Anatomy of a N-Forcer (industrial gas spring)

[QVTom]

I've been working with the N-Forcer gas springs for a while now.  After discharging the gas from this example I decided to take it apart to see what makes it tick or extend in our case.

First, the colored dust shield that covers the retaining ring must be removed.  A small jewelers screwdriver picks it out easily.  Next the C-Shaped wire retaining ring must be removed, this proved challenging but I won in the end   Once the dust shied is removed you need to push down on the bushing cartridge( scroll to next pict) to make some room under the ring, then using a small blade screwdriver,  pry the end of the ring out of its groove and another to lift the end up, the ring will pop out with some authority.  Wear glasses! Once the ring is off just pull on the shaft and the guts will easily come out.


The following are the bushing and seal cartridge.  The fist pic is the Shaft end and the second is the internal end.  As you can see both the shaft wiper and the shaft seal are separately replaceable.  There is also an O-ring that seals the cartridge to the spring casing.



end of the spring casing.  The other end that houses the fill port seems to permanently installed, maybe friction welded.


Piston.  As you can see there is no sealing ring, force is manifested by the differential of the front/rear piston face area.


This is the fill valve.  I expected some kind of check ball but it is much simpler. To discharge a allen wrench is used to turn the valve body clockwise until the gas escapes. To recharge the valve is seated counter-clockwise effectively compressing the seal slightly, when pressure is added the gas flows thru  a small port in the side of the valve body pushing or blowing by the seal to get to the inside of the chamber and when the fill pressure is removed the static pressure in the chamber now presses the seal tight.  Smart!



General parts arrangement.


A fair amount of hydraulic oil inside.  These things are messy!


Time to put it back together!


Edit:

As I was reassembling this thing I had some additional thoughts.  I remember a post by scotchmo about a smaller than 19 mm gas spring that seemed to be speed limited or lacked the snap needed for spring guns.  The clearance between the piston head and the chamber wall may have something to do with it.  This spring has 0.03 clearance, I can see a scenario where the oil is dragging or bypassing the piston  and/or the pressure is not equalizing on both sides of the piston fast enough with a tight clearance.  Maybe more is better for a spring gun.

Tom

[Paul68]

NIce breakdown. I didn't expect there to be any appreciable amount of oil in there.

[Scotchmo]

QVTom,

I just saw this post. I don't know if you still have the gas spring apart, but if you do:

Would you weigh the two sets of parts that move in relation to each other? I have speculated that the weight of the piston/rod is about the same as the body/seal. In which case the direction that it is installed would not matter as far as total piston mass.

That large amount of fluid is surprising. That explains why mine continues to weep small amounts even after thousands of shots. I installed mine body forward and am now thinking that shaft forward might minimize the weeping of fluid.

Does the fluid appear to have any other purpose besides lubrication?

[QVTom]

scotchmo,

I measured the weight of the components when it was disassembled for exactly that reason.  This spring has an extend piston shaft length of 12mm compared to a catalog part # so the differential of a standard unit will even larger.  Spring body 138 grams and piston shaft 70 grams, almost 2-1!

I too was surprised by the fluid volume.  It was described as a teaspoon or so by the the engineer at Automation Die Mold.  I believe its sole purpose is to keep the seals conditioned for the life of the spring.  As you said, these things weep and it doesn't seem related to the number of cycles.  I have weepers that have never been cycled.  I can see the remote possibility of the oil transferring from one side of the piston to the other in certain orientations and possibly slowing the rate of extension.

Tom

[willbird]

Piston.  As you can see there is no sealing ring, force is manifested by the differential of the front/rear piston face area

I don't understand this part of your post.

If you restrained the piston in the cylinder, say in a large vice, with the piston installed BACKWARDS as I understand it the force it would exert once pressurised would be exactly  the same as if the piston was installed correctly, the "head" on the piston if I "get it" correctly is just to prevent the piston from escaping the assembly at the end of the stroke ??

The force it exerts would be equal to the charge pressure times the square inches of the piston "rod"....in effect it is more of a ram than a piston in a cylinder ?? The force can only work upon the dia of the piston rod for lack of a better term...because the "head" of the piston is not sealed to the bore the gas pressure cannot work upon it ??

It may also act to some degree to provide  a cushion to prevent the piston rod from moving so quickly that it destroys important stuff...it may slow the maximum velocity that the piston rod can move ??

To make a short point longer as far as force exerted I do not see how the differential in diameters comes into play at all ??

:-).

Really cheap gas springs, like used in automotive stuff I often times could see no fill ports, I assumed they assembled them inside some kind of pressurised fixture.

Bill

[QVTom]

If you restrained the piston in the cylinder, say in a large vice, with the piston installed BACKWARDS as I understand it the force it would exert once pressurised would be exactly  the same as if the piston was installed correctly, the "head" on the piston if I "get it" correctly is just to prevent the piston from escaping the assembly at the end of the stroke ??

If installed backwards the force would be = pressure * PI *Rod radius^2 . 2500*0.2227^2*3.14 ~ 404 lbs.   

Standard installation the force is pressure * (Piston area - (Piston area - Rod area)) ~ 405 lbs.

So the force is basically the same either way and with out doing the math I'd bet the volume of the cylinder will expand a similar amount as the rod extends. You are correct but function as I described is still accurate.  Thinking of the head as a retention device only is not quite precise because it does present its two normal faces to the gas pressure and effectively creates force by the differential of its front and rear areas.

Good reply.  I did have to think some but  not too much thankfully.



Tom

[Scotchmo]

The fact that it is acting on the shaft area rather than a piston explains why some brands/models of gas springs respond slowly. Maybe I could vent the face of the pseudo "piston" by drilling a couple of small holes. Thereby allowing it to respond at a faster rate.

[willbird]

If you restrained the piston in the cylinder, say in a large vice, with the piston installed BACKWARDS as I understand it the force it would exert once pressurised would be exactly  the same as if the piston was installed correctly, the "head" on the piston if I "get it" correctly is just to prevent the piston from escaping the assembly at the end of the stroke ??

If installed backwards the force would be = pressure * PI *Rod radius^2 . 2500*0.2227^2*3.14 ~ 404 lbs.   

Standard installation the force is pressure * (Piston area - (Piston area - Rod area)) ~ 405 lbs.

So the force is basically the same either way and with out doing the math I'd bet the volume of the cylinder will expand a similar amount as the rod extends. You are correct but function as I described is still accurate.  Thinking of the head as a retention device only is not quite precise because it does present its two normal faces to the gas pressure and effectively creates force by the differential of its front and rear areas.

Good reply.  I did have to think some but  not too much thankfully.



Tom

I'm not understanding how any force can be exerted upon the head if it is entirely within the pressure vessel ??

The pressure inside the vessel presses against all sides of the head equally from all directions ?? The head would act dynamically maybe to control motion.

But I am not seeing where the .025% difference in force comes in.

Probably I am just thick headed, but you can explain maybe :-).

The easiest way I can think of to explore it would be to make the head a slip fit on the rod, but with an O-ring sealing it to the shaft, and see if it creates any force, if it did it would end up on the exit end of the gas spring.

[QVTom]

But I am not seeing where the .025% difference in force comes in.

 0.025% Really?  Lets call it a rounding error.

[Motorhead]

Just finding this post and wanted to add something ....

Just as in Race car/motorcycle suspensions they address "Sprung Weight" by mounting shocks / dampers so the bulk of the there weight is carried by the chassis and the lower mass center rod travels with whats stroking/traveling.

So by flipping the ram end for end, your also changing the total piston weight and therefor also altering it's momentum.

[QVTom]

Yes, and the case with the N-Forcer, the difference in mass is greater than 2-1.  Very significant.  It's nice to have options

Tom

[rsterne]

I'm going to borrow a drawing of a PCP valve to try and explain the force generated by the piston.... Consider the exhaust port blocked and follow along....



If Pr = Pe (ie the same pressure on both sides of the piston), there is NO net force on the red portion of the piston.... ZERO....
The force on the piston rod A is the pressure times the area of the black piston rod.... Force = (Pr - Pa) times Area A....

Only if the red portion of the piston is a close enough fit to the sides of the cylinder to generate a DIFFERENTIAL in pressure on one side of the piston during movement will the force not be as desribed above.... If the piston moves rapidly (as in an airgun) and the clearance is small (and even more so if there is fluid in the way), Pressure Pe will be higher while the piston is extending, actually REDUCING the total force available.... Scotchmo indicates this may occur with some Gas Rams, in which case drilling holes through the piston to eliminate that (temporary) pressure difference would eliminate (or nearly so) that "braking effect" which would be the greatest at the end of the stroke (full extension)....

Bob

[QVTom]

Bob, I think misinterpreted my statement about differential force.

Piston.  As you can see there is no sealing ring, force is manifested by the differential of the front/rear piston face area.

 
I was referring to the force of the gas pressure normal to front and rear side of the piston, not the dynamic pressure differential on each side of the face during extension or compression for that matter.

Dynamically  the pressure differential is working against the static force for sure.  Hence the conclusion of the OP.

Tom

[ericw]

Hmmm just to point out a math equality that was missed:

If installed backwards the force would be = pressure * PI *Rod radius^2 . 2500*0.2227^2*3.14 ~ 404 lbs.   

Standard installation the force is pressure * (Piston area - (Piston area - Rod area)) ~ 405 lbs.

So the force is basically the same either way

piston area-(piston area - rod area)) = rod area,    so one equation is identical to the other anyway.

[rsterne]

Tom, as I tried to point out, there is no net force on the piston head, only the force on the rod area.... What Scotchmo posted about installing the piston backwards, with only the rod inside and no piston head at all, will produce EXACTLY the same force.... Take away the red area on my drawing, and the STATIC results are the same.... As Eric pointed out, it amounts to the same thing anyway....

Bob

[QVTom]

Thought we had all that worked out already 

[Scotchmo]

Tom, as I tried to point out, there is no net force on the piston head, only the force on the rod area.... What Scotchmo posted about installing the piston backwards, with only the rod inside and no piston head at all, will produce EXACTLY the same force.... Take away the red area on my drawing, and the STATIC results are the same.... As Eric pointed out, it amounts to the same thing anyway....

Bob

I posted about installing the entire gas spring forward (shaft forward) vs backwards (body forward). That only affects the total piston mass of the gun. It is obvious that it has no affect on pressure.

Your instance is not so obvious unless you know the internal configuration of the gas spring.

Otherwise, I agree with what you said about the force being the same regardless of which direction the shaft was installed in this type of cylinder. Though backwards would not work for practical purposes, as a gas spring assembly would come apart as soon as you removed it from a preloaded condition.

[rsterne]

If you restrained the piston in the cylinder, say in a large vice, with the piston installed BACKWARDS as I understand it the force it would exert once pressurised would be exactly  the same as if the piston was installed correctly, the "head" on the piston if I "get it" correctly is just to prevent the piston from escaping the assembly at the end of the stroke ??

Sorry, it was willbird that stated that, not Scotchmo.... I was just trying to point out that he was correct, since the piston head doesn't seal into the cylinder it has nothing to do with the static force generated....

Anyway, we all agree.... the force is the pressure times the area of the piston rod....

Bob

[QVTom]

[willbird]

Tom, as I tried to point out, there is no net force on the piston head, only the force on the rod area.... What Scotchmo posted about installing the piston backwards, with only the rod inside and no piston head at all, will produce EXACTLY the same force.... Take away the red area on my drawing, and the STATIC results are the same.... As Eric pointed out, it amounts to the same thing anyway....

Bob

I thought about this several times last night, glad some others chimed in. I was going to go all in with the final hypothetical construct I came up with.

Take an automotive intake or exhaust valve, make a pressure vessel with a sealed hole for the valve stem to slide in. Put the valve with the head inside the pressure vesel, apply 500 psi, measure force exerted by the valve stem, reverse the valve, again apply 500 psi to the vessel, again measure force exerted by the valve stem, it will be exactly equal. All in :-).

Bill