Question about Belleville washers in regulators.

[happymecanic]

Hi guys,

Is there a way to determine the number, dimensions and disposition - ()()()()( - or - (())(())(( - etc  of the Belleville washers (spring discs) in the internal regulator of a PCP air rifle? If for example you have one that’s messed up from the start (not by me). Assuming I know the specs of some spring washers I have bought – 10 mm x 5,2 mm x 0,5 mm - ( https://spaenaur.com/wp-content/uploads/2018/02/Disc-Spring-Washers-D39-D64.pdf ) , the dimensions of the regulator and the high/low pressure I have/want (3000/850 psi), would it be feasible to calculate the quantity and strength of necessary spring washers? What about preload, if necessary, how much? Thanks in advance.

[rsterne]

If you have all the dimensions, including the assembled height of the Bellevilles, the specifications on them, and the space available when the regulator is closed, plus the diameter of both ends of the piston, then it can be calculated to some approximate value.... However I don't know of any simple formula that is readily available to do the job, it would be a matter of figuring out the force on the piston at 850 psi, and then coming up with a Belleville stack where the force equals that when the stack is compressed to the installed height with the seat just closed....

Bob

[happymecanic]

If you have all the dimensions, including the assembled height of the Bellevilles, the specifications on them, and the space available when the regulator is closed, plus the diameter of both ends of the piston, then it can be calculated to some approximate value.... However I don't know of any simple formula that is readily available to do the job, it would be a matter of figuring out the force on the piston at 850 psi, and then coming up with a Belleville stack where the force equals that when the stack is compressed to the installed height with the seat just closed....

Bob

Thank you Bob, I was so hoping you'd chime in . I guess I need to calculate the surface area of the piston head (regged pressure) and of the piston stem tip (high pressure) to be able to determine the difference of pressure between the two, which will therefore be useful to determine the force of the spring washers pack? I made this prehistoric hand-drawing  , hope it can be of some help:

[CableStop]

It would help if you can measure the height of the washer when decompressed, that way you can get a more accurate "h" dimension which is roughly "overall height - thickness of the washer".  Knowing "h" would help to calculate the load at max deflection and find the correct constant to use for calculating load at different compressions.  Measuring some similar belleville washers I have from a different regulator I got an "h" value of 0.011 inches so that's the distance the washer has to compress to be completely flat.  Using that number along with the dimensions of your washers and the properties given in the document, I found the max load to be 135 pounds and an h/t of 0.56.  Table 1 in the document has the constants you multiply against the max load to find load at different compressions based on the h/t value, 0.56 falls right between 0.5 and 0.6 in the table so I don't know which would be best to use.  Again this is where having a more accurate overall height measurement would help.

As for finding a layout that will work in this application the document more or less says "idk figure it out yourself", they give formulas for parallel and series configurations but in a regulator you most likely need a hybrid set up which they don't give any formula for.  I think Bob is right that there is no good way to calculate this in one formula, it's basically just trial and error based on height of the stack, required deflection, and the spring rate needed.  It seems like a parallel stack of washers can be treated like one thick washer but I think that will change the h/t.  Sources are conflicted as to whether or not thick washers require more complex calculations to find "h" but it seems like most think it's ok to use the "overall height - thickness" equation. 

OK so here is where my own analysis begins and where I'm not entirely sure I'm correct so ideally if someone more knowledgeable comes along and reads this they can give their input as well.  The main thing to remember here is that IF this is correct, it's only correct for linear spring stacks where the "layers" are all the same.  As in ((()))((())) and not progressive stacks like ((()))(())() where the spring rate increases with compression.  I think it will mostly come down to trial and error to get exactly the results you are looking for, calculations can only get you in the ball park.

I would start by finding the required height of the stack by measuring the distance from the face of the piston to the face of the regulator where the small piston enters.  Figure out how much deflection you need in the stack by determining the number of adjustment wheel turns you you want between max and minimum settings. Divide the adjustment distance by the height of the stack to find the percent change in deflection. Calculate the force on each side of the piston when the reg is open by subtracting the area of the small end from the area of the large end and multiplying that by 850 to find the spring rate needed at the lowest setting.  Find the constant in table 1 that corresponds to the amount of deflection you want, then divide 850 by the constant and see if it gets you to 3000, if it doesn't then find a new arrangement and repeat. 

To calculate the spring rate of a given arrangement you have two options.  In a parallel arrangement you can just multiply the number of washers against the spring rate of one washer to get the total force required to flatten that stack, a series arrangement will have the same spring rate as a single washer with more deflection.  In combination you would multiply the max deflection load of one washer by the number of washers in a parallel stack to get the max load, and multiply the distance of deflection by the number of parallel stacks to get max deflection. 

Hopefully this is useful, it got way longer than I initially planned on it being but the first half is really just a summation of the document you linked plus me plugging those numbers into a belleville calculator.  The second half is my own ideation on the matter and may or may not be right, YMMV, good luck.

[rsterne]

That regulator is difficult to assess, because the force on the piston from the HP air would appear to change significantly, depending on the input pressure.... unless I am not understanding the operation.... When the reg. is seated, it would appear that the HP air is pushing against an area of (0.196^2xPI/4) - (0.118^2xPI/4) = (0.0302) - (0.0109) = 0.0193 sq.in.... At 3000 psi, that would be 58 lbs., but at 850 psi only 16 lbs.... If the output pressure is 850 psi, the force trying to collapse the Bellevilles would change, because on the large end we have (0.411^2xPI/4) = 0.133 x 850 = 113 lbs.... so it would swing from (113 - 58) = 55 lbs. when the tank is full to (113 - 16) = 97 lbs. at setpoint.... 

Bob

[CableStop]

That regulator is difficult to assess, because the force on the piston from the HP air would appear to change significantly, depending on the input pressure.... unless I am not understanding the operation.... When the reg. is seated, it would appear that the HP air is pushing against an area of (0.196^2xPI/4) - (0.118^2xPI/4) = (0.0302) - (0.0109) = 0.0193 sq.in.... At 3000 psi, that would be 58 lbs., but at 850 psi only 16 lbs.... If the output pressure is 850 psi, the force trying to collapse the Bellevilles would change, because on the large end we have (0.411^2xPI/4) = 0.133 x 850 = 113 lbs.... so it would swing from (113 - 58) = 55 lbs. when the tank is full to (113 - 16) = 97 lbs. at setpoint.... 

Bob

I'm glad I'm not the only person confused about that. The drawing was posted while I was in the process of writing the response so I didn't get to see it until after I posted. The original picture was confusing but I assumed that there was just something I wasn't seeing and approached it like it was a normal reg. Like if the small end of the spool went in and out of an o-ringed chamber that would work despite being a weird design. As it stands from the drawing it does look like you'd have a wild variation over the range of tank pressure.  What gun is this reg for?

[rsterne]

Well there IS an O-ring on the small end of the piston, it's inside the chamber instead of on the spool (which slides through it).... I think it would work fine without the step down in diameter to the knife-edge seal on the end.... I have NO idea why it has that feature.... 

Bob

[CableStop]

I think that one at the end of the spool is there to seal the spring chamber and the knifes edge seals against a flat seat that threads into the end.  The way I understood the drawing was that it was the housing and the internals side by side.

[happymecanic]

Gents I forgot to include the high-pressure-side Oring locking ring in my drawing, my bad . You can see it in the picture of the parts. This regulator goes in a Listone Industry Taichi-X .22 pcp rifle.

[CableStop]

Does the small end of the spool piston fit through the brass ring to press against the white delrin?

[happymecanic]

Does the small end of the spool piston fit through the brass ring to press against the white delrin?

Yes it does. The Oring goes in first and abute on the bottom of the threaded hole, then the brass lock-ring screws in, then the white delrin piece.

[happymecanic]

That regulator is difficult to assess, because the force on the piston from the HP air would appear to change significantly, depending on the input pressure.... unless I am not understanding the operation.... When the reg. is seated, it would appear that the HP air is pushing against an area of (0.196^2xPI/4) - (0.118^2xPI/4) = (0.0302) - (0.0109) = 0.0193 sq.in.... At 3000 psi, that would be 58 lbs., but at 850 psi only 16 lbs.... If the output pressure is 850 psi, the force trying to collapse the Bellevilles would change, because on the large end we have (0.411^2xPI/4) = 0.133 x 850 = 113 lbs.... so it would swing from (113 - 58) = 55 lbs. when the tank is full to (113 - 16) = 97 lbs. at setpoint.... 

Bob

I hear you loud and clear on that, Bob. This is something that's bugging me too; if we had a constant high pressure input I understand my reg would work as expected and deliver a constant low-pressure set-point. But as a pcp rifle tank loose pressure, this cause an imbalance in the reg, in my case anyway . How do other pcp regulators work to compensate for that imbalance? Or am I missing something?

Here's a diagram of the pressure tube/regulator/receiver adapter from my owner's manual:

[CableStop]

Most regulators have the seat on the adjustment knob not the piston so when it seals no air can get through to the small end, when it opens the air pressure is allowed to pass through the the piston freely so the pressures equalize on each side.  That way the springs are the only thing holding it open since the pressure on each side of the piston cancels out the force on the small end.  That's why I said originally to subtract the area of the small side from the large side, since that difference is the force trying to close the piston.  Once it closes again there is no longer any air trying to push the piston open except for whatever surface area is exposed to the seat which is usually a virtual pin hole so the force only changes by five or six pounds over the course of a 3000 psi fill.

[sb327]

I agree with Bob and CS, unless I’m missing something, as designed, that reg would have a remarkably wide swing.

I may have missed it in the body of the thread but you must have been experiencing issues to have brought you into this rabbit hole. ??

According to the parts diagram vs your piece, they have added another oring and switched to a single atm vent hole instead of two. Neither changes anything. The extra oring probably was because they were getting leakage caused by the single oring passing by the vent hole in tube (just guessing).

Dave

[happymecanic]

I agree with Bob and CS, unless I’m missing something, as designed, that reg would have a remarkably wide swing.

I may have missed it in the body of the thread but you must have been experiencing issues to have brought you into this rabbit hole. ??

According to the parts diagram vs your piece, they have added another oring and switched to a single atm vent hole instead of two. Neither changes anything. The extra oring probably was because they were getting leakage caused by the single oring passing by the vent hole in tube (just guessing).

Dave

Yes I had quite a few issues with this rifle, the regulator being the last one I found. I don't know the original configuration of the spring washers, and I just found another major quirk: the regulator body is about 0.031'' smaller than the part of the tube it goes in. No more wonder why the Orings were extruding after a very short period . And I thought it was due to cheap OEM orings . Now I need to make a new regulator body before I can seriously hope to do some testing. Will update as soon as I can.

[sb327]

If you are going to make a new reg body, a redesign might be in order.

Dave

[happymecanic]

If you are going to make a new reg body, a redesign might be in order.

Dave

I'm all in for suggestions dear fellow .

[sb327]

Surely we can come up with a better design than they have done.

What is the od and id of the small end of piston? It needs to be plugged with ball or wafer and then the shaft cross drilled.

Then the hp end of body needs to be designed to allow for existing oring seal to function on piston but only allow hp air to enter through the center of an adjustable plug.

I can draw this out but it might be a couple days.

Dave

[CableStop]

Open source regulators courteousy of GTA, I like it.

[CableStop]

Can you measure the overall height of the washers you have?  I can't accurately find a washer arrangement without knowing that measurement. Also by 3000/850 do you mean a reg that adjusts from 850 to 3000 PSI or do you want it fixed at 850?