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The DCCD Primer Part I
by: Seth Cooper
This document is
intended to show how the Subaru STi Driver Controlled Center
Differential (DCCD) works. This will be a three-part document. Part 1
consists of a theoretical view of the DCCD as a simple planetary
differential with a simple clutch, and explains how torque gets from
the engine to the front and rear wheels. Part 2 consists of a
description of the actual operation of the DCCD, referencing the Subaru
tech manuals that explain the nitty-gritty mechanical details of the
DCCD, and Part 3 is practical; how does auto mode work and
suggests when it might be desirable to use manual mode. The DCCD is a
complicated mechanism, and it is only one part of an AWD car's
drivetrain, which is also very complex. This complexity can get in the
way of understanding the basic physics of how the engine makes the car
move. This text will strive to simplify as much as possible the
mechanics of the car to aid in understanding, without ignoring any
important concepts in the process. Much of the confusion regarding the
DCCD operation has come from sales literature that (in the interest of
not confusing potential buyers) oversimplifies the operation of the
DCCD, and LSDs in general.
The information in this article comes from several sources, including Subaru published materials, websites and various individuals. A list of links and contributors including NASIOC members is provided at the end of this document. A great deal of information is from the website howstuffworks.com, the perfect name for an excellent site. Portions of this text have been lifted directly from these sources sometimes without proper credit, for which apologies are made in advance.
Before discussing the DCCD, it is necessary to define torque, which will be mentioned over and over in this document. Torque has been discussed a lot on NASIOC, but the definition is worth reiterating since "moving torque around" is what the DCCD is designed to do, and we need to be clear on just what it is that is being moved. From howstuffworks.com comes this definition of torque:
is a force that tends to rotate or turn things. You generate a
torque any time you apply a force using a wrench. Tightening the lug
nuts on your wheels is a good example. When you use a wrench, you apply
a force to the handle. This force creates a torque on the lug nut, which
tends to turn the lug nut. The unit of torque we will use is the
lb-ft. As you can see, the lb-ft contains a unit of force (the pound)
and of distance (the foot), which is what torque is: force, applied at a
distance, in a manner to rotate or twist something. The engine in a
car generates this twisting force, and the drivetrain - including the
transmission, center, front and rear differentials, and axles - moves,
multiplies and divides this force to get it to the wheels. By twisting
the wheels, the car forces the tires' contact patches to push against
the ground. The ground pushes back, and the car moves.
The DCCD is a
just a special type of limited slip differential (LSD), so to understand
how it works, one needs to first understand how basic differentials
work, and then move on to what the "limited slip" part does, and then
finally examine the special feature of the DCCD, the fact that the
amount of "slip" in the "limited slip" portion can be adjusted, either
by onboard computer or by the driver.
The Open Differential
The differential is a device that splits the engine torque two ways, allowing each output to spin at a different speed. The differential has a special property in that the torque from the engine is divided between the output shafts in a fixed ratio, regardless of the rotational speeds of the output shafts. In the conventional differentials found in the rear and front of cars, the torque split ratio is 50:50. The DCCD is a center differential and happens to be a planetary type. In this design it is easy to select the ratio in which the engine torque is divided by varying the different sizes of the gears involved. In the STi, the Subaru designers selected the gear sizes in the DCCD to split the input torque in a ratio of 65% Rear: 35 % Front. The WRX and Mitsubishi EVO use 50:50, the BMW 3-Series AWD models and the Porsche Cayenne use 62R:38F. The term "Open Differential" is used to describe this kind of differential, where there is no limited slip component (or where that component is deactivated). Figure 2 shows this case for the STi.
[Note: It is difficult to show why a differential has this property, so
it will remain beyond the scope of this article.]
The benefit of this property (the one in bold print) is that it helps when the car goes through a turn. In the case of a turn, the front and rear driveshafts must be allowed to rotate at different speeds, because the turn causes the rear wheels to track inside the front wheels, and so the rear driveshaft will turn at a slower speed than the front driveshaft. The center differential allows this to happen while still keeping the same torque split between front and rear shafts.
Open Differential Drawback:
To understand what the limitation of an open differential is, one needs to understand some things about traction. Isaac Newton's third law of motion says "For every action there is an equal and opposite reaction", or to paraphrase: "You can't push against something that won't push back". This concept becomes important when talking about traction. Again, borrowing from howstuffworks.com:
“There are two factors that determine how much torque can be applied to the wheels: equipment and traction. In dry conditions, when there is plenty of traction, the amount of torque applied to the wheels is limited by the engine and gearing; but in a low traction situation, such as when driving on ice, the amount of torque transferred to the ground is limited to the greatest amount that will not cause a wheel to slip under those conditions. So, even though a car may be able to produce more torque, there needs to be enough traction to transmit that torque to the ground. If more throttle is applied after the wheels start to slip, the wheels will just spin faster.” It takes very little torque to rev the engine and drivetrain without the load of the pavement pushing back, so the engine produces very little additional torque before it gets to redline. It’s not possible to push against something that won't push back. So with all four wheels on ice, the amount of torque (or push) the engine can produce and transfer to the ground is severely limited. This is because the only things that are pushing back are the traction the ice gives the car and the inertia and friction of the engine and drivetrain parts, which are both negligible.
Now what happens if just the front wheels of the STi are on ice? Here is the problem with the open differential. The differential must preserve the 65:35 ratio. It is understood that the front wheels can accept very little torque without breaking traction. So the torque the engine puts out becomes limited by that fact. The total engine torque being applied is 2.8 times the amount of torque that goes to the front wheels. (1/0.35) If the maximum torque the front wheels can make use of is a very small number then the max the engine can usefully supply is 2.8* a very small number which is still very small. 65% of this total torque goes to the rear wheels, but it is probably not enough to move the car.
Applying some numbers to that example (for simplicity in all examples
assume a transmission gear ratio of 1:1):
Open differential and front wheels on ice: Assume the torque on the front driveshaft that will cause front wheels to spin is 10 ft-lbs. So the engine can put out a maximum useful torque of 10*2.8=28 ft-lbs, and when it does, the rear driveshaft sees a torque of 18 ft-lbs. 18 ft-lbs may not be enough to turn the rear wheels, but that does not matter, the engine (through the differential) is still twisting the rear shaft whether or not it is actually turning. Remember the open differential allows the shafts to move at different speeds and still keeps the same torque split. In this case, no matter how much the engine is revved, the torque it produces is limited to 28ft-lbs (to be precise we must add the amount it takes to accelerate the engine and front wheels when we rev it, and overcome internal frictions etc.).
Here is another case of the open differential that illustrates the "twisting" vs "turning."
Open differential with the handbrake pulled: Here the rear wheels are prevented from turning by the rear brake calipers, but the car is still capable of motion. Assume that the rear tires are on gravel and front tires are on pavement. If the torque required to turn the front wheels while dragging the rear wheels is 100 ft-lbs, then the engine is required to put out 2.8*100 = 280 ft-lbs. The rest of the torque (180ft-lbs) goes to twisting the rear driveshaft. This torque ends up "pushing" against the brake pads, which must "push back" to keep the rear wheels from turning.
Note that in both these cases all the engine power is going to the front wheels, but the torque is still split 65:35. (And also note that the law of conservation of energy dictates that in both of these cases the front driveshaft must be spinning 2.8x faster than the engine, and the design of the center diff ensures that it does.)
Now in the second case, the open diff is a benefit. If one wishes to do a "handbrake turn" and still move the car with the front wheels, this is possible. (This is why the DCCD unlocks when the handbrake is pulled.) But in the first case, there is a problem, because the fact that the front wheels are on ice limits the torque output of the motor and prevents the rear wheels from getting the car off the ice.
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
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:
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
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
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.
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
||Front Acutal Torque
||Rear Actual Torque
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.
Subaru STi manual section 6MT
My coworker, J. Hemmes, mechanical engineer, for his comments
The following NASIOC members:
cgroppi and STI_FFY for their reviews
Zoso for the suggestion that resulted in this document
Khaos, johnfealstead, afpdl, ToddStratton, Aquavir and others for providing crucial information that aided in my understanding.
All who participated in DCCD discussions, including but not limited to:
the above members and RafalW, stonemonkey, strangerq, Leonardo, zavigm,