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Clutch pack preload?
(137 posts, started )
Clutch pack preload?
Now with the test patch that includes clutch pack pre load i am wondering what it is? There have been some comments to it in the test patch section but between each comment there is a 100 other comments, also this is the more correct place to ask.

I have read some articles on it but they are all way above my knowledge and i dont understand them, so what it pre load?
Also does this now mean lfs will be more realistic and clutch packs will be quicker than locked diff's?

From what i can gather it seems to be something along the lines of the locking force when there is no torque being applied by the engine. But surely that is just the coast locking?
With a clutch pack there are three possible states of operation. Under load (acceleration), over run (coast) and no load (transition phase from coast to acceleration).

The no load state is were I had quite a dislike to the previous LSD behaviour in LFS as it would give you a sudden change in the cars behaviour just before the appex normally

So all preload does is set a minimum level of locking so you don't get the no locking effect in the transition phase of cornering. On doing some trial runs I've found that between 400 - 600 works well, much below 400 and you still get the drop in RPM and nasty kick on mid corner transition

Preload is great BTW I can see that the Locked Diff's days are numbered now
Does anyone know how the clutch pack operation used to work? If it was set to 80/80 - surely this wasn't really 80/0/80? 0 being the transition phase.
Quote from Blowtus :Does anyone know how the clutch pack operation used to work? If it was set to 80/80 - surely this wasn't really 80/0/80? 0 being the transition phase.

yes it was that was the whole problem with it
#5 - J.B.
Well explained Glenn. Put into a formula a clutch pack diff is actually quite simple. With an open diff each wheel always gets the same amount of torque as the other. What an LSD does is it transfers torque from on wheel to the other i.e. the torques are no longer the same for each wheel. The difference in torque is simply:

delta_torque = locking_factor * input_torque

So of you set for instance 80% locking factor then the actual torque that is trasferred from one wheel to the other is not constant but depends on how much torque the final drive is actually transferring between the wheels and the gearbox. When this torque is low (after releasing the brakes) the effect of the LSD is low. All preload does is set a minimum amount of difference in torque between the two wheels.

So while saying the diff is 80/0/80 isn't entirely correct, it is what the old LFS diff felt like. The new diff works just as advertised and makes much more stable and realistic setups possible. :up:
Can't wait for this new patch. I'm hoping it makes cars like the XFR fun to drive. At the moment they are just 'orrible
So, im a getting this correct?

an LSD set to, lets say, 70% locking. when going round a corner, the wheel with the least friction (Left for example) gets 70% of the power, while 30% is transfered to the right hand wheel.
Quote from djgizmo68 :So, im a getting this correct?

Hmm no I'd think that if you had 70% locking on power it would be saying that 70% of the available torque would be going to the wheel with most grip
Quote from djgizmo68 :So, im a getting this correct?

No, not really. By that logic adding more lock would make the wheel with less grip spin more, but it's the other way round - adding lock battles this behaviour.
70% power locking means that the diff will allow the wheels to turn at up to 30% different speeds. So, if you turn a right-hander and the inside wheel spins up, with an open diff all the power will go to that wheel because the power will take the path of least resistance. There is no diff locking so the wheels can turn at completely different speeds with no limit.

If you have a diff that locks to 70% on power then it will only allow the inside wheel to spin up to the point where it is travelling 30% faster than the left hand wheel. This means more power is going to the left hand wheel where there is more grip. This gives better acceleration than an open diff obviously.
The clutch pack diff has no limits to wheelspin. Even at 80% locked, one wheel can still spin at double the speed of the other. It just relates to the torque transfer possible.
Quote from Bob Smith :The clutch pack diff has no limits to wheelspin. Even at 80% locked, one wheel can still spin at double the speed of the other. It just relates to the torque transfer possible.

I stand corrected
#13 - J.B.
Quote from Glenn67 :Hmm no I'd think that if you had 70% locking on power it would be saying that 70% of the available torque would be going to the wheel with most grip

Almost. The deciding factor on which wheel gets more torque is wheel speed not grip. The slower wheel always gets more. So when both wheels are gripping it's the inside wheel that gets the extra torque (increasing understeer) but when it starts spinning it's the outside wheel.

And the locking factor doesn't really indicate a percentage of torque or power, it just tells you where the diff is on a scale of 0% = open diff to 100% = locked diff. As an equation:

locking_factor = (torque_left - torque_right) / (torque_left + torque_right)

So 70% locked means that the difference in torque between the faster and the slower wheel is 70% of the total torque. So if the wheels are getting 100 Nm of torque then the faster wheel will get 15 Nm and the slower wheel will get 85 Nm (85 - 15 = 70).

Quote from Gentlefoot :70% power locking means that the diff will allow the wheels to turn at up to 30% different speeds.

Sorry, that's wrong. Clutch pack diffs don't react to wheel speed differences. The slower wheel always gets more torque than the faster wheel but just how much more is not influenced by wheel speeds. See the formulas I posted above.

EDIT: argh, beaten by the Bob!
Quote from Glenn67 :With a clutch pack there are three possible states of operation. Under load (acceleration), over run (coast) and no load (transition phase from coast to acceleration)

These no load/transition situations confuse me. Isn't a simpler way of explaining it just: The diff acts like a locked diff until more torque than the preload is transferred through it? Is that correct?
Quote from geeman1 :The diff acts like a locked diff until more torque than the preload is transferred through it? Is that correct?

Quote from J.B. :When this torque is low (after releasing the brakes) the effect of the LSD is low. All preload does is set a minimum amount of torque that is always transferred from one wheel to the other.

From Wikipedea
Quote :The clutch type LSD responds to driveshaft torque. The more driveshaft input torque present, the harder the clutches are pressed together, and thus the more closely the drive wheels are coupled to each other.

What preload does is effectively put a certain amount of static pressure on the clutch plates which means there is a certain amount of static locking (resistance) between both wheels at all times i.e. it doesn't vary with driveshaft torque so its effectively the minimum locking force of the diff. When torque is applied to the drive shaft it will still increase the locking forces on the clutch plates dynamically or proportionally.

So Preload doesn't cause the diff to act as a locking diff but rather as a more stable LSD i.e. reduces the dynamic range of locking forces that the diff opperates in.
#16 - J.B.
Yep. But now there's one thing I can't quite get my head around. In a free rolling state there is no torque being transferred between the wheels and the gearbox. Now roll around a corner. The preload is now supposed to be the amount of difference in torque between the faster outside wheel and the slower inside wheel. What does this mean if there is no torque there to transfer, so to speak?

EDIT: I think I know. The LSD does indeed still make sure that the preload is the difference in torque between the two wheels. The only way it can do this when there isn't any torque there, to distibute between the wheels, is by applying a braking torque to the outside wheel and an accelerating torque to the inside wheel. --> less oversteer.
Quote from J.B. :What does this mean if there is no torque there to transfer, so to speak?

LOL I was just about to ask you that question But isn't there always torque between the wheels in a corner due to the speed diference of the wheels and the momentum of the car? I was also going to suggest that the momentum of the car in a corner would have a small influence on the way the LSD operates even in coast and power states as some forces would be acting back through the wheels on the LSD as well wouldn't they?
The only thing that defines the operation "mode" of the clutch LSD is the input torque from the engine - the difference between wheel speeds has no (or only very little?) influence.
Quote from Glenn67 :LOL I was just about to ask you that question But isn't there always torque between the wheels in a corner due to the speed diference of the wheels and the momentum of the car? I was also going to suggest that the momentum of the car in a corner would have a small influence on the way the LSD operates even in coast and power states as some forces would be acting back through the wheels on the LSD as well wouldn't they?

the situation you describe is the coast condition, and yes, the momentum of the car combined with the different distances the inside and outside wheel must travel around the corner cause the diff to load up.
Quote from Glenn67 :LOL I was just about to ask you that question But isn't there always torque between the wheels in a corner due to the speed diference of the wheels and the momentum of the car?

not with a completely open diff

Quote :I was also going to suggest that the momentum of the car in a corner would have a small influence on the way the LSD operates even in coast and power states as some forces would be acting back through the wheels on the LSD as well wouldn't they?

in terms of torque the wheels actually use there is a difference yes
#21 - Vain
I also have a question.
Why is the unit of preload Nm?
If I assume a linear clutch the torque exerted is some constant by the difference in angular velocity between the two wheels.
M = c * (omega1 - omega2) or in units:
Nm = ? * 1/s
So obviously the constant's unit must be Nms, or Nm*s/rad, if you measure angular velocity in radians/second instead of 1/second.
Or is the c in the formula calculated from normalization as in
c = M0/deltaOmega0
with M0 being the setting in the setup and deltaOmega0 an example difference in angular velocity (usually 1 1/s)?

Vain
Now, I go completely :insane: when looking at these formulas, but, what else would you want to specify preload in? Strength of the spring that presses the clutch plates together? It just specifies the minimum amount of torque the wheels have to overcome in order to rotate at different speeds (I think ).

E: Someone really needs to calculate through a few test cases of clutch LSD operation (like I failed to), otherwise I fear the confusion will never clear up.
Quote from Vain :I also have a question.
Why is the unit of preload Nm?

Vain

because IRL preload is set using a spring that pushes the clutch plates together. stiffer spring = more preload.
Quote from Vain :I also have a question.
Why is the unit of preload Nm?

because torque is measured in Nm ?
#25 - J.B.
Quote from Glenn67 :LOL I was just about to ask you that question...

hehe, you posted while I was editing my above post. Looks like we had the same thoughts twice in 6 minutes.

Quote from Vain :I also have a question.
Why is the unit of preload Nm?
...the torque exerted is some constant by the difference in angular velocity between the two wheels...

Probably this assumption is wrong for a clutch pack diff although I can't find any information on this. In fact it's correct for a viscous clutch which has Nms/rad as its unit in LFS.

Quote from AndroidXP :Someone really needs to calculate through a few test cases of clutch LSD operation (like I failed to), otherwise I fear the confusion will never clear up.

It' really quite simple.

An example for accelerating in a corner with 40% locking:

1. Find out how much torque is coming from the engine. For example the FOX has a max torque of 221 Nm so I'll just pick the case where I'm using about half throttle to accelerate --> torque_engine = 100 Nm

2. Multiply this torque with the gear ratio and final drive ratio to get the torque at the diff. Example: torque_diff = 100 Nm * 1.6 * 4 = 640 Nm

3. use

delta_torque = locking_factor * torque_diff

to find what the torque difference between the two wheels is

--> delta_torque = 0.4 * 640 Nm = 256 Nm

Now if this number is bigger than the preload then
--> torque_inner = 640/2+256/2 = 448 Nm
--> torque_outer = 640/2-256/2 = 192 Nm

If the preload is set to more than 256 Nm, say 300 Nm then
--> torque_inner = 640/2+300/2 = 470 Nm
--> torque_outer = 640/2-300/2 = 170 Nm
--> more locking, less oversteer than before

I hope that clears up some confusion.

Clutch pack preload?
(137 posts, started )
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