Hi.
Where does this come from mechanically?
I can believe what you say and try to memorize it, but I'd like to understand how it works mechanically, but technical drawings of LSDiffs are seldom.
If I press two (or however many) clutchplates together to create preload they should always resist torque between them by exerting an opposing torque. Regardless of the torque difference. So if I have set up my preload-less differential to stay locked in situation B in your diff2-picture, and then put additional load on the clutches I can't help but believe that now the differential should be able to resist a higher torque-difference between the driven wheels than before.
I think the preload always has an effect on the clutches, no matter in what state it is.
In this picture you can see the preload springs pushing the pressure rings.
If you add spring hard enough, it will keep the clutch pressed all the time.
I don't think it's possible to have a lower force, in a torque loaded state, then the preloaded one. Of course, I don't think anyone would do such a thing in RL
"To change the ramp angles, new ramps are needed – just exchange one for the other. If preload is needed add shims behind the preload springs, or change to stiffer springs. Locking rate of both sides of the ramp can be lowered by reducing preload, or by rearranging the clutches to reduce the number of active surfaces."
Here is a good illustration of the three LSD types.
You can see how the pinion, under torque, pushes the plates creating pressure on the clutches. It moves up on throttle and down on coast.
-the 2 way diff has the same locking pressure on coast and throttle
-the 1 way diff puts pressure and moves the plates only on throttle
-the 1.5 way diff has a lower pressure on coast.
The ramp angles on the cam set the locking power. You may remember that from the GPL setups.
Here's one simple explaination to clutch diff preload:
With the new preload setting, we get to set the STATIC preload of the clutch diff. The % settings are how much locking force (aka preload) the diff gets whilst experiencing engine input shaft torque (both on and and off power(aka engine braking).
So, what is this preload variable? Well, it's simply how much difference of torque required to turn a clutch diff from a clutch friction locked state (aka locked diff) to a limited open diff set.
What does this mean for the grease monkey who's working on the car? Well, here's the somple way to understand preload:
First, get a car that has a clutch pack LSD (e.g. many pre-2000 Pajero models, etc). Jack the rear wheels up with a grarage jack and hold the rear tires off the ground, using axles stands, etc to ensure safety (as long as the wheels are free off the ground for you to spin as you please).
Now, hold any wheel from any side (let's just stick to the left wheel for this example) steady so that it can't spin at all. Then try to spin the wheel on the right side whilst the left wheel remains practically unspinable. You will discover that the right wheel requires a finite amount of torque to wrench it free and spin whilst the left wheel remains static. The torque required to overcome this differential locking force is the preload.
FYI, a properly working Pajero clutch diff has a static preload of of around 90-110Nm. These static preload settings are considered pretty tight in production cars. It is a 2-way 4 pinion clutch diff, and one of the best and tightest LSDs among production 4WDs. Another production 4WD known to have similarly effective rear LSDs is the Nissan Patrol(as much as I can rememeber for now). It's also a pinoin 2-way clutch pack diff. Some other vehicles have "LSDs" so horrible (aka cheap excuse of a thing chucked in just to make the catalogue features list look longer and more impresssive with all sorts of acronyms the average non-car guy isn't too familiar with) that they have only static preload, all from springs pushing on relatively few clutch packs. No extra locking under power or coast.
What do you mean by this? As far as I can tell the input torque from the engine is in fact the variable that tells the diff what to do, assuming it is not in the locked or preload state of operation. Just like the wheelspeed difference is the controling variable for a viscous diff. In fact clutch pack LSD's are also known as "torque sensing" differentials.
Another thing I notice is that you state a diff running at a torque bias of, say, 2 (how to covert locking factor to torque bias) limits one wheel to a maximum of double the torque that is on the other wheel. The way I see it, it not only limits it, it actually makes sure that the torque on one wheel is exactly double the torque of the other wheel (as long as the clutch plates are slipping). After all if the clutch plates are slipping then the torque they are transfering from one wheel to the other is a fixed percentage of total torque. Add to this that the total torque at the wheels has to equal the torque coming from the engine (Newtons 3rd Law) then the tyres have no choice but to adapt their slip ratios to the torque and normal loads they are getting. Am I disagreeing with you here or simply misunderstanding something?
@Vain:
Just try to imagine the concept of preload with a linear coil spring. Say you have a spring that is 100mm long and needs 10N to be compressed by 10mm. To preload it you install it in such a manner that it can't fully expand and is compressed to 90mm length. Think something like this with limited travel. The mechanism that is holding the spring in this position will have to deal with 10N force that the spring is exerting. Now if you press down on the device, for the first 10N of force you apply, the spring will not actually compress further, you will only be taking the load off the preload device and moving it your thumb. Only when you get over 10N of force will the precompressed spring beginn to compress further than the 90mm it was compressed to by the preload.
How exactly this is implemented in an LSD I don't know as pictures seem to be very rare.
Thanks for that thought. This makes a lot sense when you look the pictures undoz posted (thanks a lot for those, undoz! The 'Operational Theory' section of that linked page is great!).
Yes, that's probably right. However, in order for the road reaction torques to exit at all, there will usually be engine torque of course. Mathematically you're already taking that into account by dealing with it only with the reaction torques coming in from the wheels. If LH = 300 and RH = 500 then there's probably already an engine torque there.
"Limits" is correct. If the car is in a very gentle turn the diff is locked and the slip ratios and vertical loads may be nearly the same. The outside tire force is not necessarily fixed to the ratio. It can be less or identical to the inside tire such as when running straight.
I'm not too sure about that. There should be, I suppose, but have seen no mention of it being important enough to consider anywhere. If the change is small enough it wouldn't be noticeable so might be disregarded. Add in that temperature, clamping force, sliding velocity, and perhaps other factors might be changing the friction coefficient(s) too in an unknown way and it might not be worth considering. I don't bother with it and I'm not aware of any full vehicle simulations used in engineering that do either.
I was just going to have a look at diff/tyre forces using forces view and raf data. I went to the skidpad, started out with an open diff and adjusted my driver inputs until I was keeping to a more or less constant, steady state circle.
I was expecting the longitudal tyre forces to be identical for both rear tyres in this case (fully open diff) yet the result can be seen in the attached screenshot.
Does anyone have an explanation? Is the assumption that in steady state an open diff delivers the same amount of torque to each wheel wrong? I can't find any disagreement via google.
What is the forces view actually showing you though? Is it the sum of all forces - including those counter forces acting back through the tyre, whereas what your looking for is the torque split from driveshaft only maybe
Well, I think if the wheels are not accelerating but keeping a constant speed then the reactive forces at the tyres (the ones I think we're seeing in forces view) should be exactly the same as the forces coming from the driveshafts (Mechanical equilibrium).
aren't the yellow lines showing the weight distribution and the other coloured lines showing the amount of grip the tyres have? why would they show the drive shaft forces?
Why would they be the same? The inside wheel is on a different path to the outside and there is dramatic wieght differences between the wheels. Saying torque is the same is not the same as saying that tractive force at each wheel is the same or am I missing something Afraid its out of my depth Have to wait for Todd to reappear
@csurdongulos - I thought the colour is indicating grip and the length of the line force, but what force exactly I'm not sure
If they were the potential tyre grip/traction then they wouldn't be zero at standstill. They grow longer when you accelerate harder so I think they are actual forces not grip indicators.
Hmm, I thought it is. In the case that the tractive force is not the same as the drive shaft torque the wheel will be forced to accelerate, right? Newton's Laws basically. In the screenshot case the wheel wasn't accellerating.
I do believe that the lighter green is indicating less grip than dark green or blue so maybe there is some slip happening on the wheel showing greater force
I was about to point that out. Torque cannot be transferred across an open diff, so engine braking must be equal. Tyre scrub can still be different though, which could account for the difference in force magnitudes.
Just to be sure we mean the same thing: With not accelerating I meant that the wheels had a constant speed, not that I was engine braking. In fact I was using a constant 40% throttle.
What I meant is that, from the perspectice of the wheel, if the torque coming from one side (from the driveshaft) is not exactly the same as the reactive torque coming from the other side (the tractive force between the tyre and the road) then the wheel wouldn't be able to maintain a constant speed. It would be forced to accelerate or deccelerate due to the non-zero torque sum.
The colors are relative to the slip ratio, afaik.. At max green, the slip ratio is optimal and the tire is producing as much force as it can. Go beyond that, and the line decreases length, and turns red because you're past optimum longitudinal slip ratio.
If the weight is mostly on the outside wheel, and just a tiny bit of force is being applied to keep the car at steady speed, then the inside tire is probably slipping a bit and getting closer to it's optimal slip ratio which accounts for the colors/lines.
[ot]
@Bob, in the test patch thread you called the patch compatible - instead of incompatible. I didn't want to spam the thread to point it out though [/ot]
Ah, yeah, he did specify the rear wheels, I was assuming a FWD car.
Tyre scrub is still going to affect the total force on each tyre though, thus the length of the lines. The tyre supporting the majority of the weight will have the greater rolling resistance and tyre scrub resistance, so the lines should be shorter. I just don't know if that would accout for the different in forces, it's hard to say how many times larger the force on one tyre is than the other (the left line goes off the top of the image, and I think both lines start from the centre of the tyre, so the bottom of each line is obscured by the tyre itself).
That's a good point too Bob. The color difference I believe is definitely due the slip ratio. There is very little weight on the inside wheel, so it's spinning slightly faster (not slower as you would think in this direction of turn) than the outside wheel. The outside wheel is also bearing a huge lateral load, which means it's ability for tractive effort is diminished. So in this situation, the inside wheel is actually producing a tiny bit more foward tractive effort than the outside. The difference is not very big though, and must be attributed to the slip ratio being higher due to less vertical load for a given torque. As to why the lateral vector on the inside tire is greenish also, that must be for the same reasons, but I'm not sure if that makes sense or not ... Interesting.
Have to wait for Todd to reappear 
... Interesting.