Hey guys I am still having trouble designing the valve for my hybrid gun.
I have a question here, In the image below what would happen? Will the piston stay in place if the piston seals to the valve chamber?
I am now looking to have a valve that stays open after the shot so that the piston inside my chamber can automatically exhaust the spent gas inside left over from ignition. and was thinking of using something like drawn below with a pop valve behind the piston.
Piston Valve help
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- Just a quick, crude, inproportionate drawing.
left side is the combustion chamber, right side is the valve chamber, and on the right side of the valve chamber but not shown is a pop valve. - valve.JPG (4.88 KiB) Viewed 2690 times
- Technician1002
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That design tends to open as soon as the ignition causes pressure to rise instead of providing a higher pressure for opening. A close ratio piston valve will remain closed until the chamber pressure is several times higher than the pilot pressure.
A good design will provide a piston that starts to move only after the chamber pressure is more than 4X higher than the pre ignition pressure.
A good design will provide a piston that starts to move only after the chamber pressure is more than 4X higher than the pre ignition pressure.
- jackssmirkingrevenge
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As Tech pointed out, with the size ratios illustrated, the piston can be forced open by an unpressurised 1x mix. If this is for a hybrid, you might want to consider a higher ratio, or the addition of a spring to help keep the piston shut until higher pressure is achieved.
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hectmarr wrote:You have to make many weapons, because this field is long and short life
Hmmmm. Yeah my design is no good.
But maybe I can change it a bit like JSR is saying. The only concern I have is that the piston would have to be kinda large. I will be using a combustion chamber of 1"dia. and a max ammo size of 1/2". So my piston I am thinking may have to be 1"dia. + with 1/2" exposed on the combustion chamber side to 90psi.
I was trying to design something that did not use a spring.
O, and JSR, where can I get formulas for working out the piston area and pressure forces on the exposed piston area?
But maybe I can change it a bit like JSR is saying. The only concern I have is that the piston would have to be kinda large. I will be using a combustion chamber of 1"dia. and a max ammo size of 1/2". So my piston I am thinking may have to be 1"dia. + with 1/2" exposed on the combustion chamber side to 90psi.
I was trying to design something that did not use a spring.
O, and JSR, where can I get formulas for working out the piston area and pressure forces on the exposed piston area?
- jackssmirkingrevenge
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As above, or you can actually use GGDT as a quick calculator:
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hectmarr wrote:You have to make many weapons, because this field is long and short life
O, wow thats pretty neat.
O.K. I think I have the math down now.
So to calculate the piston characteristics I came up with:
0.5" diameter of the piston wil be exposed to the combustion chamber side.
area=0.6168". Force=pressure X area. So 90psi X 0.6168" = 55.512 of force.
Now on the other side of the piston the diameter exposed to the valve chamber side is 1.25". The are of that is 3.8553" and to calculate force,
90psi X 3.8553" = 346.977 of force.
So the pressure at which the combustion chamber has to reach to equalize the force on the valve on both sides, is determined by:
Larger Force / Smaller Force X psi. or
346.977 / 55.512 = 6.25 then 6.25 X 90PSI = 562.54psi
So now the piston will not start to open until the pressure in the combustion chamber rises above 562.54psi.
Is this all correct?
Also how in the world do I know what is the unit for the calculated "force"
O.K. I think I have the math down now.
So to calculate the piston characteristics I came up with:
0.5" diameter of the piston wil be exposed to the combustion chamber side.
area=0.6168". Force=pressure X area. So 90psi X 0.6168" = 55.512 of force.
Now on the other side of the piston the diameter exposed to the valve chamber side is 1.25". The are of that is 3.8553" and to calculate force,
90psi X 3.8553" = 346.977 of force.
So the pressure at which the combustion chamber has to reach to equalize the force on the valve on both sides, is determined by:
Larger Force / Smaller Force X psi. or
346.977 / 55.512 = 6.25 then 6.25 X 90PSI = 562.54psi
So now the piston will not start to open until the pressure in the combustion chamber rises above 562.54psi.
Is this all correct?

Also how in the world do I know what is the unit for the calculated "force"

- jackssmirkingrevenge
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Looks correct at a glance 
Force = Pressure/Area
Pressure - lbs/in<sup>2</sup>
Area - in<sup>2</sup>
∴ Force = lbs/in<sup>2</sup> x 1/in<sup>2</sup>
The Area units cancel out, and you're left with lbs
In a nutshell, "55.512 of Force" means it's like you put the piston vertically and puy a weight of 55.512 lbs on it.

It's in my diagram, but you can also derive it:Also how in the world do I know what is the unit for the calculated "force"
Force = Pressure/Area
Pressure - lbs/in<sup>2</sup>
Area - in<sup>2</sup>
∴ Force = lbs/in<sup>2</sup> x 1/in<sup>2</sup>
The Area units cancel out, and you're left with lbs

In a nutshell, "55.512 of Force" means it's like you put the piston vertically and puy a weight of 55.512 lbs on it.
hectmarr wrote:You have to make many weapons, because this field is long and short life