Front/rear Weight Distribution

Hi John, do put us out of our misery! Can we assume from your last post that you have dashed out and sectioned a leg? If you haven't then I decline your kind offer to let me section it. But as Ken wrote, there are probably a lot of members who are awaiting the result.

I'm having to rethink one of my arguments. As I suggested, when the car is stationary with the suspension in its equilibrium state those holes above the hatched bit at the bottom must have allowed pressure fluid to fill that annular space. This will balance the equal area below leaving only the ram as the effective piston. But in that case, why have the hatched bit at all? Why not just have the ram? Unless it is fitted to do what John originally suggested by acting on the bump stop. So if John is wrong about his piston it looks like honours are shared, which would be a very satisfactory outcome. If he's right about the piston as well I'm right down the pan.

For those who are holding their breath over the questions I posed in my 17.20 posting, the answers are:

1. No, static load pressure generated by John's piston at the maximum 600kg load is 4684 kPa, as compared to the hydractive sphere charge pressure of 7093 kPa.

2. Yes, the 25mm ram will generate a static load pressure (same conditions) of 11,990 kPa, so the sphere will be well into its operating range.

3. Stupid me couldn't get the sums to work out for the ram because I was calculating for a 35 mm ram diameter!

Hi Aerodynamica, that's a massive posting. I'm afraid it'll take some time for me to absorb it.

Isn't this fun?

Derek

Hi Derek, no need to go through it too closely - it really just confirms what you already stated about the structure of the strut - it was taking me a while to 'see' the truth!

There's just a query about what you say about sphere volumes under static pressure etc.

-this IS enjoyable? isn't it? I'll get my coat...

Hi Derek

With a quick look this evening I could only find recovered struts I want to keep. There are some more older ones but they are at the back of that heap. Will need to shift that heap and the spare engines when I do my next car move round which should be fairly soon.

John

Hi John

I can see what u mean, and I can see on Derek's photo that the two skins are indeed apart for a significant lenght -though they do seem to directly touch each other at the bottom of the strut, but could there be some softer material between them where they touch at the top side?-.

So, does that mean that the outer and inner skins are allowed some -tiny obviously- freedom of movement between them? I.e. the axis of the piston+rod is not always totally parallel to the axis of the outer skin??? And how would this happen (though material deformation??) Otherwise I'm missing something!

cheers
George

ps. Derek, sorry to deviate from the core of your topic -I think we should change this section's name from "Hydraulic" to "Pandora's box" !!!-.

Yes, agree with you Aerodynamica, I think on minimum load the spheres' pressures should have been calculated as for the diaphragm to finding equilibrium at some point that will allow sufficient LHM volume in the sphere for a full suspension extension in the case of a hole?

This point however -I think- is not necesarily around the middle of the sphere, I think we are tricked to visualise it as such, it all depends on the volume of LHM the strut is pushing -you could find that in reality the sphere's diaphragm plays within a tiny amount (e.g. 1mm) of movement between full suspension extension and compression-. If this is the case, I imagine the actual seating point of the diaphragm, with car up to "driving" height, will be very-very near the sphere's damper-tablet.

It's a becomes more tricky on hydractives where the middle and corner spheres presssures are different (front: 70-75 vs. 45-55 bar), which means effectively that, on minimum car load, the corner spheres' diaphragm will always be seating further near the top of the sphere than the middle sphere's one. This would mean quite assymetric spring characteristics from the corner spheres once the car is locked to "hard": plenty of expansion course but little in the way of compression. This would mean that e.g. a slaloming XM would tend to be lifting the inside more than the outside is compressed. All this diference being filtered by the relative strenght of the antiroll bar. Hmmm, have to check this Topgear video.

This goes to show how very tricky is to choose sphere pressures on hydractives, as e.g. with a bigger volume or higher pressure middle sphere you are altering much more than your "soft mode" comfort...

cheers
George

Hi George

The comments about the load bearing function of the front strut outer skin and the load free inner piston tube are not mine. They are my interpretation of the relevant part of the M&S Citroen XM book. There is even a small coloured drawing of a sectioned front strut and sphere assembly just not big/clear enough to answer the queries. The reason the two skins are come together at the bottom is that this is the fixed end of the strut that fixes in the hub assembly clamp. The fixing on the outside of the strut for the drop link is shown opposite the centre of bearing on the main rod. This droplink must put tremendous torsional forces onto the strut case so I can understand why the designer wanted to avoid distorting the cylinder tube.

John

Hi John, although I think your offer to section a strut will add much to our store of knowledge on our beloved cars, there's no desperate urgency about it and it's certainly not worth destroying a good strut. When Dean comes up with his weighbridge results I can calculate for both piston and ram displacement.

Hi George, Graeme and everyone, There is no need to guestimate the displacement of the sphere diagram as it's easily calculated. First let me explain again the terms I'm using. Static load pressure is the pressure in the suspension leg that is needed to support the static load, which I defined for calculation purposes as the minimum to go driving, car weight plus 5 gallons of fuel plus 70kg for the driver; I suppose you could say kerb weight plus driver. Charge pressure is simply the pressure the sphere is charged to.

Static load pressure is the weight on that leg divided by the ram area.

Now to calculate the fluid displacement into the sphere.

I agree that the displacement in the leg spheres under static load must be enough to allow full rebound, it is this requirement that allows us to do the calculation.

Imagine the car jacked up with the legs on full extension, the legs are full of fluid, pressure is nil (actually atmospheric, but we're not talking absolute pressures) because the height control valve (HCV) opens the leg to return to reservoir, and the sphere diaphragm is hard up against the damper disc. Now lower the wheels slowly onto the ground, as soon as the first sign of compression occurs the ram will move into the cylinder and will displace fluid into the sphere chamber, the pressure will be just enough to balance the increased load. as the car is lowered until its weight is fully supported by the suspension the ram will move further into the cylinder and the diaphragm will be displaced further into the sphere until sphere pressure exactly balances the static load.

If the sphere charge pressure is too low the car will sink below normal ride height before the pressure is high enough to support it and the HCV will open to admit more fluid. This extra fluid only raises the suspension, it does not flow into the sphere which already has too much fluid displacement. Because of the large fluid displacement there is less volume available in the sphere and the ride will be hard.

If charge pressure is too high the car will be above ride height when equilibrium is reached and the HCV will open to release fluid. In this condition, there is plenty of available spare volume and the ride will be soft, but diaphragm displacement is insufficient to allow full rebound.

Perhaps this explains why I believe there is only one correct pressure.

So how to calculate this correct pressure and the amount of sphere diaphragm displacement? The answer is, I believe, surprisingly easy. If you've followed me so far, we know that ideally the suspension should go from full extension (no pressure) to normal ride height (static load pressure) with the HCV remaining inactive. So we calculate the volume displaced by the ram between full extension and normal ride height (extension x ram area). This gives the volume of fluid displaced into the sphere. We know the static load pressure (static load over ram area) and from those two figures we can calculate the initial charge pressure.

The hydractive sphere shouldn't come into it as for maximum effectiveness it should have its full volume available to absorb bumps, at least at static load. as the load increase it will of course be compressed and this is demonstrated by my calculations for maximum load (which we know from the handbook). It is this wish to avoid the complications of the effect of the hydractive system that made me decide to make my calculations on the minimum driving load condition.

I don't know about you but I'm shattered after that. Time for my two sherries.

Derek

PS As a young junior rank in the services I used to occasionally find myself in the butts on a firing range, hoisting a target above the parapet to be fired at. This is quite like old times.

This post has been edited by DerekW on November 30, 2008 12:44 am

Excellent long posting! I don't feel so bad for my massive one earlier now!

You want to determine the ideal sphere initial pressure OK - I need to confirm I'm getting you here! we are assuming that Citroen's own choice is insufficient?

Anyway I thought this was a simple statics question! I'm still trying to fully follow the raising the car on jacks model - sorry bear with me..

- so this means that the wheels are hanging at 'full droop' - am I right?

-agreed, the pressure in the sphere is what I've been calling 'rest pressure' - could be something like 40bar or 50 or something - anything.

- right with you again, some of the LHM displaced will compress the gas but most will go thru the HCV ? until it's lowered enough to shut off?

- this is the bit I don't get, when does this state occur from atmospheric pressure? I can't see how the ram will have moved enough LHM volume to squash the gas enough to reach this balance - even from full extension to the ground again (guess) 300mm just isnt enough stroke length to lift the diaphragm off the damper disks of the sphere very far at all because the sphere is much wider in its cross section than the ram.

- yes it'll move further into the cylinder compressing the gas at a much slower rate and the ram will run out of travel before the diaphragm has moved even 15mm off the damper inner side!

I can't see this detail quite like you state it. For every say 50mm the ram moves LHM into the sphere the diaphragm does not move a corresponding 50mm away - it's a lot less due to the 'control volume' of one cross section versus the other and the ram is a lot narrower than the sphere.

I still think the opposite and simpler approach gives your answer: going from minimum height and theoretically atmospheric pressure LHM and putting the car on normal height takes the sphere from full gas volume to reduced gas volume and raised gas pressure and equal LHM pressure. This is the car sitting undisturbed with the axle weight and additional payload you mentioned.

If you wanna know what the sphere is doing it's simply the force on the suspensionX cross sectional area of ram for the static pressure. This static pressure in the sphere compresses the gas to the same value reduces its volume by an amount as this is the value I think you want to know - it's simply Boyle's law: P.V= Const. So P1.V1 = P2. V2, and V2 is what you want to establish as value P1 is the sphere rest pressure, V1 is 400cc (probably more like 390cc inside?) P2 is your pressure value from your first calcs of the force and therefore pressure in the ram and the only unknown is V2, the now, compressed gas volume.

If V2 is an unsatisfactory value and turns out to be say, less than half the 'unpressurised' sphere gas vol. then you can choose different 'new' values for P1 and work out which value gives a better V2 that leaves a larger volume of gas in the sphere with the same strut LHM pressure acting on it. Once you get a value for P1 that gives a value for V2 that will leave a large volume of gas to play with then you've found the ideal sphere pressure to achieve this.

If say, you were interested in 500cc spheres then you choose value V1 as 500 not 400,

That's the only way I can see this!

This is how you can determine the volume of LHM held in your accumulator at cut out, and the rate of leakage if you're keen.

This post has been edited by Aerodynamica on November 30, 2008 02:02 am

Oh, Okay, can see now, didn't realise u reffered to the forces coming from the antirol bar drop-link fixing.

Was thinking myself of the forces that act on the strut rod other than the "ideal" forces, i.e. identical to the vector this one operates/slides. I still cannot mentally come to terms with the idea that, with the McPherson set up, the strut tube+rod set looks like it'll also be receiving forces that tend to "fold it in to two" (to bend it), rather than to just push the piston inwards, if u see what I mean.

In a traditional citroen cylinder (like at the rear of the XM) there's no such problem as the rod floats freely at the contact point with the piston and so is free to take a range of angles towards it as necessary. So with the McPherson setup, does it look like there's room for operational friction other than simply the friction of sliding the rod in/out of the tube? But perhaps I mentally tend to magnify those forces too much, the XM's front suspension doesn't seem to have any problem absorbing anything on the road? (here, Andrew comes and says how much better the CX was )

cheers
G

Hmmm, I'd think the hydractive sphere is equally -or even more than the corner ones, which are twice as damped- needed to provide suspension motion both ways, even at static load. Can't believe they would have it set as for it to be uncompressed with just one passenger in.

BTW, if it remains uncompressed, this means the pressure in the front system at static load will be 70-75 bar or less. Does this check out true for the XM?

cheers
G

Hi all

Right...........

XM 2.0 8v auto hatch with 13gals of fuel:

with driver (80kg)

FRONT AXLE, 980kg
REAR AXLE, 580kg
TOTAL, 1560kg

Without driver

FRONT AXLE, 940kg
REAR AXLE, 540kg
TOTAL, 1480kg

Hope this helps Derek keep us posted, i am very interested in this

Regards D

This post has been edited by dean on November 30, 2008 06:19 pm

- that's what I'd have thought - and if it was higher than the 70 odd bar pressure of the middle sphere but unconnected to it in sports mode then as soon as it connected the car would fall to the floor and then be replenished with more fluid from the H corrector.

I think the XM front strut structure differs so much from the previous BX to improve on the problems BXes could suffer with seized struts and such.

Hi Dean, thank you for going to the trouble to get that information, now I/we can start calculating in earnest.

I assume the driver loaded front end was a mistype and should read 980?

It looks as if the weight distribution hovers around 63/37%

Derek

Hi Derek

It did read 980 on the front you are right i have edited.

Regards D

This is like trying to knit a jumper out of cigarette smoke. Instead of trying to come up with all the answers in a single post I think I'll try one at a time.

Hi Graeme, "As a young junior rank in the services I used to occasionally find myself in the butts on a firing range, hoisting a target above the parapet to be fired at. This is quite like old times." Your first shot scored a direct hit. How could I have missed the fact that the HCV will be open to return until the car is below normal ride height!

I was trying to explain a wrong point of design principle and using a spurious argument, thank you for correcting me.

Quote (I don't know how to get the quotes in boxes as you do) "I still think the opposite and simpler approach gives your answer: going from minimum height and theoretically atmospheric pressure LHM and putting the car on normal height takes the sphere from full gas volume to reduced gas volume and raised gas pressure and equal LHM pressure. This is the car sitting undisturbed with the axle weight and additional payload you mentioned."

I agree that your method is better but with a slight amplification. The HCV will be open to system pressure supply but the car will not begin to raise until the pressure in the leg x ram area equals the weight on the leg. whilst the leg pressure is building up to that point the diaphragm will be moving, keeping the sphere pressure the same as leg pressure. When the leg starts to extend the sphere pressure and leg pressure remain constant, the leg extends because of the extra fluid being admitted.

From this we can deduce that, because of the built in delay in the HCV*, at full extension the leg pressure will equal sphere charge pressure. Another unknown put to bed.

If you'll forgive me, I think your comment about the relative sizes of the diaphragm and ram is not valid. It doesn't matter how big or small the diaphragm is, it will displace just enough to raise the sphere pressure to balance the weight of the car, and yes, Boyle's Law will determine the amount of that displacement.

Derek

PS *The original DS/ID had instantaneous HCVs, whenever the car momentarily "went light" in driving, the HCV would sense the leg extension and open to dump fluid with the result that the car would return to earth with a teeth-jolting crash.

This post has been edited by DerekW on November 30, 2008 08:24 pm