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Old 07-13-2006, 01:02 PM   #151
corey_dyck
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There is so much confusion out there, I did the following sketches that are greatly simplified to try to clear up what happens within the diff and how it affects torque delivery at the wheels.

#1 is showing how the torque is transmitted through the diff in fully open mode, with lots of traction for both front and rear tires.


#2 shows when the rear wheels begin to slip with a fully open diff. Of course, in this simplified model this condition could not occur for long. In the real center diff, the gears can spin indefinitely.


#3 shows a fully locked center diff with the rear tires on a frictionless surface with the front tires having more grip than needed.


#4 shows a fully locked center diff with the front tires on a frictionless surface with the rear tires having more grip than needed.


Diff settings between fully open and fully locked are hard to represent in this diagram. Basically these settings are like adding friction between the red and blue bars - they can still slip relative to each other but it takes effort.

I hope this helps someone!
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Old 07-13-2006, 01:39 PM   #152
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Jim -

Your distinction between "fully locked" and "actually locked" is useful and I will try to use it on this forum.

As to the original question, if you're convinced that there is no slop, then a center that is actually locked is no longer a 35/65 (or 41/59), since you are now driving a 4WD and not an AWD. Sorry I ever brought up my idea (although it was still interesting).

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Old 07-13-2006, 01:45 PM   #153
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Quote:
Originally Posted by corey_dyck
#2 shows when the rear wheels begin to slip with a fully open diff. Of course, in this simplified model this condition could not occur for long. In the real center diff, the gears can spin indefinitely.
Why are you drawing the crossbar at a tilt (and adding that green arrow) when you have the outputs still showing a 35/65 split? Is this your way of indicating that the torque split remains 35/65 (assuming an open center) even when the rear tires begin to slip?

- Jtoby
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Old 07-13-2006, 02:42 PM   #154
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Quote:
Originally Posted by jtmcinder
Why are you drawing the crossbar at a tilt (and adding that green arrow) when you have the outputs still showing a 35/65 split? Is this your way of indicating that the torque split remains 35/65 (assuming an open center) even when the rear tires begin to slip?
If the rear tires are slipping, they are turning faster than the front tires. In my super-simplified drawing, the blue crossbar would rotate as shown in that situation. Of course it would rotate 90 degrees in short order and the whole thing becomes pointless, but this is a greatly simplified version.

I should add that the drawing is not to scale. Pretty close, but I didn't spend the time to get it perfect.

Are you implying that the torque split is not 35/65 when the diff is fully open and one axle is rotating faster than the other? I believe it is 35/65 at all times (when the diff is fully open) independant of output shaft speeds due to the nature of planetary gears.
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Old 07-13-2006, 03:45 PM   #155
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Quote:
Originally Posted by corey_dyck
I believe it is 35/65 at all times (when the diff is fully open) independant of output shaft speeds due to the nature of planetary gears.
Which is what I believe, too. What changes, of course, when the rears start slipping is the power split. When the rears start spinning faster (due to the slippage), the fronts must start spinning slower, assuming no other changes (i.e., no sudden increase in engine speed). Therefore, since power = torque x speed, the power split suddenly shifts even more to the rear.

Note that the above is theoretical. In real-life, what usually happens at the moment you break the rears free (again, assuming no LS device) is that the engine does speed up, such that the speed of the fronts stays contant and the rears and the engine speed up together. However, given the constant torque split, this increasing difference betweeb the front and rear output speeds means more and more power is being sent to the (now almost useless) rears. This is particular nasty with a torque split that is biased to the rear, since it takes more of an increase in engine speed to keep up with the under-geared rears. Hence, it would be crazy to try to run something like a 35/65 split in a front-heavy car with no center LS device. (Just ask Charles Moss [national autocross driver of a DSM], who has a Cusco center.)

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Old 07-13-2006, 04:00 PM   #156
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Quote:
Originally Posted by jtmcinder
Jim -

Your distinction between "fully locked" and "actually locked" is useful and I will try to use it on this forum.
Actually, I was trying to draw the distinction between "locked" and "actually locked" (or "fully locked").

Quote:
Originally Posted by jtmcinder
As to the original question, if you're convinced that there is no slop, then a center that is actually locked is no longer a 35/65 (or 41/59), since you are now driving a 4WD and not an AWD. Sorry I ever brought up my idea (although it was still interesting).

- Jtoby
Some (many?) would argue that the distinction between AWD and 4WD lies not in the existence of a center diff (many 4WDs have them in the transfer case) but rather in the presense or absence of a separate "transfer case" which is aft - WRT torque flow - of the transmission.

I don't find myself ready to be aligned with either school of thought.

Terminology is often a sticky wicket. :-)

- Jim
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Old 07-13-2006, 04:12 PM   #157
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I use "AWD" for vehicles with center diffs, "4WD" for those that are permanently locked (when engaged at all), and "pseudo-AWD" for things like 4-Matics, 4Motions, xDrives, and those Porsche 959s (where a VC or one or more transfer-clutches are used to connect the non-gear-driven end).

And, since you brought it up: I was also a wicked spin-bowler in the British equivalent of high school.

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Old 07-13-2006, 05:12 PM   #158
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Quote:
Originally Posted by jtmcinder
And, since you brought it up: I was also a wicked spin-bowler in the British equivalent of high school.

- Jtoby
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Old 07-13-2006, 06:02 PM   #159
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Quote:
Originally Posted by jtmcinder
Which is what I believe, too. What changes, of course, when the rears start slipping is the power split. When the rears start spinning faster (due to the slippage), the fronts must start spinning slower, assuming no other changes (i.e., no sudden increase in engine speed). Therefore, since power = torque x speed, the power split suddenly shifts even more to the rear.

Note that the above is theoretical. In real-life, what usually happens at the moment you break the rears free (again, assuming no LS device) is that the engine does speed up, such that the speed of the fronts stays contant and the rears and the engine speed up together. However, given the constant torque split, this increasing difference betweeb the front and rear output speeds means more and more power is being sent to the (now almost useless) rears. This is particular nasty with a torque split that is biased to the rear, since it takes more of an increase in engine speed to keep up with the under-geared rears. Hence, it would be crazy to try to run something like a 35/65 split in a front-heavy car with no center LS device. (Just ask Charles Moss [national autocross driver of a DSM], who has a Cusco center.)
One key thing you're missing in your idea is that the force at the road surface decreases as they start to slip (the classic slip angle graph). That reduction in force (and rotational moment) results in an increase in speed. Actual power put to the spinning wheels does not increase, it actually decreases. That's why the engine spins faster almost instantly - some of the power it was delivering to the wheels is now going into accelerating the flywheel and all the other rotating masses. The loss of torque at the rear wheels means there is a corresponding loss of torque to the front wheels through the diff.

The planetaries in the diff cannot sense power, they can only sense torque. They will spin at any speed the wheels will allow them, as long as the torque split is 35/65 (assuming an open diff). This is represented in my simple diagram above, except torque values are replaced with force values. The 'teeter-totter' type motion of the bar (planetaries in real life) doesn't care how fast the assembly is moving (rotating), just that the forces (torques) are 35/65.
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Old 07-13-2006, 07:40 PM   #160
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That's a good point about the amount of grip available dropping off when you spin the rears (faster than by about 10% more than ground speed). But that doesn't really change anything for the diff. As you wrote, it still splits the torque 35/65, such that (after multiplying by speed) a huge majority of the power is being wasted turning your rears into blue smoke, leaving little for the fronts.

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Old 07-14-2006, 11:43 AM   #161
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Agreed, in principle. But the diff does not sense or manage power at all, only torque. Whether the rear output shaft is spinning at 1 RPM or 1000 RPM, it'll still send 65% of the current engine torque there. Power = RPM x torque, there's a very large difference in power being sent to the rear wheels in those two cases but the diff doesn't care.

The interesting part of this is that once the rear tires start slipping (force at contact patch dropping), the front tires lose a similar ratio of grip even though they're not slipping. Some pretty crazy dynamics!
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Old 07-14-2006, 11:53 AM   #162
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Any open diff has the ability to send 100% of the torque to whichever side slips. The 35/65 is just the initial split.
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Old 07-14-2006, 04:42 PM   #163
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Quote:
Originally Posted by Crazykev
Any open diff has the ability to send 100% of the torque to whichever side slips. The 35/65 is just the initial split.
Did you mean to say power instead of torque in your first sentence?

Did you mean to say torque instead of initial in your second sentence?
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Old 07-15-2006, 03:28 PM   #164
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Quote:
Originally Posted by Crazykev
Any open diff has the ability to send 100% of the torque to whichever side slips. The 35/65 is just the initial split.
Sorry, no. On a typical open diff (50/50 split, like in a FWD car), the torque applied to both wheels is always the same. If one wheel can only support 20 ft-lbs, the other wheel will also put 20 ft-lbs to the ground.
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Old 07-28-2006, 03:07 AM   #165
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Not a diff expert by any means, but here is my understanding of what post #4 of this thread means and how the diffs work.

http://forums.nasioc.com/forums/show...highlight=DCCD

- The mechanical LSD, consisting of conventional clutch and springs, can transfer up to 30% of torque to the slower turning output shaft (shafts leading to the front and rear diffs).

- The electric solenoid clutch can lock the remaining 70% beyond the mechanical clutch's ability. This implies that the center diff is capable of locking to 100%.

- The electric soloneid clutch is activated automatically based on inputs from sensors.

- With DCCD under manual control, the adjustment is between 30%-100% locked

- In fail safe mode, the electric solenoid clutch is disengaged, leaving only the mechanical clutch engaged.

-If the center diff is an open diff (which I'm not sure it is), then torque split would be 50/50 F/R during normal no-slip conditions. When slip occurred (relative to front and rear), the diff would be capable of locking up to 100%. So, if the rear wheels had zero traction, the front would receive an equal amount of torque (as the front) by means of locking the diff.

-The center diff does not affect how the front and and rear diffs manage torque between left and right wheels.

Please correct me if I am misunderstanding. In other words, flame-on.
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Old 07-31-2006, 09:53 PM   #166
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Quote:
Originally Posted by gharari
-If the center diff is an open diff (which I'm not sure it is), then torque split would be 50/50 F/R during normal no-slip conditions.
There is a planetary gear assembly which dictates this ratio. For the '04 the split when open is 35/65 F/R. For the '06 it is 41/59
Quote:
Originally Posted by gharari
When slip occurred (relative to front and rear), the diff would be capable of locking up to 100%. So, if the rear wheels had zero traction, the front would receive an equal amount of torque (as the front) by means of locking the diff.
You said "the front would receive an equal amount as the front". Assuming you meant the front would receive an equal amount as the rear then no, that's not correct. In your zero rear traction example, the front wheels would make use of all of the available DCCD input torque once the diff has actually locked.
Quote:
Originally Posted by gharari
-The center diff does not affect how the front and and rear diffs manage torque between left and right wheels.
Yes, you are correct - it does not affect this.
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Old 08-01-2006, 01:55 AM   #167
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Quote:
Originally Posted by STI_FFY
You said "the front would receive an equal amount as the front". Assuming you meant the front would receive an equal amount as the rear then no, that's not correct. In your zero rear traction example, the front wheels would make use of all of the available DCCD input torque once the diff has actually locked. Yes, you are correct - it does not affect this.
I thought that when a diff is locked, then both output shafts spin at the same rate/torque?
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Old 08-01-2006, 02:25 AM   #168
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Quote:
Originally Posted by gharari
I thought that when a diff is locked, then both output shafts spin at the same rate/torque?
Same rate - yes. Same torque - only if the load on the output shafts are equal.

If you have two shafts with a different load on each of them, and you accelerate both shafts at the exact same rate, then you *must* have applied different torques to them.
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Old 08-01-2006, 02:45 AM   #169
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right
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Old 08-08-2006, 12:08 AM   #170
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holy ape-***** My hair hurts after reading this!
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Old 12-21-2006, 07:11 PM   #171
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sorry to bring this thread back from the dead but.....

...what position on the DCCD wheel is best for spinning donuts in the rain/snow? Is it the orange "lock" (i seriously doubt) or the green arrows all the way back?
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Old 12-21-2006, 07:17 PM   #172
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Set it to open. Drop the clutch.

- Jtoby
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Old 12-22-2006, 04:31 AM   #173
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Quote:
Originally Posted by C J View Post
sorry to bring this thread back from the dead but.....

...what position on the DCCD wheel is best for spinning donuts in the rain/snow? Is it the orange "lock" (i seriously doubt) or the green arrows all the way back?
depends... I would say lock, but it will put much stress on the driveline.. But then again so will the open setting...
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Old 12-22-2006, 11:13 AM   #174
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Quote:
Originally Posted by Roffa View Post
depends... I would say lock, but it will put much stress on the driveline.. But then again so will the open setting...
Are you serious?
Read the manual and it even tells you no to do tight turns in lock because it could damage the drivetrain....
so use open.
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Old 12-22-2006, 11:57 AM   #175
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Quote:
Originally Posted by 19-Kilo View Post
Are you serious?
Read the manual and it even tells you no to do tight turns in lock because it could damage the drivetrain....
so use open.
Yes, but it's diffrent rules for slow speed / normal driving versus playing with your car..

Why should you not use 100% lock in tight turns? Because there will be diffrent speed on all wheels when you make a turn. This it not the case when you are spinning dounuts and playing with your car. "If" you have enought wtq to spinn all tires they will all spin at the same speed and "no" damage will be done to your diffs.
Haveing the unit in auto might lockup the center diff and it might not if it's not locked up there will be speed diffrence between front and rear wheels.Oil temp will raise and you might do damage to the center diffrential.

depending on the dounut my might acctually have less steering input than on a tight slow speed turn.

So if you can make all wheels spin my teory is that locked will do least damage to your car. Auto will, if not notisable, "damage" your car for sure.
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