How To Obtain The Maximum Power from a PCP Air Rifle (hard mag)

[shorty]

I am not seeing it the way you wrote.

I am seeing it as (with an open system-like a barrel) that the volume of air can not flow through as fast when comparing smaller to larger. Maybe the same thing.

Almost like how a small hole in a regulator works to refill the plenum. Huge droop and slow with small hole but less droop and maintained pressure with larger hole (like when coming off regulation).

There's something here that warrants investigation and all I can say is I am all in. I can't stand only having 35% to 40% of the guns potential power.

[shorty]

Bob,
I am not doing this to be a thorn in your kidney. It didn't dawn on me to run "your" max theory fpe calc with a barrel length that can adjusted to the dwell of a valve as you posted.

The numbers still don't add up (worse) and now I am thinking that I am chasing my tail due to your calculation.

Believe me, I don't have the answers but I (and a lot of people) believe/trust in your calculations to advance our knowledge of PCP's. Maybe you can explain this table for us. I used your (max theory) calculation and did some fpe/fps back calculations.

[rsterne]

Your calculations for the maximum theoretical FPE (Col.1) are correct, and include the FPE given to the mass of the air, and all losses.... They of course won't work if the mass of the projectile (and included air) goes to zero (which is impossible), as the velocity (Col.2) would be infinite.... That is why the number in Col.1 is a THEORETICAL MAXIMUM.... Of course in the real world, the air has mass, which accounts for the largest loss in a conventional PCP.... To give some consistency to the results, you should hold the velocity constant (I use 950 fps, so that the pellet mass is half the FPE)….  This "evens out" the playing field so that the pellet weight is a relatively constant percentage of the total mass and energy.... This, and other losses, are taken into account in my "lofty goal" formula which is based on 50% of the theoretical maximum, at 950 fps.... For example, with your 10" barrel, half the max. would be 61 FPE, using a 30.5 gr. pellet at 950 fps.... If you use a lighter pellet, it will be harder to reach 61 FPE because the air mass becomes a larger percentage of the total mass (leaving less energy available for the pellet).... If you use a heavier pellet, it will be easier to reach 61 FPE because the air mass is less important....

The easiest way to think of this is hold the velocity constant at 950 fps, and vary the pellet weight to get the FPE.... Since the velocity is constant, then for a given barrel length, the time spent in the barrel is constant, but more pressure can accelerate a heavier pellet and produce more FPE.... As an extreme example, my 6mm, 28" barrel at 4200 psi reached 173 FPE with a 73.4 gr. bullet.... but when I used a 1.8 gr. airsoft BB, which reached 2092 fps, the FPE dropped to only 17.5 FPE (1/10th as much)…. because the mass of the air absorbed most of the energy of the shot.... The "lofty goal" for that gun is 227 FPE, but that would require a bullet weighing 114 gr.... and I still would not likely be able to reach that.... Most of my "high powered" PCPs reach only about 45% of the theoretical maximum with the appropriate bullet weight.... If I ever reach my 50% "lofty goal", I will be a VERY happy camper....

HTHs....

Bob

[skorec]

My  (.177 /24” barrel ) Condor_clone  shots  13,43gr Monsters at 27 FPE  buy using  .22 transfer port from big air bottle reservoir at 1750 PSI .

My question is if the valve close before pellet is one half of barrel length. 

[rsterne]

I need more information to give you a better answer.... I need the reservoir or plenum volume, and the efficiency (FPE/CI)….

Bob

[skorec]

Bob,

0,45l bottle  pressure fall from 200bar to 180bar after  10 shoots   and  120 bar  reservoir/plenum is about  10cm3 .
.........

Sorry ,
I probably made mistake. There  was more shoots ( 20 ? ) . I need to check it again tommorow.

[rsterne]

Plenum is only 10 cc?.... About how long is the air path from the valve seat to the pellet base?.... and the approximate diameter of that?.... I am trying to get an idea of the "wasted volume" in the port system, that could have a big effect on performance with such a small plenum (big pressure drop)….

Bob

[skorec]

Sorry,
I have measured the plenum today but from outside only and it seams that it is about 20cm3. If I calculate energy myself/EU  whit ideal average 12Mpa and pellet career 0,3m the energy should be  3,14*2,25mm*2,25mm*12Mpa*0,3m= 57Jole/42FPE so the valve should to close a little bit sooner and/or  air transfer way from valve up to pellet (3,14 *0,35cm*0,35cm*5cm = 2cm3) is dropping  energy to 50J/37FPE.

..........
I have checked number of shoots again and calculate efficiency .

Air pressure at 0,45L bottle dropped from 2320PSI  to 2030PSI by shooting 19*37FPE  shots.
Internet calculator tell me  efficiency 0,9 FPE/cuin .
Why the result is at cuin when  I have  putted  cc   ?

[rsterne]

Your 42 FPE calculation for the "lofty goal" is very unlikely to be reached with a small plenum volume and a large transfer port system.... For example, the highest pressure the pellet can see, with a 20 cc plenum and a 2 cc "wasted volume" is not 12 MPa, but only (20 / 22) x 12 = 10.9 MPa…. In addition, the "lofty goal" calculation is for a "dump shot".... ie the maximum you might see when the valve stays open until the pellet leaves the muzzle.... even then, you are unlikely to achieve it....

I will run your numbers later today and see if I can come up with an estimate of when the valve is closing.... but I can comment right away that an efficiency of 1.86 FPE/CI indicates the valve is closing VERY early in the shot cycle, when the pellet is less than 20-25% of the distance to the muzzle....

Bob

[skorec]

I again made mistake se previous post.

soooory

[rsterne]

For your efficiency I get the following calculation....

0.45 L = 27.4 CI.... (2320 - 2030) = 290 psi = 20 bar.... Air used is (27.4 x 20) = 548 CI total

19 shots @ 27 FPE = 513 FPE total.... Therefore efficiency is (513 / 548) = 0.94 FPE/CI....

Lloyd's internal ballistics spreadsheet tells me that your valve is closing when the pellet is about 1/3 - 1/2 of the way from breech to muzzle.... If that is the case, you should be operating on the "knee" of the curve, and the plateau velocity would be just over 1000 fps.... Have you ever increased the hammer strike to find the plateau?....

Bob

[PikeP]

First calculate for pressure drop during shot. For you its (2320-2030)/19 * (450/20) = 343 psi. Divide that by half and subtract from initial shot pressure to obtain average. Which is 1578. Then calculate pressure drop from transfer volume. Using .75 cc I get 20/20.75 * 1578 and you get 1520 psi. Re run the lofty goal calculation and you get 37 fpe.  Hope this helps.

[rsterne]

Mike, I'm trying to follow your calculation, so bear with me.... The 343 psi is the total pressure drop in the plenum during a shot, correct?.... It starts at 1750 psi and ends at (1750 - 343) = 1407 psi so the average is 1578 psi.... However, at the pellet, the pressure is lower because of the transfer port volume (I agree, including the chamber volume, 0.75 cc is a reasonable estimate, the OP used diameter, not radius).... That results in a starting pressure at the pellet of (20 / 20.75) x 1750 = 1687 psi.... The pressure at the end of the pellet travel must include the barrel volume exposed by the receding pellet.... That pressure is 1407 psi.... Therefore the pressure seen by the pellet is the average of those 2 pressures, which is 1547 psi.... Other than that, I like your reasoning....

You are correct that the average pressure during the shot is the correct one to use for calculating the "lofty goal" FPE.... Using 1547 psi, we get (0.177 x 0.177 x PI/4 x 24 x 1547) / 24 = 38.0 FPE.... However, that would be for "bore size" porting, and will not be achieved, even with a dump shot, with smaller ports.... The ports in this gun are 0.138" (3.5 mm) and you can get an approximation of the likely FPE by multiplying the lofty goal FPE by (0.138 / 0.177) = 29.6 FPE.... This is quite close to what skorec has achieved....

We can approximate the pellet position at valve close by using the pressure at the end of the shot, which is 1407 psi.... If the expansion is isothermal (which it isn't), the total volume when the valve closes would be (1750 / 1407) x 20 = 24.9 cc.... However, the plenum plus porting system is 20.75 cc, meaning that (24.9 - 20.75) = 4.15 cc of that is the barrel volume behind the pellet.... The volume of the barrel is (.177 x .177 x PI/4) x 24 x 16.4 = 9.7 cc.... Therefore the pellet must be at least (4.15 / 9.7) = 43% of the way from breech to muzzle.... The reason I say "at least" is that assumes equal pressure throughout the system when the valve closes, which is not the case, there is always some pressure gradient in the barrel or the air could not flow.... also the expansion is partly adiabatic, not isothermal.... and both those factors would move the pellet closer to the muzzle at valve close.... However, it provides an interesting, if approximate, confirmation for Lloyd's spreadsheet....

I guess this method could also provide an interesting approximation of the pressure drop in the plenum that would occur if the valve closed exactly at pellet exit.... If we add the barrel volume to the plenum and transfer port volume we get (9.7 + 0.75 + 20) = 30.45 cc.... At the instant the pellet exits, with a "dump shot" the plenum pressure would be (20 / 30.45) x 1750 = 1150 psi.... That (1750 - 1150) = 600 psi drop in pressure would translate to a pressure drop of (20 / 450) x 600 = 26.7 psi in the bottle.... This compares to the current pressure drop of 15.3 psi for the gun as tuned.... If the pressure drop in the bottle exceeds about 30 psi per shot, then we can assume the pellet has left the muzzle before the valve closed.... Wasted air and a loud report would result....

These numbers are only approximations, of course.... because the expansion is likely closer to adiabatic than isothermal, and there must be some pressure gradient in the barrel.... but it provides another way of looking at what is happening.... Thanks for making me look at this from a different perspective.... always nice to learn something new first thing in the morning.... 

Bob

[skorec]

Ufff
I try to use now only   metric system whit Mike magic value 1520PSI

If I have putted  one shot  average pressure 1520PSI=10,5Mpa  to my calculation  3,14*2,25mm*2,25mm*10,5Mpa*0,3m=50J=37FPE answer is 0,3m .
or
If I have putted  one shot  average pressure 1547PSI=10,7Mpa  to my calculation  3,14*2,25mm*2,25mm*10,7Mpa*0,29m=50J=37FPE answer is 0,29m .
 My barrel has 0,635m.


Mike, Bob  Is it correct ? 

[rsterne]

I am confused.... In one place you said 27 FPE and in another 37 FPE.... For 19 shots at 27 FPE, the efficiency is 0.94 FPE/CI.... For 19 shots at 37 FPE it is much higher, 1.29 FPE/CI....

In addition, your calculation using the barrel length gives the MAXIMUM theoretical FPE, and the "lofty goal" is only half that.... This is why your calculations show a barrel length of only half what you have.... Does that make sense?....

Bob

[skorec]

Ufff my next mistakes . Really too much.

 YES 27 FPE is my ( Monster/13,4gr/950FPS lowest spring preload )  reality. Also internet calculator told me 0,93 FPE/cuin
http://www.calc.sikes.us/1/index.php

If I  put   one shot  average pressure 1547PSI=10,7Mpa  to my calculation   3,14*2,25*2,25*10,7*0,22=37J=27FPE answer is 2*0,22m=44cm . My barrel is  only 63,5cm.

May be it is the reason why my  China Condor valve  is too noisy. Much louder then SS balance valve in my Puncher at the same pover.   

Thanks for your time.

[rsterne]

I think you are looking at the problem the wrong way.... The "lofty goal" for your gun, with full bore area porting (ie 0.177" from valve to pellet), and maximum power (a dump shot) would be (0.177 x 0.177 x PI/4 x 1547 x 24) / 24 = 38.0 FPE.... However, your ports are only 0.138", so you can reduce that ESTIMATE to (0.138 / 0.177) x 38.0 = 29.6 FPE.... You are achieving 27 FPE, which is 91% of what you might reasonably expect.... IMO that is pretty good....

If you cranked the preload to get to the plateau, and managed to achieve that 29.6 FPE, you would likely be using about 26.7 psi or more per shot from the bottle, and the efficiency would be a dismal 0.58 FPE/CI or less.... That is why we don't tune guns to operate on the plateau.... and why the "lofty goal" is something you very likely will not achieve but you don't care anyways, because you would never tune a gun like that....

Bob

[skorec]

Bob,
I use next Airforce Condor valve and the  hole/port  is 5,5mm.
Unfortunatelly I do not understand ( 0.177inch x 0.177inc x PI / 4 x 1547psi x 24) / 24 = 38.0 FPE

[rsterne]

That is my "lofty goal" formula in Imperial units.... It is developed from the basic formula for work (energy)…. Energy in (ft.lb), abbreviated as (FPE) comes from the force applied and the distance it acts through....

Energy (ft.lb) = Force (lbs) x Distance (ft)

Force (lbs) = Area (sq.in) x Pressure (psi) …. and Area (sq.in) = Bore (in) ^2 x PI / 4 = Bore (in) x Bore (in) x 3.14 / 4 …. This is the same as radius ^2 x PI

Distance (ft) = Barrel length (in) / 12

Therefore Maximum Possible FPE (infinite reservoir, constant pressure, no friction or other losses) for your gun can be found from....

Max FPE = (0.177 x 0.177 x 3.14 / 4) x 1547 x 24 / 12 = 76 FPE.... However, the are many large losses, the greatest of which is that part of that energy must be used to accelerate the mass of the air itself.... and air under pressure is surprisingly heave.... HPA at 3000 psi has a mass of 60 gr/CI.... The mass of the air can be as much or even more than the mass of the bullet in some cases....

Therefore, for my "lofty goal" calculation, I use HALF of the formula for the maximum, so the divisor becomes 24 instead of 12 (in/ft)…. For your gun the "lofty goal" would be....

Lofty Goal = (0.177 x 0.177 x PI/4) x 1547 psi x 24" / 24 = 38.0 FPE

You can do the same thing in Metric if you watch your units.... which you seem to have done by using the bore in (mm), the pressure in (MPa), and the barrel length in (M)…. with the result being the energy in Joules.... You can either use the caliber (squared) and multiply by PI/4 = 0.785.... or you can use the radius of the bore (squared) and multiply by PI (3.14)…. Both formulas give the correct bore area in (sq.mm)….

Bob

PS, the effective port can never be larger than the caliber.... However, if ALL the ports before the pellet are larger than 4.5 mm (0.177") then your lofty goal would be 38.0 FPE.... Currently your ports are smaller, so the goal goes down because of that....

[skorec]

Thanks for explanation I didn`t know that:
1.     24”/12 = 2ft = full length of barrel
2.   Mass of pressure is similar as mass of pellet

Yes I need to do serious inspection of next item.