Complicated pneumatic/hydraulic question.

[miniwally]

More to come but this is simple hydraulics

Force is created by the area of the piston and the pressure of teh fluid behind it.

putting kid to bed and will be back

 

[CherryK5]

by the volume, small to big and vice versa..... pressure is always the same thruout a hydro system..

But 1psi on a 5 square inches should yield 5 psi on the smaller 1 square inch piston.

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Edit: assuming my tank will act as a hydraulic piston like i'm thinking

 

[miniwally]

I always like to seperate things into the individual components

So provided there is no way for the fluid and air to swap places in your tank setup and you have your 1 PSI air over an area of 5in^2 you have 5 psi in the exit line of that tank.

Since you have 5 psi (edit 5lbs not PSI) leaving that tank you now have converted your 1 psi air to 5 psi fluid.

Now your 1in^2 ram will have 5psi (edit 5lbs not PSI) pushing on that ram and thus by default in this example the rod of that ram will exert 5 pounds of push on something.

However, air over hydro is going to make a piss poor setup for steering, the air is too compressible.

 

[CherryK5]

I always like to seperate things into the individual components

So provided there is no way for the fluid and air to swap places in your tank setup and you have your 1 PSI air over an area of 5in^2 you have 5 psi in the exit line of that tank.

Since you have 5 psi leaving that tank you now have converted your 1 psi air to 5 psi fluid.

Now your 1in^2 ram will have 5psi pushing on that ram and thus by default in this example the rod of that ram will exert 5 pounds of push on something.

However, air over hydro is going to make a piss poor setup for steering, the air is too compressible.

So it will work!

This is for suspensiin , not steering btw. And thank you! You reinforced what I was thinking.

 

[miniwally]

So it will work!

This is for suspensiin , not steering btw. And thank you! You reinforced what I was thinking.

So your making an overly complicated air shock?

 

[CherryK5]

So your making an overly complicated air shock?

Just something to keep my mind and hands busy. And not a shock. A spring.
Although... some restrictors and different weight fluids... hmmm

 

[miniwally]

Just something to keep my mind and hands busy. And not a shock. A spring.
Although... some restrictors and different weight fluids... hmmm

It's still the spring part of the air shock you are creating, add in a piston, valves, etc. and you have an air shock and theres some people that are doing that really well.

 

[Fordum]

Well, the answer is simple, lets see if I can explain it.........

Use a single tank, no outside lines. Fill it half full of oil, put two pressure gauges in it. One in the air, one in the oil.
Put 10PSI of air pressure in the top. Do you expect the gauge in the oil to have more pressure than the air?

If you can make that work, you have solved the energy crisis...........

Here is your problem. You are confusing Pounds per Square Inch with Pounds.

Take a 10 square inch piston against oil. Put 1 pound of weight on it. You do not get 10 pounds per square inch of oil pressure, you get .1 PSI, because that one pound of pressure is spread out over 10 square inches.

To get 10 PSI, you would have to apply 100 pounds of weight.

With air, you trade volume for pressure. If you put 10 PSI on top instead of an actual weight, you will get your desired 10PSI on the oil, but it will take lots of air volume to do it.

Its hard to explain without diagrams.

But air over hydraulic can act as a reserve source of power for the system, letting the tank act as a spring.

I have been up a long time today, so I am starting to fade. I'll try to be clearer tomorrow.

 

[miniwally]

It's well known that with our steering systems specifically hydro steering that if you need more power out of your system you either up the pressure or ram size. Just because we're discussing air doesn't change anything. If you need to provide more power from a hydraulic ram you increase the size of the piston/rod and you get more power, your not doing anything but providing a larger area for a set PSi to push on.

In this case he is doing both but still it's psi. 1 psi means there is 1lb pushing on every square inch. 5 square inches with 1 lb pushing on every square inch = 5 lbs. of force, translate this into the outgoing line from the large tank and run that line into the smaller ram that has an area of 1 square inch. The ram is now pushing with 5lbs of force. 1 psi air pressure has become 5 lbs of force. I think where this is confusing is that even if you increase the size of the rod in the smaller ram you don't get more than 5lbs of force.

SO yes you can "multiply" the power of the 1psi of air but it doesn't matter what you put on the other end of the bigger tank you only have the set out put that is dictated by the area of the tank and the PSI of the air.

FYI we have to lower our spring rate on the the race car to take into account the amount of force that the 200PSI gas charge puts on the 7/8" dia shock shaft. We actually get the vehicle to rise three+ inches just with the gas charge. Same damn thing here he's just running it through a few more hoses and assemblies.

 

[FrostBite]

So it will work!

This is for suspensiin , not steering btw. And thank you! You reinforced what I was thinking.

He made a typo, the exit line will have 5 lbs on it not 5 PSI. Every component in the system will have the same pressure on it. Think about it, PSI is pounds per square inch, so pounds (of pressure) over square inch (area). If you put 1 PSI in, there is 1 pound of pressure on every square inch of everything, whether it be the tank walls or the surface of the fluid in the tank. The area in the term PSI and the area of the fluid cancel each other out if you work the units out leaving you with just pressure, so just 5 pounds.

Well, the answer is simple, lets see if I can explain it.........

Use a single tank, no outside lines. Fill it half full of oil, put two pressure gauges in it. One in the air, one in the oil.
Put 10PSI of air pressure in the top. Do you expect the gauge in the oil to have more pressure than the air?

If you can make that work, you have solved the energy crisis...........

Here is your problem. You are confusing Pounds per Square Inch with Pounds.

Take a 10 square inch piston against oil. Put 1 pound of weight on it. You do not get 10 pounds per square inch of oil pressure, you get .1 PSI, because that one pound of pressure is spread out over 10 square inches.

To get 10 PSI, you would have to apply 100 pounds of weight.

With air, you trade volume for pressure. If you put 10 PSI on top instead of an actual weight, you will get your desired 10PSI on the oil, but it will take lots of air volume to do it.

Its hard to explain without diagrams.

But air over hydraulic can act as a reserve source of power for the system, letting the tank act as a spring.

I have been up a long time today, so I am starting to fade. I'll try to be clearer tomorrow.

Like this guy says.

 

[FrostBite]

Another example. Say you have a tank large enough to have a 3ft diameter access door on the side. If you put 5 PSI in the tank what would happen if you unbolted the door with out releasing all the pressure? I mean its only 5 PSI. Since the door is a 3 ft circle that means it has a 1.5 ft radius, which is 18 inches. Since the area of a circle is pi x radius^2 that means the door has an area of 1017.4 inches ^2. With 5 pounds of pressure for every square inch of door that means the door has over 5000 pounds of force pushing on it. When you undo the last bolt the door is going to fly off like a missile.

Now if you wanted to drain the air out of the tank through a half inch pipe that means you could stop the flow with your thumb because the force would only be 0.2 pounds through the half inch pipe. Pi x .25^2 = 0.2

That's how hydraulics work, by keeping the pressure constant but varying the amount of area it is acting on.

 

[miniwally]

My post was a bit misleading

Your "ram" will have 5lbs of force acting on it via the tank with air over fluid. It isn't 5 PSI it's 5lbs. You can't change that or multiply it.

Up the air pressure and the force coming out of the tank rises just like a normal simple hydro setup. OR up the area for a set PSI and the force rises.

 

[DEMON44]

Like has been mentioned. Not a complicated question. The struts on our big haul trucks are set up nitrogen over oil. 4 struts, Thats literally what carries the 600ton of truck and dirt around the mine.

 

[DEMON44]

Your just talking about making the gas, remote reservoir right? Ya it works.

First one is of rear struts on Cat 797B. Second one is rear struts on Komatsu 930E electric truck. The box is anchored with 2 slings, is what you sort of see in the way. Ignore that.

 

[CherryK5]

I was trying to find a way to use a smaller ram and keep a low enough pressure for my oba ssystem.. A 2.5" diameter ram will need close to 200psi in the upper chamber and 80-110 in the lower for sway control.

 

[DEMON44]

Ya man. That's **** all, you just need to figure out the ratios and pressure you will need specificly for your weights and application. But if I'm pretty sure I'm understanding what you want to do, it'll work just like our charged struts on the haul trucks. Though I'm not really sure why you wAnt to scratch build hydraulic struts like that. You wanna offroader suspension? Or street cruiser?

 

[CherryK5]

Ya man. That's **** all, you just need to figure out the ratios and pressure you will need specificly for your weights and application. But if I'm pretty sure I'm understanding what you want to do, it'll work just like our charged struts on the haul trucks. Though I'm not really sure why you wAnt to scratch build hydraulic struts like that. You wanna offroader suspension? Or street cruiser?

It will be in the rear of my trazer

Just a fun project honestly. Something to help me learn and understand the characteristics of air springs and air struts. And play with hydraulics and pneumatics a little.

 

[DEMON44]

Fair enough. They will be hydraulic struts, not air springs, not air struts. Basically the load rides on the oil and the nitrogen is the cushion. For our cat trucks, every service they get drained fully and reset with 7" of new oil, and 475psi nitrogen on top of that, for an overall of around 16"

 

[CherryK5]

Fair enough. They will be hydraulic struts, not air springs, not air struts. Basically the load rides on the oil and the nitrogen is the cushion. For our cat trucks, every service they get drained fully and reset with 7" of new oil, and 475psi nitrogen on top of that, for an overall of around 16"

Haha, technically yes. The main reason for the oil is to save the seals. Since my theory has been disproven I will just run a ram slightly longer than i need, fill it with enough oil to keep it well lubed and run the ram capped so that at the end of the stroke it acts as a rudimentary internal bump stop as the pressure spikes.

 

[miniwally]

I was trying to find a way to use a smaller ram and keep a low enough pressure for my oba ssystem.. A 2.5" diameter ram will need close to 200psi in the upper chamber and 80-110 in the lower for sway control.

Use more than one, dual "rams" per corner? Up the size of the "piston" so you are pushing on a larger area.

It will work but honestly I think your going through some serious headaches and pointless math when you can probably buy used air shocks or ORI's for cheap.