What if some water left or entered the vacuum barrel

[popgun pete]

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.

 

[Zahar]

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.

The energy spent on undocking the harpoon and piston will be the same in both cases, both with slow undocking and with shock!

 

[tromic]

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.

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.

 

[tromic]

It remains to calculate the force with which the closed water will push the harpoon out of the piston and the duration of this force! As the pressure increases, the spear will break away from the piston before the pressure reaches a critical value for the passage of water through the O-ring into the receiver! The force of disengaging the harpoon from the piston can be measured by scales!

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.

 

[tromic]

I made a measurement, how deep the shaft tail end (the cone) goes into the piston in my Mares Cyrano.
After inserting the shaft into the piston, at the end of loading the speargun the shaft goes into the piston additional 0.8 mm deeper.
That means that the water can not replace the shaft cone in a piston before the shaft leaves the piston for at least 0.8 mm.
After that water can enter behind the shaft into the newly created empty space and the pressure of water in front of the piston will drop.

View attachment 58257

Same task but with not used - new STC piston

 

[tromic]

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.

 

[popgun pete]

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.

 

[Zahar]

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.

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!

 

[tromic]

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.

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?

 

[Zahar]

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 !

 

[tromic]

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 !

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.

 

[popgun pete]

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.

 

[tromic]

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.

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.

 

[popgun pete]

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.

 

[tromic]

Here is what would happen if shooting under angle from the bottom having some water left in a muzzle.

 

[tromic]

To calculate the water pressure (p1) in front of the piston, when it hits the shock absorber, if there is some water left, there is a very simple formula derived from Bernoulli's equation:
p1 = p2 + 1/2 q v2^2 - 1/2 q v1^2
p1 = p2 + 1/2 q (v2^2 - v1^2)
p2 - ambient pressure of water where spearfishing
q - specific density of water (1020 - 1030 kg/m3 for salt water)
v1 - speed of the piston/water being pushed by the piston
v2 - speed of the water being pushed between the shaft and the shock absorber - through the annular gap
For the example given at the beginning v1 = 30 m/s and v2 = 207.27 m/s
After calculation we get the result for p1 = 218.66 kgf/cm2 or 214.4 bar. 1kgf/cm2 = 0.98 bar

Even a sealing gasket is OK if water would not be empty from the speargun over pressure would happen, for a very short time but enough to push some water behind the piston into the barrel.
I case of Mares shock absorber during the first 1 mm of compression of shock absorber actually only the hydro damper is working.
Regular mechanic dumper can absorb only 0.59 J because for 1 mm compression the max force is only 70 kgf. Peak force for the water is 199 kgf and after 1 mm 159 kgf.

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

 

[droomdrrom]

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.

 

[droomdrrom]

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..

 

[popgun pete]

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..

An echo, is that all you want to say?