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Calculating efficiency???

Michael

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Staff member
Alright airgun geeks, I need your help with something. What is the equation to figure out how efficient an airgun is?

I've been using this equation but I believe it only gives me how much energy is produced per CC.

Reservoir: 300cc
Total air used: 250b starting pressure – 150b ending pressure = 100b
Total air used x total reservoir: 100b x 300cc = 30,000

Projectile: 25.4g @ 960fps for 47 shots
Ave energy per shot: 52fpe (70.5 joules)
Total energy: 70.5j x 47 shots = 3313j

Total air used x total reservoir / Total energy = Energy per CC
30,000 / 3313 = 9 joules per CC


Take for example another airgun with a larger reservoir, a smaller usable air range, lower shot count, and lower FPE. This results in more energy per CC.

Reservoir: 480cc
Total air used: 250b starting pressure – 180b ending pressure = 70b
Total air used x total reservoir: 70b x 480cc = 33,600

Projectile: 25.4g @ 940fps for 40 shots
Ave energy per shot: 48fpe (65 joules)
Total energy: 65j x 40 shots = 2600j

Total air used x total reservoir / Total energy = Energy per CC
33600 / 2600 = 12.9j per CC


But in my mind the 2nd airgun is less efficient because I'm getting less energy, less shots, less FPE, out of a larger air reservoir.

There must be a different equation to figure out "efficiency".
 
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Solution
In your example can you define what 12.76 "efficiency" means? I understand it's (Pressure used * reservoir size / total FPE). What does 12.76 mean in layman's terms?
It's a measure of how much air is consumed to produce one foot*pound of energy. In that particular format of "bar*cc/fpe", a smaller number is better. Meaning the gun is using less air.

I'm more in the habit of using fpe/ci instead. In this format, we're talking about how many foot*pounds of energy is being produced per unit of air consumed. So a bigger number is better which is a little more intuitive.
The second gun is less efficient because the air used had more energy. Not full formula but easy comparison. Multiply cc by reg set pressure.
Ahhh maybe I was just looking at the results backwards. So the lower the number, the more efficient it is. It should read "Total energy used per CC"?
 
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What is backwards is your units - you are dividing the air usage by the total energy, so the answers are in cc-bars / joule . . . . so the second gun IS less efficient as it is using more air per unit of energy . . .

You could flip the equation around and then you would have an answer in energy per unit of air, and that might "feel" better as a lower number would be less efficient . . . .
 
What is backwards is your units - you are dividing the air usage by the total energy, so the answers are in cc-bars / joule . . . . so the second gun IS less efficient as it is using more air per unit of energy . . .

You could flip the equation around and then you would have an answer in energy per unit of air, and that might "feel" better as a lower number would be less efficient . . . .
Can you copy, paste, and change the above to show that?
 
Here's a spreadsheet I've shared here before...you can back out the equations if you like.
https://www.dropbox.com/s/dwyfyxgj2lniwbi/efficiency.zip?dl=0

Screenshot:
efficiency.gif
 
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Maybe I'm not understanding your question but the equations are right there in the spreadsheet. The way it's laid out, the calculations are broken up into small pieces...no bewildering deeply nested parenthetical calculations.
Thanks, I didn't see the download link before. The cells have equations built in already!

In your example can you define what 12.76 "efficiency" means? I understand it's (Pressure used * reservoir size / total FPE). What does 12.76 mean in layman's terms?

@AlanMcD mentioned "air per unit of energy".
 
In your example can you define what 12.76 "efficiency" means? I understand it's (Pressure used * reservoir size / total FPE). What does 12.76 mean in layman's terms?
It's a measure of how much air is consumed to produce one foot*pound of energy. In that particular format of "bar*cc/fpe", a smaller number is better. Meaning the gun is using less air.

I'm more in the habit of using fpe/ci instead. In this format, we're talking about how many foot*pounds of energy is being produced per unit of air consumed. So a bigger number is better which is a little more intuitive.
 
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Solution
Whenever doing math, we have to remember to "follow the units." Efficiency can often be identified by most division equations with the word "per" in the description - such as Miles "per" Gallon: we measure the miles and the fuel used, and the result is an efficiency answer of MPG.

When I said "air per unit of energy", the "air" means air usage (or consumption in this case) - that is the Bar-cc, or bars of air "consumed", and the unit of energy is either FPE or joule, depending on what you are using. Total consumption divided by total energy.

One thing that often trips people up is the difference between what we could call "pure efficiency" metrics, like MPG, as compared to "pure consumption" metrics, like Gallons per 100 Miles (or liters/100 km in Europe), for any "efficiency metric" on power or energy utilization. If we work in what I will call "consumption space" then we can directly compare performance: a vehicle that uses half the gallons to travel 100 miles than another vehicle is clearly twice as efficient; but in MPG space that relationship is reversed - so for a vehicle to be twice as efficient, it has to double the MPG, leading to diminishing and non-linear returns. For example a 10 MPG improvement on a vehicle that gets 20MPG will save twice the fuel that the same improvement is to a 40MPG vehicle, even though it is 10MPG on both, but a savings of 1 gallon per 100 miles is the same savings on both.

I hope that helped, but "efficiency" can be complicated so you have to think through what is being divided by what . . . and what the answer means.
 
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It's a measure of how much air is consumed to produce one foot*pound of energy. In that particular format of "bar*cc/fpe", a smaller number is better. Meaning the gun is using less air.

I'm more in the habit of using fpe/ci instead. In this format, we're talking about how many foot*pounds of energy is being produced per unit of air consumed. So a bigger number is better which is a little more intuitive.
Not up on the whole calculation above...but...

The correct term (in red above) would be Pound-Feet (or foot).
For...whatever it may be worth.

Mike
 
Are these accurate descriptions:
fpe/ci translates to FPE produced per 1 cc of air
bar*cc/fpe translates to CC's of air consumed to produce 1 FPE

And, can y'all share some of your results so I can gauge efficient and inefficient?
bar*cc/fpe: how many cubic centimeters (cc) of air at standard atmospheric pressure (bar) it takes to produce 1 FPE
fpe/ci: how many FPE is produced per cubic inch (ci) of air at standard atmospheric pressure

I regard something like 1.3fpe/ci as being my basic expectation for most any PCP with a rifle-length barrel (say, 18 - 22")...meaning without having to do much to mitigate hammer bounce. Frequently 1.5 - 1.7fpe/ci with some effort. And for the most part, a gun hovering around 1.0fpe/ci will not be tolerated.
 
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I agree with nervoustrig's guidance, but would add that one also needs to consider the "class" of power a gun is working in as that affects basic efficiency - for example, if you are tuning for very high power levels in a given caliber, that will likely "hit" efficiency some. For example, if If you take a .22 caliber gun that has the flexibility to tune in a range of say 20-50 FPE, a good tune at 25 FPE will almost certainly be more efficient than a good tune at 45 FPE. Kind of like comparing sports cars to small sedans - the small sedans are almost always going to be more fuel efficient than the sports cars, but that by itself does not make them better (just more efficient).

Remember, efficiency is basically a ratio, and you can measure efficiency any way that works best for you. In my case, I measure all my guns and tunes with the standard efficiency values, but what I care about more is what I calculate as "Pellets shot per minute of Shoebox runtime," since I use a Shoebox compressor to fill my tanks. My higher power guns/tunes run about 5 pellets per minute, and my "most efficient" gun is my 10 M pistol at 89 pellets per minute. Interestingly, the FPE/CI of the two extremes are not massively different, with the 45 FPE .22 caliber B-Rod at 1.1 and the 3.5 FPE .117 pistol at 1.3. I have other mid power .22s that are up in the 1.6 range . . .
 
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Michael,

One might also calculate the tube for size.
Often the manufacturer is optimistic about the total cc. of only the tank size.
Shoot till the regulated pressure is reached, otherwise a big plenum wil give a different result then a smaller one.
Reaching regulated presure is different than the point of dropping speed.
Try to shoot the different guns in the same delta bar, 250 to 150 for example. (https://en.wikipedia.org/wiki/Van_der_Waals_equation)

It is in the details

Regards, Robert.
 
As @Veerkracht mentions about plenum volume and shooting below it...

Nervoustrig's calculator/spreadsheet doesn't account for shots off regulator, as for regulated guns, the spreadsheet/formula needs to include your plenum volume if your string falls below the set point, so its easy to get higher results when not adding this into the equation. So using that spreadsheet make sure to stay above your setpoint!

There is also volumetric and thermal efficiency. I personally prefer thermal efficiency! One can extrapolate here between volumetric and thermal efficiency...including where the energy is lost thermally. For those interested, the following information is provided courtesy of Domingo Tavella here https://www.researchgate.net/publication/326191398_Internal_Ballistics_of_PCP_Airguns and is something I was very vocal about (thermal efficiency and air's limited available energy) years before his wonderful contribution!

1699549229136.png


1699549311359.png
 
Agreed, for regulated guns, be mindful that a well-tuned gun will continue producing a stable velocity after the pressure has fallen below the regulator’s setpoint, at which point the plenum is contributing to the volume. You can avoid this source of error by stopping somewhere above the regulator’s setpoint. You just want to have enough shots recorded that the pressure has fallen by a meaningful amount (say, at least 500 psi), otherwise the poor resolution and accuracy of the typical gauge may represent another significant source of error.
 
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