Subaru has an permanent AWD so the Power cant be fully cut.

You can only adjust the center diff. a little.

The center diff works like the normal axle diffs.

maybe:lol:

 

The thing with the normal front and rear axles are that I can see where the power is coming from. Take the rear, it's driven by the driveshaft into the diff, and divided into left/right. But I can't or don't understand how's that happening for the center. The front wheels/axles are so far front vs the start of the driveshaft to the rears.

 

The front and center diff are part of the tranny. it is called a transaxle, fwiw.

 

This is an old Vid. but its nice!

YouTube - Subaru AWD vs. competitors

 

This is an old Vid. but its nice!

YouTube - Subaru AWD vs. competitors

That was a neat video. It gets a "Yay Subaru" from me.

I'm thinking an STI in the "full lock" mode will just cruise right up that ramp and then proceed to rip it in half and eat it :devil:. You know, not that I'm biased or anything...

 

There is a thread that you can search for the explains all this.

To make a short few comments:

•The engine connects tot he tranny which connects to the center diff which connects to the f&r difs which connect to the wheels.
•The torque split is permanently set. The DCCD wheel does not change this.

 

You sure?

"All-wheel drive performance cars are rare enough. But the WRX STi is even more unique because the driver can choose the value of the torque split between the front and rear wheels!"

Source: Subaru Drive Performance Magazine : DCCD - Driver Controlled Center Differential

 

Positive.

 

The only thing we can change is the amount of locking force.

 

Exactly. With the '04-'05 diff, we had an electromagnetic clutch to adjust the lockup. This changed for the '06+ centre diff, as there is always some form of mechanical LSD action available, in addition to whatever lockup we or the computer sets but the geared split is always 41:59.

So to answer your question, I believe an '06+ STI should be able to move the rear wheels, even if the fronts have extremely limited traction.

 

I still don't understand, but I'll try watching that video later to see if it helps.

 

Oh I definitely still don't understand either, that's why I'm lurking around here waiting for some more explanation.

How is torque split related to the amount of power routed to either end? Cause I keep reading the torque split is defined by the gears (41:59) and is not variable. Okay, fine, but then when one end slips the power is routed successfully to the other end. Those two sentences seem contradictory to me, someone please explain.

:tup:

 

if your front tires use 10ft-lbs of tq to break traction, then that is 49% of your available torque. the other 51% is present at your rear wheels, i.e. 10.X ft-lbs.

 

That was OUTSTANDING! Here I am getting hung up on engine torque when it is the torque at the wheels that is relevant! 2 sentences. Very nice.

Now, to help me even further, how does the DCCD play into this?

 

this has all been covered before. i really don't feel like searching for the thread, but there is one that answers this and every other question you're bound to have.

 

100% of the APPLIED torque, yes, but it is always split at the fixed ratio.

 

That fixed ratio is with the diff open. When it's locked, it's another story.

 

There are a ton of them.

My explanation is in this thread http://www.iwsti.com/forums/drivetrain-components/83357-dccd-clarification-3.html

I'm actually studying for my M.E. degree and this little math problem has me kinda stumped. Feel free to completely call me out on this.

Stargazer all that math works out but I think one thing we left out is that Force=Mass*acceleration. Im pretty sure Mass in this case would include a portion of the cars Mass. On the other hand, if a wheel if off the ground(or cant apply force to the ground) then it is not contributing to the acceleration of the body.(only the acceleration of the wheel, rotor etc...plus we know with a locked diff, front and rear acceleration are equal to eachother)

But if that wheel can apply force to the ground, than the Mass must include a portion of the car's mass. The force applied to the ground, or rather the reaction to it, is what actually causes the body(car) to accelerate.

So if we dont have force applied to the ground then the mass is only the Mass of the wheel/rotor etc..(For simplicity, lets consider this Zero right now)

Since we dont have Mass we dont have any Force either.(Force =Mass*acceleration)



Imagine that the rear end is off the ground with the center diff locked. Acceleration is directly porportional front and rear(with our locked diff) so let "1" represent acceleration to make this even more simple.

Using Force=Mass*acceleration,

The front end has 100% traction so the mass is the mass of the car. Lets say 2010kg.
The rear end is not accelerating any Mass.(realistically it is but we called it zero because it is very small compared to the mass of the car).

Front end----Force=2010*1= 2010
Rear end-----Force=0*1= 0
If we dont have force than we dont have torque. (Torque= Force*Length)

I'm positive actually.

With a perfectly locked center diff, the front and rear is forced to spin at the same angular velocity. It's determined by the gear ratio.(as long as that baby stays locked)

If we must keep equal angular velocity front and rear, but at the same time have varying loads, then torque is inevitabley 'transfered'(proportionally) to the side with a greater load.

Torque will be 'transfered' appropriately in order to turn the two out-put shafts at equal angular velocities.

The torque 'transfer' I mentioned above has NOTHING to do with the DCCD. My example was with a completely locked center diff. The torque 'transfer' I described is soley the product of physics.

 

Thanks, the thread you linked has been quite helpful.

 

If you have the 35/65 torque split in the centre diff you have an open diff, meaning no LSD function. If you have a 41/59 torque split then you have a centre diff with LSD function.

Both of these were available with DCCD.

If you have DCCD you have an electrically operated clutch which acts as a brake on the shaft that is spinning fastest and it brakes it against the shaft that is spinning slower. In other words it tries to stop one shaft from spinning faster than the other by connecting them together through a clutch that can stop or allow slip.

The shafts I'm talking about; one goes to the front diff and one goes to the rear diff.

How hard it tries to stop one shaft spinning faster than the other is dependant on the setting you have set in manual mode or the setting that the computer has applied when it is in auto mode. Both can set anywhere from no assistance to almost locked.

The 41/59 torque split diffs also have a LSD function which will resist the shafts spinning at a different speed once the shafts start to do so - not before, this is a safe guard against suddenly loosing traction to one end and
the computer making a sudden unexpected change that may cause an accident.

The torque split is confusing for most. It has nothing to do with the speed of the shafts. Torque is a twisting force, it can be applied to a stationary shaft or one that is spinning so it has nothing to do with speed or revolutions. It can only be applied where there is resistance, it can not exist if there is no force working against it.

So if you have the front wheels on ice, there is not much resistance so you cant apply much torque. If there is 10flbs of torque required to accelerate the front diff and make the front wheels spin on the ice then (10/41) * 59 = 14.39. Where 10 is the torque in flbs 41 is the front % split then multiply that answer by the % of torque split which goes to the back and you have how much torque in flbs available at the back assuming there is more resistance at the back wheels. Add the amount of torque at the front and at the rear together and you have how much the engine can make.

So if this were the case you engine will only be able to make 24.39flbs of torque because thats all the resistance there is available because the front has poor grip and you are probably stuck.

EXCEPT

You have a LSD and a DCCD, if the DCCD is at maximum lock then you have stopped the front wheels from spinning by connecting them through the brake (or clutch if you like) to the rear wheels so they will only spin if the rear wheels spin. So in this case you can apply as much torque as there is total resistance of all wheels (assuming you have not slip across each axel) and you may have 300flbs of resistance so you will just drive away.

Remember the DCCD can apply any amount of braking from none to full.

 

If you have the 35/65 torque split in the centre diff you have an open diff, meaning no LSD function. If you have a 41/59 torque split then you have a centre diff with LSD function.

The DCCD's lockup performs the LSD function so this statement is not quite true. It is true that the 41/59 split diffs also have mechanical LSD functionality that operates completely independently from the DCCD.

 

I'm actually studying for my M.E. degree and this little math problem has me kinda stumped. Feel free to completely call me out on this.

Stargazer all that math works out but I think one thing we left out is that Force=Mass*acceleration. Im pretty sure Mass in this case would include a portion of the cars Mass. On the other hand, if a wheel if off the ground(or cant apply force to the ground) then it is not contributing to the acceleration of the body.(only the acceleration of the wheel, rotor etc...plus we know with a locked diff, front and rear acceleration are equal to eachother)

But if that wheel can apply force to the ground, than the Mass must include a portion of the car's mass. The force applied to the ground, or rather the reaction to it, is what actually causes the body(car) to accelerate.

So if we dont have force applied to the ground then the mass is only the Mass of the wheel/rotor etc..(For simplicity, lets consider this Zero right now)

Since we dont have Mass we dont have any Force either.(Force =Mass*acceleration)



Imagine that the rear end is off the ground with the center diff locked. Acceleration is directly porportional front and rear(with our locked diff) so let "1" represent acceleration to make this even more simple.

Using Force=Mass*acceleration,

The front end has 100% traction so the mass is the mass of the car. Lets say 2010kg.
The rear end is not accelerating any Mass.(realistically it is but we called it zero because it is very small compared to the mass of the car).

Front end----Force=2010*1= 2010
Rear end-----Force=0*1= 0
If we dont have force than we dont have torque. (Torque= Force*Length)

This is a calculation of force at the front tyres contact patch required to accelerate the car at 1g. The engine will not have this at the flywheel. The gearbox, diff ratio's, and wheel diameters are all torque multipliers or dividers, but these are not effected by the torque split of the centre diff, as you say this is only relevant with a locked centre diff. This is different than a DCCD in locked mode but the outcome will be the same. Although a DCCD in locked will slip if subjected to this much torque variation between front and back.

The mass is everything that is moved, BUT they are all being moved at different rates because of the force is applied through different ratios.

 

Although a DCCD in locked will slip if subjected to this much torque variation between front and back.

The mass is everything that is moved, BUT they are all being moved at different rates because of the force is applied through different ratios.

I'm not sure if you directly responded to my post or not.

How do you know the DCCD will slip?

Yes, ofcourse the Mass should include a portion of all the mass on the car. But for simplicity, I concentrated on the central mass of the car.

The point is, with a totally locked center diff, the front or rear end can put down ALL the torque in the drivetrain.

 

Good explanations. I also found another post in the big NASIOC thread which my brain can understand -

NASIOC - View Single Post - Dccd

 

I think wdb was just saying that the DCCD is able to limit slip, regardless of the mechanical diff. So if you have a 35/65 split w/ an open diff w/ DCCD, saying it has no LSD isn't really true. It just has no mechanical LSD.