[WRC_Obsessed]

The Limited Slip Differential

The solution to this problem is the limited slip part of the DCCD. To sum up what a limited slip differential does in as few words as possible: A limited slip differential transfers torque. The DCCD has a complex clutch style mechanism that will be discussed in detail in Part 2 of this article, but for now it is convenient to think of it as a clutch with one plate on the front driveshaft and the other plate on the rear. Engaging the clutch connects the two driveshafts and so will tend to try to bring them to the same speed. But what happens if the two driveshafts are already traveling the same speed? A lot of discussion in these forums centered on that question, which can also be asked: What is the torque split when the LSD clutch is fully engaged and the driveshafts are turning at the same speed? Note: This question only applies to a clutch style LS mechanism. A viscous LS mechanism cannot transfer torque without a speed difference.

The answer to this question is: It depends. The answer is neither “35:65” nor “always 50:50”. Often literature (Subaru’s and others) will mention something like "equal torque distribution" but this is an inaccurate simplification of a much more complex process. A more considered answer would be:


It depends on:


The front/rear weight distribution of the car

Any dynamic loading of the axles (due to acceleration or deceleration)

The available friction at the tires, which depends on the surface under the tire (ice snow gravel tarmac) and the tire itself

The amount of lock (the clutch engagement force)

The torque being output by the engine

Varying any of these may (or, with the last two, may not) cause the torque split between front and rear driveshafts to change. Now what does this mean? Well, not much. As long as both the front and rear driveshafts are turning at the same rate, who really cares which one is doing all the work? That is the beauty of the limited slip clutch; because the driveshafts are forced to rotate together, it does not matter which shaft is doing what part of the work (is accepting what % of the engine torque), the net result is all four wheels turning at the same speed and the car moving right ahead. But to better understand exactly what the clutch is doing and what forces are acting on the driveshafts and tires, let us examine this more closely.


Let us look at our problem scenario again but this time with the limited slip engaged so the two driveshafts are forced to move together:

Locked differential and front wheels on ice: As in the previous example, the torque on the front driveshaft that will cause the wheels to spin is 10ft-lbs. Just like in the first example, that is the max amount of torque that can be applied to the front driveshaft. Why? Because the conditions under the front wheels are still the same, the ice will only "push back" with 10 ft-lbs, so the front wheels can only "push" with 10 ft-lbs. This is very important to understand. You can't push against something if it won't push back. But since the driveshafts are locked together, the engine is free to apply more than 10 ft-lbs to its connection at front driveshaft. In fact, think of it as just one large driveshaft, with the engine twisting the shaft at two places (side by side), one with 35% of its effort, the other 65%. Fig 3, the locked case, shows this concept. If the engine applies 100 ft-lbs, what happens? 35 ft-lbs on the "front twist", 65 on the "back twist", but remember the front axle can only accept 10 ft-lbs. If the rear can take 90 ft-lbs before slip, then rest of the engine torque (90 ft-lbs) goes to the rear axle, which is (let us assume) enough to move the car. In this case the torque split is 10%F/90%R (I ignored the case where the rear can take more than 90 ft-lbs we’ll get to that below.)



In this case, what determined what the torque split turned out to be? It was the amount of traction at the axles. But what determined that value? It was the icy surface and the type of tire and the weight the axles. With more weight on the axle, more traction could be achieved (which is why pickup owners put sandbags in their truck beds in the winter). With winter tires, more traction could be achieved as well. So you see how torque split depends on how much torque each driveshaft "can accept" which goes to the first three items in that list: weight, dynamic loading, and friction. Here is another example to make that point:


Locked differential and all wheels on tarmac, rear weighted: Assume that the car has a whole trunk full of monkeys that have given it a 40F/60R weight distribution. Let us also assume that when perfectly balanced the force necessary to spin either front or rear tires was 200 ft-lbs. Well now in this unbalanced state the force to spin the front will be reduced by 20% and the rear increased by 20% so the max torque at either axle is 160ft-lbs F/240ft-lbs R A key point to realize is that even when applying less than max torque, the torque at the axles is still in proportion. So if the engine was applying 100 ft-lbs total, again using our “virtual single” driveshaft, the torque split would be 40ft-lbs F/60 ft-lbs R. But if we put bad tires on the front, or put the front tires on dirt, lets say the max torque per axle changes to 100 ft-lbs F/240 ft-lbs R. Now at a 100 ft-lbs engine torque, the distribution would be 29 ft-lbs F/71 ft-lbs R. With a locked diff the torque split is always proportional to the relative maximum amounts of traction at the tires. And this always depends on load and friction.


But how do the last two numbered items figure in? How does the amount of lock and engine torque output affect torque split? They affect it because the DCCD clutch does not turn the front and rear driveshafts into one perfect single driveshaft. This “virtual single” driveshaft idea only works until the torque difference between front and rear driveshafts becomes greater than the force required to slip the clutch. After that point the clutch cannot transfer any more torque, regardless of whether or not this limit will result in the clutch slipping and the driveshafts moving at different speeds.


Here is a paraphrase from howstuffworks.com: "The torque supplied to the wheels not on the ice is equal to the amount of torque it takes to overpower the clutch" Again what this is saying is that the maximum amount of torque that can be moved from the engine's "twist" of the front driveshaft to the rear driveshaft is the amount of torque that would cause the clutch to slip. This clutch slip amount is determined by the lock %, and even when at full “lock” the DCCD clutch will slip if enough torque is applied. Here is an example that demonstrates this:


Locked differential and front wheels on ice and rear wheels on tarmac: Say in this case it will take 3 ft-lbs to spin the front wheels and 200 ft-lbs to spin the rear wheels, and these numbers take into account both the available traction and load distribution. Initially we will view the locked diff as creating one long driveshaft.

Assume that it takes 50 ft-lbs to slip the clutch in the DCCD at 100% full lock. What this means is the maximum torque the engine can twist the front driveshaft with is 3 ft-lbs plus up to a max of 50 ft-lbs. Any more torque on the front driveshaft will cause the clutch to slip and the front wheels to spin. Working these numbers out shows the following: If the engine is supplying 151 ft-lbs of torque, then the front driveshaft "twist" is 35% or 53 ft-lbs, and the rear is 65% or 98 ft-lbs. At this point any more torque applied by the engine will overcome both the ice friction and the limited slip clutch friction, and the front tires will start to spin, and any more effort to increase engine torque after that will just cause the front driveshaft to spin faster and rev the engine. The split at the axles at that point will be 3 ft-lbs F/148 ft-lbs R.


Front wheels on dirt and rear wheels on tarmac DCCD at 10% lock: Same premises as before but here it will take 50 ft-lbs to spin the front wheels and 200 ft-lbs to spin the rear wheels. Also let us set the DCCD so that it is only 10% engaged and so will slip at 5lbs. In this situation the point at which the clutch can’t transfer torque will occur before the point at which the front tires will slip. After that point any additional torque is divided in the 35/65 ratio, same as the open diff. That changeover point comes at an engine torque of 33.5 ft/lbs, at which point the actual torque at the axles is 6.7ft-lbs F/28.7ft-lbs R. If we increase engine torque to 100 ft-lbs, then the rest (66.5 ft-lbs) is split up 35/65 so we end up with a total actual torque split of 30 ft-lbs F/70 ft-lbs rear. If we then change the lock to 100%, the clutch can transfer all the torque difference (15 ft-lbs) and our “virtual driveshaft” analogy holds and so the torque difference is 20 ft-lbs F/80 ft-lbs R. So by changing the lock % we can change the torque distribution, even though there was no slip in the system.


Here is a table that summarizes this:


Front Dirt Rear Tarmac
Engine Torque
Front Acutal Torque
Rear Actual Torque

Open Diff
100 ft.lb.
35 ft.lb.
65 ft.lb.

10% Lock
100 ft.lb.
30 ft.lb.
70 ft.lb.

100% Lock
100 ft.lb
20 ft.lb.
80 ft.lb.




So what should you remember from Part 1 of this article?


With an open diff the torque split is always fixed by the gearing of the planetary gearset. With the STi, it is 65R/35F.

When the diff is locked the answer to “what is the torque split” is: It depends. It is not always 50:50.