To learn and share about dampers / shock absorbers

[747heavy]

Gas force increase due to stroke or why a damper has a spring rate
Let´s look at the physics and put some numbers down:
To have a reference, I "borrowed" the following illustration from the JRZ presentation. I hope they don´t mind.




so we have:

shaft dia (D_Sh): 22 mm
stroke (S): ~ 100 mm
internal reservoir dia/separation piston dia (D_Sp): 44 mm
effective reservoir length (L_res): ~100 mm
Gas Pressue: 5 bar (Case 1) 25 bar /Case 2)

what else do we need?

formula to calculate the peload:
F = p*A

F = Force [N]
p = Pressure [bar]*10
A = Shaft area [cm^2]

F case_1 = 5*10*PI/4*2,2^2 = 190 N [42,7 lbs]
F case_2 =25*10*PI/4*2,2^2 = 950 N [214 lbs]
difference = 760 N [171 lbs]

If we use a 100 N/mm [570 lbs/in)] spring on our damper, the change in gaspressure will extend the damper by 7.6 mm [0.29"]

now to calculate the increase in pressure due to the decrease in volume in the canister/reservoir when the damper compresses, we use the ideal gas law (not accounting for temperature effects yet)

Formula
po*Vo = ps*Vs

po (bar) = initial pressure
Vo (cm^3) = initial volume
ps (bar) = pressure at stroke S
Vs (cm^3) = volume at stroke S

to start with, shaft completely extented, we have a volume in the canister/reservoir gas chamber of:

PI/4*D_Sp^2*L_res = PI/4*4,4^2*10 = ~152 cm^3
@ an pressure of 5 bar
that makes 760 barcm^3

our shaft volume is:
PI/4*D_Sh^2*S = PI/4*2,2^2*10 = ~38 cm^3

our new volume in the resevoir/canister gas chamber is
Vs = Vo-V_shaft = 152-38 = 114 cm^3

our pressure inside the chamber now is:
ps=po*Vo/Vs = 760/114 = 6.666 bar

our new force now is:
F_new = ps*10*PI/4*D_Sh^2 = 6.666*PI/4*2,2^2 = ~253 N

which makes our internal spring rate
253N - 190N = 63N --> 63N/100mm = 0.63 N/mm (~3,6 lbs/in)

for Case_2 it is:
1267N - 950N =317N --> 317N/100mm = 3,17 N/mm (~18 lbs/in)

so, for the damper in our example the change in "spring rate" due to the gas pressure change is 2,54 N/mm [14.5 lbs/in]
It´s up to you, if you consider this worth, writing home about, or if you worry about it.

[RacingManiac]

You're a champ 747heavy, takes a lot of time and effort to write all those...

[747heavy]

I did some example calculations, so that we can see, how much the preload / ride height of a given car would change, if we changed the gas pressure in the dampers.

the first example is a Dallara F3 car (all values for springs and motion ratio´s are out of the Dallara handbook, they may change a bit in the real world, but it still helps to illustrate the point. At least I hope so )

Let´s see how much the preload changes, and how many turns (if any) we need to compensate on the spring perches/plateform to compensate.
How much would the ride height change, if we don´t compensate for the change in preload?

F3 car with "normal" dampers such as Penske or Ohlins with 5/8" (16mm) shaft
what happens for a moderate pressure change from 150-250 psi (100psi difference)


what happen is we loose all the gas pressure in one damper or change the car to TRD dampers?


100 psi change with 22mm shaft damper such as Moton or JRZ


250 psi change with 22mm shaft damper, changing the car to TRD (Ohlins TTX for example)



Touring car example

100 psi change with 16 mm shaft (Penske, Ohlins TT44 etc.)


250 psi change with 16 mm shaft (Penske, Ohlins TT44 etc.) - change to TRD or loose all the gas pressure


100 psi change with 22 mm shaft (Moton, JRZ etc.)


250 psi change with 22 mm shaft (Moton, JRZ etc.) - change to TRD or loose all the gas pressure


Radical (small amateur sports car, light springs)

100 psi change with 16 mm shaft (Penske, Ohlins TT44 etc.)


250 psi change with 16 mm shaft (Penske, Ohlins TT44 etc.) - change to TRD or loose all the gas pressure


100 psi change with 22 mm shaft (Moton, JRZ etc.)


290 psi change with 22 mm shaft (Moton, JRZ) - this represents all the adjustment range in damper gas pressure these dampers offer/recommend.


As we can see, in all these cases, the change in preload and ride height is perceptable. As moew lightly spung the car is, as greater the effect on ride height, and as more compensation is needed.

For a standard 16 mm shaft damper such as the Penske and Ohlins TT44 and a moderate change/or difference in gaspressure of up to 100psi, the change in ride height and preload is resonable small for cars such as an F3 o similar.
This may be the cases Speedsense has refered to.
The change in preload of the main spring by changing the plateform/perch height is between 1/2 to 1 turn, and the ride height effect is ~1-2 mm in this cases.
Which even for a Touring car, is something, most race engineers would want to know about, and consider in there set-up.

The changes are more pronouced with dampers using larger shafts, and cars using light main springs, such as Formula Fords, Formula SAE and small sports cars.
The chnage in actual spring rate is negliglibale, provided, that the damper has still a large enough gas volume, and that strokes are < 100mm.
I will show some possible exeptions later, but in general, we can conclude, that the change in actual ride spring rate is either insignificant or a secondary consideration.
This agrees with speedsense´s assessment, from earlier in the thread.

.... to be continued

[747heavy]

You're a champ 747heavy, takes a lot of time and effort to write all those...

Thanks for the kind words RacingManiac !! - it´s very much appriciated
I hope that one or the other interested reader finds some of it useful or
at least interesting.
PLease feel free to point out any mistakes - I may have made.
Nobody is perfect

[xavier111]

great explanation 747!!! continue in this way!!.....

Gracias amigo!!

[speedsense]

Awsome 747, what a report! We should be paying you for this work , I for one greatly appreciate the investment of time.

With most of the "gas spring" calculations that been done, aren't the gas spring numbers generated from a stationary point for the piston position? Whether at a fully open position or some distance of insertion into the shock?
At this stationary point the valving is closed on both sides of the piston (no flow through it) and if the measurement of the gas spring was taken while in this "state" the gas spring effect would be at it's highest. As seen by 747's numbers, the larger shaft size increases the gas spring, as the volumetric displacement of the fluid by the solid rod (assuming it's solid as it is in JRZ,Moton) provides more force towards the gas. It would stand to reason that the highest gas spring rate would be a solid rod, solid piston with ample seals and no bleed holes open.
In use (running on a car) the shock has varying fluid amounts flowing through it at the different velocities and the highest flow is at the shock's highest attainable velocity, would contribute the least amount of force/resistance (of the assembly of rod and pistons) towards the gas spring having an effect.
This would have a reduction effect on the stationary gas spring number, especially with a very high flow piston in use...

Though even to this day, Ive seen engineers who "try" to use the shock as a spring (over dampen, way beyond critical). Maybe that's why their drivers seem to have twitches and seem uncomfortable driving that type of setup, especially when they hit a bump at a critical point in cornering.....the gas spring left them for a moment and got inserted into the drivers bowels
AA IMHO

[speedsense]

Hi speedsense,

Sorry mate, I did not forgot about you, but needed to get some othe things out of the way first.
I will answer or comment to your questions/comments in seperate replies. I feel that way it is easier to follow, and we don´t need to requote the whole lot, when replying.

What do I mean with hydraulic lock out?

For me it´s the same as for RacingManiac.
In a conventional damper (e.g. no constant volume dampers such as TRD or Rotational dampers) you need a reservoir (volume) to take up the volume displaced by the shaft
(PI/4*d(shaft)^2*stoke) plus the volume increase due to the expansion of the oil volume with temperature (~0,00064*Volume_Oil*delta_T).
If the volume in the reservoir can´t accomodate this increased volume, the damper will "lock out" and the force is approaching infinite (assuming oil to be incompressible (~0,07%/Mpa-for pure oil), and nothing in the damper to yield).
If this happens on a dyno, it either triggers the overload protection, or something is going to give way.
If happen in the car, it may have a similar effect to the damper bottoming out - instant loss of grip is a very likely outcome. But I have seen dampers in rally and off-road applications explode under these conditions, after jumps.

About Moton and JRZ dampers.
I think the are the same, as far as te working principle and philosophy behind is concerned. I don´t know the finer details, but it probably goes back to the same man beeing involved at different times - Jan Zuijdijk
This is quite common within the dutch damper industry, that they are all somehow "connected" with each other, in one way or the other.

some info for the interested reader can be found here:

http://www.janzsuspension.com/
http://www.vehicledynamics-expousa.com/ ... ijdijk.pdf
http://www.jrzsuspension.com/uploads/20 ... og2010.pdf
http://www.motonsuspension.com/index.ph ... ilosophyms

As far as the function goes, they are the same damper. If you read there product discription, it´s like a carbon copy of each other.
I post that only as reference, it does not mean, that I agree with everything which is said in there publications, but we will come to that later.
They both use a 22mm shaft (not 25mm, as I initialy thought), but the underlying concept remains the same nonetheless.

This is "second hand" information on JRZ, Moton. Jan Zuijdijk was the designer of Koni shocks (racing division) in the early 90's, he left and formed JRZ when Koni was not accepting of his theory of larger shock shafts. He chose to go on his own with his designs.
Ever use a quick connect on a shock can? That was one of Jan's inventions..
While he was a great shock engineer, JRZ as a business had problems (apparently he was much better at engineering). One of the engineers working for Jan, left JRZ during troubled times and headed over the pond and was involved in starting Moton, which is a very similar shock to the JRZ.
Not sure if Jan is involved at Moton though. All of this is second knowledge and is the story as I heard it.

[speedsense]

... the author is truly mistaken. Especially were he states that some people actually lower their spring rate because of the spring rate of the shock. It's just beyond me that someone with experience with race cars would think this way (IMHO, wrongly so)

Mmm.. Rockers on race cars frequently have a "rising rate" characteristic. Hence reducing damper preload will, before adjustment, cause an increase in average spring rate. I have watched a race team recover the change in static ride height by adjusting push rod lengths, rather than adjusting spring platforms. I wonder....

If I had car with a 100lb spring, with 1/4" preload (25 pounds of preload) (regular shock perch spring mounting) and a 1:1 rocker, progressive to 1:5 (for simplicity) and it was sitting on a set up table with 400 lbs showing on the scale (at the wheel) as supported wheel weight.
If I remove the the preload (at the spring perch) and make it 1/8" (12.5 lbs) preload on the spring perch , will I see:
A) An increase or decrease in wheel load weight? (on the scale)
B) An increase,a decrease or no change in ride height? (measured at the chassis)

[747heavy]

Thanks for the reply speedsense - it´s appriciated
Unfortunately you have a habbit (and I don´t mean it in any disrespectful way), to mix different things together.
This makes it a bit difficult at times, to follow and to seperate the issues and principles involved. I don´t say that you are wrong with what you observe, just that you introduce another variable, which needs seperate attention. IMHO

Your last example is a perfect example for this.
I understand perfectly what you want to say (at least I hope so), but we need to consider, that a 4 wheeled vehicle is staticly over defined (it´s the 3-leged chair vs. 4-leged chair discussion).

This is why you see a load split when you take of load from one wheel, in an extreme case, you can take all the load off one wheel, the other three will share the load, and depending on the position of your CoG, you may be even able to still drive the car around the track.


Nevertheless, this is not helpful when we consider our problem at hand here.
This is we onlny look at a quarter car model at the moment (one wheel at the time).

If you change gaspressure in your damper you change preload (a force).
The ratio will only depend on your shaft/piston rod diameter. The relation which governs this is P=F/A --> F=P*A.
Which means for a given pressure you get a larger force (preload change) with a larger shaft.
Now we have a force, which we take off our spring.
Assuming that we had equlimbrium before, the spring will extend now, as we have taken load off her.
How much the spring will extend/relax in length is defined by the spring rate.
k=F/s --> s=F/k
This means a softer spring will relax much more then stiffer spring for a given load change. As an example if we take 1000 N (~100 kg) off an 100N/mm spring it will relax 10 mm if we to the same with an 500N/mm spring it will relax only 2 mm.

This change is totaly independent from your motion ratio, because we have a force equlimbrium. THe corner of our car still has the same weight, so the spring still need to support the same load.
How much the height of this coner will change is defined by your motion ratio, but in a purely geometrical way (no MR^2 only MR).
If we don´t compensate for it the car will be higher or lower by the amount of s*MR (or s/MR this depends on your definition of MR).
If your wheel moves 2 mm for any 1 mm of spring, then it will be now 2 mm higher, if our spring has extended by 1 mm.
If we have a motion ratio of 1:1 then it will be 1 mm higher for the same change.
So the amount of ride height change depents from your motion ratio.

Think about the example you have given to Dave and consider what happens if you make the change on all 4 wheels at the same time.
What will happen? In my world (the car has still the same weight) it will be x- amount higher or lower.

That the load changes different when you do it only at one wheel, has other reasons, as the car with 4 wheels on a set up plate is staticly overdefined.
If you want to make a test for yourself.

Put a resonable stable table on 4 corner weight scales, which you level nicely before. Zero the scales --> now put a weight (lets say 25 kg) on the table.
This is our car weight in question (25kg) if you at the readings of your 4 scales together it will allways be 25 kg.

But chances are that only 3 of your scales will show a weight.
Move the weight around, the individual scales will record a different weight, but the overall weight remains the same.
In this example the "Ride Height" of your table (if we consider the distance between table surface and scale surface as such) does not change.
Therefore the weight has to change.

Put the table on 4 springs and do the same.
Put the weight in the centre of the table (x and y) measure the "ride height" put a 2 mm spacer between 1 leg and one spring (motion ratio 1:1) and see what happens.
If you like, do the same with different springs and with the weight not in the centre, and you vary the preload (spacer) under the different springs.

I think this little experiment (which does not include dampers or friction) shows the challenges of corner weight balancing a car and trying to keep the ride height constant quite nicely, especially if the weight is off-centre.

Hope this makes some sense.

[DaveW]

I think I understand your argument, speedsense, but I hope you can accept that the effective wheel rate of a spring is the actual spring rate divided by (MR)^2, where MR is the motion ratio defined as wheel movement per unit spring movement. The effective wheel rate is a property of the suspension that helps to define how the suspension will react to disturbances. I will assume that you agree, & repeat the comment you queried:

Mmm.. Rockers on race cars frequently have a "rising rate" characteristic. Hence reducing damper preload will, before adjustment, cause an increase in average spring rate. I have watched a race team recover the change in static ride height by adjusting push rod lengths, rather than adjusting spring platforms. I wonder....

It should, perhaps, be noted that a "rising rate" rocker is one that, when coupled with a constant rate spring, will cause the wheel rate to increase as the spring is compressed. This has implications when the spring is required to carry an increased proportion of the static load, as it will if the damper preload is reduced. In that case the spring will be compressed more, the rocker will rotate, the wheel will retract & static ride height will be reduced. The change will cause an increase in the wheel rate if the rocker has a "rising rate" characteristic.

There are, of course, two ways of recovering the loss in static ride height. The position of the spring platform can be adjusted (increasing spring "preload"), or the push rod length can be adjusted (lengthened). The first method will return the rocker to its original position, whilst the second will leave the rocker in its new position. The first method will cause the change to have no direct effect on the way the suspension responds subsequently to a disturbance, whilst the second method may, depending upon the design of the rocker (it will if the MR varies with rocker position).

BTW, your comment on a damper "locking" when static is possible theoretically, but it rarely happens in my experience. A leakage path usually exists across a piston, & is even inserted deliberately on occasion by drilling a small bleed hole through the piston.

p.s. Great stuff again, 747.

[marcush.]

It may or may not have been mentioned ,but raising the gas pressure is not only adding a preload but will cause considerable friction and stickslip effects to dampermovement...it may as well compromise the seallife at the shaft..
At Ohlins it was always paramount consideration to keep gas pressure as low as possible ,just enough to prevent cavitation with the non TTX dampers.

[747heavy]

With most of the "gas spring" calculations that been done, aren't the gas spring numbers generated from a stationary point for the piston position?
Whether at a fully open position or some distance of insertion into the shock?
At this stationary point the valving is closed on both sides of the piston (no flow through it) and if the measurement of the gas spring was taken while in this "state" the gas spring effect would be at it's highest.
As seen by 747's numbers, the larger shaft size increases the gas spring, as the volumetric displacement of the fluid by the solid rod (assuming it's solid as it is in JRZ,Moton) provides more force towards the gas.
It would stand to reason that the highest gas spring rate would be a solid rod, solid piston with ample seals and no bleed holes open.
In use (running on a car) the shock has varying fluid amounts flowing through it at the different velocities and the highest flow is at the shock's highest attainable velocity, would contribute the least amount of force/resistance (of the assembly of rod and pistons) towards the gas spring having an effect.
This would have a reduction effect on the stationary gas spring number, especially with a very high flow piston in use...

Hi Speedsense,

It seems to me, that you still have a slight misconception in regards to the gas spring in a damper - with appologize and the upmost respect.
Let´s see if this drawings help, to get my point of view across.

Imagine this to be our damper, in a stationary position, as you say all valves in the piston o closed.



As you can see, we have the same amount of pressure above and below the piston, as it´s allways the case, based on the fact, that within a fluid under pressure, the pressure is equal at any point.

As you can see, we have equal pressure acting on both sides of our piston.
The only difference is, that said pressure has a larger area to act on in one chamber (bump chamber) and as we know that F=P*A, we can see that the extension force (preload), does not result from a pressure difference from above and below the piston.

It purely results from the different surface areas on both sides of the piston.
This difference will allways be there, whatever the pressure, it does not matter if valves in the piston are open or closed.
The force would be there even if we have no piston at all.
(please see the following drawing)



We have a pressure of 20 bar inside the damper, this pressure acts all around inside the damper. And it acts on the shaft. The only place where it can´t act, is the cross section area of the shaft (PI/4*d(shaft)^2)which is outside the damper.
This part has 21 (absolute) bar on one side and ~1 bar (atmospheric pressure) on the other.
This difference * the shaft cross section area gives you the force, you feel and measure as preload on your dyno or spring tester.

If you like you can make the test for yourself, if you have access to a damper and a dyno or spring tester.
Take all the shims of the piston and test the damper, you will still measure the same gas force/preload.
You can even take the piston off, but please be careful, it will shoot the shaft straight out, when you pressurize the damper, and you will have a mess, with oil everywhere .
Therefore, if you really want to do this, you should clamp the damper into your dyno or spring tester first, so that it can´t blow apart -->
Zero, your loadcell, and now pressurize your damper.
The force you will read on your loadcell is P*A, give or take ~10-20 N to account for friction.

The other small glitch in your conception, is about the hollow shaft.
It does not matter, as far as the volume change is concerned, if the shaft is steel or filled with oil.
Both mediums are incompressible, so both will displace the same amount of fluid. The damper with the fluid filled shaft, has a slightliy higher oil volume to start with, is maybe a bit lighter and provides maybe a bit more cooling, as the oil can dissipate heat via the shaft now.
But as far as displacing fluid goes, and as far as our area for the gaspressure to act on is concerned, it does not matter.





Hope this makes some sense, and that the gaspressure vs. preload argument is a bit easier to understand now.

[Arunas]

Just one thing is wrong - shaft has capacity, and different shaft capacity will take different amount of space inside shock by movement, and the only medium able to be compressed inside it is pressurized gas. As gas capacity will change differently (according to different space occupied by full or hollow shaft), gas pressure will also change accordingly.
Sorry for my English

[747heavy]

your english is fine Arunas,and thanks for your reply.
Let´s see if this is the case.

Here I made a quick sketch, to show the difference:



Let´s consider the following:
In case[A] I have a damper body with internal diameter of 40 mm and a height
of 100 mm, and a shaft with an diameter of 20 mm and a max stroke of 50 mm.

this gives me a oil volume (green) of (PI/4 * 40^2mm * 100 mm) = ~125644 mm^3
my shaft has a volume (red) of (PI/4 * 20^2mm * 50 mm) = ~15708 mm^3

if I insert the shaft into my damper body, I have a volume left for the oil of
125644 mm^3 - 15708 mm^3 = ~109596 mm^3
This means, as by damper body does not growth, the volume of the shaft (15708 mm^3)
has to be taken up by the gas volume and assuming I have a seperation piston of 40 mm diameter, the piston will move 12,5 mm

Now in case I have the same damper body but a hollow shaft. ( For the ease of explaination, I assume I have a "magic shaft" with "0" wallthickness, so that all the volume is represented by oil)
This means my total oil volume is now 125644 mm^3 +15708 mm^3 = 141372 mm^3.
When I now insert the shaft in to the damper the volume at the end (geen area) is 125644 mm^3. And the difference is a again 15708 mm^3.
Overall, I have more oil in the damper,yes, but the volume change if the same.

You have to keep in mind, that your shaft has to be closed at one end !!!!
If it is not, well, then we have no pressure --> no force and a lot of oil on the floor.

Hope that makes some sense

[Arunas]

Very good explanation! You are right, thank you. That means in both shaft cases we will end with different gas pressures (over shaft travel). Then this difference can be reduced only by reducing shaft diameter (e.g. volume), or is it not desirable?