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Old 12-29-2003, 01:45 PM #22
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Alright, being the type of person I am (lazy and a programmer) I wanted to try to automate the whole process of figuring out shots/fill, so I made an Excel spreadsheet. It looks like the math is right based on what you guys gave me. A 68/3000 tank with a full 3000 fill should get (at 100% efficiency) 1706 shots... with an realistic approx of 1499 when running at 195psi @ 300fps. If they're shooting at 280, it's about 1959/1721


The spreadsheet will be uploaded soon.

*edit*
the spreadsheet is up at: http://s90767649.onlinehome.us/downl...otsPerTank.xls or go to www.autocockerfaq.tk and look in the downloads section.
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Old 12-29-2003, 03:32 PM #23
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Very nice, thanks for taking the mystery out of the whole process.

Uh... Sticky????
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Old 12-29-2003, 03:40 PM #24
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The spreadsheet is great!!! Thanks for taking the initiative!
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Old 12-29-2003, 03:44 PM #25
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Hey, you might go post that xls sheet at the Tinker's Guild, it'd be a handy addition to the FAT.

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Old 12-29-2003, 03:51 PM #26
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Granted this is theoretical, but the process doesnt include the energy to cock the marker. What we have here is only the numbers for how many times a ball can be expelled from a marker with 100% eff. What it really is, as we all know, is much lower. So people reading this, dont think that you will still get very close to these numbers if you go and fine tune your cocker. It just doesnt hold. Unless CQ somehow figured that into his equations without telling us...
Anyways, good work boys.
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Old 12-29-2003, 04:01 PM #27
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you guys may want to download the excel file again. The original file I posted had a math problem in it where the lower you set the "Tank Fill PSI", the higher your Approx shots went. It's been fixed and is now correct.

If you'd rather just make the change yourself, you'll want to change cell B8 to "=((((B3-B5)*B1)/12)/0.73756)/F4" (without the quotes of course)

Dank: You're right, this does not include the energy required to cock the marker, only fire it. The excel file was never meant to tell you what you should be getting. This is the absolute best 100% efficiency, even then it's still going to be off. If I can figure out how much energy is required by the cocking mechanism, that would make it a bit more accurate.
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Old 12-29-2003, 11:15 PM #28
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Man, I go out of town to work for a little while, I come back and feel dumb. I've read something like 5 posts so far and 4 of them have math in them I'm a gasfitter for a living and I understand the gas laws and so forth. Doesn't mean I remember them. Or rather, want to remember them. Interesting facts.
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Old 12-30-2003, 01:02 AM #29
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You think this is bad, you should read some posts on AO in the Deep Blue forum. Some very smart people there IMO.
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Old 12-30-2003, 09:18 AM #30
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Quote:
Originally posted by MobiusMorte
Uh... Sticky????
No, but he's free to add it to the FAQ.

I did not add the pneumatics into my actual math, but they are part of the reason I said I was so sure that a case on those tanks was impossible. And it's actually a fairly easy process to figure how much air the pneus use... it's simple to calculate the internal volume of the pneumatics (to a relatively close approximation anyway, since the internal volume of the 3-way is tricky but small) and, given the cocking pressure, one can calculate how many cubic inches of gas at ambient pressure are used for each shot to recock the marker. Subtract that from the available tank air every shot, and shot count goes down accordingly.

I'll do it someday when I'm more motivated.

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Old 01-01-2004, 01:54 AM #31
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Quote:
given the cocking pressure, one can calculate how many cubic inches of gas at ambient pressure are used for each shot to recock the marker.
Did you mean ambient temperature here CQ? Since you already are assuming you know the cocking pressure and volume to calculate the actual moles of gas and temperature would play an important role. I'm basing this off of the usual

PV = nRT formula. So if we're trying to figure n (number of moles) we'd get....

PV/RT = n

Let me know if I totally misunderstood you
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Old 01-01-2004, 02:49 AM #32
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Ok, let me just say this, only guys from the 'cocker forum could bring in JOULES, and the physical state of nitrogen and compressed air under pressure in a confined area effectively to figure out something related to paintball. My Chemistry class HAS come into use because I understood what Tanner and Conqueror were talking about!!! The closest ANY other forum will come to this discussion is "psh, my setup runs 175psi and mows muppets down at 34bps with debounce at 4" good job guys
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Old 01-01-2004, 11:40 AM #33
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lol.

Gotta love the cocker forums. Not as much flaming in here either.
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Old 01-01-2004, 01:28 PM #34
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Quote:
Originally posted by Hatebreed13
The closest ANY other forum will come to this discussion is "psh, my setup runs 175psi and mows muppets down at 34bps with debounce at 4" good job guys
The closest discussion I've seen compared to this is in the Bushmaster forum where we were talking about lightening the valve spring to lower LPR pressure to pinch and increase efficiency. On the other hand, I don't spend tons of time in other forums.
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Old 01-01-2004, 03:17 PM #35
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Quote:
Originally posted by daishi
Did you mean ambient temperature here CQ? Since you already are assuming you know the cocking pressure and volume to calculate the actual moles of gas and temperature would play an important role. I'm basing this off of the usual

PV = nRT formula. So if we're trying to figure n (number of moles) we'd get....

PV/RT = n

Let me know if I totally misunderstood you
Good eyes, but in this case I actually did mean ambient pressure. You could also use tank pressure, but that's not a constant as the tank empties. Basically, the idea is this: 1) Tank pressure and cocking pressure are different, thus if we convert both into volumes of gas at ambient pressure we have an "equal playing field" to work with, and 2) Until now we've only been working with the JOULES of energy in the tank, and it's impossible to directly compare joules in the tank with cubic inches of air at a certain cocking pressure in the pneumatics. By converting BOTH into ci or L or whatever at ambient pressure, we can appropriately determine the relative amount of gas used by the pneumatics against how much air volume is available in the tank.

Make any more sense?

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Old 01-01-2004, 06:41 PM #36
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Ah, I see what you're getting at. I totally missed the fact that we were still discussing the energy instead of calculating concentration of gas in the tank. Once that is done then my post would make a bit more sense I guess. Thanks for clearing that up.
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Old 01-01-2004, 10:24 PM #37
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Sure thing. I was just trying to find an appropriate way to compare the "amount" of air used per pneumatic recock cycle to the total "amount" of air in the tank.

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Old 01-01-2004, 10:26 PM #38
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Oh, and btw (as I hinted before) - you're right to think of using PV=nRT, but that is the ideal gas law, which breaks down far below the tank pressures used for paintball.

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Old 01-01-2004, 11:30 PM #39
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Yeah, I wasn't sure at what stage the gas in a HPA tank would be at (near super critical or not), so I just broke out the handy dandy ideal gas law that absolutely never works in real life Out of curiosity, which equation would you use to figure out the relationship then? I don't want to fumble back through anyh of my general chem or pchem books unless I have obsolutely have too.
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Old 01-02-2004, 10:37 AM #40
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I'm not positive, but I don't think there's one equation that could neatly describe the situation. The critical point of nitrogen is 126K and about 492psi, so it's interesting to note that a blowback and even some mags running on nitro would actually have supercritical fluid in the VALVE CHAMBER, whereas a cocker would not. In any case, I think the proper way to calculate this would be to use Van der Waal's equation on the tank to calculate the number of moles of gas given the gauge pressure, ambient temp, and tank volume. That would give us the true number of moles in the tank, n. Once we have n, we can calculate the volume that n moles of gas would occupy at the firing pressure of the marker (say 200psi for a hypothetical cocker). Then we'd need a way of figuring out the volume of 200psi gas needed to fire a paintball ASSUMING 100% efficiency (not as hard as it sounds). The volume of n moles at 200psi divided by the volume-per-shot number would yield (I think) a more accurate idea of shots possible on that particular tank. Then we factor in the volume of air used per shot by the pneumatics per shot and try to find the point where the number of shots X would yield the equation X(shot volume) + X(pneus volume) = total tank volume. THAT is our theoretical max.

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Old 01-02-2004, 05:08 PM #41
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Another question to think about... how does the barrel affect this? I mean you said "Kinetic Energy at the muzzle = potential energy from the gas", right? If a 10" barrel with no porting shoots 280fps, then you stick on a 16" with lots of porting, you'll have to increase the air usage, but the equation doesn't factor this in, correct?
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Old 01-02-2004, 06:36 PM #42
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Quote:
Originally posted by FallNAngel
correct
lol. there is a point when the nerd in me wants to stop doing math. here it is. This is when engineers stop doing equations to figure out everything in a perfect world and start applying ideas to see how they work.

This may be my favorite thread ever. Right up there with the hot girls thread in small talk.
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