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Shock absorber

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tromic

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Aug 13, 2007
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I made an estimation how high is a force on the shock absorber after the shoot.
I made assumption that the shock absorber can move 8 mm under the impact of the piston. Also supposed that the loading effort of the gun is 30 kg and the speed of the piston is 30 m/s. If the weight of the piston were 15 g, than the force on the shock absorber just after the shot would be F = 86 + 30 = 116 kg. Normally the force would be 30 kg. But, if for some reason the absorber could move just 4 mm. than the force on the shock absorber (and the piston) would be 2 x 86 + 30 = 202 kg! That might happen if the shock absorber tubing were thicker then should be.
I measured the OD of the Mares and Asso shock absorber (tubing OD). Mares was 16.9 mm and Asso was 17.4 mm. I suppose Mares would be safer for the gun. Absorber housing ID in the muzzle was in both cases about 17.7 mm.

 
popgun pete

popgun pete

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I made an estimation how high is a force on the shock absorber after the shoot.
I made assumption that the shock absorber can move 8 mm under the impact of the piston. Also supposed that the loading effort of the gun is 30 kg and the speed of the piston is 30 m/s. If the weight of the piston were 15 g, than the force on the shock absorber just after the shot would be F = 86 + 30 = 116 kg. Normally the force would be 30 kg. But, if for some reason the absorber could move just 4 mm. than the force on the shock absorber (and the piston) would be 2 x 86 + 30 = 202 kg! That might happen if the shock absorber tubing were thicker then should be.
I measured the OD of the Mares and Asso shock absorber (tubing OD). Mares was 16.9 mm and Asso was 17.4 mm. I suppose Mares would be safer for the gun. Absorber housing ID in the muzzle was in both cases about 17.7 mm.



The force on the shock absorber should include the effort to jerk the spear free of the piston as when the piston hits the shock absorber spear and piston are travelling as one. The piston is stopped, but the spear carries on.
 
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tromic

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.Right, Pete! It is additional about 2 J (my estimation). So the force might be about 110 kg instead o 86 kg, or 220 kg in case of 4 mm breaking distance. If some water would be trapped in a muzzle (broken vacuum system) it is clear that under such a high pressure some water would be injected behind the piston into the barrel!


This is how mares shock absorber works in asso muzzle. I made measurement using a vice and a dynamometer, so it is not very precise but might be interesting. 5 mm of compressing of shock absorber requires close to 1400 kg. If measured with normal measuring equipment the result would be less, but that would be a big value too!


 
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tromic

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I made a calculation of energy from above graph. For energy of the piston of about 8 J the compression of shock absorber tubing is less than 4 mm. It would be impossible to reach compression close to 8 mm or more. So the efficiency of the shock absorber is not the best and the stress to the piston and shock absorber is higher than should be.

 
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popgun pete

popgun pete

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The effort to pull the spear out of the piston seems easier on a plastic piston than it is on a metal one. Sometimes it requires a sliding hammer blow to free the spear from the latter when wiggling the spear around fails to release it, whereas the plastic pistons tend to let go when you do this while pulling on the spear shaft at the same time. Of course some guns don't have any shock absorber at all, principally because there is no room to fit one as is the case on most of the forward latching pneumatic guns. This is due to the release mechanism being installed in the muzzle which takes up space where the shock absorber would normally be. Generally these guns have a lighter piston as it has no metal tail for a sear tooth to latch onto and the piston face only pushes on the rear of the spear tail instead of tightly gripping it. The ball tail spears on pneumatic guns such as the GSD models and the earlier Technisub guns aim to release the spear tail without much mechanical effort by employing a ball gripping device that opens up as the piston hits rather than using a frictional spear to piston coupling, however these guns had a shock absorber as well, on the body of the piston for the GSD guns (rubber sleeve) and in the muzzle and on the piston for the Technisub guns (metal coil springs used on both elements). The Technisub pistons carried two springs, a heavier spring for the shock absorbing function and a lighter spring to operate the ball retaining device when you loaded the gun so that the ball was trapped in the piston and then released when the device hit the shock absorber anvil.

Hence lots of ideas have been used, but I imagine trial and error long ago worked out the dimensions of shock absorbers, however that was for wet barrel guns. The loss of hydraulic damping with dry or vacuum barrels means that if the piston is travelling faster at impact then the gun's shock absorbing capacity needs to be increased, however if the speed is the same (using lower charge pressure in the gun) then this capacity can probably remain unchanged.
 
popgun pete

popgun pete

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Besides reducing the stress of an energetic impact on the large screw threads that hold the muzzle onto the inner barrel, the shock absorber has to limit the forward travel of the piston seals so that they do not move beyond the confines of the inner barrel tube or compressed air may escape the gun's reservoir. On a dry or vacuum barrel gun which has no muzzle relief ports the inner barrel could extend right up to the rear face of the shock absorber anvil. Now Tromic has previously mentioned the possibility of shortening the piston as it would no longer require the section that bridges the length of the internally opened out area which contains the relief ports on a standard pneumatic gun (in commentary on the forthcoming Sporasub "One Air") and a shorter muzzle length would be a consequence (with no relief ports section). However the possibility of trapped water forcing its way into the air reservoir when the piston hits the end of travel with nowhere to go due to a seal leak at the muzzle of a vacuum barrel gun that has unexpectedly allowed some water to enter presents a potential problem for internal gun corrosion. How would you know that it had even happened? Maybe a slightly weaker shot than expected.

Some time back I mentioned the benefit of using flexibly covered muzzle relief ports as a "blow-off" valve if the vacuum barrel did leak water as the hydraulic pressure would then open the ring of muzzle valves rather than water pushing past the piston seals going back inside the gun. Another possibility would be to use a relief valve in conjunction with the muzzle shock absorber's movement under impact, however such ports would be small and located forward of the rear outer sealing edge of the shock absorber body (it then being an integral part of the muzzle sealing system) which would expose the ports if it underwent excessive forward travel. "O" rings sliding past small ports located in a ring around their periphery to open them up can damage the sealing edge of the ring, so any damage here would then allow water to penetrate the vacuum barrel system via the shock absorber actuated relief valve system. Not so good!

I guess the only way to know if this "trapped water due to slow spear to muzzle seal leaks" will ever be a problem is to find out if vacuum barrel gun users have been finding water in the oil of their guns during servicing. If the muzzle/spear shaft seals have been replaced on a preventative maintenance basis then they will have had no wet "dry shots", so no water should get inside the gun unless the spear surface itself is damaged or there has been off-axis loading of the shaft, but the number of shots put through the gun will also be a factor as only a small amount of water may transfer each time.

The stroke of the shock absorber body movement under impact in a dry barrel gun no longer having any bearing on the forward placement of the seals on the piston body means that the seals could be placed very close to the front of the piston body and immediately behind the rear termination of the spear tail socket in the piston's front face. However that minimizes the volume available for accommodation of water inside the gun above and beyond that which is in there when the spear tail is first inserted in the muzzle, there always being some water there which assists in inner barrel lubrication during the shot. This then poses the question as to whether water gets past the shock absorber anvil pressing firmly on the front face of the piston with the gun discharged. On metal pistons it does as they have a small radial hole in the side of the spear tail piston socket to stop hydraulic lock on the shaft tail, but plastic pistons don't have this hole. Over time some water must get inside there, but does it ever fill up? I posted a thread about washing this area out on vacuum barrel guns to eliminate saltwater corrosion in the muzzle, but is the annular volume accommodation ever filled up or does it have some capacity to swallow the effects of minor water leaks? The answer will have a bearing on how much piston length can be used in front of the outermost piston seal.
 
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tromic

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Pete, that is why I prefer the simplest Tomba, with rubber tubing. It serves as a valve and it is easy to switch to water barrel operation, if necessary.
 
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tromic

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Piston has an energy of approximately 10 J. In my opinion better shock absorber is that, which absorbs 10 J on a longer distance.
4 mm is better than 2 mm. So, from the results, STC (2 mm) was worse on a test than Mares (Cyrano, Sten 13) shock absorber (4 mm). That means, with Cyrano, Sten 13 shock absorber, both piston and shock absorber would last longer. Sten 11 shock absorber (2,5 mm) was worse (?) than older Cyrano or Sten 13 shock absorber.



 
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tromic

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Additional test with O-rings!

 
popgun pete

popgun pete

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I have not seen a "Sten 11" except in photos, so was surprised to see the smaller diameter alloy muzzle body. Does this mean the plastic nose cone it backs onto is also smaller at the front end? The cross-sectional area of the shock absorber sleeve is reduced, so there is less material in the sleeve unless it is thicker and mounted on a smaller diameter plastic body, which looks like that may be the case. That smaller diameter may make the rubber sleeve stiffer and thus harder to compress and hence you get the 2.5 mm travel result on the shock absorber body.

I assume two things happen when the rubber sleeve is compressed during shooting. First it progressively fills up the available space which is probably already filled with water in the small gap around it inside the muzzle body. That water is rapidly squeezed out and is completely replaced by rubber, so that is the rubber moving from a thinner wall sleeve which is longer, to a shorter one which is now thicker. After that the rubber itself has to compress to a smaller volume if the shock absorber body moves any further forwards. There would probably a significant increase in the force required from this point on, but I don't know when or if it occurs. It could probably be calculated from the respective volumes as to when the rubber sleeve shortens in length sufficiently to fill up all the available space in the muzzle body. I know that the rubber sleeves bear marks near where they are squeezed, but only at one end, next to the shock absorber body rear shoulder. I have never found a broken sleeve, but then I don't usually open the muzzles up. The marks on the rubber outer surface may be from abrasion caused by it touching the surrounding muzzle body interior, in which case the rubber sleeve never is compressed fully outwards to touch all along its length.

One way to find out is to temporarily fill the annular gap in front of the shock absorber with something that will easily crush (wax?) and not recover its shape and then shoot the gun, that will tell you what the energy absorption stroke of the shock absorber body is. Either that or something that will leave a scratch mark on the shock absorber body to see how far forwards it actually moves in the muzzle. Of course a more powerful shot will move it further forwards, but you need to establish a starting point. Another possibility would be to paint the inside of the alloy muzzle with a thin sticky coating and see if the rubber sleeve picks up this coating all along its outer length after a shot, or alternately the coating inside the muzzle is marked by being rubbed against the rubber sleeve.
 
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tromic

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Sten 11 shock absorber (shortened because of damage) with additional 2 orings 6 x 2 mm. Braking distance is 3 mm at 10 J of energy of the piston.



 
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tromic

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I have not seen a "Sten 11" except in photos, so was surprised to see the smaller diameter alloy muzzle body.
...
That smaller diameter may make the rubber sleeve stiffer and thus harder to compress and hence you get the 2.5 mm travel result on the shock absorber body.
...
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The previous test with 2.5 mm result had been made with stiffer, replacement rubber sleeve, not the original. That is why I made a new test, the latest, but I had to shorten the rubber sleeve for about 2 mm because it was damaged on the edge. I couldn't obtain the original spare part from my supplier shop. The original sleeve was 16 mm long. Shortened with two additional O-ring it was 17,5 mm. With original, new sleeve the braking distance might be between 2,5 and 3 mm, for 10 J piston energy. Before measurement I supposed it would be more than 5 mm.
 
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popgun pete

popgun pete

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Here is another calculation using a "Work-Energy" approach. If the piston is travelling at 30 m/s when it hits the muzzle and it weighs 15 grams (0.015 Kg) then its kinetic energy is given by 0.5 x mass x velocity^2 which equals 0.5 x 0.015 x 30^2 = 6.75 joules. For the moment we will neglect the mass of the spear and the air mass travelling inside the gun. So to bring the piston velocity down to zero that energy has to be dissipated as noise and heat in the muzzle. Deforming the rubber absorbs energy, some of which turns into heat, but in water this heat is rapidly lost, so we do not notice it, but we do hear the sound of piston impact. Some of the energy is stored in the rubber sleeve and then returns the shock absorber back to the discharged gun position. Work and Energy are the same measurement units and work is defined as the force applied over a certain distance, say X, which let us say is the incremental movement of the shock absorber during the impact. With the gun discharged the piston is already leaning on the shock absorber with force 30 kg (using your figures), so initially the shock absorber is moved in by a distance D where there is a balance achieved between the force applied by the piston (under the reservoir's air pressure) and the resistance of the rubber sleeve to being crushed forwards to this distance D. When the shock absorber has moved to distance (X + D), during the shot, the piston has stopped moving forwards and the force on the piston is now 30 kg again (maybe a tiny bit less due to volume change with the piston slightly further forwards in the inner barrel), so then the rubber sleeve recovers and the stored energy pushes the shock absorber back to distance D where it always is when the gun is not cocked. The rubber sleeve is now in an equilibrium, so to speak, between rubber sleeve compression and the air pressure inside the gun.

If we add the spear to the equation then the kinetic energy is much greater as it is spear mass and piston mass combined moving at 30 m/s, but as the rapid deceleration commences the spear tail breaks free of the piston. So at some instant the spear and piston combined energy is going into the rubber sleeve, then it drops to only that of the piston. Now we know that spearguns of the same model have exactly the same muzzle even though they shoot different length spears, so either the muzzles are designed for the longest models with spare absorption capacity for the shorter ones, or absorption of any of the spear's kinetic energy does not really come into it. If it did then the spear would slow slightly while leaving the muzzle. Instead it is just the force to pull the spear free of the socket in the piston face which I think would be the same for all of the guns in that model series regardless of their length. The energy or work to pull the spear free is the force to do it times the distance that the force is applied over, however the distance is very short due to the tapered cone friction connection between spear tail and piston socket, so the energy used will be small as the distance will be small, maybe only a few millimetres.

Let us say that it takes 5 kg to pull the spear out, which we need to turn into a force of 5 x 9.8 Newton (9.8 conversion or 9.8 m/s^2). Say the force is exerted over 2 mm of travel distance Z inside the piston socket which is 0.002 metre, then the work done is F x Z = 0.098 Joule. Add that work-energy to the 6.75 Joule for the piston obtained earlier and we have 6.848 Joule.

Hence I think the energy absorbed by the shock absorber may be somewhat less than 10 joule, for this example anyway. If the gun was fired without a spear then the piston would travel much faster, the energy stored in the gun working only on accelerating the piston mass and not on the combined spear and piston mass, therefore with much greater terminal velocity. That is when the energy absorption capacity of the muzzle shock absorber may be exceeded and I know that such a shot will destroy the shock absorber in a "Seabear" breaking the polyurethane bush into tiny fragments that acts as the shock absorber in those guns. The plastic pistons and shock absorber bodies in Mares guns (and others) may be cracked by the experience, so I don't suggest anyone try it to find out what happens, in air anyway.
 
popgun pete

popgun pete

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Just to continue the above example, if the spear has a mass of 800 g (0.8 kg) and is travelling at 30 m/s then its kinetic energy will be 0.5 x 0.8 x 30^2, which equals 360 Joules. So the moving spear has plenty of energy to break free of the piston and will lose about 0.1 Joule in doing so, based on the figures assumed previously. The spear will not fall out of the gun with the muzzle pointed downwards due to gravity if it takes 5 kgf to pull it free from the piston. When metal pistons wear out, or more likely the spear tail's taper jamming into them wears down thus losing most of that taper, then the spear can fall out, but it is a marginal situation. I replaced one of my "Sten" 90 cm gun's spears for this very reason, although fine surface rust on it had totally replaced the cadmium plated coating by then, it had never been bent in all those years, unlike some other spears. Most of my Mares spears were bought when a local dive shop quit their old pneumatic gun parts stock, so all the spears were cadmium plate finish which is a dull golden colour, but this plating method has been largely discarded due to heavy metal poisoning issues with the liquid wastes discharged after the electroplating process.
 
popgun pete

popgun pete

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We can use the numbers here to work out a few more things. At distance D of shock absorber travel the piston is stationary and pushing against the shock absorber anvil or stopper, so the rubber sleeve must be producing a resistive force of 30 kgf on the front piston face to balance the air pressure producing 30 kgf of force going in the opposite direction. A force of 30 kgf is equal to 294 Newton (multiply 30 x 9.8). (Note when using these Energy and Work formulae you have to use the correct measurement units for the calculations). That force, say Fi, is going to be the initial contact force on the shock absorber sleeve which will rapidly increase as it is being axially squeezed, otherwise to absorb 6.848 Joules of energy it would take a shock absorber stroke (X) of 23.3 mm to stop the piston and jettison the spear (6.848/294 x 1000 = 23.3) with only 294 Newton of force being continually applied over that distance. We used the formula (W = F x X) here by transposing it to read (X = W/F), that way as we know W (which is the same as energy E) and we have assumed F, so we can obtain the value of X. No practical shock absorber travel stroke is going to be that long using rubber elements, however with metal coil springs it may be a different matter!

Let us say that the force increase is linear with the sleeve being progressively crushed and that it has to move 4 mm to stop the piston, so X = 0.004 metres. Now we can use the "area under the graph" calculation to work out what the final force Fs has to be which is similar to the calculation used for estimating the energy of a pneumatic gun given the cocked and discharge pressures and the working course of the piston and its cross-sectional area. This is the familiar rectangle with a right angled triangle sitting on top. The area of the lower rectangle will be (Fi x X) which is 294 x 0.004 = 1.176 Joules. The area of the triangle above it will be (Fs - Fi) x 0.004/2 = Fs x 0.004/2 - 294 x 0.004/2 as Fi = 294 Newton. The triangle's area is half of a rectangle of the same dimensions, hence we divide by 2.

So summing the two areas, rectangle plus triangle, we have [1.176] + [Fs x 0.002 - 1.176/2] = 1.176/2 + Fs x 0.002 which will equal 6.848. Therefore we can calculate Fs as being equal to 3,130 Newton or 319 kgf. [using Fs = (6.848 - 1.176/2)/0.002] Now the force increase is not linear, so the upper triangle will not have the straight line hypotenuse we have assumed here, it will be a curve instead as Tromic has shown, but this calculation gives us a ball park figure. Now you will understand why undoing the muzzle on a pressurized gun is potentially lethal if your body has to emulate the shock absorber by opposing the piston with 319 kgf to finally stop it with your head!

Note that I have not included the initial compression of the rubber sleeve to distance D in the absorption of the piston energy, principally because I don't know what that distance D is, however the work done in that travel section will be (294 x D)/2 as we go from zero force to 30 kgf or 294 Newton. If D was say 2 mm or 0.002 metres then we would already have 0.294 Joules absorbed, so the calculation in the paragraphs above would be for the remainder (6.848 - 0.294) = 6.554 Joules or a difference of 4%. By starting from zero compression the graphical calculation of work/energy is much simpler being only the area under a triangle. Having originally assumed that the force increase is linear (which it probably is not) the gradient from zero to 294 Newton would have to be the same as that from 294 Newton to 3,130 Newton. Now I have assumed that X = 0.004 metres, so what will D be on that linear increase basis? 3130 - 294 over 4 mm is 2836/4 or 709 per mm, hence to be at a value of 294 Newton, D will be 0.415 mm.

Then we would have ((Fs - 0) x (X + D))/2 = 6.848. If (X + D) = 0.004415 then Fs = 2 x 6.848/(X + D) = 2 x 6.848/0.004415 = 3,102 Newton or 316 kgf. Not much difference, for these assumptions that is, as before we obtained 319 kgf.
 
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tromic

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Peter that was an interesting calculation of shock absorber. Here is another subject for estimation for a physicist. I made a rough assumption that impact energy of the piston of 10 J should be same to the kinetic energy of the water that leaves the gun though the holes of 4 mm and the gap between the shaft and the muzzle boring. I made a calculation having on mind that S1*v1=S2*v2. (cross section * speed).


 
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tromic

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Aug 13, 2007
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Maybe would be better to make a design of shock absorber using a real energy of the piston. Mares Sten 13 is 10 g. The mass of shock absorber is 4,5 g. At 30 m/s the energy of the piston hitting the shock absorber is 4,5 J. After hitting the shock absorber the speed of both piston and shock absorber is 25 m/s. So I should calculate the water escaping cross section to have such a speed of water that kinetic energy equals to 4,5 - 5 J. This could be calculated from S1*v1=S2*v2. v1 is 25 m/s. S1 is a cross section inside muzzle filed with water that serves to slow down the piston on the length of 8 mm. The amount of water inside a muzzle has mass "m" that would have 5 J kinetic energy at some speed v2.
 
popgun pete

popgun pete

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A hydro shock absorber has to use some physical resistance to the flow of the water contained inside it so that this water is ejected at a lesser rate than it would be if the water's exit was totally unrestricted. The water exiting via the muzzle shock absorber ports moves at a higher speed than the water travelling inside the shock absorber, but the restriction of the ports changes the mass flow rate which is lower than it would if the muzzle was unrestricted. At any given instant the mass flow rate inside the shock absorber body and that going out the (combined) ports is the same, so that is where the cross-sectional areas times the velocity for each location equivalence equation comes in (the one you mentioned above), but that will not tell you about the magnitude of the decrease in the mass flow rate inside the shock absorber body. To achieve a considerable drop in mass flow rate, with water as the transfer medium inside the shock absorber, the exit ports have to be very small, so the ports you have indicated are too big, in fact they are about the size of muzzle relief ports in the older model pneumatic guns. Although you have two hydro shock absorber sets in a sense, using the front ring of four ports and the rear ring of four ports, they represent a large exit for a relatively small volume of water in each section being pumped out by the respective advancing faces of the shock absorber body, namely the rubber sleeve edge (rear set) and the shock absorber nose (front set).

The "Doroganich" pneumatic gun uses a hydro shock absorber muzzle and there are no ports at all in the shock absorber's water pumping section, the water has to flow out around the small annular gap between the extended piston nose and the muzzle's outermost bore which closely surrounds it. The cylindrical step formed by the larger diameter piston body immediately behind the extended piston nose is essentially the face of the water pumping piston that pushes the water out of the section of muzzle which is nearly the same diameter as the piston body is before it hits the shoulder of the smaller bore that is just large enough to allow the extended piston nose to pass through it which acts to block off most of the exit pathway for the water coming out of the pumping section. I think for your design to work you need to make the exit pathways much smaller by just using the gaps around the concentric parts where they pass through each other rather than adding radial ports. Those gaps cannot be too small as they may jam with the introduction of sand and grit.

The advantages of the design, maybe a quieter piston impact; the disadvantages are more piston travel to stop it moving with the seals being located further back on the piston in order to keep them inside the inner barrel tube, but here through using two muzzle pumping sets you have halved the stopping distance at least compared to a single pumping unit in the muzzle.
 
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tromic

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Thanks Pete for your observation! Those 4 ports for water exit (not 4 + 4) were calculated based on 10 J. They would be smaller for 5 J. But my problem is I suppose something was wrong with my calculation. I took for calculation of energy the speed when piston hits the shock absorber. But over the distance of 8 mm that speed drops down to 0 m/s. Also, as you mentioned, the mass of water in muzzle also drops down to 0. I am not sure how to take this facts in calculation.
A guy on russian forum (Alex Chebotarev-HaNTeR) made his own gun design with hydro shock absorber but he was using water barrel. He sent me some pictures of his design. I think he was working on that for long time. It is easier to make this kind of damper with water barrel than with vacuum barrel, but the principle is same.

I made a calculation, something to start with (probably wrong, or maybe not?)
Mp = 10 g _____Mass of the piston
Ma = 5 g ______Mass of the shock absorber+water
S1 = 2.16 cm2 __ Cross section in muzzle pushing the water
S2 = 0.65 cm2 __ Cross section of all water escaping ports (gap around shaft + 4 x 4.5 mm boring in muzzle)
M = 1.7 g _____ Mass of water in a muzzle
v1 = 25 m/s ____Initial speed of piston+shock absorber, after piston hits the shock absorber with speed 30 m/s
v2 = 83.5 m/s __Speed of water escaping muzzle ports and shaft gap
Epa = 4.68 J __ Initial energy of piston+shock absorber at 25 m/s
Ev = 6 J ______Energy of water escaping from muzzle ports with speed 83.5 m/s
Fw = 585 N __ Force of the water that slows down piston and shock absorber on length 8 mm.
 
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popgun pete

popgun pete

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The viscosity of the fluid has a bearing on the result, using just the changing cross-sectional areas and mass of the fluid and the velocity is only part of the calculation. That is why shock absorbers use different viscosity oils to change their damping performance and that determines the degree of slowing of the mass flow rate in their pumping action. The fork oil we use in pneumatic guns comes in different grades for that reason. So the calculation of performance is not so straightforward as it was for a rubber sleeve damping unit.
 
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