regards hydraulic circuits, can you do the following

[scott33]

New here, hello, a google on my topic brought up hydromech's thread here somewhere. hydromech seemed the right person for the job. I would like to know if you could have the following hydraulic setup.

A 10-15mm diam and say 100mm long cyclinder and piston (piston 1, lever A) provide pressure like say a motorcycle levered hydraulic brake circuit.

The exit pressure/fluid line is a tube exiting perpindicular to the said cylinder. At the other end of the cylinder, opposite end to the the levered piston, is another piston 2, its lighter spring loaded function is fluid pressure release/moderation.

So I'm imagining in response to force on the lever A which pushes the piston 1, there is then an exit pressure which climbs and then plateaus (as piston 2 retracts in response to pressure).

The force on the piston 1, from the lever A, is spring resisted, but this lever, if given enough force, will move sufficiently to mechanically (via more levers ...'B') exert negative pressure on the pressurised circuit. (say this levers B system connects to a piston 3, acting on the cicuit)

So the system is imagined to have an exit pressure, in response to increasing input lever pressure, that initially increases in pressure, plateaus, and then drops off to zero .

I imagine it compact, which is why try to have one cyclinder providing two components to minimise weight and space, If it weighed a lot more more than say a motorcycle brake front brake reservior it wouldn't be so viable.

Apart from it being potentially completely unworkable (I dont know)

Any thoughts on that very much appreciated, not sure if I should continue to imagine this one or not.

Thanks
Scott.

 

[Hydromechdude]

Perhaps I'm not thinking well, but drawing a schematic of what you're envisioning would probably be quite helpful.

 

[scott33]

Thanks Hydromechdude for taking the time and Ok attached is a (rudimentary and rough) sketch of the imagined circuit.

[

http://files.engineering.com/getfil...1-a7da-f87f3cbb3f0f&file=hydraulic_diag_2.jpg

]

I should have added previously that the input force on lever 1 may cycle haphazardly through min to max or part thereof for periods from say 0.5sec to a few seconds and then have no force acting for some further period of a few seconds.
The no. 2 levers would not function as is I am sure but the gist of that setup is there, and the piston 3 and its cylinder could be somewhere else but the requirement for compactness makes as it is drawn attractive (altough perhaps not workable ...of course I am not sure).

The graph at the bottom of the diagram shows the required exit pressure as a function of force on lever 1.

You may guess its function, I'm reluctant to make it explicit, for all I know it may be a good (applicable) idea, but I hope the idea is examinable to some basic extent at this level of abstract.

Again thank you for your time, if it is way off the mark (in theory or workability) ..I can take the hit.

Scott.

 

[desertfox]

Hi scott33

If piston 3 is mechanically linked to lever arm 1 in such a way it should retract it won't have a choice.
Whilst I realise its only a sketch, altough piston 3 is mechanically linked to piston 1 lever, the mechanical advantage doesn't look right.
Piston 2 won't retract if fluid can flow down the exit tube due to it having a lower pressure at the far end of the tube.
We need more information in terms of your spring forces etc.

desertfox

 

[scott33]

Hi desertfox,
Thanks for taking a look at this.

ahh yes about the levering (as drawn) onto piston 3 ...it doesn't work, couldn't quite figure it out exactly in short order. But what I wish to imply is that essentially as lever 1 reaches the inner range of its movment (ie as input force - F1 increases) lever system 2 comes into play and is used to provide an outwards force on piston 3, ie. a negative pressure is created and I imagine this to act on the exit pressure ie bring it down to zero 'ish.

The 'max pressure' spring (behind piston 2) is imagined to be preloaded sufficiently to provide a pressure plateau with increasing input force via lever 1. (I think pressure values would be around 10-40% of that found in a typical motorvehicle brake line...if that's helpful..I don't know what such a pressure value may be). So if spring rates are kg/mm and pressure is N/m*m ..I'd presumably pick the plateau max pressure and match the spring rate and its preload to that... at least I assume that's how you would do it.

And there's a little graph sketch in the picture showing relationship required between input and output force and pressure.

With further increasing input force then the piston 3 would have a withdrawing force bringing exit pressure to zero.

The exit pressure is directed / applied to a hydraulic piston (not shown) ..ok a brake calliper.

I imagine that as input force increases or decreases there is a smoothish transition between the various zones of the exit pressure. (ie see graph, initially a linear increase then plateau, then zero or the reverse).

Thanks
Scott.

 

[desertfox]

Hi scott33

Look at piston 1 for a minute, the force applied to get lever 1 moving needs to overcome both the lever 1 return spring and the piston 1 return spring as they are both working against you, I am assuming at this time that fluid can move through the exit pipe, at some point the pressure on piston 1 will result in a force that is equal and opposite to the force on piston 2 given by its spring, but that will only occur if whats at the far end of the exit tube cannot move any more,in other words no flow out the exit tube.
Now because piston 3 is mechanically coupled to lever 1, piston 3 will move down automatically as lever 1 moves and will stop when lever 1 stops, it can't move after that unless lever 1 moves again.
While piston 3 is moving all its doing is increasing the volume for a short while of the main cylinder, but the cylinder will just fill with oil close the void, I can't see that its going to reduce the exit pressure to zero.

desertfox

 

[scott33]

Ahh, I see, thank you, it seems I have neglected something in the picture (at least), namely the stop for piston 2 (and piston 3 but I'll get to that). So when piston 2 is at rest it is forced, by the spring, upon the stop some distance up the cylinder. Pressure build up in the cylinder causes it to displace towards and be opposed by the spring.

If the exit pressure line goes to a brake calliper (not shown) with usual pistons etc, then the swept volume of piston 1 will equate to the volume then swept of the calliper pistons (up to the point of the piston 2 release), which is to also say I'm imaginig it as a closed circuit, if that is to say anything useful.

Yes I've been a bit hazy (because I am a bit hazy) about the lever 2 setup. The piston 3 and lever 2 coupling should have some ...slack built in for most of the lever 1 inwards travel. Maybe like a slotted connection for one of the pivot arms. (then I think you need a return spring to return it to seated position)

In which case piston 3 will be sprung the same as piston 2, and it will require a stop too, with its piston position withdrawn by the lever and returned to seated by a spring with rate, at least in excess, of piston 2 spring.

Regard your last thought about piston 3 increasing cylinder volume and having oil fill the void, ... not sure. I'm assuming that, give or take, a fixed amount of incompressible, and presumably unstrechable fluid will make for no pressure in the line if a force acts to increase volume.

I'll draw those changes (ie include the stops / lever slack for pistons 2 and 3) and see if that can make a more likely case of the idea.

Thanks
Scott.

 

[desertfox]

hi scott33

The best thing to do is draw free body diagrams for each component assuming for instance that the brake is applied.
If you do that I think you will probably see clearly whats going on.

regards

desertfox

 

[scott33]

THanks desertfox, yes I have approached the matter in step wise fashion and have a better appreciation because of it. Intersting how ones thoughts have to step up a notch or two (six) when voicing them publically ...

I think I have solved the problems you have highlighted ...including the one regarding piston 3 if it is being withdrawn and then simply fluid will fill void.

If piston 1 is actually hard bound to the lever 1 and piston 3 is large enough its swept volume (on withdrawal ...even with a small displacement) will overcome the swept volume of piston 2 (it will effectively suck piston 2 back in prior to achieving a pressure drop) and counteract/overwhelm the slight attempted increase in pressure consequent to the continued inward displacement of piston 1. That is, a net negative (presumably to zero) can be achieved with large piston 3....I think.

Oh I might add, if you happened to be apraising that circuit that the force applied to lever 1 is fairly large and it appears that the spring required to counteract it is in the region/order of 20kg/mm (which seems is not normal units for springs but made sense to me ) hopefully there are springs in the 20-30mm diameter range which can do that.

Again thank you for your time, still tentatively plan to figure out who I might approach to build such a thing...not so easy perhaps in New Zealand.

Scott.

 

[desertfox]

Hi scott33

Have you got a better sketch of your device now, or freebody diagrams as I mentioned previous, if so post them up, in the meantime I'll try and do freebody diagrams for what you posted previously, because I still don't think it will work but I could be wrong "Grin"

regards

desertfox

 

[btrueblood]

Um, trying to imagine the output requirement. Do you want the length of the pressure "plateau" and point of subsequent drop-off to be related to piston stroke, or time, or ...?

Would a pressure regulator on the output line do what you want? I.e. a simple piston, retraction stroke draws from a reservoir via a check valve. Pressure stroke closes check valve, output line feeds to (brake caliper or whatever), with a pressure relief valve plumbed somewhere along the output line. If output pressure rises to setpoint, relief valve opens, and fluid is passed back to reservoir.

 

[scott33]

Desertfox, I'll attach a new diagram now. It's a bit different, but essentially the same - no need for messy lever 2 business. Places piston 3 parallel to piston 1.
[

http://files.engineering.com/getfile.aspx?folder=ff4609ce-ec84-4316-b2ca-8db6f7da9c41&file=v2.jpg

]


I should note, The secondary cyclinder (housing piston 3) is, on the quick calc I did, meant to be at least 2.5 times the primary cyclinder (not just slightly larger as I have drawn it). IN that case even with the relatively small displacement of piston 3 the swept volume (SV) of piston 3 will be greater than the SV of piston 2 + SV piston 1.

Further as piston 3 is displaced imparting increasing volume on the system, piston 1 will have continued to attempt to decrease it... to no avail, or so is my thinking since the net effect is increasing volume - decreasing pressure...with luck it can go to zero

My imaginings have displacement of piston 3 less than 1/4 of piston 1 displacment. Pistons 3's displacement occurs only at last 1/8 - 1/4 of piston 1 / lever 1 displacement. Return spring of piston 3 is greater than piston 2 return spring. (so it can't act as pressure relief)

Ignore the implied effect in the diagram of the lever ratios of spring 1 as compared with the other 2 springs (ie implied by spring 1 being nearer the lever 1 pivot), since in actuality I imagine them to be all acting near enough at the same point on main lever.

Would it work?? having not even got the 'in principle' bit figured out I start to see the working detail as easilly getting beyond me...


Hi btrueblood, thanks for looking at this.

The length of the pressure plateau and drop off points are dependant (begin consequent to) the input force, so in effect the lever 1/piston 1 position.

The input force is large and will be effectively resisted only by spring 1. So the piston 1 will be whereever the input force places it (given spring 1 resistance). ie I don't imagine springs 2 or 3 will significantly effect piston 1 position. If I make useful sense.

regards your thoughts I'll need to google a few terms to understand the idea. I'm no hydraulics engineer..

Cheers
Scott.

 

[desertfox]

Hi scott33

Thanks for the updated picture, I'll study it and come back.
My thoughts initially is that your going have some mechanical binding problems because of the force on the bar driving the pistons generated by the tension spring and the input force.
If I understand correctly what your trying to do:-

Move piston 1 to generate pressure so that a brake is applied? Then any further increase is pressure via piston 1 is to be prevented by piston 3 relieving it?
So I am not sure what piston 2 is doing?
Please confirm what I am thinking is correct or not.
If you can put a sequence of what you expect to happen but without writing it in paragraphs it will be easier to understand for instance:-

1 link 1 drives piston 1 untill ------

2 piston 2 then does ------

2 piston 3 then ----------

regards

desertfox

 

[scott33]

Hi desertfox.
ok
1. intial move piston 1 causes linear increase in Exit pressure (acts like a simple piston hydraulic circuit)

2. At some point (specified by spring rate of spring 2) with piston 1 continuing to move in -
2. piston 2 moves out, acts as pressure release. Creates constant exit pressure...this constant pressure is a continued with increasing inward displacement of piston 1.

3. As the lever 1 and therefore piston 1 reach near their inward limit of travel (say last 1/8)
3. piston 3 is displaced outward (which because it is larger diameter than other pistons) effectively creates net increase in volume and thereby decreases exit pressure to zero.
3. So piston 2 is sucked back in..yet net (exit) pressure is decreased).
3. piston 1 continues to move inwards (however the swept volumn of the piston 1 displacement, in this last phase, is much less than the swept volume of piston 3...thus countering/overwhelming piston 1 effect)

4 On reverse, ie piston 1 withdraws from max inward displacement, the exit pressure will go from zero'ish to constant pressure, and then back to linear decrease.



Many thanks again for your time

Scott.

 

[desertfox]

Hi scott33

Okay thanks for the sequence, however it isn't going to work, consider:-

1/ piston 1 moving inward.

2/ At some point(specified by spring force 2 and not the spring rate) piston 2 moves out, it cannot act as a pressure release and create constant exit pressure, because as piston 2 moves out the force on piston 2 increases, due to spring 2 being compressed.

3/ Piston 3 will only move if the pressure inside face of the piston stop generates enough force to overcome spring 3 or it is driven by the beam connected to piston 1.

4/ If the spring on piston 3 is a stronger spring than that of piston 2 then its possible that piston 2 will continue to move out while piston 1 is still moving.

In addition as mentioned in my earlier post your going have some trouble with binding forces generated at the beam connected to piston 1.
I'm sorry but I think it is simply not going to work, your design will suffer from spring tolerance issues not to mention alignment issues.

regards

desertfox

 

[gerhardl]

Coming in late, and got lost in the mechanical logic, but interesting problem, and a 'contrary' thought:

This seems to be trying to construct a completly self-sustained, repetitive mechanical system giving a variable output force (for one acting rod? or two?), again repetitive and depending on the rods mechanical position.

If this is the case would it be of interest to try to obtain the same result by magnetic strip on the sylinder piston, activating limit switches placed in adjusted distances on the outside of the cylinder wall placed, that again operates inlet/outlet solenoide valves with different sized throttle and check valves, leading hydraulic oil back and from cylinders/mechanically pressuriced reservoirs?

Eg electro/mechanical solution?

Would this in case work without ooutside power force other than springs (and electrical power (from battery?) to solenoids?

Ref ABS vehicle brake systems.

(As said, just a thought, I do not need an answer....)

 

[scott33]

desertfox, excellent figurings thank you. ...oh ye of little faith

Do let me though persist one more round.

The point you make that there will be an increasing force exerted by spring 2 on piston 2 as piston 2 displaces is well taken. Didn't think of that and is a bit of a hitch.

However it could be that the F = kx relationship of springs (specifically for spring 2) would provide, if not a flat plateau (of exit pressure) then a slope significantly less steep than that provided without it. That is, the rate of increase in pressure would be reduced ...and it may be that such a slowing of increase (rather than creating a constant pressure) will in practice suffice.

The dead set crucial element is the full release that piston 3 is imagined to provide. It may be that I have not drawn that mechanism clearly. I'll describe it as follows.

Spring 3 is a return spring, Piston 3 is displaced/withdrawn only due to lever 1 movement - the last bit of lever 1's movement. The gap in the rod (to piston 3), is meant to effect that mechanically.

Would not then piston 3 reduce pressure? That is, if piston 3's swept volume (SV) was large enough (larger than SV of piston 1 + SV piston 2) would not its displacement get pressure to zero? And do it (get to zero P) proportionally (proportional to its displacement) ...even given the small displacement available.

Yes the binding issues I can appreciate, Is there a brick wall problem translating a pivoted lever force to a piston. Is there a semi-floating connection or something solution...maybe.

Regards spring tolerance issues - I guess you're referring to functioning being compromised because of spring rates being required to be so close to one another (and yet definitely different) that the values will be closer than the tolerances of spring rates manufacture ?

If that is your thinking then I'm imagining spring 1 is an order of magnitude (if not 2 or 3) different from the others, while spring 2 need be carfully chosen, spring 3 need only be say 2 or 3 times that of spring 2.

Anyhow how say you Sir desertfox. Have I dodged any of those bullets?


Hi gerhrdl = thanks for your thought - I apologise for not mentioning it - but yes I would love the elctro mechnical solution, I've experienced a couple and they are good, but I was specifically going for no power, no battery etc. Hopelessly perhaps...;-)

 

[desertfox]

Hi scott33

Okay I agree that if piston 3 is driven by the beam attached to piston 1, it is possible that you can reduce the pressure by increasing the volume within the vessel, but that depends on the forces acting on the pistons and force and stiffness generated by the respective springs.
The pressure would drop back to its original pressure if piston 3 sweeps a larger volume then piston 1 + piston 2.
Now where are the fliud seals, it seems to me that they would be mounted on the pistons for number 2 and 3 and on the piston rod of number 1 piston.
I still think your going to have alignment and binding problems which will hinder manufacture, also the diameter of piston 3 will be much larger than pistons 1 and 2 so you need to ensure that the pressure within the vessel cannot move piston 3 before its driven open by the beam attached to piston 1.

desertfox

 

[scott33]

Thanks again desertfox, your thoughts are very much appreciated. Where would you might expect the binding to occur? (on the misaligned piston in the cylinder or the rod guides...or maybe both??)
last question - to avoid binding is the ideal design require that the force acting on the piston rod act purely inline with the rod?
Or is there another typical method (lever rod junction) to minimise such problems?

Cheers
Scott.

 

[scott33]

And the other last question, regards the seal mounted on the piston 1 rod, what or how does that work?