the piston moves back. Ideally this chamber would have a piston of its own so it can maintain pressure as it fills the VC fully. Now we vent the other half of the air in that chamber. This cuts the waste of air in half again. Compared to filling and dumping from a full piston stroke for the mid-chamber with a single piece piston and 3 chamber system like Bob described, we are only filling the mid chamber with 2/3 the air then conserving half of that by using it to fill the VC. We've got the efficiency improved by 66%.
Now, if we don't dump that extra 1/3 into the atmosphere but instead combine the above 2 ideas, we can have 1/2 of the mid chamber air go into the VC and half into a second VC that dumps into the barrel with the shot. There is zero wasted air, other than the inefficiency of the 2nd dump valve.
We have a VC of X quantity of air at 6000psi, a 2k psi piston following the air released from the VC, and VC2 with X air at 2000psi that behaves like a regular dump valve. I would probably have this release into the opening of VC1 with a valve that opens when the the pressure in the chamber drops below 2000psi. That way there's only one valve to open and close and VC2 is self timing.
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Sorry, once again, you totally lost me.... You need to draw it out and figure all the forces on all the moving parts, and if you are convinced it will work build one to see how well....
At stage 3, the middle chamber? at 0 psi isn't pushing back against the piston, so there is still 1/3 of the 2K force on the piston pushing it towards the VC....
The efficiency was never better than about 1.33 times what we can get conventionally at 2K, let alone at 3K.... so basically ANY wasted air makes it not worth the bother, IMO....
Bob
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Sorry, once again, you totally lost me.... You need to draw it out and figure all the forces on all the moving parts, and if you are convinced it will work build one to see how well....
At stage 3, the middle chamber? at 0 psi isn't pushing back against the piston, so there is still 1/3 of the 2K force on the piston pushing it towards the VC....
The efficiency was never better than about 1.33 times what we can get conventionally at 2K, let alone at 3K.... so basically ANY wasted air makes it not worth the bother, IMO....
Bob
Regardless of the mechanism, can you run the numbers for what we could expect out of X volume of air at 6000psi followed by X at 2000psi? You'd need to know where the projectile will be when the pressure gets down to 2000psi to figure out from there what doubling the volume of air behind the pellet is going to do. If it takes a 45" barrel to get get full velocity out of that much air, then that would be good to know before pushing the design any further. You could use the numbers you quoted for the .308 with a 10.3cc valve at 6000psi and a 30cc valve at 2000psi for the dump.
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As an aside, where can I find Lloyd's internal ballistics spreadsheet you used to do some of your math?
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Here is a sketch of the basic idea using a stepped piston like that in a regulator, only acting backwards like a typical pressure booster.... Black lines are the stepped cylinder, blue the stepped piston, and red an airflow passage and check valve (located in the cone shape in the small end of the piston).... The 2K chamber on the left is supplied from the regulator.... The 6K chamber on the right is the firing chamber, vented by a dump valve to the barrel....
(http://i378.photobucket.com/albums/oo221/rsterne/Parts%20for%20Sale/BoosterPiston_zpsb07f14ef.jpg) (http://s378.photobucket.com/user/rsterne/media/Parts%20for%20Sale/BoosterPiston_zpsb07f14ef.jpg.html)
The dashed blue line is the piston in the compressed (firing) position (ie to the right), it is sitting up against the shoulder in the cylinder, (actually against a light return spring in the middle portion of the cylinder, not shown).... When the gun is fired, the 6K chamber drops to zero (guage), and the 2K cylinder at the left is full (up to the dashed line) with 2K air.... Because of the extremely short response time of the shot, I do not want the piston to move during the shot to cause any variations in velocity.... Since the pressure is so high (6K), even a small movement would make a big difference....
In order to move the piston back to the cocked (decompressed) state (ie to the left), three things have to happen....
1. The volume of 2K air in the low pressure chamber must decrease (note air cannot flow backwards into the regulator to accomplish this)
2. The pressure in the firing chamber must reach 2K
3. The pressure in the middle chamber must also reach 2K
Conditions 2 & 3 are required to equalize the pressure across the piston.... Condition 1 would be achieved by venting it to the port where the red arrow is on the top of the cylinder.... The air would flow into the middle portion (where the spring is) and through the piston and check valve into the firing chamber.... With equal pressure on both sides of the piston, the spring would recock it.... NO AIR IS WASTED IN THIS COCKING STROKE (the air simply moves from the left side of the piston to the right side, into the middle and firing chambers).... The passage between the low pressure chamber and the middle chamber would then have to be closed, and the middle chamber vented to the atmosphere.... The check valve in the piston would close and as the pressure in the middle chamber bleeds off, the regulator would top up the low pressure chamber, moving the piston to the right and compressing the air in the firing chamber back to 6K.....
The problem with this design is that the volume of 2K air used for every shot is the volume of the low pressure (left) chamber over the entire stroke of the piston.... which is 3 times what actually goes out the barrel.... 2/3rds of the air, all of the 2K air used to cock the piston in the middle chamber, is vented to atmosphere on the compression stroke.... Obviously, that is terribly inefficient, and renders the whole exercise useless....
Your task, should your choose to accept it, is to eliminate that wasted air.... *grin*
Lloyd's spreadsheet is proprietary, he was nice enough to give me a copy.... I don't have any way to superimpose two shot cycles as you suggest....
Bob
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Here is a sketch of the basic idea using a stepped piston like that in a regulator, only acting backwards like a typical pressure booster.... Black lines are the stepped cylinder, blue the stepped piston, and red an airflow passage and check valve (located in the cone shape in the small end of the piston).... The 2K chamber on the left is supplied from the regulator.... The 6K chamber on the right is the firing chamber, vented by a dump valve to the barrel....
(http://i378.photobucket.com/albums/oo221/rsterne/Parts%20for%20Sale/BoosterPiston_zpsb07f14ef.jpg) (http://s378.photobucket.com/user/rsterne/media/Parts%20for%20Sale/BoosterPiston_zpsb07f14ef.jpg.html)
The dashed blue line is the piston in the compressed (firing) position (ie to the right), it is sitting up against the shoulder in the cylinder, (actually against a light return spring in the middle portion of the cylinder, not shown).... When the gun is fired, the 6K chamber drops to zero (guage), and the 2K cylinder at the left is full (up to the dashed line) with 2K air.... Because of the extremely short response time of the shot, I do not want the piston to move during the shot to cause any variations in velocity.... Since the pressure is so high (6K), even a small movement would make a big difference....
In order to move the piston back to the cocked (decompressed) state (ie to the left), three things have to happen....
1. The volume of 2K air in the low pressure chamber must decrease (note air cannot flow backwards into the regulator to accomplish this)
2. The pressure in the firing chamber must reach 2K
3. The pressure in the middle chamber must also reach 2K
Conditions 2 & 3 are required to equalize the pressure across the piston.... Condition 1 would be achieved by venting it to the port where the red arrow is on the top of the cylinder.... The air would flow into the middle portion (where the spring is) and through the piston and check valve into the firing chamber.... With equal pressure on both sides of the piston, the spring would recock it.... NO AIR IS WASTED IN THIS COCKING STROKE (the air simply moves from the left side of the piston to the right side, into the middle and firing chambers).... The passage between the low pressure chamber and the middle chamber would then have to be closed, and the middle chamber vented to the atmosphere.... The check valve in the piston would close and as the pressure in the middle chamber bleeds off, the regulator would top up the low pressure chamber, moving the piston to the right and compressing the air in the firing chamber back to 6K.....
The problem with this design is that the volume of 2K air used for every shot is the volume of the low pressure (left) chamber over the entire stroke of the piston.... which is 3 times what actually goes out the barrel.... 2/3rds of the air, all of the 2K air used to cock the piston in the middle chamber, is vented to atmosphere on the compression stroke.... Obviously, that is terribly inefficient, and renders the whole exercise useless....
Your task, should your choose to accept it, is to eliminate that wasted air.... *grin*
Lloyd's spreadsheet is proprietary, he was nice enough to give me a copy.... I don't have any way to superimpose two shot cycles as you suggest....
Bob
Can you point me to a link where I can get a copy through appropriate channels or at least price it out, or is it proprietary as in secret? I certainly don't want to cheat him his due if he put the amount of work I suspect he put into that spreadsheet.
I'm curious if the superimposed shots would act like a single shot from a 60cc plenum that dumps at 4000psi or if it behaves completely differently.
Though I'm not good enough at the math to do the numbers, I suspect strongly that this can be done without venting the air. If half is boosted to 6k and half is left at 2k, so long as it all gets applied to the rear of the pellet I don't consider that a waste. Whether it is practical for a rifle or for an artillery battery is a different question.
The main practical question I have as to whether this is feasible or not is whether I can force 2k of air through a check valve with minimal external force. If that is possible, my design will work given a competent machinist and the right materials.
I will try to put all the pieces together in my design to draw and better explain.
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Lloyd's spreadsheet is basically not available, you would have to convince him you really needed it, and that he could trust you 110%.... I am not aware of anyone else besides me that has a copy.... As with all such things, it is a work in progress and a labour of love, and I think it is the feedback I provide him which made it available to me.... something I am grateful for nearly every day....
Using the example we are working with, 10.3 cc of 6000 psi air through a dump valve, the residual muzzle pressure is over 1300 psi (so the gun will be incredibly LOUD), and the pressure doesn't drop to 2000 psi until the bullet has already gone 17" down the 30" barrel.... By that time, the bullet has already reached 87% of it's muzzle velocity.... Although I can't use the spreadsheet to calculate what additional velocity one might gain by increasing the volume available at that point, all it could do is add more push during that last 13" of the bullet's travel down the barrel, by reducing the rate at which the pressure declines (ie increasing the muzzle pressure and muzzle blast).... That is assuming, of course, that you could open that second chamber completely and instantly, trying to flow it through a check valve in the remaining millisecond before the bullet exits the barrel ain't gonna cut it, IMO.... The critical part of the acceleration is the initial push of the 6000 psi, the part that could be influenced by a second chamber opening at 2000 psi (and 17" down the barrel) is VERY small.... The pressure as seen by the bullet would still be 6000 psi initially, 2000 psi at 17" down the barrel, and instead of 1300 psi at the muzzle, some value greater than that, depending on how large the second chamber was.... If that second chamber was 30cc, then the decay in pressure for that last 13" of travel would be from 2000 psi (in both cases) down to about 1560 psi instead of 1300+.... The greatest difference in pressure would occur in the last part of the barrel, where virtually no gains are to be made anyway.... I don't think that is a very efficient way to use air.... and in fact I'm not sure you would even see a consistent, measurable difference in velocity....
Instead of using a dump valve, closing the valve on the 6K chamber at the 17" point in the bullet travel would only lose 7 fps (~3.5 FPE) and yet would save fully 33% of the air inside the 6K chamber.... Once again, you can see why dump valves are a poor choice....
BTW, once the second chamber is dumped, the two valves together would act like a 40.3 cc valve at 3022 psi.... ( (30 x 2000) + (10.3 x 6000) ) / 40.3 = 3022....
Bob
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With the the amount of air doubled, couldn't we use smaller chambers for the same size projectile?
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As I tried to explain, doubling the amount of air when the bullet has already achieved 87% of it's velocity at 17" down the barrel will do virtually nothing.... However, cutting the volume of the 6000 psi chamber by 50% would drop the energy down from 224 FPE to less than 180 FPE, still using the 30" barrel....
Bob
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As I tried to explain, doubling the amount of air when the bullet has already achieved 87% of it's velocity at 17" down the barrel will do virtually nothing.... However, cutting the volume of the 6000 psi chamber by 50% would drop the energy down from 224 FPE to less than 180 FPE, still using the 30" barrel....
Bob
Is that 180fpe for just the 6000psi or both 2000 and 6000? Is it still going to be at 87% 17" with the smaller chamber?
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Regarding the mid chamber air, can we just move it back and forth between 2 chambers? If we had a check valve that could switch directions or two check valves going opposite directions that could be turned on/off depending which way you want the air to move. For instance if the mid chamber were compressed and allowed to vent through a check valve into a chamber with 0 psi, the next time we need to refill the mid chamber we just vent the air back from the other chamber through a check valve going the other direction. These could be controlled by a cover that slides back and forth between the two. The other chamber would also need a piston working on it, but that's not difficult.
Edit to add picture. Pardon the bad art work.
Edit2: Looking at it on paper the issue seems to be that the rear chamber changes volume without changing pressure. Half of the difference is made up by filling the valve chamber from the rear chamber. That still leaves some displacement unaccounted for. This might work better if we only tried to compress 2 to 1. That means the torus of air in the mid chamber is equal in volume to the air in the valve chamber. As the piston moves to the rear, it forces air from the rear through a check valve into the valve chamber. As the piston moves forward it can pull more air from the regulator.
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Upon further review, this idea seems to work using 2 to 1 as the compression. I don't even think you have to have check valves between the two sides of the mid chamber. The divider just has to be slightly perforated. Because the mid chamber never changes volume, it's pressure doesn't have to change, just the relative position of the piston. The pressure pushing back on the piston also includes the area around the valve chamber, but this is vented to the atmosphere and doesn't add anything. So long as the regulator keeps filling up the rear chamber, we should still get effortless 2:1 compression. If we regulate that to 2500psi and compress up to 5000psi in the valve chamber, we should still get some considerable force and hopefully some good efficiency. Or if we want to run a 4500psi tank, we can regulate to 3k and still get 6k psi in the VC.
I'm not certain about perforating the divider or using reversible check valves. I almost wonder if the mid chamber is even necessary. With a check valve from the rear to the front chamber, if the piston moves back it just displaces the air from back into the front. Once you hit the rear of the stroke and close the check valve, as you push forward, the air from the regulator replenishes the rear and sends the piston forward. This could be completely automated with a light spring pushing the piston to the rear. The regulator may need to be blocked while the piston moves to the rear then opened going forward again.
Am I missing something here?
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Yep you keep missing that you have to have the same force on both sides of the piston to reset it for the next compression stroke.... Force = pressure x area.... If you just open a valve between two chambers of equal volume, one at 2K and one at zero, you get both at 1K.... the air doesn't move completely from one to the other, only half of it does....
The 180 FPE was just for the 6000 psi x 5.15 cc and the 2000 psi point would now occur earlier at about 8.5" down the barrel.... If you have the 2nd chamber dump at that point, you will get more out of it than before, the two chambers will work (from that point on) as if it was 20.15 cc at 3022 psi.... You will gain a few FPE instead of basically nothing, but you are still using a LOT of air to do it.... and your velocity is down 100 fps from your goal.... None of these dual chamber setups are as efficient as a straight 3000 psi conventional PCP with the valve closing at 50% (15" of travel)....
I have to tell you, I'm growing a bit weary of this.... You need to do some drawings and work out the math for each step and figure out how to move the piston without losing any air.... or it's a waste of time.... I'm going to take a break from this thread for a while and let you ponder it....
Bob
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Yep you keep missing that you have to have the same force on both sides of the piston to reset it for the next compression stroke.... Force = pressure x area.... If you just open a valve between two chambers of equal volume, one at 2K and one at zero, you get both at 1K.... the air doesn't move completely from one to the other, only half of it does....
The 180 FPE was just for the 6000 psi x 5.15 cc and the 2000 psi point would now occur earlier at about 8.5" down the barrel.... If you have the 2nd chamber dump at that point, you will get more out of it than before, the two chambers will work (from that point on) as if it was 20.15 cc at 3022 psi.... You will gain a few FPE instead of basically nothing, but you are still using a LOT of air to do it.... and your velocity is down 100 fps from your goal.... None of these dual chamber setups are as efficient as a straight 3000 psi conventional PCP with the valve closing at 50% (15" of travel)....
I have to tell you, I'm growing a bit weary of this.... You need to do some drawings and work out the math for each step and figure out how to move the piston without losing any air.... or it's a waste of time.... I'm going to take a break from this thread for a while and let you ponder it....
Bob
Thanks Bob, sorry for being a bother. I'll have to work on it some more as you say.
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Perhaps a totally different track?.... Don't worry about the efficiency.... Build the gun to use a 6K dump valve with a volume of 50% of the barrel volume.... Carry a 4500 psi Pony tank into the field, regulated down to 3000 psi and equipped with a 2:1 pressure booster (or the two combined into a regulated pressure booster if such a thing exists) so that it has a 6K output.... Use hoses and fittings, and a gun, designed to hold 6K in the valve.... Fill, shoot, and repeat....
The gun is a single shot, but it gets incredibly light, even with a 30" barrel, because it has no reservoir.... Using an 18cc dump valve at 6K should be able to push a 150 gr., .308 cal bullet about 950 fps (300 FPE).... The efficiency of the shot is 0.66 FPE/CI (better than most big bores) but when you include the air vented from the booster, it drops to about 0.44 FPE/CI, still in the range for many big bores.... You will lose some 6K air from bleeding the hose, though....
If instead of the dump valve you can control the dwell accurately using a balanced valve (so that the hammer energy doesn't get overwhelming) you can, of course, drastically increase the efficiency.... If you can close that same 18 cc valve when the bullet exits the muzzle you would have the same power, and the efficiency would be 0.98 FPE/CI.... Close the valve at 50%, the velocity only drops 20 fps and the efficiency jumps to 1.26 FPE/CI.... However, you have the added complication of filling the gun to exactly 6K for each shot.... although a smaller amount of air would be wasted by the (smaller) booster....
Bob
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Perhaps a totally different track?.... Don't worry about the efficiency.... Build the gun to use a 6K dump valve with a volume of 50% of the barrel volume.... Carry a 4500 psi Pony tank into the field, regulated down to 3000 psi and equipped with a 2:1 pressure booster (or the two combined into a regulated pressure booster if such a thing exists) so that it has a 6K output.... Use hoses and fittings, and a gun, designed to hold 6K in the valve.... Fill, shoot, and repeat....
The gun is a single shot, but it gets incredibly light, even with a 30" barrel, because it has no reservoir.... Using an 18cc dump valve at 6K should be able to push a 150 gr., .308 cal bullet about 950 fps (300 FPE).... The efficiency of the shot is 0.66 FPE/CI (better than most big bores) but when you include the air vented from the booster, it drops to about 0.44 FPE/CI, still in the range for many big bores.... You will lose some 6K air from bleeding the hose, though....
If instead of the dump valve you can control the dwell accurately using a balanced valve (so that the hammer energy doesn't get overwhelming) you can, of course, drastically increase the efficiency.... If you can close that same 18 cc valve when the bullet exits the muzzle you would have the same power, and the efficiency would be 0.98 FPE/CI.... Close the valve at 50%, the velocity only drops 20 fps and the efficiency jumps to 1.26 FPE/CI.... However, you have the added complication of filling the gun to exactly 6K for each shot.... although a smaller amount of air would be wasted by the (smaller) booster....
Bob
I think that is a very sensible solution that totally negates all the fun of figuring out the mechanics. I think using air to run the pressure booster can't be done without wasting at least the same as what you dump down the barrel. It just can't be done. It took your previous post and an epiphany to get that. What could be done is to combine the first idea of a lever and a pressure booster. Assuming a 2:1 booster with 1" diameter, 3000psi doubled to 6000psi, the pressure to lift the ring of the bigger diameter section is 2356lb. That's better than the 4712lb to compress the front chamber directly. If we use a hydraulic chamber instead of the air chamber to open the piston and fill the front chamber, when you release the hydraulic the piston will have its full 2:1 force to apply. What's the math on a 2356lb lever?
Edit:
Nix that. It's worse than on page one because the chamber volume is so much more. If we apply this back to the math for that first lever it would bring the 100lb weight to 50lb. Use 18" instead of 12", and it's 37.5lb. It's thus nonviable for big bore but a contender for 50fpe.
On your last post, I definitely want to understand balanced valves better but I'm nowhere close to that.
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OK, so before you start jumping off the walls, that is FAR from a dump valve, in fact the dwell is so short the valve is closing when the pellet has only moved less than an inch.... To get back to about where we were for power and efficiency, and with a dump valve, we need to reduce the plenum (in this case valve) volume.... I had to do some trial and error, but here is what I came up with.... It turns out you only need about 1/10th the volume, 1.67 cc of air at 6000 psi to get back to 50 FPE with a dump valve....
Regulated, compressed to 6000 psi, 1.67 cc valve with dump shot.... 50 FPE @ 1.29 FPE/CI.... So the answer is yes, what you are proposing should work.... IF you can build it....
Unfortunately, that's a big IF.... You need to start with 5 cc of air (0.305 CI) at 2000 psi and then compress it 3:1 to 6000 psi.... Let's say that you use a piston 1/2" in diameter, which is 0.196 sq.in.... you need a cylinder 1.56" long, and a piston with a 1.04" stroke to increase the pressure 3:1.... At the end of the compression stroke, the force on the piston would be 1178 lbs. so you would need quite a leverage system to achieve that.... If the lever was a foot long, you would still have a 100 lb. load.... You could trade off smaller diameter for longer stroke, or larger diameter for less stroke on the piston, but you still end up with a 100 lb. load on the end of a 1 foot lever to compress that 5 cc of 2000 psi air to 6000 psi....
So the answer to your question is, yes, by using a much higher pressure you can make a dump valve as efficient as a conventional PCP.... The question then becomes, can you build and operate it?....
Bob
Going back to this post. Substitute 3000psi and 2:1 compression. Put a pressure booster between the lever and the chamber. Use 18" of lever. 37.5lb load on the lever is the end result. My umarex octane had a 50lb lever. If we work backwards from what amount of lever strength is tolerable we can figure out a max FPE for this system. 75FPE aught to get us back to 50lb on the lever.
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Starting with 3000 psi, you only need 3.34 cc of air, and compress it 2:1 to get that 1.67 cc @ 6000.... Assuming the same diameter (1/2") the load at the end of the stroke is still the same, but you only need half the stroke, cylinder is 1.04" long, stroke is 0.52".... Half the stroke means twice the MA, so half the force on the same length lever.... However, your starting load is greater than before.... Alternately, you can decrease the diameter and the loads, and increase the piston stroke, getting back to the same load at the end of the lever....
This is do-able, but of course you are talking wayyyyyyyy less power than the 224 FPE of the .308 cal.... 50 FPE using the numbers above.... Plus you need a source of 3K air, of course....
There is some info on balanced valves here in the Geek Gate.... they are in their infancy at the moment.... but show promise.... Basically a knock-open design with a balancing piston that removes the additional load of the air pressure holding the valve closed, just leaving the pressure times the stem area for the opening and closing forces.... requiring much less hammer strike....
Bob
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I want to correct my earlier statement that running the pressure booster can't be done without wasting the air. The only instance where I think you could get away with that mid chamber run by air is if you ran two pressure boosters side by side. They would need to alternate with one compressing as the other releases. Because there are equal forces in the two rear chambers they can push the mid chamber air back and forth. This is obviously not practical. You either run a double barreled gun with twice the weight or two pressure boosters for one chamber. It is quick on the follow up shot either way though.
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Sorry, I still don't see it.... When one booster is on the compression stroke, ALL of it's force is required to raise the pressure of the firing chamber to 6K.... There can be NO pressure in the middle chamber at max. stroke.... However, in the other booster, returning the piston to it's uncompressed position (against 2K pressure) requires 2K of pressure on both it's firing chamber (1/3rd of the area) and middle chamber (2/3rd of the area).... Connecting the two middle chambers together would see the volume move back and forth, but there would still be 2K of pressure in both middle chambers.... so no compression could occur....
In addition, the large (low pressure) side of the chamber that is on the cocking stroke is full of 2K air, which in my design would flow into the middle and firing chambers.... If the middle chamber is being filled by the air from the middle chamber on the other booster, then that excess air has nowhere to go.... unless you vent it....
BTW, in my suggestion of having the booster external to the gun, I realize that there is an air loss during recocking the booster, and mentioned that as a loss of efficiency.... The other alterative, of course, is to build a PCP that runs on 6K and figure out how to fill it safely.... There are big bores safely running 4500 psi now, it's only a matter of time before they go higher.... The problem currently is a safe way to store air at an even higher pressure to fill them.... or use a booster....
ob
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Sorry, I still don't see it.... When one booster is on the compression stroke, ALL of it's force is required to raise the pressure of the firing chamber to 6K.... There can be NO pressure in the middle chamber at max. stroke.... However, in the other booster, returning the piston to it's uncompressed position (against 2K pressure) requires 2K of pressure on both it's firing chamber (1/3rd of the area) and middle chamber (2/3rd of the area).... Connecting the two middle chambers together would see the volume move back and forth, but there would still be 2K of pressure in both middle chambers.... so no compression could occur....
In addition, the large (low pressure) side of the chamber that is on the cocking stroke is full of 2K air, which in my design would flow into the middle and firing chambers.... If the middle chamber is being filled by the air from the middle chamber on the other booster, then that excess air has nowhere to go.... unless you vent it....
BTW, in my suggestion of having the booster external to the gun, I realize that there is an air loss during recocking the booster, and mentioned that as a loss of efficiency.... The other alterative, of course, is to build a PCP that runs on 6K and figure out how to fill it safely.... There are big bores safely running 4500 psi now, it's only a matter of time before they go higher.... The problem currently is a safe way to store air at an even higher pressure to fill them.... or use a booster....
ob
Yup. I had to picture the two boosters back to back and balance the forces out that way to make sense of it in my head. You end up with 3k in one chamber trying to compress the other chamber to 6k, because the two middle chambers at 3k just balance each other out. Only if both chambers and both mid-chambers were open to each other could the piston move back and forth without effort. Because we're sealing one off and trying to double its pressure, we have to vent its mid chamber to balance the two sides. Bah.
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Since it seems I have to keep combining ideas to get better results, let's combine a double piston system and a lever to make up the difference in force. Instead of moving air back and forth, lets vent the mid chamber of one into the front chamber of the other and vice versa. Here is a simplified representation of back to back pistons and the extrapolation to how it might actually work side by side.
1) Start ready to fire one chamber. Open and close that valve.
2) One chamber is 0psi. Vent the other's mid chamber to this and they'll equalize at 2000psi between the compressed chamber and the other mid chamber.
3) The piston hasn't moved yet. Run 3000psi to the mid chamber that needs to expand and apply force on piston to compress the other mid chamber's air into the expanding firing chamber.
4) You're ready to fire the other chamber now.
For a 1" diameter piston, 3000 to 6000 psi boost, I get 356lb 785lb of force required to compress the mid-chamber and firing chamber air from 2000psi to 3000psi. That's low enough to run with a 1' 18" side lever. This is for the 230fpe .308 at that.
Thoughts?
Edit: correction, the force difference should be 785lb.
(.25 x 3.14159 x 3000psi) - (.25 x 3.14159 x 2000psi).
Not as awesome but still doable with a longer lever.
Chris
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It looks like you may have something, but without all the volumes shown it's hard to visualize.... You need to redo the drawings, using your squares, but show volume on BOTH sides of the center (large) portion of the piston (there has to be some) and the check valves and supply passages to the regulator.... I get that the piston is moved by a lever to overcome the pressure difference.... However, doesn't that lever have to operate both ways?....
On second look, there appears to be nowhere for the 2K to go in the middle chamber when you apply 3K on the other side?....
Bob
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That air gets pushed into the firing chamber. The volume in the middle chamber is equal to that of the firing chamber at each of their respective maximums. We're going from zero psig after opening and closing the valve up to 2000psig when we open the other mid-chamber. The connection stays open as the piston compresses the mid air out and into the firing chamber. The firing chamber doubles in size as the pressure reaches 3000psi and it's piston is opened up.
I have an idea for how to operate the lever. I can draw later once your satisfied with how it operates.
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Look at diagram #3, the right middle chamber is at 2K and the firing chamber yet to be compressed at 3K.... You show air flowing to the left side of the middle chamber piston from the regulator.... When you move the piston to the right using the lever, the 3K becomes 6K in the firing chamber.... Where does the 2K air in the right side of the middle chamber go?.... It must be vented, or as it compresses it will prevent you from moving the piston to the right with the lever....
Bob
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Coming in late here Ron, curious how the displaced piston movement occurs at such low pressure.
If it is precharged to 925psi then when it moves 50% the pressure is double so 1850 psi. At 67% travel the pressure must be triple so 2775psi.
If the pump generates 1007psi then the displaced piston has only moved 9% of its available stroke.
I will keep reading this thread looking for more info might have missed.
Cheers
Walter....
For purposes of illustration and to ensure that we're all on the same page, the Olafsson patent
drawings # 6 and 7 work well (shown partway down in the old original post).
-- the chamber to the right of the piston is precharged to 925 psi
-- air from the pump is introduced at the corner thru passage 40
-- once the pump pressure reaches 925+ the piston is forced to the right........
for a 10 inch stroke the final pressure is ~ 1007 (both sides of the piston)
-- the valve is opened, air flows directly into the barrel
Bob.........." the piston is in a precharged chamber, so in fact working like a gas ram?"
Not really. It merely separates the precharge from the pump charge.
..........." I'm not sure what the active volume of the valve is"
Could you rephrase please?
............" with the pressure staying above 925 psi until the piston completes its travel"
Due to the speed of the process (therefore at least "partly" adiabatic) I'd say
it drops below that value.
Hope this helps.
Ron
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Welcome, Walter.... yes, Ron said the piston displacement was only 0.186 CI.... I see what you are saying, though, that is the only air available for the shot.... That's what I used in my calculations, which shows the valve volume is 39% of the barrel volume, so the gun ACTS like a PCP with a 36cc valve filled to 1007 psi and then dumping 3 cc of that during the shot.... It's a clever way to get the gun to ACT like a PCP yet use a dump valve.... Without the piston and preloaded chamber, the valve would just be a chamber of 3 cc, filled to 1007 psi, and the pressure would drop throughout the shot, so that when the pellet was 39% of the way down the barrel (barrel volume = valve volume) the pressure would be 1007/2 = 504 psi.... With the floating piston, at that point the pressure is still 925 psi and the valve ACTS like a PCP valve (that held 36 cc @ 1007 psi) that closes at that point (except it is empty instead of closing).... It's a way of using what amounts to an air spring to store and release energy to change the pressure profile of the shot cycle.... which improves the efficiency....
Bob
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Look at diagram #3, the right middle chamber is at 2K and the firing chamber yet to be compressed at 3K.... You show air flowing to the left side of the middle chamber piston from the regulator.... When you move the piston to the right using the lever, the 3K becomes 6K in the firing chamber.... Where does the 2K air in the right side of the middle chamber go?.... It must be vented, or as it compresses it will prevent you from moving the piston to the right with the lever....
Bob
It fills up the other firing chamber.
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Not according to your drawing.... At point 3, both firing chambers have air in them, one is at 2K and the other at 3K.... Neither could be further filled from a chamber at 2K....
The left chamber was ready to fire at point 1, fired at point 2, refilled from the middle chamber and the regulator at point 3, and then compresses the opposite chamber to arrive at point 4, which is the mirror image of point 1....
Bob
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Not according to your drawing.... At point 3, both firing chambers have air in them, one is at 2K and the other at 3K.... Neither could be further filled from a chamber at 2K....
The left chamber was ready to fire at point 1, fired at point 2, refilled from the middle chamber and the regulator at point 3, and then compresses the opposite chamber to arrive at point 4, which is the mirror image of point 1....
Bob
Regulator doesn't flow into the firing chamber. It flows from regulator to mid-chamber to opposite firing chamber and out the barrel. If you take 3k in the mid-chamber and vent to the still compressed firing chamber, the pressure goes up to 2k in the firing chamber and down to 2k in the mid-chamber. Now is where the work begins for the lever. 785lb is the force it takes to move the pistons and squeeze the air out of the midchamber into the firing chamber. I'll do better drawing after I get some more work done and can spare the time to do it right.
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Sorry, still don't get it....
Bob
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Here is a drawing, ready to fire from booster A. Please pardon the bad drafting skills. Do not mistake arrows for anything that is supposed to be taking place. This drawing is static, everything is in balance. All airflow is one way through check valves (not drawn).
Drawing is not to scale.
Chamber diameter is 1in.
Larger booster diameter is 1.414in.
The larger booster diameter will need to be adjusted up to make up for where the lever mechanism attaches to be sure adequate area is exposed to 3000psi in the rear chamber.
From this point:
1) Open the valve to A, release 6000psi, and close the valve. Chamber A is now 0psig.
2) Turn the regulator on to mid-chamber A and off to B.
3) Open mid-chamber B to firing chamber A, and the pressure will equalize at 2000psi in both.
4) Press up on lever to pull up on A and press down on B. 758lb of force will overcome the imbalance and get the pistons moving. This forces the rest of the air out of mid-chamber B into firing chamber A.
5) Chamber B will now be 6000psi and A will be 3000psi. We are ready to fire again.
I hope this is more clear. Short of doing an animated gif, I'm not sure how else to explain it.
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OK, let me have a go, based on your drawing and explanation.... I'm figuring this out as I type, no idea what the results will be.... I am assuming the CR is 2:1, as the area of the firing chamber is (1/2)^2 x PI = 0.785 sq.in.... The top of the piston is twice that area (1.414/2)^2 x PI = 1.57 sq.in.... The middle chamber is the difference in those areas, which equals the area of the firing chamber, 0.785 sq.in.... All three have the same stroke.... All further calculations are based on those areas....
Top side of drawing (B) force on piston is zero (3K top and bottom, same areas top and bottom)....
Bottom of drawing (A) force on piston is also zero PROVIDING the pressure in middle chamber A is zero.... as follows:
force on top of piston is 3K x 1.57 = 4,710 lbs.
force on piston from firing chamber is 6K x 0.785 = 4710 lbs. (balancing the top of the piston)
If middle chamber A is at zero, the system is in balance, no force on the lever....
Now fire the gun, firing chamber A drops to zero, it will take 4710 lbs. to raise the lever....
Pressurize middle chamber A to 3K from regulator, stop regulator flow to middle chamber B but it is still at 3K, it will take 2355 lbs. to raise the lever....
Vent middle chamber B to firing chamber A, pressure will equalize to 2K.... at this point, the following applies....
Upper chamber B, top of piston 3K = 4710 lbs.
Middle chamber B, 2K x 0.785 = 1570 lbs.
Firing chamber B, 3K x 0.785 = 2355 lbs.
Net force side B = 4710 - 2355 - 1570 = 785 lbs. away from the lever (ie compressing the firing chamber)
Lower chamber A, top of piston 3K = 4710 lbs.
Middle chamber A, 3K x 0.785 = 2355 lbs.
Firing chamber A, 2K x 0.785 = 1570 lbs
Net force side A = 4710 - 2355 - 1570 = 785 lbs. away from the lever (balancing the force on the upper chamber)
As I see it, there is no force on the lever at this point, it should start to move easily, displacing air from Middle chamber B into Firing chamber A as it completes the stoke.... However, Firing chamber A is now sealed off, so to compress it takes whatever force is required to compress the air from 3K to 6K.... Once the stroke is completed, we have:
Upper chamber B, top of piston 3K x 1.57 = 4710 lbs.
Middle chamber B, 2K x 0.785 = 1570 lbs.
Firing chamber B, 6K x 0.785 = 4710 lbs. (balances the force from the top of the piston)
Force required from lever = 1570 lbs. (remaining force from the middle chamber towards the lever)
Lower chamber A, top of piston 3K x 1.57 = 4710 lbs.
Middle chamber A, 3K x 0.785 = 2355 lbs.
Firing chamber A, 2K x 0.785 = 1570 lbs.
Force required from lever = 4710 - 2355 - 1570 = 785 lbs. (away from the lever)
Total force required from lever to compress Firing chamber A = 1570 + 785 = 2355 lbs.
Look at it another way.... Since both pistons at 1.57 sq.in. and the pressure in the upper chamber is 3K, the force on the top always balances out, creating no load on the lever.... Therefore we only need to look at the middle and firing chambers....
At your starting point, with no pressure in mid-A, the system is balanced
After firing A, load to lift lever is 4710 lbs.
Open regulator to mid-A, off to mid-B, load to lift lever is 2355 lbs.
Open mid-B to firing-A, system is balanced, both are at 2K
Move lever to compress B, at the end of the stroke, the following applies:
Air has moved from Mid-A to Firing-B, but both are still at 2K, this force cancels out (same area)
Firing-B is at 6K
Mid-A is at 3K
Net force which must be supplied by the lever is (6K x 0.785) - (3K x 0.785) = 4710 - 2355 = 2355 lbs.
I get the same answer both ways.... In addition, Mid-B is still at 2K, whereas your premise at the beginning of the cycle was that Mid-A was at zero.... Therefore to get back to that condition, Mid-B must be vented to atmosphere.... In addition, Firing-A is only at 2K, but needs to be at 3K before the next compression stroke starts.... Not sure how you accomplish that....
Once again, this seems like a lot of effort.... This post took well over an hour....
Bob
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Bob,
Give me a day to process that. I don't think well when I'm tired. I'll process what you have done and see if I get the same results. I appreciate your scrutiny.
Chris