Looking for a great 1st regulator!! HELP :)

[halocline]

For easy math, increase the diaphram area to 2 square inches. Decend 2 ft so we have another 1 psi increase in ambient pressure. The increase is now 2 in sq x 1 psi = 2 lbs of additional force on the diapharm and an additiona 2 psi of IP.

Herman, I understand the forces at work. It doesn't matter how big the diaphragm is, if it's the same size on both sides, the pressure will equalize. It has to. In your example the increase in IP is not 2 PSI, it's 1 PSI. 2lbs of additional force spread over a 2 sq" surface equals 1 PSI. You'd need 4 lbs total additional force to sustain a pressure of 2 PSI on a surface area of 2 sq". I believe you're confusing total pressure with PSI.

I think that's right, anyway.

 

[herman]

You are missing the complete picture, the diaphram is only half of it, all of the force applied by it is concentrated into the push pin. The surface area of the push pin, which does the actual work of opening the HP seat does not change, if you increase the surface area of the diaphram and apply the same force to it, the force on the push pin must go up .....think about it, if increasing the surface area of a diaphram did not increase the force it applies second stage diaphrams could be made tiny-a 1 inch one would do the same work as a 6 inch one. Go back and read (I did last night ) the sections on force in Reg Savvy.


This is way off the OPs topic....hope we are not boring the others to death.

 

[Zung]

Don't worry, we're not reading - Please carry on

 

[halocline]

The surface area of the push pin, which does the actual work of opening the HP seat does not change, if you increase the surface area of the diaphram and apply the same force to it, the force on the push pin must go up .....think about it, if increasing the surface area of a diaphram did not increase the force it applies second stage diaphrams could be made tiny-a 1 inch one would do the same work as a 6 inch one. Go back and read (I did last night ) the sections on force in Reg Savvy.

Please keep this civil. I am not missing the complete picture. Of course the total pressure on the diaphragm increases with it's size. That's basic. The thing that I can't seem to get you to understand is that if the interior size is the same as the exterior size, then the pressure must equalize on each side. I'm frustrated that you can't seem to agree with this. This can only mean that any increase on the exterior of the diaphragm is met by an equal increase on the interior. This, in fact, is how the increase in ambient is equalized with an equal increase in IP. The force on the pin is the exterior ambient force (plus spring) MINUS the interior force (IP).

Go back and look at the very simple numbers that you presented. If a diaphragm has a surface area of 2 square inches, and you increase the depth so that you get an increase of ambient of 1 PSI, the total force on the diaphragm's exterior is 2 lbs. That's right from your example.

Okay, so the interior surface of the diaphragm must produce 2 lbs of pressure to keep the diaphragm from collapsing and pushing the pin which opens the valve. Right? That's IP, you already know this. What you seem to refuse to accept is that the interior surface of that same diaphragm is also 2 square inches, and in order to exert 2 lbs of force on a surface area of 2 square inches, you need to raise the pressure 1 PSI, which is what happens to IP to keep the valve closed.

Lets try 10 square inches. Descend to get your 1 PSI increase in ambient, you get 10lbs of force pushing this big diaphragm in. But, since the interior surfaced is also 10 sq", guess how much pressure increase you need...1 PSI, applied to that 10 sq", is 10 lbs total force, and the diaphragm stays put.

Look, we've both been around this forum a long time, and I hope you realize that this argument is nothing personal; I'd like to keep it that way.

 

[herman]

:blush: SORRY Matt, I did not intend to sound like I was taking it personally. I apologize for my poor choice of words, I really was enjoying a debate (I sure hope that comes across like I intend. ). I learn a lot from them....including this one.

I reread your last post and after some pencil time on paper I realized you were right but there had to be something I was missing. After a couple of hours of confusion I finally figured it out.....the diaphragm is different size, at least in effect anyway. The OB does require 2 diaphragms to be "overbalanced" with the outer one being larger than the internal one. Because the 2 are connected together with a solid rod and pads they end up acting like a diaphragm with different surface areas on each side. In effect it becomes a solid piston with different end sizes. All of which you plainly stated in post 21 and I totally missed.....I have got to learn to read slower, think more and type less.

 

[Luis H]

Hi Herman

I just found this thread. I hope your confusion is now cleared.

The only over balanced (over compensated) regulators are the ones with the solid push rod between the two diaphragms.

The old Conshelf Supreme’s that are filled with silicone oil are not over balanced. The fluid transmits the outside pressure directly to the chamber, which in turns it transmits that same pressure directly to the IP. Therefore the IP increases directly with the outside pressure. It doesn’t matter what size are either diaphragm since there is pressure working on both sides of both diaphragm.

If you remember the old Cyklon 300 didn’t use an outer diaphragm, it just used a large rubber boot filled with fluid (often the preferred fluid was vodka).

With the dry chamber the pressure in the chamber becomes irrelevant as long as it is less or equal to the surrounding. As you mentioned, the solid rod becomes a piston between the two diaphragms.