by B.B. Pelletier
This report is for Herb, who requested it several weeks ago. I’ll present empirical (observed) data on this topic–not theory. I have been testing airgun velocities since the first day I got my first used chronograph in 1994, and I’ve learned quite a lot just by watching.
Can pellets go supersonic?
There have been several arguments over the years that, since they are driven by compressed air, pellets cannot go supersonic. Let me lay that one to rest right now. I’ve shot thousands of pellets faster than sound–the most recent was during the test of the Career Infinity a couple weeks back. Yes, pellets do go faster than sound.
When I first started chronographing pellets, I was curious to see if the .177 RWS 48 sidelever could drive light pellets to 1,100 f.p.s. Those were the days before trick lightweight pellets, so we relied on RWS Hobbys that were lighter than they are today. I saw many shots in the 1,050 f.p.s. region, but other than obvious detonations, I didn’t see any that went 1,100 f.p.s. When you think about it, though, 1,050 f.p.s. is pretty fast.
Then came the 1250 Hurricane
The next big leap forward came when Gamo brought out the 1250 Hurricane, a breakbarrel springer they claimed could shoot 1,250 f.p.s. I was fully prepared for this gun to also shoot less than the claim, but to my surprise, it didn’t! In the first test, I got a velocity of 1,257 f.p.s. with a Hobby pellet. The average speed for the string was less than that–about 1,220 f.p.s. as I recall, but the gun really did achieve its advertised speed. And it was so much faster than the speed of sound that nobody who witnessed a shot could doubt it and keep a straight face. The supersonic crack was convincing evidence, because it always came well after the sound of the discharge.
And then the AirForce Condor
Then I went to work for AirForce. For the first 18 months, nothing big happened in the supersonic realm, but then AirForce owner, John McCaslin, came out with the Condor–a rifle that exceeded 1,200 f.p.s. in .22 caliber. In fact, I test-fired each one of the first 100 guns to ensure they were going faster than 1,250 f.p.s. with Crosman Premiers. We didn’t ship them unless they were because we knew there would be an army of new owners sitting out there with chronographs waiting to expose us if the guns didn’t perform.
We kept a ledger of those first 100 guns, some of which exceeded 1,280 f.p.s. with .22 Premiers. They were the fastest .22 caliber air rifles the world had ever seen, though by this time I knew that certain Korean rifles like the Career 707 could sometimes get a first shot out the spout at 1,270 f.p.s. But the Korean rifles couldn’t keep shooting that fast. They declined in velocity with each shot, while the Condor kept the first 20 shots faster than 1,200 f.p.s.
Believe it or not, it was nearly a complete year before anyone thought to test the Condor for the top velocity in .177 caliber. Heck–we were already over 1,200 f.p.s. in .22 with medium-weight pellets. How much faster did we need to go? However, the day finally came when AirForce wanted to know how absolutely fast the .177 could go, so I got tapped to test it. I already knew it would go over 1,300 f.p.s. with a .177 Hobby, but Skenco Hyper-Veloocity pellets were available then and they weighed over a full grain less than Hobbys, plus their bodies were plastic, not lead. Surely they would be the fastest.
And they were. I fired several 5.4-grain Skencos, and the fastest recorded shot went 1,486 f.p.s. Had I known at the time how big an issue this would become, I would have photographed the chronograph readout, but I didn’t. As of today’s date, 1,486 f.p.s. is still the fastest velocity I’ve seen from a pellet being driven by air, alone. I’ve seen plenty of shots powered by a fuel/air detonation that have wandered into the 1,700 f.p.s. realm, but those are shots from a firearm, by definition. When just air is the propulsive force, 1,486 f.p.s. is the best I have seen.
Gamo, again
A year went by and then Gamo brought out their Performance Ballistic Alloy (PBA) pellet, made from a non-lead casting metal plated with 18-carat gold. The Raptor pellet was the first PBA to hit the market, and it increased the velocity of almost everything it was used in. A year later, the Hunter Extreme came to market with the claim that it could shoot a pellet over 1,600 f.p.s. Actually, the gun bears a sticker that claims 1,650 f.p.s. Gamo even showcased the gun and pellet on Jim Scoutten’s Shooting USA television program, where the camera recorded the rifle shooting a Raptor through the chrono at over 1,630 f.p.s.
So, I tested a Hunter Extreme. To my surprise, however, it failed to achieve even 1,400 f.p.s. The best I saw was 1,395 f.p.s. I asked other Hunter Extreme owners to report their top velocities to me, and the best I got was a third-party report of 1,425 f.p.s. What was happening?
What happened is very clear. The rifle shown on TV was detonating! I can get the Hunter Extreme over 1,600 f.p.s. by introducing a light oil through the air transfer port. Anyone can. When the piston compresses the air, the heat of compression ignites the oil droplets, causing an explosion. But I challenge anyone to shoot the rifle for 50 shots AND THEN chronograph it! I’m pretty sure they won’t see velocities above 1,400 f.ps. with PBA pellets. I don’t know if I’ve convinced you about this or not, but I am satisfied that the Gamo Hunter Extreme is not capable of achieving 1,600 f.ps. with PBA pellets in a supervised and honest test.
Before we move on, I have two more comments. First, 1,395 f.p.s. (the fastest I have ever been able to get a Raptor to shoot in a Hunter Extreme on air, alone) is not a poor performance! It’s stunning! But it’s not 1,600 f.p.s. and people need to know that, so when some gun really does shoot that fast it can get credit for doing it honestly.
The second thing is that Gamo now advertises their .22 caliber Hunter Extreme as shooting 1,300 f.p.s. in .22 caliber, which they say makes it the fastest .22-caliber air rifle on the planet. That statement is untrue–plain and simple. I have shot several shots from .22-caliber Condors faster than 1,300 f.p.s. I did it years ago, before Raptor pellets existed in .22 caliber. I always tested Condors with 14.3-grain Premiers, so all it took to pass 1,300 f.p.s. (with some guns–not with all of them) was to drop down to a lighter lead pellet. It was no trick; the gun was always that fast. Who knows how fast it would be with a trick pellet?
Enter, Dennis Quackenbush
I got a call from Dennis Quackenbush late last year. He wanted to test maximum pellet velocities because of things people were saying on the Yellow Forum. He constructed a special testbed smoothbore gun in .25 caliber and .375 caliber. We tested both versions at the 2008 LASSO big bore airgun shoot in Texas, and I filmed the test. That video is provided for you in a new article on the Pyramyd AIR website. That article and video is the summation of this report.
So Herb, that’s my report. It’s as far as I’ve seen pellet velocity testing go to date. These numbers have all been witnessed–sometimes by over a million viewers, so nobody can argue they aren’t real.
My goodness… Gamo had the poor taste to wack a good-sized pig with their .177 Hunter Extreme – thereby implying that the gun could realistically be used for something it couldn’t.
What are they gonna go after with the .22 version? Deer? Water Buffalo? Bin Laden? I’m quaking in anticipation…
BB, I’m surprised you didn’t mention two things: 1. that detonation is BAD for the rifle (although many of us know this), and 2. Supersonic tends to throw accuracy out the window with a pellet, right? I’m curious when the companies are going to quit adding their “industry standard” to the velocity advertisements. Let me know if that will happen. JP
Morning B.B.,
Isn’t it primarly the shock of going supersonic and leaving supersonic that adversly affects the accuracy? My question is, how accurate are the pellets if they’re shot at a target before the leave supersonic? I hope this makes sense to you. Mr B.
Mr B
Drag is what stabilizes pellets. Air must flow over the pellet the right way to produce the right amount of drag . The shape of the pellet also influences how air flows over the pellet. We are not even talking about weight distribution yet.
There is a great deal of turbulence near the sound barrier. The air flow over a pellet changes and becomes less effective at producing the proper amount of drag. The pellet becomes less stable.
Keeping a pellet above the sound barrier for a short distance (they lose velocity fast) is not going to help much because because they are not properly stabilized at that time and will become destabilized even more by the turbulence as they drop back through the sound barrier.
You need to keep in mind that pellets have more in common with darts, arrows , and shotgun slugs than they have with bullets.
Rifling does not stabilize pellets…drag does. Drag does not stabilize bullets….rifling does. And you need to keep bullets away from the sound barrier too.
twotalon
Mr. B.,
You understand the accuracy problem with supersonic pellets. But the speed issue is not about accuracy. It’s about a number that sells guns to unknowing buyers–just as hypervelocity .22 long rifles sell to the general public.
B.B.
B.B.
Did you say hyperinaccuracy long rifles?
No kidding. Junk.
twotalon
twotalon,
School was a 2 hr delay this AM, snow, I was on first cup of coffee, not thinking and my brain missed change in air flow over pellet at supersonic speed.
Since a pellet is stabilized by it’s shape,is a smooth bore as inherently accurate as a rifled bore?
Mr B….
A rifled bore is inherently more accurate ….provided the rate of twist helps instead of hurts.
There are always flaws in the surface of a pellet that affect the symmetry of the air flow over a pellet. This causes the pellet to plane away from the original direction of travel. Rifling causes the pellet to rotate…therefore constantly changing the planing direction so that a tighter helical path to the target will be produced rather than having it sail off in one direction.
This is much like shooting two arrows…
One with straight fletch that is damaged, or the shaft is bent. It does not rotate, and the lack of symmetry causes it to plane off in one direction because of uneven air flow over it (drag).
Another with helical fletch..
Fletch may be damaged, or shaft not straight…but the air flow over the fletch causes the arrow to rotate and plane on a tight helical course to target.
twotalon
twotalon,
Thanks. The muddy stream is now a clear flowing brook sparkling in the sunlight.
Vince,
The video about Gamo shooting a pig is EXACTLY why I don’t own and NEVER will own any Gamo product. I agree that it is in the poorest taste and they should have never shown such a video to the public and give airgunners a bad reputation. I do agree that they will probably try a Safari hunt with their .22 version soon .
B.B.,
With all due respect, and I have a lot of respect for you, I must say that your velocity tests are interesting but without mention of accuracy I think they offer credibility to the misplaced emphasis on velocity alone especially for new airgunners doing research.
JP’s reaction to your article and the linked article (and video) are the same as mine. Total emphasis on velocity without any mention of accuracy.
It’s going to take an enormous effort to undo the “velocity without accuracy brainwashing” that so many advertising dollars have created in so many new airgunners minds.
If I understood your linked article correctly, Dennis Quackenbush’s motivation for the velocity testing was to dispell all the misinformation shared on the yellow forum about velocity of air. The purpose of home made delrin “pellets” was speed (velocity) alone to create a scientific basis to state fact to these VETERAN airgunners sharing misinformation on the yellow (misinformation on the yellow,surprise). I think velocity alone will be glorified in the minds of newbie airgunners and become THE priority for their purchase. IF I wanted to buy my first airgun and didn’t know better, after watching the video with all those veteran airgunners, with their custom big guns and all their fancy electronic gadgetry singing praises of a new record in velocity, I would run to the nearest store and buy a gamo hunter, 1,650 FEET PER SECOND!!!
This in turn financially rewards and validates the priorities of those manufacturers that design and build guns with only higher velocity as their ultimate goal.
Just my two cents.
kevin
b.b., so if I use the Gamo PBA in my low powered Slavia 630 it would seem I can expect to raise it from 510fps to 625fps or so?
In your opinion would I be able to retain the same accurace? At 10m the Slavia will put 5 shots through one ragged hole with the stock open sights, pretty impressive in my opinion.
And the $60,0000 question. I am plagued by magpies in my back yard. I live in an area where the nearest neighbour out back is 1/2 mile away. I’m only worried about the small garden I have which has a couple of birdhouses for songbirds whose little ones constantly go missing because of the magpies.
No more than a measured 40 feet. Would the Slavia with the PBS’s be enough to do them in. I know that with the existing accuracy of the gun I could easily make head shots at this range.
Or am I dreaming.
Thanks, CowBoyStar Dad
CowBoyStar Dad,
If you can put head shots on the magpies at 40 feet everytime with a lead pellet why change to pba pellets?
kevin
Kevin, just because I know that at 500fps, even at only 40′, they are not carrying much energy.
People have mentioned how thick skulled certain small animals are, I don’t wish these animals to suffer, but want a clean kill.
I’m confident on my shooting abilities…do mostly 10m shooting and do indeed have no trouble putting 5 shots in the 10 ring of a Crossman 10m target (about 1/2″).
Is it just me or is anyone else not able to view the video at the end of the linked article? Could be my computer I guess.
-Aaron
Aaron,
It’s just you cause it worked fine for me.
CowBoyStar Dad,
If you connect with the head shot, by by magpie. My Crosman 1377 kills crows with three pumps at 10 yards when I hit them in the head, every time. Trust me you will drop them dead on the ground every time!
Mr B.
Thanks Mr. B…exactly what I need to hear.
CoeBoyStar Dad
CowboyStar Dad,
No offense intended.
I should have elaborated. If you shoot .177 caliber gamo pba 5.4 grains with an ESTIMATED velocity of 625 fps you have 4.69 ft. lbs. of energy. If you shoot .177 caliber jsb 10.2 grains with a known velocity of 500 fps you have 5.66 ft. lbs. of energy. If you shoot the jsb 8.4 grains you have 4.66 ft. lbs. of energy.
The point I was trying to make and did I very poor job of is why buy the expensive, untested (for velocity AND accuracy) gamo pba’s if your lead pellets have relatively the same energy in a light pellet and more energy in a middle of the road weight pellet like the jsb 10.2 grain.
Sorry.
kevin
Well this could be considered to be part 2 of “How fast can an airgun shoot?”
/blog/2009/1/how-fast-can-an-airgun-shoot-part-1/
Think I’ve finally sorted out part of the conundrum.
“Ideal” gas means that the gas is governed exactly by the ideal gas law:
PV = nRT
Where P is the pressure of the gas, V is the volume, n the number of moles of the gas, R is a constant, and T is the absolute temperature. No gas is ideal. But over limited ranges the gas laws work well. Essentially you let R vary instead of being a true constant.
Now let’s consider a “pneumatic” type gun. How much work is done on the pellet? Assume:
* The rifle has a pressure reservoir connected through a transfer port to the barrel.
* There is a StarTrek force field in the transfer port separating the pressure reservoir and the barrel section.
* No air loss around pellet, thus pellet seals barrel perfectly.
* Vr is the volume of the reservoir (up to force field), and Vb is the volume of the barrel (on the other side of the force field).
* We assume that the gas expansion happens isothermally.
* Pi is the initial pressure inside the pressure reservoir; Pf is the final pressure in both the pressure reservoir and the barrel.
* We ignore any friction or viscosity of air.
* When the gun “fires” the force field is dropped and the air flows out of the reservoir and pushes the pellet down the barrel.
* When the pellet reaches the end of the barrel, the force field goes back up in the transfer port.
The force field in the transfer port thus acts like an ideal valve. If we assume that volume of the reservoir, Vr, is much greater than the volume of the barrel, Vb, then the final pressure in the barrel and reservoir would be almost Pi.
Now for a Benjamin Discovery the starting reservoir pressure is roughly 2000 lbs/inch^2
A 0.22 pellet has a cross-section of 3.14*(0.22/2)^2 = 0.038 inches^2
Barrel length 24 inches = 2 feet
So work done on pellet = 2000lbs/inch^2 * 0.038 inches^2 * 2 foot = 152 foot-lbs
Now let:
Wp = weight of pellet in grains
Sp = Speed of pellet in feet / second
Wp*Sp^2/ 450,400 = Ft*Lbs
– or –
Sp = Sqrt(ft*Lbs * 450,400/Wp)
So the speed of a 16 grain pellet would be about
Sp = SQRT(152*450,400/16) = 2,070 feet/sec
That is of course a “bit” faster than a real Benjamin Discovery shoots. Why is this wrong, or how do we need to refine the model? Using the “real” number for Vr and Vb to calculate the “exact” value for Pf would drop Sp a small amount, but nothing nearly enough to get the value in a realistic range.
Obviously under the initial conditions there is no momentum overall for the gas. It isn’t until the force field is released that momentum is created. The pellet/piston goes down the barrel and the “gun” recoils. Some small amount of energy is lost in the recoil of the gun, but that is very small too.
It would seem that there are a couple of “obvious” contenders for a much lower limit.
* It is doubtful that the valve in a real Discovery stays open long enough for the pellet to exit the barrel.
* The final speed of the pellet is much greater than the speed of sound. It seems like there should be some limit imposed by how fast the air molecules can travel down the barrel. First the air molecules have to go through the transfer port (which chokes the flow?). Second virtually none of the air molecules in the pressure reservoir are going as fast as 2,070 fps. In order to get the gas molecules moving faster, you would have to heat them. In fact the expansion of the gas would cool the gas, the exact opposite.
It seems to me that there is a very real limit imposed by how fast the gas molecules can travel down the barrel. Faster than the speed of sound would seem possible by a small factor, but 2,070 feet/sec would seem beyond what would be possible at normal room temperatures. There seems to be an interesting fluid dynamics problem here. My calculus is too rusty to go through all the exact derivations; hence I’ll not even try a mathematical derivation. Also if you did use calculus I don’t think the equations would be integratable, so numerical methods would have to be used anyway.
Herb
CowBoyStar Dad,
I can vouch for what Mr. B said. When I had my 2100b, 5 pumps and a head shot at 10 – 35 yards was always fatal. Some times, the crow, starling or dove would start to fly off only to crash down after a few wing beats. So, if you’re good with head shots at 10 – 25 yds and 500 FPS, go for it, no need for those buzzer PBA’s. The CP’s or Jsb ‘s will do just fine.
Bob C NJ
Herb – I just read your entire comment twice and the only part I understood was “StarTrek Force Field.” 🙂
-Aaron
All in all the point was that for a pneumatic rifle there are a number of ways to get a higher velocity for the pellet:
(1)making a longer barrel
(2)keeping the valve open longer
(3)increasing the reservoir pressure
(4)decreasing the weight (mass) of the pellet
The question is “Do you get to a point where the pellet is traveling so fast that the air molecules can’t ‘push’ on the back of the pellet to keep accelerating it?”
I think the answer is yes. Now the question is what is that limit?
The point of the Discovery calculation was to show that ignoring fluid dynamics (in other words ignoring fact that air molecules can only travel so fast…) the calculations predict an absurdly high velocity. Since that velocity isn’t observed, the reverse must be true. Fluid dynamics is a real factor in the ultimate limit for an air gun.
Herb
Herb,
I’m not sure I follow: You modeled the maximum power potential of the system and translated it to the output. What confuses me is whether your question (about why the real Discovery is slower than the ideal) is real or rhetorical. It seem obvious to me at least that your present formulation does not take into consideration any of the actual parameters affecting the system, e.g. valve duration, friction loss, energy to rotate the pellet, etc. By your calculations, the Discovery is perhaps 40% efficient compared to a system with no friction or other losses and maximum valve duration? That is about all we can conclude. Doubtless, fluid dynamics enters the picture for a perfect model, but until we see the results of an imperfect model integrating more known parameters, we don’t know how important it is.
RE: Discovery reservoir volume
Went looking for the “real” numbers. Jim Chapman reports that the Discovery reservoir is 8.3 cubic inches.
https://www.americanairgunhunter.com/benji_discovery.html
Ignoring any “dead space” in the transfer port.
24 inch barrel, 0.22 inches in diameter has a volume of:
24*3.14159*(0.22/2)^2 = 0.912 cubic inches
So:
Pf = 2000 * 8.3 /(8.3+0.912)= 1800 psi
So if valve did stay open until pellet exists gun, and gas pressure was always at equilibrium throughout the system, then the final pressure would be 1800 psi not the 2000 psi with which the Discovery had initially in the reservoir.
I’ll have to do the calculation, but I think this would drop the velocity something like 5% from 2,070 fps to something like 1960 fps.
Herb
PS – Think the limit imposed by fluid dynamics was something that BB recognized because of his experience without being able to do the “exact” calculations. If the Discovery had been designed for the fill pressure to 3000 psi, it wouldn’t have gotten (3000/2000) or 150% more velocity. Thus designing for the lower pressure evidently makes the gun a whole lot easier to use, without much of a sacrifice in the ultimate velocity.
This is also reason not to keep the valve open until pellet gets to the end of the barrel. You trade off a little bit of velocity increase for a whole lot more shots.
BB,
I’m reluctant to see airguns go the way high speed leads. Put a boat-tailed bullet into a high-powered pellet rifle and accuracy will not be the same issue at supersonic speeds as it is with a diabolo pellet. Unfortunately, the safety inherent in current pellet rifles also goes bye-bye. I guess this is why I, at least, am not excited by big-bores and bullet-like projectiles used in PCP’s. The whole speed race seems designed to attract idiots who shouldn’t even have BB guns, anyway.
Okay Herb, are you saying in a pneumatic rifle there has two be a definate ceiling as to how fast you can “push” a pellet because there is a definate ceiling how fast you can “push” air? That’s interesting.
-Aaron
P.S. I hope you don’t think I was being rude, I thought the Star Trek Force Field was very clever on your part!!
BG_Farmer
RE: other efficiency factors
I agree, but I think any of those factors are small compared to overall discrepancy. They do all add up the “wrong” way, in that they all reduce the overall velocity that can be obtained. So none of them cancel each other out.
Hopefully we can entice Jane into the conversation and she can shed some light on how to model the overall problem. I’m stumped. In chemistry you generally calculate based on equilibrium conditions. This isn’t an equilibrium problem, it is a problem in dynamics.
Herb
PS – Aaron, rude? absolutely not. My skin is thicker than that! I’m fascinated by such discussions. I always learn something which is the goal.
Herb,
The type of model that you are attempting to build will require a *lot* more math than you’ve so far included, and I speak as an aerospace engineer by training. Compressible fluids are notoriously hard to model accurately because they’re, well, compressible, so things break down pretty quickly. Also, the values that you have used change throughout the impulse event, meaning for example that the pressure behind the pellet changes as it moves down the barrel. That’s just to name one variable that changes over time. Don’t forget that the properties of the gas behind the pellet also change as it expands.
It’s been years since I attempted such a modeling challenge, so I’m probably too rusty for even back of the envelope type calculations any longer. The last time that I did do something of this sort though (calculating burn time needed to push a satellite from one orbit into another orbit), we simplified using lots of assumptions and still ended up writing brute-force computer programs to solve the many equations involved.
This is not meant to be a knock on your figures so far, which are pretty fascinating just an encouragement not to get too frustrated by the answers you’ve gotten. Shoot me an email if you’d like to discuss the work necessary to do a thorough job on answering modeling the theoretical maximum speed of a pellet.
RE: Different way to look at problem
Let’s say we have no pellet, but our Star Trek force fields at transfer port and end of the barrel. Now we release the force field at the transfer port. How long does it take the overall system to get to equilibrium pressure? It is going to take some finite amount of time for the gas molecules to travel down the barrel.
If the overall time to reach equilibrium is much much faster than the pellet exits the barrel, then the problem is not one of dynamics. If the time is roughly the same or longer, then the problem does involve dynamics.
Herb
Herb,
You can’t take the volume of the reservoir, you have to take the volume of the valve. I know from looking at them that the flow into the Discovery and RWS 850 valves are restricted (the 850 worse than the Discovery). The valve volume is less (a lot less) than the barrel volume.
I have had the same thought as you to try to at least see what the simple formula would work out to be without taking into account fluid dynamics – just to see what it looks like. Your calculations are interesting although I don’t have time to look at them closely right now. Perhaps I could give you some real numbers for the valve volume of the two valves I have if you are interested (that would have to be this evening).
Thanks,
.22 multi-shot
Bobby,
Yes, I’d really like to figure out roughly how the “right” calculations could be done. I absolutely was trying to do some “back of the envelope” calculations to “prove” that fluid dynamics was necessary to solve the problem.
There seem to be two significant factors. First just the speed of the air molecules, and second the transfer port in a real rifle chokes the gas flow. The transfer port thus serves as sort of a secondary regulator. Keeps you from wasting too much air. Without the transfer port (smaller in diameter than the barrel) you use a lot more gas but get a very very small increase in pellet velocity.
The first limitation is inherent in that, just starting with compressed gas, the temperature of the gas can’t be increased. The gas molecules are bouncing around only so fast.
The reason that BB correctly eliminated detonations is that they do raise the temperature of the gas. Thus the pellet can go faster.
It really seems to me that something in the range of 20% faster than the speed of sound would be the practical limit starting off at room temperature. About 68 degrees F, or 20 degrees C.
I don’t have an e-mail address for you, but my yahoo e-mail ID is hcacree if you want to discuss this off line.
I studied rocket motors since Jane and I discussed this in the past, and I think the correct way is to note that the pressure inside the barrel can be separated into two kinds of pressure. Dynamic pressure is the pressure of the gas going down the barrel. Static pressure is the pressure of the gas molecules bouncing off the barrel itself. Total pressure is given by the addition of the two, and in the case posed would be 2000 psi.
For a “real” rifle the ideal would be if the transfer port directed the flow of gas straight down the barrel so that the gas molecules didn’t bounce around on the barrel. This maximizes dynamic pressure. Even so, once the gas molecules get to the pellet, then they would bounce back hitting gas molecules coming down the bore. (not to mention bumping into gas molecules already in barrel behind pellet.) So there are probably pressure zones that are created as the pellet goes down the barrel.
I find it fascinating that such a “simple” problem can require exotic math to solve.
Herb
.22 multi-shot,
RE: Valve volume
Yes, I’d be curious. It isn’t going to change the numbers much. from BB’s third report on the discovery:
/blog/2008/1/the-benjamin-discovery-part-3-velocity/
I’d guess that a 16 grain 0.22 pellet would have a velocity of something like 840 fps. That is a long way from the 2070 fps of the initial calculation.
Herb
Herb, BobbyNations, et al,
Herb,
Your calcs of a “perfect” 100% effecient 16gn 2000psi 24″bbl Benji doing 2070 are right on. I calculated it a different way and got the same answer. So we’re either both right or both wrong, LOL.
I also found that in that “perfect” case, the pellet reaches supersonic (say, 1,180fps) about 8 inches down the barrel. But then, the air is compressed and therefor denser on both sides of the pellet. The speed of sound is faster in denser air. Way too many variables.
I’ll have to check with some of the other engineers here and see if anyone has fluid modeling software that will handle at least the flow of the air thru the barrel behind the pellet. The drag of the pellet in the barrel is probably a hopeless variable. I’m very curious.
Lloyd
Kevin,
The intent of this report, and of the article and video it refers to, is to explore just the velocity aspect of an air-driven projectile. Accuracy was not a consideration.
What we were trying to discover is how fast it is possible to propel a pellet of any kind with just air. That knowledge has no bearing on the usefulness of an airgun. It’s just raw scientific data.
But we have learned this from the experiment. That Gamo probably has never been able to shoot a pellet of any kind from an airgun to 1,600 f.p.s. without resorting to a chemical explosion. Also that Benjamin has never been able to propel a pellet to 1,500 f.p.s. with air power, alone.
Now, in the future when we all have those technical discussions about airgun speed and somebody brings up 1,600 f.p.s. and the Gamo Hunter Extreme, everyone who reads this blog will know it’s not possible and they will have this report to back them up.
It will be of no value in a discussion about what good useful airgun to buy, but it will serve as a cautionary safeguard against the blatherings of the lunatic fringe in our hobby.
That was why I reported it. Because someday I will be able to say, “Of course you know they are lying about the velocity. Airguns don’t really shoot that fast.”
B.B.
CowboyStar Dad,
As long as you remain at that low velocity the only way to know if the Raptor is accurate is to shoot them. Remember they were quite accurate in the Marksman 1010 I tested.
B.B.
RE: Friction of pellet
A first approximation would seem to be that the friction is independent of velocity. Thus the friction takes the same amount of work per inch of the barrel.
All things being equal it would seem that you could extrapolate the velocity/mass curve to zero mass for the pellet and get the “ultimate” speed that the gas molecules could create. For real pellets, there are other factors besides just the mass of course. So not sure if this would be practical or not.
I don’t have a PCP so I can’t get a pellet moving near fast enough to make an extrapolation meaningful for the data I can collect.
Herb
I just want to know where to get one of them there Star Trek force fields.
B.B.
RE: Energy lost to recoil of rifle
For recoil, The pellet goes one way, the rifle in exactly the opposite direction. Momentum is conserved, not kinetic energy.
so:
Wp – weight of pellet
Vp – velocity of pellet
Wr – weight of rifle
Vr – velocity of rifle
Wp * Vp = Wr * Vr
or
Vr = Wp * Vp / Wr
Wr = 5.13 pounds or 35,909 grains
Vr = 16 * 2070 /35,909 = 0.92 fps
Using calculator at:
/article/What_is_Muzzle_Energy_August_2003/5
35,909 grains at 0.92 fps is 0.07 foot-pounds of energy.
Insignificant loss…
Herb
Herb, BobbyNations, Lloyd, et.al,
If you don’t mind I find your conversations mind stretching and fascinating.Please keep them going here so we all can learn from them.
B.B.
I’d settle for the beam me up gizmo–take a hike OPEC.
RE: Obtaining Force field
When they make the next Star Trek movie maybe we can ask to use one of the holosuites to get data for our airgun experiment….
😉
Herb
RE: How far down barrel to reach 500 fps?
Seems like at velocities of less than 500 fps that the gas would act without any significant dynamic considerations.
16 grain pellet, 500 fps is 8.88 ft-lbs.
Distance = 8.88/(2000*0.038) = 0.117 feet
or about 1.4 inches!
So it would seem that in a pneumatic gun that the pellet gets moving fast very quickly.
Herb
Herb,
Don’t forget the math paper we saw (from CowboyStarDad). I know you were skeptical, but he came pretty close on multiple springers (a more difficult problem) without any real consideration of fluid dynamics and with fairly simple math. If that method holds up with errors less than 10%, I would say who needs fluid dynamics:).
BB,
I have a chroniton emitter in the barn somewhere, if you wan’t it its yours — but you have to move the old wood stove that’s on top of it:).
BG_Farmer,
I store a lot of dilithium crystals here at the house. Wouldn’t they become unstable in the presence of a chroniton emitter if there is any flux in the magnetic envelope?
B.B.
BG_Farmer
RE: Who needs fluid dynamics?
Yes, but…as I remember the kid assumed that the temperature of the gas in the springer was a constant. At that point I immediately stopped believing anything in the paper.
The kid was absolutely right in that the calculations get nasty, and an exact integration is not possible. So you end up using numerical methods. He introduced a lot of the calculations that would be needed.
I don’t want to disparage the kid, but I just don’t believe him. I’d want to see a lot more data plugged through his model before I’d believe it.
Herb
Herb,
Sorry about that, I’ve enabled my google profile so that you can see my email address now. Just in case, it’s bobby.nations@gmail.com. Oy, I may have signed up for more work than anticipated here, but what the heck, new hobbies are always fun 😉 I’ll give this more thought tonight and see if I can add to the conversation meaningfully. Now, where did I put all of my old textbooks?
BG_Farmer,
I missed the math paper that you reference. Could you point me towards a link for it?
RE: Paper
http://home1.fvcc.edu/~dhicketh/Math222/spring07projects/StephenCompton/SpringAirModel.pdf
Took me a while to find this again…
Herb
Bobby,
Herb got it (thanks, Herb). Its a school project, but I thought he took a good stab at a thorny problem.
Herb,
Perhaps so, although I think too much is made of the temperature spike inside a springer (in the absence of combustibles)…wouldn’t a PCP go the other way?
Herb,I have a couple questions:if a pellet starts moving in the barrel,won’t the friction cause heat,and won’t that heat have to be accounted for in pressure against the barrel? and won’t the flow through a valve greatly mitigate the ability of the air pressure in the resevoir to exert force on the pellet? also,how was recoil measured? I only ask because the duration is short,the distance is too,and the velocity is caught{by the shoulder or a pad} I only ask out of curiosity. no formal education,but eager to learn… FrankB
RE: the paper
Agreed that the author took a interesting stab at a thorny problem. I just think that there are some significant flaws in the modeling that negate the conclusions.
Page 4.
“The first assumption will now be made in this model. The assumption is that the temperature of the compressed gas within the internal mechanism remains constant.”
Well no…when you compress the gas with the spring piston it gets hot. He should have assumed that the gas was changed adiabatically – without heat loss/gain from outside.
page 6
He introduces 4 different kinds of friction which is plain overkill. What is does do is throw in more fudge factors that you can hide in the calculations.
Page 8
“Finally, only half of the output energy will actually by imparted to the pellet. Newton’s third law of motion clearly state that for every action there is an equal and opposite reaction – in the case of firearms this is commonly known as recoil. The summation of these two equal energy outputs, the pellet energy and the recoil energy, represent the total output energy from the system at any given moment during the internal energy conversion process also known as the firing sequence.”
Well no…momentum is conserved not kinetic energy. I’ve shown in the calculations for the Discovery that the recoil energy of the rifle is insignificant.
He also added factors that are probably insignificant. For instance, I don’t think that the rotational energy of the pellet amounts to much and could just be ignored.
How the author got the agreement that he did is a mystery. In general you make a mathematical model of the right “form,” then use some experimental data to calibrate it. You can then use the model’s calculations to make predictions.
All in all the discussion did make me realize more about the complexity of analyzing a spring piston gun. That was why I switched to analyzing a pneumatic. Thought the analysis would be simpler.
Herb
Herb,
assuming a “perfect anything” is the best way to solve a real problem.
I think it is possible that friction is a tremendously important factor. The pressure behind the skirt will tend to push it outward. It only takes a minuscule tendency of the skirt to deform outward before a significant pressure is developed between the rim of the skirt and the barrel.
If you assume a coefficient of pellet-barrel friction of 0.4, and you estimate the contact pressure between the skirt and the barrel, I think you will easily find that friction is very significant. Thus, that is another term in the energy equation.
There are other assumptions that are not totally accurate. One of them is the system being isothermic. Oh no! there is a lot of heat expelled out of the system.
Efficiency values in typical pneumatic or gas driven machinery are typically lower than 50% (they include friction and heat loss). Thus, I think that velocities less than 1,000 fps would be reasonable considering your over-2,000 estimate and some efficiency value.
Said this, I think it would be a pretty safe bet to assume there is a number of people that have developed thorough theoretical models for air guns that are calibrated based on real tests.
Ditch Digger…
FrankB,
RE: If a pellet starts moving in the barrel,won’t the friction cause heat,and won’t that heat have to be accounted for in pressure against the barrel?
Yes, friction causes an energy loss. Not sure exactly how you think about pressure against the barrel. The pellet would get hotter, and heat would expand it, thus more friction. All in all I think the overall energy loss to friction would be small. Pellets are not molten blobs coming out of the barrel.
RE: … won’t the flow through a valve greatly mitigate the ability of the air pressure in the reservoir to exert force on the pellet?
Yes, the design of the valve and things which affect air flow would have a significant effect. The turbulence would tend to create a choke point. It is a mystery if such factors could be ignored or not. Probably not in a “real” mathematical rifle model.
Herb
BB,
I spoke to 7 of 9 (not easy to get permission from my wife for that, and I don’t even ask about Captain Janeway anymore!)…this particular unit has a flux radius of only two meters, so you could store it fairly safely.
tunnel engineer
RE: Friction
I’d have to agree that friction should be considered. It is surely a factor, but I don’t think that it would cut the velocity in half. I’ll have to look and see if I can figure out what a reasonable loss would be.
Herb
BG_Farmer,
Okay, I’ll take it. I just have to get rid of a few hundred tribbles to make some room.
B.B.
There is a formula in the plans for the caselman automatic rifle that can be used to reasonably predict pellet velocity assuming that your volume of the gas you use is enough to push the pellet right up untill its exit (ideal). The formula is:
V=sqrt(A*P*W*G*2B)
A=bore area in inches
P=pressure in PSI
W=number of bullets per pound
G= acceleration constant of 32.2
B= barrel length in feet
The example that Caselman used for that carbine:
sqrt((pi*(.31/2)^2)*3000*(7000/100)*32.2*(2*2))*.7
Where Pi*(.31/2)^2 is the area of the bore of the .31 caliber weapon
3000 is the PSI of the weapon
(7000/100) is the number of 100 grain bullets in a pound
32.2 is an acceleration constant
(2*2) is the constant two times the length of the barrel in feet (2 feet)
and .7 is used to account for losses of the system (70% efficient)
I found playing with this equation gives numbers that match up pretty well with known quantities given by various setups.
Enjoy,
Ben Benson
B.B.,
Glad you are thick skinned and took my comment in the light it was intended. Re-read my comment from early this morning and it seemed a little prickly. Not intended.
What an amazing discussion today. I didn’t realize the depth of your audience. You certainly have an educated, passionate and loyal following (with one exception). I don’t understand half of it so I guess I’ll go shoot and not worry about it.
kevin
Ben
RE: Caselman Formula
Exactly what I did…the terms just arranged somewhat differently.
Herb
Kevin,
You and me, both!
B.B.
RE: Friction
Coming at this sort of backwards again…
For the calculation of the energy of the Discovery, there is a tremendous amount of energy involved. To get some idea of just how much, let’s look at how much energy is required to MELT the lead pellet. The 16 grain pellet at a velocity of 2070 fps has 206.45 Joules of energy. At a velocity of 840 fps the energy is 34.0 Joules. So we’re “missing” 172 joules of energy.
— Lead —
Specific heat capacity 26.650 J/mole/Degree Kelvin
Melting point 600 degrees Kelvin
Heat of fusion 4.77 kJ/mole
Let’s convert this to grams…
Lead atomic weight 207.2 g/mole
Specific heat (26.650 J/mole/Degree K)/(207.2 g/mole)
= 0.1286 J/g/Degree Kelvin
Heat of fusion = 4.77 kJ/mole
= 4770 J/mole /(207.2 g/mole)
= 23.02 J/g
So assuming we starting at 300 degrees Kelvin, we have to heat the lead to 600 degrees kelvin (300 degree difference) and then melt it.
Energy = 0.1286*300+23.02 = 61.60 J/g
Now there are 15.43 grains in a gram so the 16 grain pellet weighs a bit more than a gram. So it would require 63.9 J to heat pellet from room temperature and melt it. Since the pellets don’t melt, I just don’t think that the overall amount of energy lost to friction could be great enough to account for a significant amount of the inefficiency.
Herb
And I do realize that we could try to figure out how much of the heat goes to the barrel and how much to the pellet. Also as pellet gets hot, then more of the heat would go to the barrel.
Just think we’re “missing” way to much energy for friction to be the most significant cause.
Herb
Can anyone help me with this darn flux capacitor? I can’t get back to the future.
I can hardly wait for you guys to tackle a springer and try to model how the turbulence caused by the piston compressing the air affects the ultimate pressure able to be placed behind the pellet! I think you folks will soon be onto Chaos Theory. I just don’t have the patience or mental capabilities to do this type of stuff anymore and marvel at the folks that do!
Anyway, I just wanted to give BB another kudo on the Leapers scope base. I finally assembled everything back together on the RWS350, mounted the scope and prepared to sight it in again. None of the settings needed to be adjusted thanks to the precision manufacturing of that base together with the Weaver rings. You and Leapers really did a first class job here.
OK,
Looking at the caselman equation with the discovery:
sqrt((pi*(.22/2)^2)*2000*(7000/14.5)*32.2*(2*2))*.7 = 1522 FPS.
~81 FPE
That is with and IDEAL valve, but also taking into accouOnt losses due to friction, etc.
I don’t know about the Discovery valve, but judging by the pictures they appear to be based on the 22xx series valves. If the 22xx series are any indication, that that valve is releasing less a maximum useable volume for the barrel length of the gun (this is why you can almost double the power of a 2260 with valve mods – 11.5 FPE to ~22 FPE)
Taking that into account, I think this equation is very close to a maximum number on this gun – at least it is not completely out of the question. You might only be able to get 1300-1400, but still you can chalk that up that difference to that 70% fudge factor for inefficiency.
Ben Benson
Ben,
Maybe I’m dense, but what is the acceleration constant G in your equation? The similarity to local acceleration due to gravity (g) threw me for a loop. I like the “efficiency coefficient” — I thought of that earlier today, when I first saw Herb’s post, but its the kind of thing you have to be an “authority” to do:).
Herb,
1) abdiatic heating: he did make that assumption and disregarded the impact. I haven’t calculated the impact, either, so I can’t say whether or not that simplification has any significant effect.
2) 4 kinds of friction: really just three, as rotational friction with rifling is shown to be zero. Normal gravitational friction in this case is probably also not worth worrying about, but he did.
3) 1/2 on page 8: I’m still thinking. I see your point, but I’m not sure he’s wrong, maybe just not stating it clearly. Oh, how it hurts to think:).
Herb,
I agree that friction losses and energy conversion to heat are only a small percentage of the inefficiencies. I think the bulk of the “apparent” losses have more to do with airflow through the valve and transfer port, and with the air simply not being able to keep up and maintain full pressure behind the pellet as it accelerates down the barrel.
Most of the energy seems to be imparted to the pellet within the first 12 or so inches, but after that, the velocity gains drop rapidly. I won’t say that the efficiency drops off with the square of the MV, but it is more drastic than a straight linear decrease.
Back to your original 2070fps theoretical from the .22 Benji, the pellet reaches 500fps in about 1.4″, 1000fps in 5.8″, 1500fps in 12.6″, and then uses the last half of the barrel to get the additional 500fps. BB has often said that, at least with springers, barrel lengths over 12″ don’t add much additional velocity.
In reality, lets say the Benji reaches 950fps, I think the pellet reaches 500fps in about 6″, 700fps in 12″, and gains the last 250fps in the final 12″ of the barrel.
My testing has always shown, given a specific pressure and bore dia, the heavier, slower pellets produce more energy than the fast light pellets. Theoretically, given a dia and pressure, the energy (MV-squared) should be the same despite pellet weight, but I have never found that to be the case. The heavier pellet invariably yields higher ME.
Just a couple more cents into the pot.
Lloyd
B.B.
Excellent report sir!! As always, I enjoyed reading it.. Take care…
Cheers,
Jony
Fred,
Quite a road you’ve traveled with that RWS 350.
I had an RWS 54 (scope killer) and until B.B.’s base came out I was almost ready to run over that gun. The UTG leapers base turned was instrumental in making that gun usable.
You’re fortunate that your follow up “no droop” base came out as quickly as it did. I waited 4 months beyond the original target date to get mine last summer.
I’m glad everything worked so well for you. Keep those mount screws, rings screws and stock screws tight and let us know how well that gun can group when you get to know it better.
Good/safe shooting to you.
kevin
Fred,
The word “turned” does not belong in my post. Should have re-read. Sorry if it caused confusion.
kevin
BB and Kevin
DUUUHHH gee,I thought I could just manipulate a lever to produce compression of air then trip the trigger lever to release the pressure through a ported valve and that would accelerate my chosen projectile through the bore into my intended target.didn’t know I had to be a math whiz to do it right:)
Guess I’d better shoot for more enlightenment.
Sorry I couldn’t resist cause this threads got me so lost I’ll need a shovel to dig my way through:)!
Happy formulations everyone
JTinAL
P.S shouldn’t we hire Mr. Whorf to guard the Transport from Ky. to Tx.?
Last word veri. was CANDUNC
must have been aimed at me cause I read it as CAN DUNCE SEE
JTinAL
Herb,
Lloyd reminded me of something else. When pressure goes down (due to valve being closed and volume of barrel expanding), it might be possible to reach a point where the pellet is decelerating due to bore friction. A similar thing happens with powder rifles: velocity increases with barrel length up to a point, then declines. A simple point, I know, but a real world possibility.
JTinAL,
After reading 100% of the comments and comprehending 20% I feel like I’ve already been transported to another, more evolved, planet. Guess I need to get out more.
kevin
Ben,
We’re having a violent agreement…
You had:
V=sqrt(A*P*W*G*2B)*Efficiency
you then substituted 7000/Wp to get number of pellets in a pound.
I had:
V=sqrt(A*P*B*450,400/Wp)
Thus the biggest difference is the 70% efficiency factor which I did not have. Looking at the equations quickly It appears that you used 14.5 grains for the pellet weight, where I had used 16.
Herb
Herb,
I meant he seemed to know about adiabatic heating, but chose to disregard its impact on the calculation. I thought his reasoning was sensible for doing it that way, but that is of course up for debate. Shouldn’t affect a PCP, anyway, right, or do you want to assume adiabatic cooling…way beyond me:).
PS. I will write out “adiabatic” 500 times as penance for my mispelling:).
BG_Farmer:
RE: “…but chose to disregard its impact on the calculation.”
That is my point. I think he made the wrong decision. The compressed air will get hot. Hot enough to significantly raise the air pressure.
Herb
Great, now that my curiosity’s piqued looks like its time to get out the integral calculus textbook and the machinery’s handbook……heheheh
(and maybe a webapp later…..mmmmm)
Ben Benson
JT,
Wish we had transporters up, but they haven’t functioned since the ice storm:). Imagine: we could all just pop in to talk/shoot airguns with BB during breaks. I have a feeling Mrs. BB wouldn’t let me wear my boots in the house though:). My friend in Huntsville (ironically he’s a rocket scientist) said you all made it through the cold spell, but did you get any of that snow?
BG_Farmer
The little snowstorm that could,Didn’t:)
Don’t know how but we only got ~2in.
while my uncle 2mi. down the road got 4 and my nephew farther so. near Birmingham got over 6 in.strange but O.K with me:)
Maybe if we brought tea and cookies and lots of ammo Mrs.BB would let us loiter on the porch while BB gathered up the toys for us to play with.THat would be great to be able to pop in for a while when we could arrange a little time then be home before bedtime.Makes the imagined future almost as appealing as the past.
OH well at least we’ve still got this great site to play on:)
JTinAL
Kevin & B.B.
Can I come shoot with you?
I'm afraid I just read 80 "odd" posts (and I don't mean the count) and I'm thinking I don't know how to think..
I wish I could say I found it interesting or enlightening..
Only about 8 miles over my head!!
But 80 comments is a good thing, and maybe someday I'll get what it was all about..
I second the idea we focus on telling the story of accuracy and where it happens and where it doesn't happen… and it doesn't happen very often over 900fps with a pellet..
It's good to disprove their claims about speed, but more important to tell the truth about accuracy!! IMHO..
And someday will have that discussion too, I hope..
Wacky Wayne
Ashland Air Rifle Range
Herb,
From your calculation, the barrel has a volume of about 0.912 inches^3 (0.038 inches^2 * 24 inches). The approximate volume of the Discovery valve is 0.1357 inches^3.
What blocks the flow to the Discovery valve is the small hole through the pressure gauge adapter.
.22 multi-shot
Ok, here is the CLOSEST thing I could find in an old physics book.
The following uses Bernoulli’s equation which doesn’t fit our case because we have a compressible fluid and there is NOT a steady state flow. Anyway, here it goes…
Bernoulli’s equation is
P1 + p·g·y1 + 1/2(p·v1^2) = P2 + p·g·y2 + 1/2(p·v2^2)
where
P – is pressure in Newtons/m^2
p – is fluid density in grams/cm^3
g – is the amount of work done by gravity in Newtons
y – is y position in meters
v – is the velocity in meters/second
This equation is for a two ended system, each end represented by one side of the equation. This describes the case of steady-state, incompressible, nonviscous fluid flow.
Example:
A cylindrical tank 1.2 m in diameter is filled to a depth of 0.3 m with water (p = 1000 kg/m^3). The space above the water is occupied by air, compressed to a gauge pressure of 1.00 x 10^5 N/m^2. A plug is removed from an orifice in the bottom of the tank having an area of 2.5 cm^2. What is the initial velocity of the stream which flows through this orifice? What is the upward force experienced by the tank when the plug is removed?
From Bernoulli’s equation,
P1 + p·g·y1 + 1/2(p·v1^2) = P2 + p·g·y2 + 1/2(p·v2^2) (equation 11.5.32)
In this example, the velocity v1 may be neglected because the area a2 is so much smaller than a1. The pressure P1 is equal to 1.00 x 10^5 N/m^2, while P2 is zero; also y1 = h, while y2 = 0. Putting all this into equation 11.5.32, we find
P1 + p·g·h + 0 = 0 + 0 + 1/2(p·v2^2)
v2^2 = 2/p(P1 + p·g·h) = 2/10^3[10^5 + (10^3)(9.8)(0.3)]
= 205 m^2/sec^2
v2 = 14.35 m/sec
To find the initial upward thrust on the tank, we may note that before the plug is removed the initial momentum of the system is zero. Since there are no exteranl forces acting on the system (tank plus water), the momentum remains zero after the plug is removed. But in time interval dt, the liquid acquires a negative y-component of momentum equal to
d·p”liquid” = -v2·dm (equation 11.5.34)
where dm is the mass which squirts from the orifice in time dt. This quantity can be expressed as the product of the density times the volume emitted in time dt, the latter in turn being equal to the area a2 times the distance the stream travels in time dt, which is v2·dt. We may write, therefore,
dm = p·a2·v2·dt
Substituting this into equation 11.5.34, we find
d·p”liquid” = -p·a2·v2^2·dt
The momentum imparted to the tank is just the negative of this, in view of the fact that the total final momentum must be zero. But the momentum imparted to the tank in time dt may also be equated to the impulse of the force acting on the tank, according to the impulse-momentum theorem. We may write, therefore,
d·p”tank” = p·a2·v2^2·dt = F·dt
whereupon it is evident that the force F must be given by
F = p·a2·v2^2 = (10^3)(2.5 x 10^-4)(14.35)^2 = 51.5 N
I’m afraid the rocket scientist’s are correct – we’re up a creek without a paddle.
.22 multi-shot
Howdy Bobby,
Nice to find another ol’ Tennessee boy here. I see you’re near Memphis. I grew up in and near Cleveland, about 30 mile NE of Chattanooga…near the base of the Smokies. Great place to grow up for a boy who loved to hunt…mostly upland game.
Regards,
Joe
Wayne,
I had no idea when I posted this report that it would become viral. But in retrospect, it’s always a fundamental topic that sets everyone off, isn’t it?
Gotta do more of them.
B.B.
Trying to tie up some loose ends…
(1) BG_Farmer said…
Ben,
Maybe I’m dense, but what is the acceleration constant G in your equation?
Accelaeration due to gravity – 32 ft/sec/sec
Comes into equation because most physical relationships are dependent on somethings mass not its weight. You weigh less on moon because the moon has lower gravity, but your mass stays the same.
March 03, 2009 8:14 PM, Blogger Lloyd said…
…My testing has always shown, given a specific pressure and bore dia, the heavier, slower pellets produce more energy than the fast light pellets. Theoretically, given a dia and pressure, the energy (MV-squared) should be the same despite pellet weight, but I have never found that to be the case. The heavier pellet invariably yields higher ME.
Agreed my modeling doesn’t account for this observation. Seems like this is where friction would come into play. The extra mass plus the friction allows more pressure to build up behind the pellet before it starts to move.
In the article that the kid did, he was right to divide the friction into two parts. Not sure if I’d call forming the pellet to the barrel friction, but the effect is there. It does take work to mark the lands into the pellet.
The second part of the friction is the work to push the pellet through the barrel.
March 03, 2009 8:43 PM, Anonymous BG_Farmer said…
…velocity increases with barrel length up to a point, then declines.
Agreed, that should happen. This is another factor where friction comes into play. With a barrel a light-year in length friction wins and the pellet doesn’t make it out of the barrel. Because of the long length of the barrel (and now very very large volume), the gas pressure drops to near nothing.
March 04, 2009 1:25 AM, Anonymous .22 multi-shot said…
Thanks for the volume of the valve. The extra dead space has to be added to the equations. Drops the pressure just a bit.
—
Been fun guys!
Herb
Herb,
Got it. I thought it looked like g.
“
From your calculation, the barrel has a volume of about 0.912 inches^3 (0.038 inches^2 * 24 inches). The approximate volume of the Discovery valve is 0.1357 inches^3.
What blocks the flow to the Discovery valve is the small hole through the pressure gauge adapter.
.22 multi-shot”
Ok, then fire up your drill press and bore that puppy out!!
😀
Ben Benson
RE: How fast?
Stumbled across this thread in yellow…
"So what IS the maximum theoretical muzzle velocity for an airgun?"
http://www.network54.com/Forum/79537/thread/1226513246/1226699539/So+what+IS+the+maximum+theoretical+muzzle+velocity+for+an+airgun-
Which lead to this thread…
Gas Flow through a tube
http://www.physicsforums.com/archive/index.php/t-200607.html
and this paper by Seigel, 1965
http://www.dtic.mil/cgi-bin/GetTRDoc?AD=AD475660&Location=U2&doc=GetTRDoc.pdf
The answer is 1640 fps for air. The gist is that ultimately for an infinitely small (weight) pellet that the gas has to push itself.
Trying to wade through all this now, but thought the find was interesting…
Herb
Herb,
That’s not true for spring piston air guns, because the air is compressed so rapidly that it heats the air and changes the dynamics completely.
Ben Benson
Ben
Absolutely correct.
But isn’t it amazing that there is a fundamental limit for pneumatics and it is tied to the speed of sound? And Seigel figured it out in 1965 !!
And the limit does exist for springers. The higher temperature of the air means that the speed of sound is faster. You’d have to pick a temperature.
Herb
Herb: right
It stands to reason if you had powerful enough force driving a piston, you could compress the air to pressures seen in firearms, and subsequently get firearm power. Now, such a gun would likely be unreasonably large or mechanically complicated (huge mechanical advantage required to cock the gun)
Determining the pressures here would require some calculus to determine the pressures between two moving pistons (the one generating the pressure, and the pellet acts as a piston).
Ben Benson
Ben,
RE: “if you had powerful enough force driving a piston, you could compress the air to pressures seen in firearms…”
Yes and no. The 1640 fps is an absolute limit. Just pressure won’t get you any faster. If you had Jane’s car spring in your springer, then when the gun fires not only does the pressure go up when the piston comes slamming through, the temperature is also increased. Since sound travels faster at a higher temperature, the springer could theoretically go faster than 1640 fps.
I think this also kills the notion of a PCP stably pushing a diabolo pellet going faster than the speed of sound. You can’t get far enough above the sound barrier to have any reasonable range. A 1640 fps diabolo pellet is just at the upper end of the transonic region. It’s velocity would very quickly drop lower.
Herb
Ugh…
RE my comment: You can’t get far enough above the sound barrier to have any reasonable range.
Drag out the wet noodle and flog me. Make that “You can’t get far enough above the speed of sound to have any reasonable range.”
Herb
RE: Yellow forum equation
The other formula that is in the yellow forum is:
v = 172 * sqrt(Q/(m+Z*Q))
where:
v = velocity of pellet
m is pellet wt in grains.
Z = gas density grains per cubic inch = 0.011 grains for air at room temp
Q = PSI x barrel_length x caliber x caliber
This equation would allow you to calculate from first principles he maximum velocity for any pneumatic gun from some basic parameters for the gun and pellet.
On yellow forum ray Carter noted:
Obviously, a max is reached when m=0. When that happens, we get
v = 172 *sqrt(1/Z)
doing the simple arithmetic, that comes out to 1640 fps.
Steve in NC contributed:
Since 1640 is a theoretical limit, you can never get all the way there.
However, to achieve a (slightly) more practical (?) 1600fps in .177 with CPLs (7.9gr) and a 24″ barrel would require 19,000psi.
Meanwhile…
1550fps would require 8000psi.
1500 would need 4900psi.
1450 could be done with 3400psi.
But 1400fps would need only 2600psi.
You could go faster than 1640 fps if you used a different gas other than air. For example Helium. The speed of sound in Helium is nearly three times as fast as the speed in normal air.
Herb
RE: Quackenbush 0.375 rifle
Couldn’t help myself.
Plugged numbers from BB’s new article:
/article/Do_changes_in_air_pressure_affect_pellet_velocity/51
into equation.
Assume fill pressure of 3000 psi for the rifle.
Delrin
V = 172 * Srt( (3000*24*0.375^2) / (15.5 + 0.11*(3000*24*0.375^2)))
V = 1536
for PB using 74 grains the calculation is 1271
V(observed-delrin) = 1381
V(observed-lead)= 1041
All in all that seems like pretty good agreement. We’ve neglected friction, and the fact that a “real” valve was used without dumping the whole reservoir pressure.
Herb
Herb,
Looking at your original formulation, I started to think you can’t assume that 152 fp are transferred to the pellet, only 76fp, since equal force is exerted on the pellet and rifle. Note this does not mean that recoil energy is equal to pellet energy, but they are proportional according to the ratio of their masses. Anyway, I’m sure you can tell me why not, but the result comes out to a much more reasonable figure for Discovery.
EG:
Now let:
Wp = weight of pellet in grains
Sp = Speed of pellet in feet / second
Wp*Sp^2/ 450,400 = Ft*Lbs*0.5
– or –
Sp = Sqrt(ft*Lbs*0.5 * 450,400/Wp)
So the speed of a 16 grain pellet would be about
Sp = SQRT(76*450,400/16) = 1460 feet/sec
Trying the same thing for QB’s speed trial:
ft*lbs = 294
;3000 psi
;.25 pellet, 74 gr. lead
;barrel length assumed to be 24″
sp = sqrt(ft*Lbs*0.5*450,400/Wp)
sp = sqrt(147*450,400/74)
sp = 945
Maybe the barrel is longer — BB might know — than 24″ in this case:
ft*lbs = 367.5
;3000 psi
;.25 pellet, 74 gr. lead
;barrel length assumed to be 30″
sp = sqrt(ft*Lbs*0.5*450,400/Wp)
sp = sqrt(147*450,400/74)
sp = 1057
It breaks down entirely for the 7 gr. pellet, which is not surprising.
Anyway, I hope you don’t mind my monkeying around with it. I think the YF formula incorporating the density of air is more likely to be useful also.
Herb,
Not that it matters, but in the 30 inch QB barrel calculation, I forgot to change 147 to the proper value when I cut and pasted.
BG_Farmer
RE: “I started to think you can’t assume that 152 fp are transferred to the pellet, only 76fp, since equal force is exerted on the pellet and rifle.”
No!!! You’re splitting the kinetic energy. Momentum is conserved, not kinetic energy. The pellet goes one way, the rifle the other as you noted. Momentum is the product of mass times velocity.
I did the appropriate calculations in post:
At March 03, 2009 3:00
RE: Energy lost to recoil of rifle
I fudged a bit and used weight instead of mass. But since I used weight on both sides, it works out fine. What my calculations ignore is that pressurized gas is ejected out of the barrel too. It has some momentum. So the rifle will actually pick up a bit more momentum. Even doubling the 0.07 ft-pounds of energy amounts to an insignificant amount of energy compared to the 152 ft-pounds.
Herb
RE: Calculation on Discovery
The “problem” is that the energy can be increased beyond limit. Boost pressure from 2000 psi to 4000 psi, to 10,000psi and so on. Obviously the “real” Discovery couldn’t handle 10,000 psi, but no upper limit is the problem.
Using the yellow formula:
V = 172 * SQRT( (2000*24*0.22^2)/(16 + 0.11*2000*24*0.22^2))
= 1286
Herb
Herb,
I think that when you calculated foot-lbs., you calculated a force, not kinetic energy. Force is equal on pellet and rifle, I’m confident of that. I’m also confident that this doesn’t violate conservation of momentum or disagree with your recoil calculations. The kinetic energy imparted to the pellet is proportional to the kinetic energy imparted to the rifle according to the ratio of their masses.
I agree that no upper limit is the problem.
BG_Farmer said…
RE: "I think that when you calculated foot-lbs., you calculated a force, not kinetic energy. Force is equal on pellet and rifle, I'm confident of that…"
I've done so many calculations that I'm a little lost about which calculation you think was in error. I'm assuming that you think the problem was in the momentum calculation.
At 3 pm on Tuesday I had:
Wp – weight of pellet
Vp – velocity of pellet
Wr – weight of rifle
Vr – velocity of rifle
Wp * Vp = Wr * Vr
or
Vr = Wp * Vp / Wr
Wr = 5.13 pounds or 35,909 grains
Vr = 16 * 2070 /35,909 = 0.92 fps
Correct?
Now I swizzled things around a bit and used the calculator at the PA website to calculate foot-pounds of energy for the rifle.
KE = 1/2 * Mr * Vr^2
I had velocity, but to use KE equation I'd have to convert weight in grains to mass using the gravitational constant. Thought that would be more confusing than helpful. So I just "cheated" and used PA calculator which hid the details of the calculation.
As I said, I "fudged" a bit in that I used weight not mass. But since the gravity factors is on both sides of the equation, the result for the rifle's recoil velocity is correct.
Wr & Mr are weight and mass of rifle
Wp & Mp are weight and mass of rifle
g = gravitational constant
Wr = Mr * g
Wp = Mp * g
So if you substitute in my "momentum" equation you get
(Mp * g) * Vp = (Mr * g) * Vr
or after dividing both side by g:
MP * Vp = Mr * Vr
which is in fact the "right" form of the equation.
RE: "Force is equal on pellet and rifle, I'm confident of that…"
Not really. Pressure*Area is a unit of force. Let's plug the barrel and now shoot the gun. The pressure becomes homogeneous very quickly. The barrel has a lot more surface area than the plug so the barrel has a lot more FORCE on it than the plug. In this situation force isn't like the gravitational force where both objects experience the same force, but in opposite directions. The force comes from the gas which isn't part of the rifle or the pellet.
Now the pellet has essentially all the work done on it – Force*Distance. The rifle doesn't recoil from the force, but from the momentum of pellet.
Does this make sense now?
Herb
Wp & Mp are weight and mass of PELLET (not rifle)
Herb
Herb,
I’ve tried to wade back through the sources for the 1640fps max air velocity in a tube, and have not found anything very convincing. Bits and pieces, but no real derivation. Where did that constant of 172 come from?
Did you find anything that really convinced you that 1640fps is the theoretical maximum speed? And with a zero weight pellet, that 1640 is independent of bore, pressure and barrel length??
Big bore Bob Dean got 1474fps pushing a Delrin slug of 15.5gn(I think) thru a .375 cal x30 or33″ barrel. The yellow forum formula says that no matter how much more pressure you add, the max vel OF AIR will only go up another 166fps. Hard to believe.
Think about it.
Lloyd
The following information was added to today's blog (Thursday) but I'm adding it here so the basic information is in one place.
I wasn't smart enough to figure the relationship out. The researcher was Dr(?) Arnold E. Seigel of the United States Naval Ordnance Laboratory.
http://www.dtic.mil/cgi-bin/GetTRDoc?AD=AD475660&Location=U2&doc=GetTRDoc.pdf
The most pertinent part is section 11, which starts on page 32 of the pdf file, page 17 of the text. The limiting case is presented in equation 11-8.
—–
Basically the limit comes about because gas molecules have to get their energy from the gas itself. There isn't anything pumping additional energy into the system.
See simulation of gas molecules at:
https://en.wikipedia.org/wiki/Kinetic_theory
Maxwell-Boltzmann Distribution
https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution
see section on distribution of speeds.
Air is about a "molecular weight" of 29. So in between Ne and Argon in the figure.
Also section on typical speeds. note relationship between the most probable speed, the average speed, and the rms speed.
See also Speed of Sound:
https://en.wikipedia.org/wiki/Speed_of_Sound
Note section "Speed in Ideal Gas and in air"
C(ideal) given in terms of SQRT(Gamma*R*T/M)
where gamma is 1.4 for air.
This makes speed of sound smaller than all of the three other quantities – the most probable speed, the average speed, and the rms speed.
The 172 comes from grinding constants for air into a more basic equation. I'm looking into this now.
I am convinced that this is a real limit.
Herb
RE: Equation
March 05, 2009 2:52 PM, Anonymous Herb said…
RE: Yellow forum equation
The other formula that is in the yellow forum is:
v = 172 * sqrt(Q/(m+Z*Q))
I pointed out before that on the yellow forum Ray Carter noted:
Obviously, a max is reached when m=0. When that happens, we get
v = 172 *sqrt(1/Z)
But the same thing happens when the pressure containing term , Q, moves towards infinity.
v = 172 * sqrt(Q/(m+Z*Q))
divide numerator and denomination in sqrt term by Q
v = 172*sqrt(1/(m/Q+z))
as Q -> infinity, then m/Q goes to 0 and drops out.
Herb
Herb,
Thanks for including the link to Seigel’s work again. When i tried to download it earlier today I thought it was a bad link because of the slowness. This time, it was definitely worth the wait. Incredibly fascinating, but I will have to be more awake to appreciate it.
I see how there is a definite velocity upper limit to a precharged system, although I have yet to convince myself of the number of variables that can be canceled out. As you said, all of the energy must come from the compressed gas itself, because no additional energy is being introduced after the projectile starts moving. That begs the question of whether a properly designed spring gun would be able to obtain higher velocities by effectively “chasing” the projectile out the barrel in a manner that a precharged gun cannot do. The window of opportunity for chasing the projectile is only about 5 milliseconds so the timing would have to be perfect.
He also mentions (I’ll paraphrase) that lower velocity guns will always be more efficient than higher velocity guns. By efficient, I/he mean/s being able to attain a greater percentage of their theoretical velocity and projectile energy. I have definitely found that to be true with PCPs.
Good stuff.
Lloyd
Lloyd
RE: “That begs the question of whether a properly designed spring gun would be able to obtain higher velocities by effectively “chasing” the projectile out the barrel in a manner that a precharged gun cannot do.”
A springer could go faster because a springer heats up the gas as well as compresses it. Sound travels faster in hot gas.
Herb
RE: Yellow Forum thread
The Yellow forum thread that I found also lead to the following paper:
A Two-Stage Light Gas Gun for the Study of High Speed Impact in Propellants” By Dr. Con Doolan, Weapons Systems Division, Aeronautical and Maritime Research Laboratory,
DSTO-TR-1092
http://dspace.dsto.defence.gov.au/dspace/bitstream/1947/4048/1/DSTO-TR-1092%20PR.pdf
It is about a two stage light gas gun. Basically a springer using helium as the gas. Did some say KA-POW!!
Herb
Herb,
Thanks for the additional lead. With complex technical topics like this, finding the research is the first hard part, and then finding the research that is written at the right level is the second problem.
I imagine you’ve already worked on this, but in a magnum springer, what percentage of the power comes from heating the air. The spring has a fixed number of foot pounds to offer, and it goes to accelerating the piston, friction, compressing the air, heating the air, other?. I imagine that the heat is responsible for only a few percent.
What do you think?
Lloyd
Lloyd,
RE: “In a magnum springer, what percentage of the power comes from heating the air.”
Ok, we’re switching gears here to a springer.
At the instant the piston comes to the end of its travel as it plunges down the bore for the first time (ie before rebound), the engery released from the spring compresses the gas and heats it. The sum of the two is equal to the energy “lost” by the spring.
It has been a long time since I took Physical Chemistry. I’m at a loss to immediately put a finger on the right equations and data to correctly show how much energy goes into compression of the gas and how much goes into heating the gas. But I very much doubt that the heating portion is small. I’d guess that heating consumes a significant portion of the energy. I’ll try to figure this out exactly later.
Herb
RE: Regrouping…
The paper by http://www.dtic.mil/cgi-bin/GetTRDoc?AD=AD475660&Location=U2&doc=GetTRDoc.pdf
His limiting case is presented in equation 11-8 on page 34 of pdf file, or page 19 of the text.
** Equation 1 **
Mu(escape) = 2*A(0)/(Gamma-1)
were:
Mu(escape) = velocity where pressure drops to zero
A(0) = speed of sound
Gamma = specfic heat ratio of gas
Gamma for air is 1.4, meaning that the escape velocity is 5 times the speed of sound.
This is same as equation 1 on page 2 in Dr. Con Doolan's paper.
Although this does give an absolute theoretical limit, it seems much too high.
I'm trying to figure out where/how the "better" formula, Equation 2, on the yellow was derived.
** Equation 2 **
v = 172 * sqrt(Q/(m+Z*Q))
I can derive a much lower limit than Equation 1 from the gas laws, and the Maxwell-Boltzmann distribution.
** Equation 3 **
A(o) = C(ideal) = V(sound) = speed of sound
V(sound) = SQRT(Gamma * (R*T/M))
See:
https://en.wikipedia.org/wiki/Speed_of_Sound
** Equation 4 **
V(mean) = SQRT( (8/pi) * (R*T/M) )
See:
https://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution#Distribution_of_speeds
So combining 3 and 4 we get
V(mean) = SQRT(8/pi)/SQRT(gamma) * V(sound)
or
** Equation 5 **
V(mean) = 1.35 * V(sound)
Given that sound is about 1125 fps at standard temperature and pressure (STP), then the limit from equation 5 is about 1516 fps.
Essentially combining Equations 3 and 4 into 5 assumes that the "slug" of gas going down the barrel stays at the pressure of the reservoir, but that "magically" all the gas molecules average their kinetic energy when passing into the barrel. Furthermore the gas molecules all obtain a vector directly down the bore of the gun so that none of the gas molecules are bouncing into each other. Since on average the gas molecules have not lost any kinetic energy, the gas has done no work on anything else (the pellet).
If I understand the whole thing right, this basically infers that the gas molecules down the bore have no 'static' pressure and that all the pressure is 'dynamic.' If the gas molecules start bumping into each other (and thus the barrel also) then some of the 'dynamic' pressure is lost, the velocity of the flow drops, and the 'static' pressure rises. This model in essence turns the "transfer port" into a perfect rocket nozzle and calculates the rifle recoil as the result of the "thrust" of the "rocket." So mass of gas in barrel times its velocity down the barrel is essentially the "thrust."
The 1516 fps that I calculated, and the 1640 fps from Equation 1 are in the same ballpark. How/Why the difference I will work on. I first have to figure out how to derive Equation 1.
Sorry if my jumping the gun on all of this has created much confusion. I just got really excited when I stumbled on the thread in the yellow forum. Equation 1 just seemed of the right form and gave calculations which seemed reasonable.
Herb
Re: regrouping…
!@#$%^&*(
Think I finally have a handle on this.
The above calculation of 1516 fps is wrong. You have to look at it one way or the other.
If the work is done s-l-o-w-l-y then the reservoir pressure (assuming an infinitely large reservoir) is maintained throughout the journey of the pellet through the bore. This maximizes the work done by the gas, and the gas in the bore has no 'dynamic' pressure, it is all 'static' pressure. In other words, the gas in the barrel itself has no overall kinetic energy directed down the barrel.
On the other hand, when the air comes out of the reservoir to just atmospheric pressure, the gas can convert its internal energy (due to high pressure) to kinetic energy of the low pressure gas. So a cubic inch of air at 2000 psi (above room pressure) would expand out to to about 140 cubic inches of air at room pressure with kinetic energy. For this situation the "thrust" of the gas is maximized and the gas molecules have the most velocity. In this case, the energy lost depends on the starting pressure, so the velocity of the gas molecules increases as the pressure drop increases.
Again, I'm sorry for all the twists and turns as I figure this out.
The gist is that either extreme is wrong. The "true situation" is something in between these two. At this point I don't understand how to model the gas flow to figure this out…
Herb
RE: Have figured out constant of 172 in yellow formula
v = 172 * sqrt(Q/(m+Z*Q))
where:
v = velocity of pellet
m is pellet wt in grains.
Z = gas density grains per cubic inch = 0.011 grains for air at room temp
Q = PSI x barrel_length x caliber x caliber
Before I had:
Wp = weight of pellet in grains
Sp = Speed of pellet in feet / second
PSI = pressure in pounds per square inch
Bc =caliber of Barrel, diameter of barrel in inches
BLi = length of barrel in inches
Pi = geometric constant 3.14159…
work = work done on pellet in ft*pounds
** Equation 1 **
Work = PSI * (Pi/4) caliber * caliber * BLi/12
** Equation 2a **
Wp*Sp^2/ 450,400 = work
rearrange equation 2a to solve for Sp
** Equation 2b **
Sp = Sqrt(work * 450,400/Wp)
Substitute expression for work into 2b
** Equation 2c **
Sp = Sqrt( (PSI * (Pi/4) caliber * caliber * Bli/12) * 450,400/Wp)
Pull out constants in equation 2c
** Equation 2d **
Sp = Sqrt((Pi/4)*450,400/12) * Sqrt(PSI*caliber*caliber*Bli)/Wp)
** Equation 2e **
Sp = Sqrt((Pi/4)*450,400/12) * Sqrt(PSI*caliber*caliber*Bli)/Wp)
Simplify constants
** Equation 2f **
Sp = 171.7 * Sqrt(PSI*caliber*caliber*Bli)/Wp)
So that is where 172 number comes from in yellow equation.
The numerator inside the SQRT is simply the term Q from the yellow equation. Wp is same as term m in the yellow equiation.
The extra term z*Q in the denominator is still a mystery. In order to “fit” with Wp, then z*Q has to have units of grains.
Herb
Herb,
Just a guess, but Q would appear to be a function of volume and pressure, therefore z*Q would be a weight?
Herb,
Yes, I think z*Q is simply the weight (implicitly converted elsewhere to mass) of air in the barrel.
BG_Farmer,
Yes z*Q would have units of grains.
Just can’t figure out how/why it is the formula besides “to make it work.” 😉
Herb
Herb
Herb,
Like I said, it seems to calculate not just the weight of the pellet, in grains, that is being “worked” on, but the weight of the air also. This may be its magic limit. Anyway, I’m not a rocket scientist.
Herb,
Have you tried the YF formula applied to Discovery? How well did it work for you. Could you just plug Discovery numbers into it and let me know what you get?
Herb,
One more thing: I think I see the problem with both your formula and the YF formula. The force on the pellet in the direction out of the barrel can be assumed to be constant, but there is an opposing force in the opposite direction as the air in front of the pellet is compressed (both formulae seem to assume no pressure in that direction) and is limited in how fast it can flow out the end of the barrel. I think you might have been on the right track when you suggested putting a “force field” at the end of the barrel. What do you think?
RE: Yellow equation
See:
http://www.network54.com/Forum/79537/message/1236402420/Formula+for+maximum+theoretical+muzzle+velocity+for+an+airgun-
for more discussion. The term was basically just “thrown” into the equation as an empirical correction.
Herb
Herb,
I think the term is OK, but negligible: for .22 it comes out to about 2 gr… For all practical purposes, its the same as your equation. I ran numbers for the Discovery with the YF formula and came out pretty close to what you calculate, more than 1640 fps, by the way, which leads me to believe someone missed something (perhaps an implicit conversion) when they performed that math trick. That’s why I was hoping you could do it; I’ve had a cold lately and may not be functioning 100%. At this point, I think my 1/2ing the work is closer to correct, but for the wrong reasons.
I think it might be possible to set up the problem with the pellet as a piston with force on it from the reservoir, and a force of opposite sign from air resistance (for lack of a better term). If the force of air resistance can be approximated, it can be subtracted from the force driving the piston. Trick is, the force on the piston from air resistance varies with velocity. Does this make sense? Just trying to help, not torment you:).
Herb,
I can’t give up on this:). The force of air resistance on the pellet solves a lot of problems in the results, unless you are shooting the pellet in a vacuum.
Counteracting the force of the air from the reservoir pushing the pellet “out”, there is a small, constant force due to atmospheric pressure and quickly larger one that varies by the square of the velocity. This, for me, explains why your formula and the YF one hugely over-estimate the velocity of the pellet. Also, the opposing force does not vary by the mass of the pellet, only the area, which means light or “zero” weight pellets are also limited in velocity, depending on the air pressure.
If you don’t want to try to set it up, I’ll try myself.
BG_Farmer March 07, 2009 10:11 AM
RE: opposing air pressure & force field at the end of the barrel
Yes, need another force force field at he end of the barrel so that the pellet is essentially expanding into a vacuum. In order to make the calculations come out right you'd have to subtract 14.4 psi from reservoir pressure since the pressure is above room pressure. On 2000 psi 14.4 psi is a twit of course.
Since we are trying to calculate a maximum velocity, it is permissible to ignore any factor which decreases the velocity – such as friction in the barrel, choked air flow, and the opposing air resistance.
I'm pretty sure that I see how to improve the yellow formula. I pretty sure that the "limit" will be more than 1640 fps.
Herb
BG_Farmer
Your comment on:
RE:March 07, 2009 9:30 AM
“Could you just plug Discovery numbers into it and let me know what you get?”
— and —
RE: March 07, 2009 1:34 PM
“I ran numbers for the Discovery with the YF formula and came out pretty close to what you calculate, more than 1640 fps, by the way, which leads me to believe someone missed something (perhaps an implicit conversion) when they performed that math trick.”
————-
I ran the calculations on the Discovery at:
March 05, 2009 5:04 PM
Using the yellow formula:
V = 172 * SQRT( (2000*24*0.22^2)/(16 + 0.11*2000*24*0.22^2))
= 1286
————-
The 1286 is way below 1640.
I don’t think that the yellow formula is right, but it is very interesting…
Herb
Herb,
Check your calculation above carefully — if you use that value of z, you won’t get 1286, I think:). I get 1286 using z = 0.011, now, though, although I had a problem that undervalued z*Q the other way. If you eliminate z*Q (for example by making z = 0 or very small), the Yellow Forum formula and yours yield identical results.
I have a spreadsheet with your formula, the yf and a scratch formula of my own evil design.
I’m even more confused about your purpose if you dont care about modeling performance in atmosphere…not much to hunt in space. 2000:14 is not much to worry about, but if you don’t incorporate air resistance, the “answer” will never even come close to reality, I think. Since you started by wondering why your velocity of 2070 fps diverged so much from the actual 840 fps, I assumed you wanted to make your model real.
BG_Farmer
RE: “Check your calculation above carefully — if you use that value of z, you won’t get 1286, I think:). I get 1286 using z = 0.011,…”
Huh??
My calculations above are using z =0.011, and with that value the prediction for the Discovery is indeed 1286 fps.
Yes, I am trying to make a “real” model. But I’m looking for where the “big” problem is first rather than trying to tweak small corrections into the model. I think that there is a big problem in how much work is done on the pellet. I don’t think all the small tweaks would overall close the gap.
The whole point of the yellow formula was to give a “reasonable” absolute maximum. So you can ignore things that make the pellet go slower. For instance the friction of the pellet in the barrel is ignored.
The speed of light is an absolute maximum, but useless for any practical air rifle.
In science the whole model thing is done interactively with calculations. You calculate and get a number roughly right. Create a model then do calculations based on model. Change assumptions in model and calculate some more. Hopefully you get to a point where the model works.
Herb,
Sorry, just meant you have 0.11 for z in the calculation you show, although the result reflects that you must have used 0.011; very confusing for us simple clodhoppers trying to work along.
After a little thought, I think I understand what you mean by “there is a big problem in how much work is done on the pellet”. My opinion is that that may be premature when the model does not incorporate all the known significant factors, but I’ll be interested to see what you come up with and be proven wrong. Also, I think if you don’t incorporate air resistance in some way, you will never have access to real world results to corollate.
BG_Farmer
Hanging my head. You are absolutely right. I had 0.11 in equation but used 0.011 in calculations. The constant of 0.011 is the correct one for the original yellow formula.
Thanks for pointing out the error.
Herb
BB,
Why is velocity seemingly an issue that arouse more exponential excitement than accuracy? To me the holy grail in this hobby is nothing but accuracy.
If we have two stations discussing velocity and accuracy, the former will have the most listeners.
Just thinking out loud, my apologies.
Abe
Abe,
I don’t know why this topic is so fascinating for some people, but as you and every other reader can see–it is! If you recall, I mentioned all the discussion on the various forums. What you see here is a microcosm of that.
I agree with you, that accuracy is the main attraction, but this is also a legitimate interest, as the responses bear out. And I’m sure this discussion is not over.
B.B.
RE: Recap on Yellow formula
Started a new thread in yellow to discuss how the formula was dervived.
http://www.network54.com/Forum/79537/message/1236402420/Formula+for+maximum+theoretical+muzzle+velocity+for+an+airgun-
After finding out some of the details I posed a formal derivation of he formula
http://www.network54.com/Forum/79537/message/1236478670/More+formal+derviation+of+formula…with+formal+model
and got same form:
v = 172 * sqrt(Q/(m+Z*Q))
but a different constant for z. Using the model proposed z=0.0173 (careful with zeros this time!).
Now When m = 0, then theoretical limit for the model is:
v = 172 *sqrt(1/0.0173) = 1308 fps
This is obviously too low. See yellow for what I think problem is.
To just arbitrarily change z to make it smaller, and hence make the maximum velocity bigger stinks in my opinion. Need to refine model.
Herb
RE: Why maximum velocity is so fascinating to me
I’m a geek first. Shooter second.
The fact that velocity might has ultimate limits is very fascinating to me. The speed of light is the first limit. Much too high for any real airgun to achieve and hence of no practical value.
THhe second limit for using presurized air seems to be 5 * speed of sound as derived by Seigel, 1965. Now that is still higher than any airgun shoots, but it is getting warmer. The limit is high enough beyond the current technology to indicate that there is ample room for improvement.
The underlying factor in all of this is the transonic region. I think that there is ample experimental evidence to so that this is a “no-fly” zone. Pellets are not accurate in this region.
All in all though I’d like about 800 fps for a 0.22 to whack the squirrels that are causing problems for me. I don’t want to be shooting bullet shaped projectiles at 2500 fps with an airgun. Even if I could control the noise, I’m shooting inside a city and that is just too dangerous.
Herb
RE: March 03, 2009 11:04 AM, Anonymous Herb said…
I should have pointed out that the calculation in that post, that the derived calculation does give an ABSOLUTE limit. Given a starting pressure of 2000 psi and a 24 inch 0.22 barrel and a 16 grain pellet, you ABSOLUTELY can’t make the pellet go any faster than 2070 fps.
At March 03, 2009 11:04 AM,
I used the volume of the reservoir, the volume of the barrel, and the starting pressure to calculate what the final pressure of the system would be if the perfect valve was opened fully and then closed just as the pellet exited the barrel. The pressure drops to 1800 psi which is obviously way to low for how the Discovery really preforms. For the configuration of the discovery this would lower the absolute limit a bit to probably something like 1800 fps. That would be a somewhat better limit.
Essentially, what really happens, is closer to this – the valve pops open to pressurize only a few inches of the barrel to 2000 psi. Then as that slug of gas expands the pellet is pushed to its final velocity.
I now think the way to model the airgun would be to assume that the pressure does equalize between the reservoir and the barrel for some distance X in the barrel. Then let that volume of gas expand to the end of the barrel. Summing work as pressure times distance for each increment of distance you can calculate overall work. Then from kinetic energy equation you can calculate the final velocity of the pellet.
You could calculate the pressure drop per shot by simply shooting a number of shots and seeing how much the pressure drops.
Overall though there is another factor here – the pellet. There is enough variation between pellets to really gum up and modeling. Real world experiences show that indeed this would be the situation. You have to include at least the weight, some term to indicate when the pellet starts to move, and probably a term for friction.
The term for when the pellet starts to move would be an offset for the initial pressure. Basically if the pellet starts to move easily then the overall pressure is reduced.
The extra “calibration” factors for each pellet are the rub. A PCP operates on a narrow window of pressure anyway. If I have to throw in several “fudge factors” for each pellet, what good is the model? I just shoot the pellet at the pressure and measure its velocity directly.
Herb
RE: March 03, 2009 11:04 AM,
Guess what really bothers me in the calculation that I did in that post is the conversion of the energy to velocity.
Using:
KE = 1/2 * M * V^2
for any KE value as M goes to 0, then V goes to infinity. Obviously that doesn’t happen.
It just seems that there should be some bound on how fast the gas molecules can move. From a practical stand point the question is – does any real pellet get moving fast enough for this to make any difference. The answer to that question may be in fact – no.
On the other end there are problems as well. For a given mass of a pellet M, as KE goes to infinity so does the velocity. That doesn’t happen either. Nothing goes faster than the speed of light. But the power needed to approach the speed of light is so far away from the power of any real airgun, that the limit is irrelevant. The speed of sound is 1,125 ft/sec or 767 mph. The speed of light is 186,000 miles per second, or 669,600,000 miles per hour. So relativity won’t be a factor until we improve airgun power plant to create a billion times more power. With that sort of energy, the recoil would be enough to put the rifle in earth orbit!
Herb
RE: Seigel Factor
For special relativity there is the Lorentz factor
Gamma = 1 / SQRT(1- v^2/c^c)
where v is an objects velocity and c is the speed of light.
How about the Siegel factor?
SF = 1 / SQRT(1-v^2/(5*SS)^2)
V velocity of pellet, SS is speed of sound. Wonder if this would give us some idea if velocity of gas molecules is important. If pellet is going roughly the speed of sound, then the factor is about 1.02. Probably indicates that we have nothing to worry about.
Herb
Herb,
I think you've done about what you can without resorting to calculus, fluid flow, etc. I currently slightly prefer the YF to your original formulation simply because it incorporates a factor that seems like it ought to be considered and provides at least some limit even for light pellets:). However, the valuation of that factor is tricky: I'm not so sure the full mass of air in the barrel is accelerated to v. It seems like you have the same concern, expressed in a different manner. Ironically, the accidental undervaluation of z*Q in the original YF equation seems to be about right. I agree its repulsive to play with the value of a physical constant, but the way the formula is set up currently makes that the only option.
I vote for assuming full pressure from breech to muzzle, even if it takes a full tank of gas. If you start playing with the valve, you might as well do a full model with friction, air resistance, etc., in which case its a model of a specific airgun in atmosphere, not a maximum attainable value.
I started work on a different approach, calculating v_max from the force of air resistance limited by the force exerted on the pellet. There's too little difference in real-world pellet velocities (i.e., lead versus delrin) to attribute the difference only to the kinetic energy imparted to the pellet. I'm hoping that internally, air resistance is easy to calculate because the Cd of pellet in barrel is >= 1, but we'll see. I also tried adding the mass of STP air in the barrel to the weight of the pellet and z*Q in the YF formula, and it doesn't make much difference, so I'm not sure air resistance will give me a useful value, either. Don't take any of this too seriously, as I'm still just thinking.
BG_Farmer
RE: Calculus
Next step. My first calculations for the Discovery do give an ABSOLUTE limit given basic parameters about the rifle.
Basically I didn’t like the fact that velocity could be increased unbounded by increasing fill pressure for the rifle. Don’t think that happens.
However, going back to the REAL rifle. By shooting a string of shots and looking at the REAL pressure difference you can get an idea of how much of a drop in PSI/shot occurs. Nothing theoretical about that at all. Now from pressure drop calculate how much air is blown into barrel. For that volume of air assume it starts out at fill pressure. Work at that pressure is pressure*distance that it pushes pellet down the barrel. Now valve closes, and let the “slug” of emitted gas expand down the barrel pushing the pellet. As the gas goes down the barrel, the pressure drops because the volume is increasing. Here you need calculus to determine how much work is being done.
But this model does actually give the ABSOLUTE maximum amount of work that can be done on the pellet. It ignores friction, gas flow problems, and anything else that in the real world would slow the pellet.
My earlier calculations didn’t take into account the “dead” volume in the valve. But I showed that if the valve stayed open on the Discovery until the whole barrel filled with gas and the pressure in the barrel and the reservoir were to equalize, then the pressure in the reservoir drops from 2000 psi to less than 1800 psi (need to add valve volume). The Discovery doesn’t drop nearly this much. So each shot uses much much less gas than a “barrel full” at 1800 psi.
There is a lot of air in the barrel. A barrel full of air at 2000 psi weighs more than the pellet! The problem with the yellow calculation is that ultimately you don’t have to push all of that air. If you just left the valve open till all the gas is gone, then the pellet would be out of the barrel long before the amount of gas in a “barrel full” of gas at 1800 psi was released.
Of course the other extreme is wrong too. Let’s say that the valve just dumps enough gas into the barrel to fill the barrel with gas at environmental conditions (STP in chemistry…). That is too little air compared to what the Discovery really uses. It is also too little air to get to “real” velocities. In other words, I’m sure that the 800 fps of the 0.22 caliper Discovery requires much more air than that. You actually need to dump more than a barrel full of air at STP to get a “better” velocity. Of course having the valve open after the pellet leaves the barrel is just a waste. None of that air is going to make the pellet’s velocity higher.
Make sense?
Herb
Herb,
Maybe we ought to sing Cumbaya(sp?), because I think we’re in harmonious agreement:).
Couldn’t help myself. Plugged numbers from BB’s new article:
/article/How_fast_do_pellets_go_March_2009/59
into equation. Assume fill pressure of 3000 psi for the rifle.
Delrin
V = 172 * Srt( (3000*24*0.375^2) / (15.5 + 0.11*(3000*24*0.375^2)))
V = 1536
for PB using 74 grains the calculation is 1271
V(observed-delrin) = 1381
V(observed-lead)= 1041
All in all that seems like pretty good agreement. We’ve neglected friction, and the fact that a “real” valve was used without dumping the whole reservoir pressure.
http://www.lmpenterprises.co.uk
RE: March 09, 2009 6:44 PM, Blogger lee said…
The equation for Delrin was calculated correctly, but the equation was written wrong. As was pointed out to me it was not 0.11 but 0.011
So the original yellow equation should have been written;
V = 172 * Srt( (3000*24*0.375^2) / (15.5 + 0.011*(3000*24*0.375^2)))
The problem is that there is no basis for the Z factor of 0.011. Nothing wrong with using an empirical number to fit the data.
But we’d need to have a whole bunch of data in the “local vicinity” of the data for which the constant was calibrated. Then you calculate error limits for data set for the fit. Then you say that based on a pressure of X to Y, for pellets of weight from A to B, for a fill pressure of C to D, then for the model XYZ rifle the following equation gives the velocity +/- error%.
That isn’t quite the same as saying the absolute limit is 1560 fps.
See yellow thread for more discussion on Z.
Herb
I recently purchased a hunter extreme and a Gamo Whisper for my wife.. I’m a newbie to air rifles admittedly, had some as a kid just the crossmans (im 42) I think 650 pfs was advertised. We didn’t have chronographs to see if was that fast or not.
With that said, I use to shoot Combat (firearm) and competition on the circuit, reloaded obviously a must for doing that.
One Rifle I loved was the 22o swift, a 22 cal. bullet real flat trajectory. But it had it’s downfalls.
Yes some new to airgunning will buy the fastest thing just to hear the crack of supersonic, but the fastest gun in the world is in a lab, it shoots a tiny ball and displaces a great deal of sand.. They use this gun to simulate an asteroid hitting earth.
What I’m getting at is at real close range at supersonic speed that real light pellet may do a lot of damage. Like the swift, wind a blade of grass, tumbling ect. have an effect.. But my 22 swift would explode on impact on a coyote tearing him a new one.
I picked the fastest thing for my buck because i know I can put the heaviest pellet in it and it will still be arguably the fastest gun.
Can we make a 177cal that is longer and heavier for this gun? Perhaps as air guns increase velocity the pellet itself will be made even heavier for better punch.
But for me I’m happy for now, I’ll put in that pba just to hear a pellet gun sound like a 22 rifle and think about how far these toys have came. Then I’ll put in the heaviest led there is and still be as fast as my Wifes air gun.
Time to become familiar with airgun ballistics. Diabolo pellets will not be accurate above 1,000 f.p.s. The top field target competitors try to keep their guns around 900 f.p.s.
Longer, heavier pellets are not good for the powerplant of a spring gun. Springers like lighter pellets. Your rifles will do best with 7.9-grain Crosman Premiers (from the cardboard box) and 8.4-grain JSB Exacts.
You’re right that a supersonic pellet is better at close range than a subsonic one, but since they open to groups of 3 inches at 50 yards and your rifle can hold half that size with 900 f.p.s. pellets, they are only good at VERY close ranges! Most airgunners are pushing for groups of less than one inch at 50 yards, these days. That takes either a very accurate springer like an RWS Diana 54 or an Air Arms TX200, or else a precharged pneumatic rifle.
At 42, you are on the young side of the average airgunner. 5-0-55 is about average today. And you make an interesting analogy. That .220 Swift of yours is one of the hardest of all .22 centerfires to be accurate with, because of the extreme velocity. But drop 1,000 f.p.s. and the .222 Remington is well-known for superior accuracy. That’s the cartridge the U.S. Army initially used in the M1 Carbine (not including the failed 5.7 Johnson experiment) when they were trying to create what Eugene Stoner finally turned into the M16. And unfortunately, they did it with another accuracy failure–the .223.
Your Swift had bundles on hydrostatic shock. But no conventional pellet gun will ever have ANY. The velocity is too slow–even at supersonic velocities.
Your Gamo Hunter Extreme has now been brought out in both .22 and .25 caliber in recognition of that fact. With a slow-mover like a pellet it takes size and mass to do the job–not velocity.
The fastest I have ever witnessed a .177 Raptor pellet leave a Hunter Extreme was 1.395 f.p.s. That’s fast for an airgun, but about half of what is needed for hydrostatic shock. The tales of 1,600 f.p.s. for that rifle are just that–tales. My column in Shotgun News this month details the study I have done on this subject, and the Hunter Extreme is one of the guns I have been studying.
But topics like this are what make airguns interesting, don’t you think?
B.B.
I have broken Quackenbush's top speed. 1558fps with 6.0gr lead free pellet, with cold air PCP of my own design. Other shots were 1557, 1530 etc. video will go up soon. Comparing the top velocity of springers (hot air) and PCP's is apples and oranges. A springer with enough working volume can break the 1640fps limit of cold air PCP's.
Of course cold air is denser than hot air, so it would produce the highest velocities under similar conditions.
Good job.
B.B.