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Author Topic: What is a Reasonable FPE Goal for PCPs ?  (Read 394 times)

rsterne

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What is a Reasonable FPE Goal for PCPs ?
« on: August 26, 2017, 05:12:20 PM »
To be able to figure out what your new "Whiz-Bang" PCP can do, first of all you need to understand what the MAXIMUM POSSIBLE FPE could possibly be, with no losses.... That can be calculated relatively simply, as the "Energy Input", based on the available force, acting through the available distance.... The force is the bore area times the pressure... and the distance is the barrel length.... As always, we must watch the units....

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

The Force (lbs.) is the bore area (sq.in.) times the pressure (psi).... and the Distance (ft.) is the barrel length (in.) divided by 12 (in/ft)....

For example, let's use a .25 cal running on 3000 psi, with a 24" barrel....

Bore area = (0.25)^2 x PI/4 = 0.0491 sq.in. times 3000 psi = 147 lbs. force.... 24" barrel is 24 / 12 = 2.00 ft.... The maximum possible FPE this combination can produce is therefore 147 x 2.00 = 294 FPE....

Now before you jump all over me, remember that this would be the MAXIMUM POSSIBLE.... To achieve this, the pressure would have to stay at 3000 psi throughout the shot, which means the reservoir volume must be infinite.... In addition, there could be NO frictional losses, or any other form of loss.... That 294 FPE total would be divided into the energy in the bullet, and the energy in the air exiting the barrel.... and you would be shocked how much air weighs at 3000 psi, it is significant.... and any energy in the air is not available to accelerate the bullet.... To summarize, the maximum theoretical energy is calculated from....

Maximum FPE = Bore area (sq.in) x Pressure (psi) x Barrel length (in) / 12 (in/ft)

Pressure is pressure, regardless of the gas used.... This maximum applies for air, Nitrogen, or Helium....

Bob
« Last Edit: August 26, 2017, 09:33:35 PM by rsterne »


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rsterne

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Re: What is a Reasonable FPE Goal for PCPs ?
« Reply #1 on: August 26, 2017, 06:02:47 PM »
The next interesting point is that all PCPs with the same Barrel VOLUME will have the same potential FPE at the same pressure.... Look at the whole formula, and what happens when we rearrange it....

Maximum FPE = Bore area (sq.in) x Pressure (psi) x Barrel length (ft.) = Pressure x [ Bore area (sq.in) x Barrel length (in) ] / 12 (in/ft)

Since Barrel volume (cu.in) = [ Bore area (sq.in) x Barrel length (in) ].... we can substitute the Barrel volume in the above to get....

Maximum FPE = Barrel volume (cu.in) x Pressure (psi) / 12 (in/ft)

It doesn't matter if you have a .25 cal with a 4 foot barrel, or a .50 cal with a 1 foot barrel.... They have the same barrel volume, so the same potential FPE.... It doesn't matter which formula you use, they will both produce the same results.... Another, simple way, of looking at this, is the more air can be contained in the barrel, the greater the potential FPE....

Bob
« Last Edit: August 26, 2017, 09:35:33 PM by rsterne »
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rsterne

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Re: What is a Reasonable FPE Goal for PCPs ?
« Reply #2 on: August 26, 2017, 07:09:04 PM »
OK, so now you know what the theoretical MAXIMUM FPE possible is.... The big question is, what are the losses, and what percentage of that number might you actually be able to obtain, if you do a really good job?.... I'll give you the simple answer first, and then try and explain why.... If you can get 50% of the theoretical maximum FPE on air.... or 70-75% on Helium.... you have done a great job !!!

The biggest single loss in the system is accelerating the mass of the air (gas) itself.... Let's go back to our first example, a .25 cal with a 24" barrel at 3000 psi.... The mass of the air in that barrel is the barrel volume times the air density at 3000 psi.... The volume is (.25)^2 x PI/4 x 24 = 1.18 CI, and air at 3000 psi has a density of 238.6 kg/m^3.... That converts to 60.3 gr./in^3, so the air in the barrel weighs 71 grains, and somewhere about half that total mass will be travelling at the same muzzle velocity as the bullet....  :o .... Helium at 3000 psi has a density of 31.0 kg/m^3, which is only 7.84 gr./in^3, so the same barrel full of 3000 psi Helium would only weigh 9.2 grains.... only 13% the weight of the same amount of air.... This is the primary reason that you can get a LOT more power by using Helium.... Much less of the available FPE is lost accelerating the mass of the gas, so more is available to accelerate the bullet.... It's as simple as that....

If you have been following along so far, it may have occurred to you that the heavier the bullet, the smaller the percentage of that is the mass of the air.... This means that heavy bullets can extract a greater percentage of the maximum available FPE.... Therefore, to be able to come up with a percentage number that has a chance of working, we have to "fix" the weight of the bullet, in relationship to the FPE available to accelerate it.... Since if a bullet is travelling 950 fps, the FPE is twice the weight in grains, I chose that as a standard.... After looking at a LOT of high performance PCPs, I have found that the best of them can approach 45-50% of the maximum theoretical FPE on air, when tuned to shoot flat-out.... I don't have enough data on Helium to give you a firm rule of thumb, but about 50% higher FPE with Helium than with air seems reachable, from the small sampling of PCPs we have using Helium.... which would make the percentage using Helium about 70-75% of the maximum....

Let's go back to our example, where we calculated the maximum FPE as 294 FPE.... 50% of that is 147 FPE, achieved by pushing a ( 147 / 2 ) = 73.5 gr. bullet at 950 fps.... If you can do that with a 24" barrel, on 3000 psi, you have done a great job designing and building the gun.... I have never quite made it.... It always takes me a few inches extra barrel length, a bit more pressure, or a heavier bullet, to get to that 50% level.... So, if you manage it, pat yourself on the back for a job well done.... If you think about where the extra (50% of the) energy has gone, remember that about 71 FPE went into accelerating the air, so that leaves us with about ( 294 - 147 - 71 ) = 76 FPE (roughly 25%) of losses to friction, etc.etc.... I guess if you assume the same 76 FPE of losses in a gun running Helium, that might leave you with (294 - 9 - 76 ) = 209 FPE available for the bullet.... which is 71% of the theoretical maximum FPE.... as good a number as any to use for an estimate for using Helium....

OK, to summarize.... For our example, a .25 cal with a 24" barrel running on 3000 psi of air, the theoretical maximum energy is 294 FPE.... About 25% of that is lost to friction and other losses, another 25% goes into accelerating the air, and there is about 50% remaining to propel the bullet.... You shouldn't consider this as a "hard number", but as a "lofty goal".... one that if you can achieve, you can be very proud of.... Taking this 50% into account, changes the formula to the following....

FPE Goal (air) = Bore area (sq.in) x Pressure (psi) x Barrel length (in.) / 24 .... where the constant you divide by, 24, changes inches to feet, and includes the 50% factor....

Bob
« Last Edit: August 26, 2017, 09:37:57 PM by rsterne »
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rsterne

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Re: What is a Reasonable FPE Goal for PCPs ?
« Reply #3 on: August 26, 2017, 07:27:55 PM »
Now we get into the "conditions" I put on the above equation to predict your "lofty FPE goal".... One of those conditions is that the bullet weight be 50% of your resulting calculation, which means the velocity will be 950 fps.... For our example, 50% of 294 FPE = 147 FPE.... the bullet weight would be (147 /2 ) = 74 gr.... If you use a lighter bullet, at a higher velocity, you will have great difficulty approaching your 50% goal.... On the other hand, if you use a heavier bullet, that 50% goal will be easier to achieve.... Perhaps that means we can take that into account in our formula?.... Let's have a look (I'm going to do a bit of creative rounding)....

Off the top, we have about 25% losses, so that 294 FPE is reduced to ( 294 x .75 ) = 220 FPE available to accelerate the bullet and the air.... The air in the barrel is expanding, and being replaced from the reservoir, so for simplicity we use only half the mass of air in the barrel in our calculations, which is 36 grains.... The bullet weighs 74 grains, so the total is 110 grains, travelling at 950 fps at the muzzle, which is our 220 FPE.... OK so far.... Now what happens if we double the weight of the bullet to 148 gr.... We still have 36 gr. of air, so the total is now 184 gr.... That weight would have 220 FPE of energy at 734 fps.... However, the 148 gr. bullet would now have 177 FPE of that energy, and the 36 gr. of air only 43 FPE.... Assuming the losses don't change, the bullet could now have up to ( 177 / 294 ) = 60% of the theoretical maximum FPE.... This again, is not a hard and fast formula.... but it shows you what MAY be happening if you use a really heavy bullet.... If you exceed our 50% of maximum "lofty goal" by using a bullet twice as heavy as what the formula was intended for.... that is because that "your" lofty goal should really be 60% of the maximum instead of 50%.... OK, so now I just threw a monkey-wrench into those guys who claim they have exceeded my "lofty goal" formula.... Yeah, you know who you are....  ::)

Bob


« Last Edit: August 26, 2017, 09:40:11 PM by rsterne »
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rsterne

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Re: What is a Reasonable FPE Goal for PCPs ?
« Reply #4 on: August 26, 2017, 08:00:41 PM »
In order to have any chance of getting to the "lofty FPE goal", you will have to eliminate every possible source of lost energy.... The first thing is port size.... You will need full, bore-area porting at every point in the system.... The object is to have whatever pressure is in the reservoir available at the base of the bullet.... You need to eliminate every possible restriction to achieve your goal.... If a port is smaller in area than the bore, the flow will be reduced, and there will be a pressure drop.... Every time you turn the airflow, you introduce a resistance to flow, which is why "inline" valves (actually they are axial flow, there is no such thing as a "straight through" valve, at best they present a slalom course for the air) can produce more FPE from the same reservoir and barrel length.... What they are actually doing is delivering a bit more pressure to the base of the bullet, particularly late in the cycle when the air is rushing down the barrel towards the muzzle.... The valve throat, even in an inline valve, makes the air turn past some corners, so valve throats actually benefit from being about 10% larger in area than the bore, to reduce that effect.... Don't forget, all the porting INTO the valve needs to be even larger than that going out, to insure that full reservoir pressure is always available right at the valve seat, throughout the shot....

The second thing is to use a large reservoir or plenum, to minimize the pressure drop during the shot.... Using our example, where the barrel volume was 1.18 CI (19.3 cc).... Let's say our reservoir was 10 times that big, at 193 cc.... We fill it to 3000 psi, and fire the gun.... By the time the bullet gets to the muzzle, that 193 cc of 3000 psi air has expanded to fill the barrel too, and the total volume is now ( 193 + 19.3 ) = 212 cc.... and the pressure has dropped to just ( 193 / 212 ) x 3000 = 2730 psi.... and the average pressure to power the shot was only ( 3000 + 2730 ) / 2 = 2865 psi.... There goes 4.5% of your power out the window.... See how easy it is to not reach your "lofty goal"....

The third requirement for maximum power is valve dwell.... In order to maintain maximum pressure and acceleration all the way to the muzzle, the valve must stay wide open until after the bullet leaves the gun.... I call this a "dump shot", because it is VERY wasteful of air.... but it is the only way to get maximum FPE.... If you close the valve early to save air, and improve efficiency.... you will NOT reach your lofty goal.... So, in reality, this entire thread is taking about power you will not only find VERY difficult to achieve.... but you will not use in real life because to do so wastes an incredible amount of air....  ::)

Bob
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rsterne

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Re: What is a Reasonable FPE Goal for PCPs ?
« Reply #5 on: August 26, 2017, 08:40:03 PM »
OK, so what effect does a smaller port have in how much power your PCP can produce?.... After all, not many use full bore-area porting, that is pretty much reserved for bullet shooters.... For those shooting pellets, 75-80% of bore diameter is much more typical.... and sometimes much smaller.... Now in reality it probably doesn't matter what the "lofty goal" is if you are using smaller ports, because you are probably also using much less dwell to conserve air.... but it would be handy to be able to calculate a lofty goal for a PCP using less than bore-size ports.... Here is where hours of experience and testing, and keeping good records, pays off.... I can't justify this "rule of thumb" using mathematics.... all I can tell you is that I use this formula all the time, and IT WORKS.... or at least it predicts a maximum that is hard to obtain (the whole point of a "lofty goal")....

Instead of using the caliber squared to calculate the bore area.... I use the caliber times the smallest port diameter.... and then multiply that by PI/4 to get a "port compensated area" that the pressure can operate on to accelerate the bullet.... Yeah, I know the bullet is still the same size, and what is happening is that the pressure is dropping across the smaller port as the flow velocity increases.... but all I can tell you is that by using this formula, the results are amazingly close to reality.... So, here is the formula, make sure you use the smallest port diameter....

FPE Goal = Caliber (in.) x Port (in.) x PI / 4 x Pressure (psi) x Barrel length (in.) / 24 .... which simplifies to....

FPE Goal = Caliber (in.) x Port (in.) x Pressure (psi) x Barrel length (in.) / 30.6

If you prefer to simply calculate the "lofty goal" for a gun using bore-area porting, you can simply multiply that by the percentage of port size.... An 80% port will reduce the FPE to 80% of the goal for bore area porting etc.etc.... For many PCPs, that changes the 50% used in the lofty FPE goal formula to 40%.... For our sample gun, a .25 cal with a 24" barrel, operating on 3000 psi.... if you install a 3/16" port, the lofty goal would drop from 147 FPE to 110 FPE.... Yeah it will really take some doing to get that, but that is what your "lofty goal" is....  :o

Bob
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rsterne

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Re: What is a Reasonable FPE Goal for PCPs ?
« Reply #6 on: August 26, 2017, 09:08:06 PM »
As I write this, I realize that when you start compounding all the effects.... of smaller ports, lower pressures (less mass of air), and other things, it is possible that guys may be able to exceed the simple "lofty goal" formulas given above.... Maybe we should look at that.... Let's look at our example gun, a .25 cal with 24" barrel, at 3000 psi.... The FPE goal would be 147 FPE.... What happens if we reduce the pressure to 1500 psi instead of 3000?.... Well, for a start, the maximum FPE drops from 294 FPE to just 147 FPE because we only have half the pressure, so half the force available.... So, our "lofty goal" is now just 50% of that, or 74 FPE.... But now have only half the mass of air being accelerated as well.... Let's assume we still have the same 25% losses.... and the mass of air in the barrel at 1500 psi is 36 gr., so half that is 18 gr.... How much power is left to propel the pellet?....

Maximum FPE = 147 FPE.... 25% losses leaves ( 147 x .75 ) = 110 FPE.... at 950 fps, accelerating the air uses up 36 FPE of that, leaving ( 110 - 36 ) = 74 FPE for the bullet.... Well, the math looks like it is still working out.... pretty amazing, actually....  8)

Now let's drop the port size down to 3/16".... That leaves us with a goal of ( 74 x .75 ) = 56 FPE.... Reduce the barrel length from 24" to 20", (corrected for the lower mass of air) that drops again to 49 FPE.... which would be a 24.5 gr. pellet at 950 fps.... How does that stack up against a 25 cal. MRod with 3/16" ports, regulated to 1500 psi?.... Can you get 49 FPE from that, even with a slightly overweight pellet of 25.4 gr.?.... I think it would be tough.... but maybe attainable.... Using the simple 50% formula, 1500 psi and 20" barrel with 3/16" ports, gives 46 FPE.... still pretty tough....

How about a .25 cal MRod (20" barrel) at 2000 psi?.... Max. FPE = 164.... less 25% losses = 123 FPE.... Mass of air = 39 gr., that uses up 39 FPE, so we are left with ( 123 - 39 ) = 84 FPE for the bullet.... Let's use bore-size ports throughout.... Can you get 84 FPE at 2000 psi (a 42 gr. bullet at 950 fps) even with bore-size porting?.... I'm betting it will be pretty tough.... Even using the simple formula, 50% of the 164 FPE maximum - 82 FPE might be challenging with a 41 gr. bullet....

Anyways, there are some examples of how I would use these formulas to set a goal for a new PCP.... I guess goals are made to be broken.... but I think these are realistic, providing you follow the logic and guidelines.... Hope you have enjoyed the journey.... I am now opening the thread for questions, but PLEASE stay on topic....

Bob
« Last Edit: August 27, 2017, 11:06:51 AM by rsterne »
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