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What if some water left or entered the vacuum barrel

I know from practical experience it is much easier to jerk the spear free of the piston than just pulling it out with a steady pull, especially with metal pistons. To do this after testing the gun on land by working the piston up and down in the barrel without latching all you need to do is push the spear tip into soft ground and stand on the opened out flopper. A sudden upward pull on the tank and the spear disengages easily. If a piece of wood is used with a hole drilled through it then you can stand on the piece of wood instead.
 
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The energy spent on undocking the harpoon and piston will be the same in both cases, both with slow undocking and with shock!

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I used to load the shaft into the gun using hand T loader, for about 10 - 20 cm and than holding the spear let the inertia to do the job of disengaging the shaft from the piston.
 
Try to measure and calculate it.. The initial force to the spear from the water is 1079 N (for numbers at the beginning). This force might be present for the first 0.2 - 0.4 mm of braking distance. For the same braking distance the spear travels 2 x further after detaching from the piston.
 
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Same task but with not used - new STC piston
 
Today I did a measurement of energy necessary to detach a new STC piston from the shaft tail end cone.
After initial inserting the cone into the piston it was possible to push it deeper for additional 3.5 mm applying force.
I used two different methods to see how the results would be.
First method:



Second method:




Required energy is higher than I thought to be. But it depends on the force required to load the gun or of the pressure in a speargun.
It also depends on the brand and type of the piston used. I suppose that with my Mares piston currently in my gun that energy is lower.
 
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Remember when a speargun shoots the spear has reached its maximum velocity as the piston hits the shock absorber, so its momentum will be as high as it ever will be, momentum being mass times the velocity of the shaft. The impulse required to jerk the shaft free of the piston is given by the detachment force times the time it operates over or force times its duration. Impulse and momentum have the same units, MLT^-1. The impulse required should be a lot smaller than the momentum of the shaft as we see every time we shoot our guns underwater.
 
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In a pneumatic vacuum gun, the harpoon breaks away from the piston before it makes direct contact with the shock-absorbing sleeve if there is enough water in the barrel for this!
 
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I know how the impulse is defined. But how would you measure the detachment force and the time it operates over? The force could not be measured using house spring dynamometer because the impulse after detaching the shaft from the piston would damage/destroy the scale. You should have a special scale for such purpose. How would you measure the time? During detaching the shaft from the piston the force is changing from the max to some lower value before separation happen. It is not a steady-constant force. And the time belonging to each force is different.
Why do you think that obtaining the energy necessary to jerk the shaft free of the piston is not enough?
 

Energy E spent on undocking = And the work done to separate the harpoon from the piston! The energy is the same in both cases! At docking A = Force kg per Distance !
 
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Zahar said:
Energy E spent on undocking = And the work done to separate the harpoon from the piston! The energy is the same in both cases! At docking A = Force kg per Distance !
Click to expand...
That is exactly what I measured: energy spent on undocking which is same to the work to separate harpoon from the piston. It was in case of measurement I did: 2.3 J in first method of measurement and 2.4 J in second method. The difference in results is acceptable small because I did the testing in discrete steps varying the height of fall for few cm until a get the separation. I made a calculation base on Potential energy Ep=mgh which is equal to kinetic energy
Ek=1/2 mv^2, at the time of impulse of jerk being applied.
 
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These calculations don't really tell you anything useful in how you can make changes, the point I was making is the energy expended in yanking the spear out is trivial compared to the energy in the rapidly moving spear. As I mentioned on an earlier post for vacuum barrel guns bar fitting a relief valve behind the shock absorber face the only other thing you can do is remove the friction fit of the spear tail in the piston and have the vacuum hold the spear in the gun instead. That is what the "Taimen" does, the hole in the piston nose is just a socket for the spear tail to sit in. You could change the tapered conical spear tails by machining them to a cylinder and achieve the same effect by putting a bung in the piston nose to drive the spear without it penetrating too far into the piston’s hollow nose.
 
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Here are some numbers:
Mass of the shaft m = 0,3 kg
Speed of the shaft before separation from the piston v = 30.68 m/s
Energy of the shaft before separation from the piston Ek = 141.69 J
After detaching from the piston, if for that is used E = 2.3 J of energy, the
energy of the shaft would be Ek1 = 139.40 J and a new
speed of the shaft would be v1 = 30.43 m/s, what is 0.25 m/s lower.
Change in momentum of the shaft is from 9.24 Ns to 9.16 Ns what is 0.08 Ns
If the shaft would be mainly detached after 2 mm separation from its place in a piston that would require a force F = 1150 N (2.3 J/ 0.002 m) and that separation would happen in time of something more than 0.065 ms. (0.08 Ns/1150 N)
For full separation shaft have to exit from the piston at least for 3.5 mm.

I made this calculation because the indicial pressure of water left in muzzle is according to:

What if some water left or entered the vacuum barrel

I know from practical experience it is much easier to jerk the spear free of the piston than just pulling it out with a steady pull, especially with metal pistons. To do this after testing the gun on land by working the piston up and down in the barrel without latching all you need to do is push...
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1079 N.
From this calculation it is obvious that with that force the shaft could not detach from the piston because it needs more than 1150 N so the high pressure of the water will bi present for at least during the beginning of braking distance.
 
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I should add some pistons in the past had claws which held a ball projection on the end of the spear tail. When the piston hit the shock absorber face the claws automatically opened and let the shaft go. The problem with this system was the ball tail could be bent on its spigot attachment and the claws would wear out or break depending on the design. Last seen in the GSD pneumatic guns using nylon pistons.
 
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Here is what would happen if shooting under angle from the bottom having some water left in a muzzle.



 
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I made a things little more clear, I hope..

p1 = p2 + 1/2 q v2^2 - 1/2 q v1^2 = 1 bar + 1/2 q (v2^2 - v1^2)
p1 - pressure of water in muzzle
p2 - outside, ambient, pressure
v2 - the speed of water leaving the muzzle
v1 - speed of the piston/water being pushed by the piston
q - (1020 - 1030) kg/m3 (for sea water)
v2 = A1/A2*v1
A1 - cross section of the barrel (piston)
A2 - cross section of gap where water can exit the muzzle forward
 
The braking distance for the piston is 5 mm. Because the length of the shock absorber body is about 18 mm, the shaft tang would be still inside the shock absorber, immediately after braking. The OD of the tang is 8 mm for 7 mm shaft, and the ID of shock absorber is 8.5 mm, so that is the only space for the water to go forward.
 
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I should add some pistons in the past had claws which held a ball projection on the end of the spear tail. When the piston hit the shock absorber face the claws automatically opened and let the shaft go..
 
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droomdrrom said:
I should add some pistons in the past had claws which held a ball projection on the end of the spear tail. When the piston hit the shock absorber face the claws automatically opened and let the shaft go..
Click to expand...
An echo, is that all you want to say?
 
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popgun pete said:
An echo, is that all you want to say?
Click to expand...
Zahar, I know you'll find this very funny too..

I tried to simplify it further using simplified Bernoulli's equation
p1 + 1/2 * q * v1^2 = p2 + 1/2 * q * v2^2
Because in this case
v2 > v1 -> v1^2 << v2^2;
p2 < p1
We can write even more simple
p1 = 1/2 * q * v2^2

p1 - pressure of water in muzzle in front of the piston
v1 - speed of the piston/water being pushed by the piston = 30 m/s
p2 - pressure in space between shock absorber ond the tang of the spear p2 < p1
v2 - the speed of water leaving the muzzle = 207,27 m/s
q - (1020 - 1030) kg/m3 (for sea water)
A1 - cross section of the barrel (piston)
A2 - cross section of gap where water can exit the muzzle forward
v2 = A1/A2*v1

p1 (Pas) = 1/2 * 1020 kg/m3 * 207.27^2 m2/s2 = 21910035 Pas = 219 bar

So the pressure of water in front of the piston would be at least 219 bar (in case the mass of the piston would be 0 g.
Becauce the mass of the piston is about 10 g, the pressure of water p1 in front of the piston would be even higher...?
I tried to take into account the mass of the piston as well.The piston I was considering had a mass of 11.7 g and a volume of 4.7 cm3.The density of the piston is 11.7/4.7 = 2.48 g/cm3. The amount of water in front of the piston could be about 1 cm3, which is 1 ml. If I imagine the piston as a liquid of the same mass and volume and add the mass and volume of the water in front of the piston, then the total mass is 11.7 + 1 = 12.7 g and the volume is 4.7 + 1 = 5.7 cm3. It is as if I were applying the formula for pressure p1 to a liquid whose density is q = 12.7/5.7 = 2.23 g/cm3 = 2230 kg/m3. Now we can calculate the new value of p1:

p1 (Pas) = 1/2 * 2230 kg/m3 * 207.27^2 m2/s2 = 47901351 Pas = 479 bar

Regardless of which result is more accurate (219 bar or 479 bar) the fact is that a pulse of enormous pressure will be created in front of the piston...
Mass of the piston:


This is to determine the volume of the piston.

.
 
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