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Air Gun Home Forum Index » General Air Gun Questions and Topics » Data on 0.177 internal, external ballistics no terminal
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Data on 0.177 internal, external ballistics no terminal 
PostPosted: Sat Aug 18, 2018 10:44 pm Reply with quote
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Joined: 15 Aug 2018
Posts: 5
Location: California
I need data for the 0.177 pellet and/or round BB (5.14 gr) Daisy and/or Crosman such as projectile type, dimensions, mass, reservoir pressure and size, regulator set point, plenum size cc/ci, type and length of barrel, ambient temperature, latitude, elevation, barometric pressure, relative humidity, etc., the more the merrier.

I'm trying to device some simple equations to get comparable rule of thumb' estimates because it is practically impossible to get the full picture from a manufacturer, or dealer without it being hidden in a myriad of adjectives such as shoots fast, hits hard, goes a long way, it looks bad-a$$, or "I shot 10 rounds through one hole!" (with no other data given such as caliber/mass/velocity/range; maybe they meant that they shot the 10 rounds through the muzzle hole?), etc.

Kinetic Energy (Ke) = 1/2 x m x v^2 (three variables, ke/m/v, need two given in order to solve for the third; else, its infinite values with only one variable given unless you use whole integers then its a little less than infinite, (a slight of hand to insult our intelligence in order to get our hard earned monies)

One tactic I see quite often is where an airgun's velocity is given only, but without the mass of the projectile, or the kinetic energy (ke), and on another line the kinetic energy is given, but not with the mass of the projectile, or the velocity. Sometimes two values are given such as foot pound energy (fpe) and velocity in feet per second (fps), but obviously not from the same shot. I find that very deceiving to the point it gives me a bloody nose reading such deceptive insults.

I then calculate both fpe and fps values given by the publishers and their mix match best values of velocity and kinetic energy from obvious two different shot scenarios where one shot to get the ke the heaviest pellet was used, and the lightest pellet to get the highest velocity, and I end up getting impossible values because lead (Pb) begins to melt in the barrel starting at around 2,000 fps (>621 oF), and the calculations are showing an excess of 2K fps.

Though I have come across a very few that will give the mass of the projectile to get the velocity, or ke they publish, but that is a very few.

I seen a video by Loyd where he obtained 2031 fps from a 47.5 inch barrel firing a 9.1 grain aluminum pellet with a seal attached to the pellet and it calculated to a little more than 8% efficient, and that's with the transfer port at 100% (caliber) right behind the pellet. That shot translates that 92% of the energy is not lost, but converted to other than kinetic energy such as frictional heat energy, light energy (internal flash), sound energy, the ke carried by the compressed, expanding gas (air; 55 cc at 4,000 psi). I've seen that video hundreds of times and every time I come up with another question and calculation. I wish I had the time and money for such experiments. Thank you Loyd.

A mistake is wrong information given believed to be correct, but a lie is information given, or not given that is meant to deceive.

I have the theoretical data down, what I lack is the empirical data from my fellow air gun enthusiasts.

For example, I want to buy a CO2 BB gun that will shoot the fastest , but below 890 feet per second (fps) as efficiency is my second order of priority.

Now, on average, what I read on the world wide web is that you can get 20 to 60 "good" shots from a 12 gram (185 grain) liquid CO2 powerlet, but no data is given on the transfer port (TP) orifice diameter dimension, so I assume that if it is set at its 100% efficiency power dimension (caliber diameter), for 20 power shots; thereafter, dropping down along with transfer port diameter, and the shot count would then begin to rise to around 60 shots. I do not know, but guess it might be from 100% down to 50% of the caliber diameter.


Area = π/4 x d^2 = π/4 x d x d = π/4 x caliber (cal.) x transfer port (TP)

12 grams (g) of liquid CO2 = 12 x 15.432358 = 185 grains (gr) (rounded)

So, for 20 power shots at 592 fps for 4 fpe (80 fpe Total), 9.26 gr liquid CO2 in vapor form is needed per shot.


For a high 60 shot count at 382 fps for 1.67 fpe (100 fpe Total), 3.09 gr CO2 in vapor form is needed per shot.

So, as you can see, as the shot count goes up, the efficiency goes up too, but at the loss and expense of the high individual power shots, which is not what I want.

I need data to calculate for the sweet spot that I'm looking for, similar to calculating a ballistic coefficient (BC) where two velocities and known ranges are given, not withstanding that BC changes every fraction of an inch of travel and BC increases at range because velocity slows down; thereby, reducing the resistance drag of air, but an average can be calculated for the intended usage range.

What would be cool would be to get the geometry dimensions of the transfer port because the shape also affects the efficiency from 50% to 100% when expanding the equation such as venturi (75%n), streamline funnel (100%n), or just square corners (50%n). The trade off is the time (t) element.

Efficiency (n) is the power out divided by the power in (n = Pout / Pin).

Here's what the formula would partly look like, excluding the nitty-gritty stuff such as the myriad of resistances and their losses, time (t), air mass, gas molecule velocity for speed of sound (SoS), etc., etc.

Vapor CO2 (Temperature at 70 oF is approximately 853 psi), so:

853 psi x π/4 x 0.177 x 0.177 = 21.0379069 pound force (lb.f)

If the transfer port is at its widest then the transfer port efficiency (TPn) is equal to 100%, or 1.00 TPn, so this is how the equation would look like then:

853 x π/4 x 0.177 x (0.177 x 1.00n) = 21.0379069 lb.f; though, it looks like the same number, it is not because this similar looking number was obtained at 100%n (TPn).

From what I've seen throughout the ages, blogs, and experienced experiments, the efficiency revolves around the caliber, in percentage form. Any efficiency above this is due to more precision part fitment and mas/gas balances, but it also leads to a requirement of better cleaning, maintaining, adjustments, and repairs for premature worn parts such as pivoting metals and seals. An analogy would be to take an Indy 500 race car, which is not used as a pizza delivery car; although, I would order pizzas just to see the car drive up to my house, with extra cheese and peporroni.

Furthermore, any efficiency less than the caliber is due to looser part fit, friction(s), imbalances of mass/gas, and loss of sealing of the compressed gas, so lets calculate for that theoretical average off the shelf air gun with an efficiency of caliber, or 17.7% for my dream BB gun.

853 x π/4 x 0.177 x (0.177 x 1.00) x 0.177 = 3.715 lb.f

Note: the "1.00" is for a 100% efficiency (n) transfer port (TP) shape.

Now, lets apply a 12 inch barrel, but it has to be converted to feet to get foot pounds of force in the unit designation, and remember that there are 12 inches in a foot, so we have to divide our barrel length of 12 inches by a factor of 12.

So, a 12 inch barrel / 12 factor = 1 foot long barrel (please don't think that this is stupid and childish because try to calculate for an infinity barrel x pi. You would need a barrel longer than our universe and past the next dimension that information and calculation will be on another blog).

I will shorten the 0.177 x 0.177 x 0.177 in the formula to 0.177^3 for ease of reading and lessen the use of digital ink; in other words, I am multiplying the base (0.177) by three (3), which is the power or exponent.

Don't forget to read the instructions on your 99 cent calculator. A scientific calculator for 99 cents? Yes! At the 99 cent store. There went my dream job in manufacturing calculators unless I learn to speak Mandarin.

853 x π/4 x 0.177^3 x 12"/12 = 3.714

Again, though it looks like the same value number "3.714", it is not because it was obtained from a 12 inch barrel, or a 1 foot barrel, notice the added "ft." in the unit.

So, I want the most power, first, then the best efficiency, second, and in my opinion based on my calculations, using the laws of physics, thermodynamics, and mechanical statistics, I get pushing the projectile velocity to the magic speed of 890 fps at Std.P (15 oC, 59 oF, 101325Pa).

Even though, 891 fps would give you more power, the air resistance increases geometrically, so the minor gain is greatly offset by the major or extra power required to push the BB 1 more fps (singular foot and not plural feet), no matter how minute (as in small and not 60 seconds of time or angle), of the finite limited source (the reservoir, or PoVo), it is still a trade-off on the balance beam.

The analogy: It takes more work to lift one brick one foot higher when stacking on a 99 foot wall than it takes to lift one brick one foot higher on a 9 foot wall. But it is still the same mass and one foot difference?

On the other hand, 889 fps would give you more efficiency, the trade-off would be in the power loss, no matter how slight, and power comes first before efficiency in my order.

The analogy: I lost my view because there is a brick in my way, and I can't see the forest because the trees are in the way?

Accuracy (on target), I will not touch in this post, nor precision (group size) because they are both a science to themselves, and you need a bare minimum of 20 shots to have your study have any statistical significance; in other words, in this post I am shooting at a barn, on the second floor with the windows closed, and I am trying to knock it down with as few BBs in the shortest time as possible, from the inside.

Who cares if my BB "gun can't hit the broadside of a barn".

So, my objective is to get 890 fps with 853 psi of CO2 vapor pressure, a 12 inch barrel, and 3.715

Reverse engineering to get the velocity isolated using the exact value of gravity at sea level and 45 degree latitude.

sqrt(3.715 x 2 x (9.80665 / 0.3048) x 7,000 x (12/12) / 5.14 gr BB) = 570 fps. Too low...

"We're going to need a longer barrel." Modified from the Jaws movie. For my friends that never watched the movie Jaws, I changed "longer" from "bigger". Since I have combat induced tinnitus that is all I can remember that I heard.

Calculation calls for a 29.2 inch barrel, but there are so many other hidden variables involved (longer barrel = more frictional heat and volumetric imbalance losses) that is why I need empirical data.

To reiterate: I need all of your data so we all do not have to reinvent the wheel again, each, but move forward.

As you can read, I am not a millennial and find it impossible to write/text in short acronyms, and I hate emoticons, especially the lower right showing his
intelligent quotient (I.Q.).

Thank you
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Data on 0.177 internal, external ballistics no terminal 
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