Front/rear Weight Distribution

If I understand the figures, and perhaps John could either confirm or refute, the front rams changed from 22mm diam. to 25mm, rear hatchback pistons changed from 35mm diam to 37, both changes about 1995. Estate rear pistons have always been 42.5mm diam.

I don't think the front suspension inclination is going to have any effect on my calcs, but I still have to slide under the car to measure the lever arm ratios on the rear suspension units.

Those distribution figures you quoted will be a reasonable basis to start my sums.

Derek

Hi Derek

With the front of Mk2's you do not have to worry as a single calculation will cover it. It was just the push rod diameter that increased with the piston diameter staying the same size. My guess is that they were worried about the long term fatigue strength so increased the rod diameter when they introduced the 2.5TD. Interesting that even the earlier V6-24 did not push the rubber on to the road violently enough to need its front suspension and steering beefed up from that of a standard XM. Will check on the other figures you mentioned and post again.

John

Hi John,

Problem is that there is a fluid passage connecting the upper and lower faces of the leg piston, shown by the lower arrows in the R.H drawing:



This means that pressure on top of the piston balances out pressure over an equal area below the piston.

Result is that the ram diameter is the onlly effective agent compressing the sphere. The piston simply centralises the ram in the leg cylinder.

Derek

Hi Derek

I can only see the arrows for the oilways that feed to the underside of the piston. There is no entry for the oil to get to the top of the piston. If it gets there at all it it is by leakage up past the piston. I think the piston is the long part over the middle half of the rod not the little bump stop around the hollow bottom end of the rod. Citroen allow 22mm and 25mm struts to be mixed on the front of a car so there cannot be any significant hydraulic difference between them.

John

Hi John,

I don't understand why we're seeing different things here. I'm assuming the piston is the cross-hatched item at the bottom of the ram. At the top of the ram are the holes which communicate with the channel down the centre of the ram and thence to the space under the piston, this allows transfer to and from the height corrector and the sphere. But the two arrowed holes communicate between the central channel and the annular space above the piston. Suspension pressure is fed therefore to both the underside area of the piston via the central channel and the annular area above the piston via those drillings. take one from t'other and you get ram area.

Am I missing something obvious?

Derek

Hi Derek

I see your crosshatched bit as the fixing on the end of the piston rod that mates with the bottom bump stop/spring if there is no fluid pressure. I see the piston top and bottom as the two thick dark black annular rings around the mid portion of the rod. These are above the arrowed lower oil holes. Engineering drawings are fun to read and a bonus if got right.

John

This post has been edited by xmexclusive on November 26, 2008 09:05 pm

Hi guys, this is fascinating! I'm interested as to how different the structure of the Xm strut is to the BX which has a much simpler structure IIRC.

I'm with John here, I feel the darkest components represent the 'piston' as the cross hatched item (with holes at either side) would not have different fluid pressures on either sides during equilibrium. If the black bits are the piston then the piston area is the cross sectional area of the inner cylinder dimension minus the cross sectional area of the thinnest channel up the center.

That represents a flow rate up the central channel that's relatively high compared to the stroke of the strut since you have flow rate of the larger area A1 x V1, velocity of fluid = the flow rate of the second smaller area A2 x V2. Simply, if the area of the thinner channel is half of the area of the cylinder then the flow rate will be double for an incompressible fluid.

Funnily enough the BX has a simple narrow piston in a narrow cylinder all encased in the rigid strut housing and it displaces a small volume per unit length which corresponds to the near zero leverage ratio of McPherson struts versus say a GS front piston of greater cross sectional area but with a ratio of leverage from the suspension arms of a smaller ratio. It means for a BX at least that the actual fluid displaced is relatively similar to the GS and so the same spheres can be fitted to both with minimal difference.

I think the Xm's high ratio here (unless I got it wrong) is to suit the hydractive system that uses more LHM per suspension deflection when in soft mode and to increase the effect of firm mode when the centre sphere is locked out.

Graeme

Just to add a question: Derek, I don't fully understand the idea of going from low to normal height to get the pressure. I thought that if you know the front axle weight and the leverage of the suspension strut (MacPhersons = negligible I believe) then simply halving it will give you the force in the strut. If you can get the piston cross sectional area for sure then you can get the pressure per unit area of the column of LHM. If you have the LHM pressure and the initial pressure of the sphere and its volume then at equilibrium the gas and fluid pressure are the same - all that remains is the volume and I think that's what you're aiming for?

Sorry I found this discussion well after it started but I'm really interested in it.

This post has been edited by Aerodynamica on November 27, 2008 05:24 pm

Sorry for the delay in answering, it takes longer to get my thoughts in order these days.

John, my interpretation of the drawing is that your "piston" is a support bearing with helical slots to assist lubrication and leaky seals top and bottom, hence the need for a runoff pipe in the gaiter. If it is a piston then there is an immediate problem, the annular space above your piston will be changing its volume, and hence its pressure, continuously under suspension movement with no means of releasing it. Taking my argument a stage further, the presence of a runoff pipe suggests that fluid will find its way into that chamber and must fill it before it reaches the runoff pipe. The virtual incompressibility of fluid means the suspension could not move.

If my interpretation is correct and it is a bearing, the annular space does not alter its volume.

I think the only way we can be sure is for some kind soul with a scrap leg to cut off the cylinder bottom and extract the ram and piston.

Aerodynamica, as my calculations are based on static load conditions flow rates to and from the sphere don't come into the equation. I'm afraid you're wrong in your calculation of piston area, for equilibrium fluid pressure must act on equal areas in the upwards and downward directions. If you deduct the area of the central channel from the "upward" area then equilibrium doesn't exist and the leg will extend to its limit. If you look at the drawing the top of the ram is closed and in static conditions the fluid in the channel is a stationary column. So if John's piston is correct the active area is the cross-sectional area of the cylinder. If I'm correct it's the cross-sectional area of the ram, difference being the cross-sectional area of the seal/support bearing which is attached to the cylinder.

Ref your last posting, I think you misunderstood the point of the exercise. The idea is to try to work out the optimum sphere charge pressure and then compare it with the spheres that are specified by Citroen. By the way, thank you for your interest and your contributions, they helped to concentrate my thoughts.

Derek

This post has been edited by DerekW on November 28, 2008 05:51 pm

I hope we can all agree that, for the hydractive sphere to work, it should be vitually uncompressed when the car is at minimum load as this will give maximum volume and hence comfort when meeting a bump. I say "virtually" because there should be slight compression to ensure that it's immediately active. So at maximum static load there should be more compression.

I have a problem and I assume it's my maths, so if you could check for me:

Taking the late model V6 we know the maximum weight on each front leg is 600kg (5886 newtons).
Piston diameter is 40 mm for John's piston, or 25 mm for my ram.

Hydractive charge pressure is 70 bar where 1 bar = 101.325 kilopascals and a pascal is 1 newton per square metre.

Question 1: Does the hydractive sphere come into operation with John's piston?

Question 2: Does it with my ram only?

Question 3: What am I missing?

Derek

Hi Derek

Have just done a check on the new rear cylinder I am going to use as a sphere tester. The piston measures correct for an early Mk1 car one at 35mm in diameter while the cylinder is 49mm od. Will see if I can find a tatty orphan front strut to sacrifice. Do you want to section it or shall I run the disc cutter through it. I have some nice thin 1 mm thick stainless steel cutting discs.

John

Hi Derek, thanks for response, I think I see where you're coming from but I do wonder that if the black rings are bearings and the structure between them is equally fixed then the central 'ram' has a lot of surface contact to rub!

Just to explain where my mistake lies:

My initial problem with the cross hatched item being the piston was that it is relatively shallow in its length dimension such that a larger potential of leak back past it could manifest (this is very much the opposite to every Citroen suspension ram previously seen - they generally have a very long piston/ram length compared to the stroke of that piston) however, looking closer, if the black annular rings are bearings machined to the very close tolerances we know of Citroen hydraulics then they would serve to 'lengthen' the pistons contact area.

My other initial difficulty is that the two arrowed holes would allow a pressure equalisation above the piston so for every increase in pressure under the piston, a portion would be lost through those holes and the piston would collapse. But! then I realised the difference in volume between the annular area above the piston (outside those holes) and the volume being changed under the piston is much greater than the space above the piston (but below the bearings) and so I think eliminates the concern I had for those holes (though fluid flow through them will occur and the difference in volume above and below the piston is governed by the difference in volume of the piston thickness and the thickness of the central ram walls - the volume that takes up) - if that makes sense. That last bit doesn't read too clearly but I'm finding it difficult to explain the consideration.

Either way I think you're right about the piston area being the central column's area

I just drew a diagram to try to clarify what I was saying and seem to have come full circle! I can't justify the two holes again despite what i said ^^ for every mm of the piston and central rod displacing a volume of LHM from the bottom, the volume of the annular space above the piston will demand an amount of LHM to justify the movement (do you agree with that? I think it's simple hydrostatics but I could be screwing up somewhere) and this travels through the arrowed holes. But now thinking again it doesn't matter! this exchange can be neglected since it simply goes relatively unhindered from one side of the piston to the other.

The central thin column of LHM right up to the top of the strut has a length that will change per stroke of the piston its cross sectional area added to the cross sectional area of the central ram's walls. So I think the amount of LHM displaced per ram movement is the length of the stroke multiplied by the cross sectional area of the central column of LHM PLUS the cross sectional area of the central ram's walls. Can you see what I mean? I think this supports what you say in your reply.

I was going to copy the tech. scans and arrow the situation I mean but can't open that .gif file in my presently available photostudio thing - sorry about that.

I am certain that the spheres of the suspension at the front at least, are set at a pressure that sees them at approximately 1/2 gas volume under static and unladen conditions. So a sphere of 55 bar rest pressure will be around 110bar with the car standing at normal height and unladen. It is perhaps slightly under half gas volume and not absolutely half. Either way it would suggest a pressure of 110bar of LHM in one strut in equilibrium for a car such as a BX for example (chosen since I know it uses 55bar spheres!)

In equilibrium condition, the LHM pressure and the spheres' gas pressure are equal (I THINK this holds for the XM's arrangement with the center sphere phasing in and out - it should do unless there's an aspect of hydrostatics I'm missing!) and so the pressure you're working out in the strut given the axle weight will tell you exactly how much the gas is compressed with the car standing unladen and then worked out for it laden with a driver (I'd imagine the volume change in the gas will be very small)

With the car standing, the pressure you've worked out for the LHM in the strut will equal the gas pressure in the spheres. I am not sure you're right that it will be hardly compressed - I think the gas will be compressed to a greater extent than you say when the car is standing unladen. I think it is going to be nearly half the volume as stated. However, further additions of load from the half compressed gas will have less of an effect since at the half way stage the amount of force required to compress further begins to raise higher with every attempt to compress the gas further - half the volume, double the pressure thereafter the pressure raises exponentially and should theoretically become infinite (until the gas is compressed into a liquid that is but that should not be achievable)

Anyway sorry to load all this thinking while typing craziness! I'm equally keen for you to get this answer!

I'd agree with Derek's support bearing notion and the effective piston area being the cross sectional area of the rod, however don't forget that the front struts (unlike the rears) don't leak into the gaiters but via the leak off connection strapped to the lower half of the struts. This must mean that internal leakage upwards past the piston seal/support must find its way into the top half of the strut and then somehow into the narrow annular space formed by the double skin of the strut.

I will be interested to see the results of John's surgery on a scrap strut!

Ken

This post has been edited by kenhall1202 on November 28, 2008 08:52 pm

Hi All

I have found out the reason for the double skin of the front struts. The outer metal tube provides the vertical structrural strength and accomodates and deformation. The inner skin is designed to carry no external loads and is the 40 mm diameter piston guide. This is particularly important because of the very high lateral load the XM front suspension sees when curving. It keeps the cylinder working without friction and within its critical design tolerances.

Another thought about 22mm/25mm strut rods. If the feed through the centre of the rod is critical then the hole may well be the same size whatever the rod thickness. If Derek is right about the piston being the bottom item the higher lateral holes may just be oil feeds to ensure the inside of the cylinder/bearings and piston are continuously lubricated. If it operates in this manner then the fluid pressure has no effective force above the piston as the piston top and the bottom of the bearing are a fixed distance apart mounted on the same rod. Derek I can now see it your way round. Sorry for the red herring.

John

Yes, I also conclude that the piston is the central part but if the thicker piston has the same inner dimension (i.e. thicker walls) then it will have a different piston area by the small difference in area between the older and newer styles - this will displace a little more LHM per stroke though it might be allowed to be negligible.