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¿Como Tomba sí hace

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tromic:

tromic:
You have to use both hands. One to hold the body and the other to rotate the locking tubing (stainless steel tubing) for 90 degre). In locking postion there is a small indent to secure the locking tubing in locked position so it can not be accsidentally unlocked. Rotation of 180 degree from locked position facilitates the insertion or removal of the spring.  First idea was to use 1,35 mm spring wire but it was very dificult to rotate the tubing so I change it for 1 mm spring wire. In next tuning step I would try it with 1.1 mm or  1.2 mm...

Tienes que usar ambas manos. Una para sujetar el cuerpo y la otra para girar el tubo de bloqueo (tubo de acero inoxidable) 90 grados. En la posición de bloqueo hay una pequeña hendidura para asegurar el tubo de bloqueo en la posición bloqueada para que no se pueda desbloquear accidentalmente. La rotación de 180 grados desde la posición bloqueada facilita la inserción o extracción del resorte. La primera idea fue usar un alambre de resorte de 1,35 mm, pero era muy difícil girar el tubo, así que lo cambié por un alambre de resorte de 1 mm. En el siguiente paso de ajuste, probaría con 1,1 mm o 1,2 mm...

tromic:

tromic:
This is an example of how this kind of device is used in Rusia/Ukraina in their lakes and rivers.
Este es un ejemplo de cómo se utiliza este tipo de dispositivo en Rusia y Ucrania en sus lagos y ríos.

https://postimg.cc/wRJr6q75

tromic:

--- Cita de: tromic en 19 de Octubre de 2022, 09:47:16 am ---What would happen when shooting from the bottom at an upward angle...
¿Qué sucedería al disparar desde abajo en un ángulo hacia arriba...

--- Fin de la cita ---

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...

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