Sonic Choke? Maybe not. 2162 FPS

[rsterne]

The starting point for this idea is the equation for the energy that is produced in a shot, E = 1/2 mv^2.... If we go back to Lloyd's post about the theoretical maximum velocity (neglecting the weight of the air), we have the following equation, with it's derivation carefully explained.... Our eventual goal is to be able to input the weight (in grains), pressure (in psi), and all other dimensions in inches.... Getting there we will deal with those pesky constants.... From this thread http://www.gatewaytoairguns.org/GTA/index.php?topic=94054.0 we have Lloyd's equation.... v = (F/m) x {sqrt ([2 x d x m]/F)}

This can be further simplilfied by squaring both sides, then rearranging, and taking the sqrt. again....

v^2 = (F/m)^2 x ([2 x d x m]/F) = F^2 / m^2 x 2 x d x m /F = 2 x F x d / m .... then take the square root of both sides and we have.... v = sqrt ( 2 x F x d / m) ....

That is EXACTLY the same equation as Herb1836 and Steve_in_NC are working with, once we add in all the constants to get the units correct....We need the force F in lbs. the distance d in feet, and the mass m in slugs (weight in lbs. / 32.174).... We also need to calculate the force F, which is the pressure in psi, times the bore area in in^2.... First, the force....

As explained by Lloyd, the force F accelerating the bullet (and the air) is the pressure P times the bore area A, which is A = cal.^2 x PI/4 .... The force F is therefore....

F = P x cal^2 x PI/4 with the caliber in inches, pressure in psi.... so our main equation now becomes....

v = sqrt ( 2 x P x cal^2 x PI/4 x d / m ) .... The distance the force is acting through, d (in feet) is the barrel length BLi (in inches), so we have d = BLi / 12 .... and our main equation now becomes....

v = sqrt ( 2 x P x cal^2 x PI/4 x BLi / 12 / m) .... We want to be able to input the pellet weight W in grains, and to convert that to the mass in slugs, we use m = W / 7000 / 32.174 = W / 225218, so now we have....

v = sqrt (2 x P x cal^2 x PI/4 x BLi / 12 / ( W / 225218 ) .... If we gather the constants at this point, we get Steve's equation....

v = sqrt { ( 2 x 225218 x PI / 4 / 12 ) x ( P x cal^2 x BLi / W ) } = sqrt {29481} x sqrt { P x cal^2 x BLi/ W } = 171.7 sqrt ( P x cal^2 x BLi / W )

This is a rearrangement of Lloyd's equation, and will give exactly the same results, which is really pretty KEWL, that two such bright guys got the same answer.... However, once we add the mass of the air, things change, and I will deal with that in my next post....

Bob

[Scotchmo]

Bob,

Looking forward to the rest of the story! And wondering what are the implications for 1650fps vs 1854fps.

FYI: grains is a mass, not really a weight(force). Technically, no reason to show the 7000/G calculation to convert to pounds(mass). You can go directly from grains to slugs.

In the mean time:

In trying to visualize the limits, I came up with this first pass at a velocity envelope for a 70F airgun at psig>0. The hatched area is the operating envelope.



Air would go from A to A' (assuming 0 friction - not possible, unless freely expanding)
Just pellet would go from B to A' (assuming a force from 0 mass air - not possible).
Just air would go from A to B' (assuming a 0 mass pellet).

A velocity curve for a pellet would go from B to B'. The curve for different setups can shift up or down but will still lie in the hatched region.  The curve for a very light pellet (magenta curve) would gravitate toward point A. The curve for a very heavy pellet (blue curve) would gravitate toward point A'. The more typical curve (green curve) would be in between. They all start at v(0) and end at v(max).

The curve could not fall outside of the hatched region without breaking some physical law.

I'm not sure if it is all that useful.

[rsterne]

Sorry, I had a appointment this morning and got interrupted just before I finished this post, so I saved it in Word and just finished it up now....

OK, so now it gets REALLY interesting.... let's include the mass of the air.... to do that, we will back up a bit, to the equation that had the mass m (in slugs) in it....

v = sqrt ( 2 x P x cal^2 x PI/4 x BLi / 12 / m )

We now want to add the mass of the air in the barrel Ma, to the mass of the pellet Mp, with everything in slugs.... We have m = Mp + Ma so the equation becomes....

v = sqrt ( 2 x  P x cal^2 x PI/4 x BLi / 12 / [ Mp + Ma ] )

The mass of the air Ma (in slugs) is the barrel volume Va (in in^3) times the density Da (in slugs/in^3), so Ma = Da x Va.... where Va (in^3) = cal^2 x PI/4 x BLi .... Hey, we have those same numbers in the numerator, so we can simplify the equation to the following....

v = sqrt ( 2 x P x Va / 12 / [Mp + Ma ] ) .... We want the pellet weight in grains, so we need the weight of the air in grains also.... Here is where it gets complicated (like it wasn't already?).... If we accept that air is an ideal gas, and use Boyle's Law, we can just calculate a value for Da at 1 psi, the way Steve did it, to get the density factor he called "Z".... He used half the air density at 1 psi, (which he calculated to be D = 0.0219 grains/in^3/PSIA).... so that he ended up with a weight in grains that could be simply added to the pellet weight W before converting to slugs.... The relationship to mass is as follows....

Ma = Da x Va where Da = Z x P so Ma = Z x P x Va.... If we substitute that for Ma in the equation above (Z to be in grains/in^3/PSIA), we get the following....

v = sqrt ( 2 x P x Va / 12 / [ Mp + Z x P x Va ] ) .... Checking the units, for Ma = Z x P x Va, we have Z in grains/in^3/PSIA x P (PSIA) x Va (in^3), the weight of air would be in grains, so we have to divide by 225218 to convert to slugs, just like for the pellet.... Hence, both the pellet weight W and the weight of air ( Z x P x Va) are divided by 225218, so we get the following....

v = sqrt ( 2 x P x Va / 12 / [W + Z x P x Va] / 225218 ) .... Now let's gather the constants....

v = sqrt { ( 2 x 225218 / 12 ) x P x Va / (W + Z x P x Va) } = sqrt { 37536 } x sqrt { P x Va / ( W + Z x P x Va ) } = v = 193.7 sqrt { P x Va / (W + Z x P x Va ) }

If we set the pellet weight W = 0, then we get the following....

v = 193.7 sqrt { P x Va / (0 + Z x P x Va) } = 193.7 sqrt { P x Va / Z x P x Va }.... the P's and Va's cancel out, so v = 193.7 sqrt ( 1 / Z )

The only variable left is the value of Z, and it drastically alters the maximum velocity.... Steve used half the air density he calculated, Z = D / 2, which gave Z a value of 0.01095 grains/in^3/PSIA.... That leads to the following calculation....

v = 193.7 sqrt ( 1 / 0.01095) = 193.7 sqrt (91.32) = 193.7 x 9.556 = 1851 fps.... We know that can't be right, because Lloyd already has achieved 2031 fps, with a pellet weighing 9.1 gr.... This calls the value of Z into doubt.... I'll look at that in the next post....

Bob

[rsterne]

OK, so if we assume that we have a problem with the air density factor, Z, let's see what we can determine.... Steve worked from the mass of air molecules and molar volume, and arrived at a value of 0.0219 grains/in^3/PSIA.... I'll try a different approach, to see if I can get the same value.... I will start from standard air density at 1 bar, at 20*C, which from this online calculator is 1.1894 kg/m^3.... http://www.peacesoftware.de/einigewerte/luft_e.html .... I double checked this with another online air density calculator.... http://wahiduddin.net/calc/calc_da_rh.htm .... which gave a value of 1.188 kg/m^3 for dry air at sea level at 20*C.... The average of these is 1.1887, so I will use 1.189 kg/m^3....

From another online conversion calculator, the conversion factor for kg/m^3 to grains/in^3 is 0.252891, so that becomes 1.189 x 0.252891 = 0.30069 grains/in^3 air density at 1 bar.... However, 1 bar is 14.504 psi, and we want the value at 1 psi, so dividing 0.30069 / 14.504 = 0.02073 grains/in^3/PSIA.... That's pretty close to what Steve got, an error of only 5%.... He was using 25*C, and using that I get 1.168 x 0.25289 / 14.504 = 0.02037, a little further from Steve's number.... Therefore, I will use the value I calculated of 0.02073 grains/in^3/PSIA.... If we use Steve's estimate that Z = D / 2 (because only half the air column is being accelerated), then Z = 0.02073 / 2 = 0.01037.... Using this value in the equation for the maximum velocity with zero pellet mass, we have....

v = 193.7 sqrt ( 1 / Z ) = 193.7 sqrt ( 1 / 0.01037 ) = 193.7 sqrt ( 96.43 ) = 193.7 x 9.820 = 1902 fps.... This is still too low a value, particularly when you consider it is for a pellet weight of zero.... and has already been exceeded with a pellet weight of 9.1 and 10.3 grains at 4000 psi in a 47.5" barrel....

If the entire air mass is used, instead of half (not unreasonable, because it is all moving), so that Z = D = 0.0207, we get the following....

v= 193.7 sqrt ( 1 / Z ) = 193.7 sqrt ( 1 / 0.0207 ) = 193.7 sqrt ( 48.309) = 193.7 x 6.955 = 1346 fps.... an even lower value.... Scott suggested that working through using integration, an increment at a time, could make a difference, but the equation we are dealing with, at zero pellet weight, has no terms for caliber, pressure, or barrel length, so we can't do that using this equation.... I would suggest that pursuing this route may be a waste of time.... I would appreciate it if someone could check through my math and make sure I have not made a mistake somewhere.... It was a long process, and that is always a possibility....

Bob

[Scotchmo]

Bob,
Good work. I want to look at that more when I'm at my desk. I'm at the range this afternoon (working, not shooting). If you are correct, at BL=0, it's all pellet. So we start at 172? At BL=infinity, it's all air. So we use 194?

Z looks right.

Bob - are you sure that D/2 represents 1/2 of the air column? I thought it was the (1/2m) from 1/2mv^2. And density is not a constant during the shot cycle. Density changes with temperature. Entropy gain is significant for supersonic shear of air. You can't ignore it. And you can't average it over the entire air column. If you did, then Z=D/4. But most of it is right behind the pellet. I'm leaning toward Z=D/8 at the limit (50% efficiency).

If 1650fps is the least length maximum limit. The most length maximum is 2x1851fps, or 3702fps.

For a very light pellet, you can end up expending almost 8x the energy to go from 1651fps to 3702fps.

I could be wrong. I'm basing on my current theory. All data and formulas support it.

[rsterne]

The D/2 is the density factor.... and I have no way of working in a temperature change without data.... The 1/2mv^2 fraction is part of the constant, it is the first "2" that you see in the formula before the constants are collected.... The constant should be 194 all the time, I only included the 172 version to show how Steve got it, by omitting the PI/4 in the numerator.... that is why I put it in red, because it really does not belong in my calculations....

I plan on working with the ACTUAL air density which decreases relative to the pressure.... ie at 6000 psi the density is only 80% of what we are using in this simplified formula.... Since air density makes a big difference, that could be very important....

Bob

[rsterne]

OK, let's back up again.... We will go back to where we still had separate terms for pressure P and the two masses, pellet Mp and air Ma....

v = sqrt ( 2 x P x Va / 12 / [Mp + Ma ] )

We will still use Ma = Da x Va.... but this time we won't accept Boyle's Law and it's simplified version of the relationship between pressure and density, but we will instead use the actual values for the air density at various pressures, as given by the Peacesoft calculator.... http://www.peacesoftware.de/einigewerte/luft_e.html

I have gone through and compiled the actual air density, in kg/m^3, at all pressures from 15 - 10000 psi, in 1000 psi increments.... Here is that information, first in a table, and then as a graph....



In the table, the middle column is the actual density from the Peacesoft calculator.... while the right column is that predicted by Boyle's Law.... all at constant temperature....



You will note that below 2000 psi, Boyle's Law applied almost perfectly, but above that the actual density is less than Boyle's would give.... The actual density at 4,000 psi is 8% less than you would get if air was an Ideal Gas, at 6,000 psi 18% less, at 8,000 psi 28% less, and at 10,000 psi 36% less.... So let's see what the equation looks like using the actual air density Da....

v = sqrt { 2 x P x Va / 12 / [Mp + (Da x Va) ] }

At this point, the masses are still in slugs, so the units for air density Da are slugs/in^3 and for barrel volume Va are in^3.... Once again, let's let the pellet mass be zero, so we have....

v = sqrt { 2 x P x Va / 12 / [0 + (Da x Va) ] } = sqrt { 2 x P x Va / 12 / ( Da x Va ) } = sqrt { 2 x P x Va / 12 / Da / Va }, the Va's cancel, and we have

v = sqrt ( 2 x P / 12 / Da ) .... We need the air density Da in slugs/in^3, and pressure P is of course in psi.... To convert kg/m^3 to slugs/in^3 we must divide by 890575, so for the air density Da in kg/m^3 the eguation becomes....

v = sqrt { 2 x P / 12 / (Da / 890575) } = sqrt (2 x 890575 / 12 ) x sqrt ( P / Da ) = sqrt (148429 ) x sqrt ( P / Da) = v = 385.3 sqrt ( P / Da )

Let's check this to see how the results compare to our previous equation, and we'll do that at 1 bar (14.5 psi), where the air density is 1.189 kg/m^3 as follows....

v = 385.3 sqrt ( 14.5 / 1.189 ) = 385.3 sqrt ( 12.195 ) = 385.3 x 3.492 = 1346 fps.... Hey, using the full air density before, we got 1346 fps, I think it works ! ....

How about at 2000 psi, where the air density is 164.4 kg/m^3....

v = 385.3 sqrt ( 2000 / 164.4 ) = 385.3 sqrt ( 12.165 ) = 385.3 x 3.488 = 1344 fps.... Still looking good over the (near) linear part of the pressure / density relationship.... What about 4000 psi?....

v = 385.3 sqrt ( 4000 / 302.4 ) = 385.3 sqrt (13.228 ) = 385.3 sqrt x 3.637 = 1401 fps.... The velocity is increasing as the density drops relative to the pressure rise, exactly what we were expecting.... Let's jump to 10,000 psi, where the air density is 524.3 kg/m^3....

v = 385.3 sqrt ( 10000 / 524.3 ) = 385.3 sqrt (19.073 ) = 385.3 x 4.367 = 1683 fps.... a huge increase.... However, we are still well below what we have already accomplished, so let's go back and use only half the air mass, instead of all of it.... This makes the equation....

v = sqrt { 2 x P / 12 / (Da / 890575 / 2) } = sqrt (2 x 890575 x 2 / 12 ) x sqrt ( P / Da ) = sqrt (296858 ) x sqrt ( P / Da) = v = 544.8 sqrt ( P / Da ) with Da in kg/m^3....

This will increase all the previous velocities by a factor of sqrt (2) = 1.414 times.... Here is a table of the calculated velocities up to 10,000 psi....



I had great hopes that we would finally have some maximum velocities that we haven't been able to reach, and looked unlikely in the foreseeable future.... Unfortunately, that is still not the case.... I do not know if having an actual pellet mass, and applying the equation to each interval as the pellet accelerates down the bore will do that or not.... However, since it still fails to predict a high enough velocity in the limiting case, where pellet mass is zero, I don't have much hope for it....

If anyone wishes to carry on from here, please do.... I would also appreciate it if someone will check my math to make sure I haven't made a mistake....

Bob

[lloyd-ss]

I'll comment on all the math work that has been done in another post, but first I need to give an update with the newest Fastest Shot.
I had to shorten the barrel to 46" to remove an oversized portion of the breech. I had not machined it correctly originally, but this time it turned out good.  It allows the use of much shorter projectiles.  The one for this shot was .335" long and weighed 7.7 gns, including the o-ring,  both before and after the shot. I was able to retrieve the projectile from the tube of grocery bags.
Still not seeing any attenuation in the velocity, in fact, this shot was 102.6% of the prediction (using the 1745fps shot as the benchmark).

Shot specs,
.278 cal
46" long barrel
7.7 gns
4600 psi air (had  a slight overfill after adjusting the compressor)
55 cc dump chamber
Velocity  2162 fps


Lloyd

[Scotchmo]

Lloyd - Wowee!

Bob,

You DO need to account for the entropy gain (heat gain in the propelling air), or you will be spinning your wheels. The inefficiency is absolutely needed.

Without it, the equation only shows the horizontal dashed line in the picture below:



The pic is a little confusing and would need a 3-D chart to correctly scale all the info (such as efficiency). And the barrel length would be some kind of  Log_x scale as I need to show infinity.

If you include an equation with entropy gain, and run a line from your equation, you will have the upper boundary line shown (cyan line). The one that the very light pellet (maganta curve) mostly follows.

For now, I'm good with it as is. Until Lloyd breaks through the boundary.

Imagine taking an airgun and using as much as half the energy in the tank to heat the air before the shot. You just raised the velocity limit by raising the molecular kinetic energy, but you reduced the overall efficiency. That's what is happening DURING the shot. The process is substituting molecular kinetic energy for mass kinetic energy. It shows up as heat and disappears as soon as it leaves the muzzle. And we can't do anything about it.

At 1854fps, the molecule temperatures in the boundary layer might be in the 1000-1200K degree range. Now check the density of the air. A simple equation is never going to predict the actual velocity. It's too complicated. I think the best we can do for accurate modeling between the limits is to get data. And then do an empirical friction formula that takes into account any pertinent variables. Put it in our spreadsheets. It will be even better than a straight efficiency multiplier.

[Buldawg76]

Lloyd
 WooHoo Getting closer to the magic number 2330 fps every shot. great work and its to late to turn back now.

Mike

[lloyd-ss]

Bob,
Thank you very much for presenting the derivation of the velocity and max velocity equations.  You have covered it quite well.  The one area that seems to be a struggle is the mass of the air, or actually, what assumption to make on how much of the mass of the air in the barrel is being accelerated.  I see several difficulties in that and I think that it varies with the barrel length, the length to diameter ratio, and the velocity (possibly just the MV) of the air or pellet.  I do not believe that just deciding to use some, somewhat arbitrary, portion of the air mass is realistic, even though that is what I do in my internal ballistics spreadsheet.  Here are a few reasons why.
We know that there is pressure drop within a pipe and that a pipe of some very long length will have almost no flow coming out the far end of the pipe.  There is friction in the barrel and friction in the air itself in a laminar flow situation, and it has the effect of (I believe) the air flowing thru more of a continually narrowing tapered barrel.  I think this is also dependent on velocity.  Slow heavy bullets are generally more efficient than light fast bullets, suggesting that there are more friction(?) losses with higher velocities. 
Could it be that at longer barrel lengths and higher velocities, the accumulation of additional air mass proceeds at a slower than expected rate, but also, at the same time, the force on the pellet is decreasing at a faster than expected rate?  It is not neat and tidy.  There is a continuum of pressures and velocities throughout the length of the and figuring that out is complex.  There could be some exponential function that determines the variable rate at which the mass of air is building up behind the pellet.  Quite complex.

Scott, I am still trying to get a handle on your entropy application in the calculations.  You seem to have it all going to heat that increases the temperature of the air and therefore the molecular velocity.  Is that correct?  But doesn't entropy also cover  non-recoverable, non-definable energy losses within the system?  Aren't some of those losses in a way  like the efficiency fudge factors that we are currently using.  Entropy losses are often empirically determined and isn't that in a way what we have been doing?  I agree that figuring the heat related energy losses (usage) helps to narrow down where the energy is going and will help to make the fudge factor  a smaller percentage number.  But there are still entropy  losses that will remain undefined, correct?  You have added a piece to the puzzle that I had not really thought hard about, but I can see that it is an important part.  Again, quite complex.

Lloyd

[rsterne]

Lloyd, first of all a well deserved, and hearty CONGRATULATIONS ! on the latest 2162 fps shot.... Just think how far you have pushed the boundaries in such a short period of time.... truly remarkable.... It shows what can be done when you ignore those who say you can't do something.... *grin*....

I think I have gone as far as I can with using a simple model of the mass of the air that is being accelerated during the shot.... and unless somebody can find an error in my math, or methodology, I don't have a solution.... At 10,000 psi, by using the actual air density, and half the mass, I have pushed the velocity to 2379 fps.... but at 5000 psi the equation spits out 2039 fps with zero projectile weight, and you are already at 2162 fps with a 7.7 grain pellet at less pressure.... Scott, I don't have the math skills, or the understanding of airflow or thermodynamics to even know how to approach going down that path.... Lloyd, it is interesting you mention the length/diameter ratio of the barrel, and it makes sense that would be very important to the real-world results.... but in the equation the barrel volume doesn't care about that, the theoretical maximum FPE is proportional to the volume only, it doesn't matter if you have 1000 lbs. force over 1 foot, or 100 lbs. force over 10 feet, it's still 1000 FPE.... Right there, the departure between theory and reality lands with a THUD !!!

I am excited about one thing, however.... It appears that Lloyd's current spreadsheet, once corrected using the VanDerWaals air density, will work over a HUGE range of situations with only a simple efficiency "fudge" factor.... and being the practical guy I am, that is good enough for me.... If I want to find out what the maximum is for that set of conditions, I will just set the factor to 100% and know we can't get there from here.... I just won't know the details of why.... *LOL*.... Incidently, to get the latest shot to balance in Lloyd's J4 Spreadsheet, using Adiabatic expansion of the air, requires just shy of 74% efficiency....

Bob

[Scotchmo]

...

Scott, I am still trying to get a handle on your entropy application in the calculations.  You seem to have it all going to heat that increases the temperature of the air and therefore the molecular velocity.  Is that correct?  But doesn't entropy also cover  non-recoverable, non-definable energy losses within the system?  Aren't some of those losses in a way  like the efficiency fudge factors that we are currently using.  Entropy losses are often empirically determined and isn't that in a way what we have been doing?  I agree that figuring the heat related energy losses (usage) helps to narrow down where the energy is going and will help to make the fudge factor  a smaller percentage number.  But there are still entropy  losses that will remain undefined, correct?  You have added a piece to the puzzle that I had not really thought hard about, but I can see that it is an important part.  Again, quite complex.

Lloyd

"You seem to have it all going to heat that increases the temperature of the air and therefore the molecular velocity.  Is that correct?"
Yes. It goes to "molecular kinetic energy" - i.e. temperature increase. I mostly ignore the small pellet friction.

"But doesn't entropy also cover non-recoverable, non-definable energy losses within the system?"
No. I'm assuming there are none (or negligible) in your straight forward system.

"Entropy losses are often empirically determined and isn't that in a way what we have been doing?"
Yes. Fluid friction, Reynolds number, I think are mostly derived from empirical data. We are working outside of the range of most previously available data.

"But there are still entropy losses that will remain undefined, correct?"
Not really. We can pretend that they are insignificant until we get there.

I'll throw another theory out here.

Once we start producing heat, we have a work-to-heat engine. At the extreme (infinity) 50% of the original energy ends up as heat (entropy gain). 50% ends up as work (KE of pellet). Maybe like a backwards engine (heat-to-work) cycle. A Carnot cycle can easily do 50% efficiency. At a Carnot cycle efficiency of 50%, we need a temperature spread of 70F to 600F (295K to 589K). The hot side is 589K.

A Carnot cycle is very slow. Ours is instantaneous. The column of air will not have time to equalize in temperature. The 589K spreads out over the air column. High density, 70F(295K) at the inlet, to low density 1660F(1178K) behind the pellet.

Plug that into the equation:

v = 193.7 sqrt ( 1 / Z )

At 1178K, air density D = .0055gr/ci/psia.
Z = 1/D = 0.00275

v = 193.7 sqrt (1 / 0.00275) = 3694fps

I know it looks like circular reasoning.
But at least it's a complete circle. 

The left graph in the picture represents the velocity limit equation that we have been trying to reconcile. But it ignores heat gain. The friction-less system is worthless as a predictor of the upper limit. It only gives you half the picture. And that picture is the lower half of the velocity envelope, which tells us very little:



We need the upper half. The right graph includes fluid friction and heat gain. It defines an upper limit that is more like what I would expect. But it follows Boyle's law, so it too will break down at pressures over 3000-4000psi. Until you use 6000psi nitrogen, we are probably OK.

Maybe the number we should be focusing on is the 50% efficiency. That is the limit. Once you drive it hard enough to generate 50% heat, it's balanced. Velocity will be maxed out. We don't really know what that velocity is. We'll just know when we are getting close. We need to test the theory:

At 50% inefficiency, velocity is maximum.

[39M]

I probably missed this and it might have been mentioned somewhere in this thread.
But I have an idea, not sure if it qualifies as theory, that by changing the length and angle of the "funnel" behind the chamber, that you could tune it to match different barrel lengths.

A long less sharply angled "funnel" might provide less turbulence and be beneficial for a longer barrel. While a wider ratio, more sharply angled "funnel" would benefit a shorter barrel.

Might mean nothing, but just a thought.

[Big Bore Bart]

OK, so if we assume that we have a problem with the air density factor, Z,   (snip)   If we use Steve's estimate that Z = D / 2 (because only half the air column is being accelerated), then Z = 0.02073 / 2 = 0.01037.... Using this value in the equation for the maximum velocity with zero pellet mass, we have....

v = 193.7 sqrt ( 1 / Z ) = 193.7 sqrt ( 1 / 0.01037 ) = 193.7 sqrt ( 96.43 ) = 193.7 x 9.820 = 1902 fps.... This is still too low a value, particularly when you consider it is for a pellet weight of zero.... and has already been exceeded with a pellet weight of 9.1 and 10.3 grains at 4000 psi in a 47.5" barrel....

If the entire air mass is used, instead of half (not unreasonable, because it is all moving), so that Z = D = 0.0207, we get the following....

v= 193.7 sqrt ( 1 / Z ) = 193.7 sqrt ( 1 / 0.0207 ) = 193.7 sqrt ( 48.309) = 193.7 x 6.955 = 1346 fps  (snippedy snip)   I would appreciate it if someone could check through my math and make sure I have not made a mistake somewhere.... It was a long process, and that is always a possibility....

Bob

  Bob; you made me break out the calculator.    
  My take is the barrel is a tube and therefore the boundary layer will cause a constriction.   I ran your equation with a divisor of 3 and got this...

       Z = 0.02073 /3=0.00691

    Plugging that in gives v = 193.7 sqrt ( 1 / Z ) = 193.7 sqrt ( 1 / 0.00691 ) = 193.7 sqrt ( 144.7178) = 193.7 x 12.0299 = 2330.186 fps....

   There is that magic number.
 
   The boundary layer acts as an orifice, due to the friction of the air column at high velocity.  The core of the air column will be moving considerably faster than the outer portions.   IMHO the constriction can be as severe as 60%+. 

   Considering the fact that the pellet is accelerated to ~60% of it's final velocity in the first half of the barrel or less, the final acceleration must be being done by the air making it through the core.   This air could be those molecules that are moving faster than 1640fps.  Either way Steve is still wrong.

[rsterne]

The problem is justifying using only 1/3rd of the air.... It could, in fact, be argued that we should be including all of the air in the barrel when we are using a dump shot, particularly if the reservoir is so large as to maintain a constant pressure.... Ultimately, the only proper way to do the calculation is by integrating the air mass used in increments as the bullet travels up the bore....

I think we are soooooooooo far past the idea of 1650 fps being the limit that we don't even need to think about justifying it any more.... I think that Scott's idea of the velocity gradient, with the air in the center moving much faster than the air on the outside (against the barrel wall).... added to my idea that one "packet" of air pushes on the next, with only the front one pushing on the bullet.... can explain velocities up to 3300 fps.... and I can pretty much state with confidence we won't get there with air.... While it would be nice to know exactly what is happening inside the barrel so we can model it accurately, I'm perfectly OK with just using an efficiency factor to balance the spreadsheets....

Bob

[Scotchmo]

The average air velocity right behind the pellet is higher than the air velocity entering the barrel. The mass flow rate is the same. Think about it!

On Lloyd's last test, the air enters at 1650fps (or 1854fps) and the first molecules to reach the muzzle were going 2162fps. What's the problem?

[rsterne]

That could very well be an accurate description of the situation, Scott.... The mass flow could be the same, but the pressure lower as the velocity increases (ala Bernoulli).... except the "narrowing" of the airstream towards the muzzle is caused by the slow molecules dragging along the barrel.... I don't suppose we will ever know for sure....

Bob

[Scotchmo]

That could very well be an accurate description of the situation, Scott.... The mass flow could be the same, but the pressure lower as the velocity increases (ala Bernoulli).... except the "narrowing" of the airstream towards the muzzle is caused by the slow molecules dragging along the barrel.... I don't suppose we will ever know for sure....

Bob

Flow rate is the same.
Pressure is the same.

Temperature is higher. Velocity is higher. That's the only difference.

[match]

...
I am excited about one thing, however.... It appears that Lloyd's current spreadsheet, once corrected using the VanDerWaals air density, will work over a HUGE range of situations with only a simple efficiency "fudge" factor.... and being the practical guy I am, that is good enough for me.... If I want to find out what the maximum is for that set of conditions, I will just set the factor to 100% and know we can't get there from here.... I just won't know the details of why.... *LOL*.... Incidently, to get the latest shot to balance in Lloyd's J4 Spreadsheet, using Adiabatic expansion of the air, requires just shy of 74% efficiency....

Bob

So using 70% in the spreadsheet would be "high end almost doable"  and 60% would be "readily doable"?