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Resolved [RESOLVED] How does AWD work? [Merged 9-7] torque split division slip drive

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To answer ther OP: The power is distributed equally to all wheels, but if one rear wheel spins more than the other rear, the rear viscous coupler will create friction, and attempt to equalize output between both rear wheels. Also, in exactly the same way, if the front or rear begins to spin faster than one another, the center viscous coupler will begin to create friction and equalize the speed of both front and rear. The front diff is open, but if one front spins, the front "as a set", would be going faster than the rear, and cause the center viscous coupler to want to make the rears spin as fast as the front. Since that would take more power than the average 200 or so horsepower DSM has, the front stops spinning, and the car just goes.

In laymans terms for the average sheep: The advanced viscous coupler system automatically transfers torque from the wheels that slip, to the wheels that grip.;)

What I am trying to decipher here, is if all wheels in an AWD vehicle see power equally. Say for instance, you could dyno each wheel individually (4 machines) at the same time, would all wheels read equal assuming no traction was lost at either of them.
 
I suppose they would, assuming each dyno applied the exact same amount of resistance to each wheel, and there were no other losses, (ie dragging brake pad, varied tire friction, varied amount of power to spin the driveshafts, axles, and bearings.)

Yes, in a perfect word, in a vcacuum devoid of all other forces, then sure, you could do that.

You could also stand pencils up on end.
 
What I am trying to decipher and state is that the amount of power the front diff sees is greater than what the rear diff sees. It makes sense in my mind, but I am open to listen to otherwise.
 
Must resist....

What I am trying to decipher and state is that the amount of power the front diff sees is greater than what the rear diff sees. It makes sense in my mind, but I am open to listen to otherwise.

OK. I'll play.

Why?

Why do you think that the front diff will "see" more power than the rear?
 
Because the center differential is independent from the front differential, meaning it is carrying more of a load.
 
must resist....



Ok. I'll play.

Why?

Why do you think that the front diff will "see" more power than the rear?

x2...

Because the center differential is independent from the front differential, meaning it is carrying more of a load.

By the same reasoning, the rear differential is independent from the center, so...

Speaking strictly based on gear ratios, the center differential sends the same amount of power to the rear differential, as it does to the front differential. There may be more drivetrain loss in the rear, but that comes with speed, so torque (as in you are holding it still, and I am applying torque at the other end) is equal. Period.

Now adversely speaking, if all the differentials were open, and you were running the car on a lift (no load) I can agree that the fronts would spin more easily than the rear, only because of the added friction losses involved with spinning the transfer case, and driveshaft bearings.

Now I'm going to fix some cars. At least they don't come up with outlandish theories.
 
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Because the center differential is independent from the front differential, meaning it is carrying more of a load.

Of course the fricken center gets more "load"! It all goes through the center while only part of it goes through the front.

But you're not answering my question, which was simply my asking you why you believe that a DSM with a 50/50 center sends more power to the front.

Again, why do you think the front diff will "see" more power than the rear?
 
By the same reasoning, the rear differential is independent from the center, so...

No, no, no! You really don't get this (probably because you have a cool little car instead of a heaving, wallowing 2G)! Because they are all independent of each other, they all get more power than any of the others! It's a perpetual-motion machine that produces infinite torque. Pay attention!

tee hee :)
 
Of course the fricken center gets more "load"! It all goes through the center while only part of it goes through the front.

But you're not answering my question, which was simply my asking you why you believe that a DSM with a 50/50 center sends more power to the front.

Again, why do you think the front diff will "see" more power than the rear?

I only said I think it gets more when the coupler is engaged, my understanding is that the rear wheels will loose traction more than the front wheels.

Supposedly it is 50/50, but in reality it is still front biased due to the nature of physics. It also depends a lot on if you have the viscous LSD. I would say it is closer to true 50/50 without it.

For reference.
 
I only said I think it gets more when the coupler is engaged, my understanding is that the rear wheels will loose traction more than the front wheels.

OK, you really need to stop posting (although this is now the all-time train-wreck thread, given the merge, so maybe you should keep going ... I dunno).

Read my two tech articles or anything else on Ferguson AWD. Read something basic on weight-transfer and learn how the tires with the most Normal force (i.e., weight) have the most grip, especially when the tires with more weight usually have less camber, too. Read a high-school physics text. Read last Sunday's New York Times. Read a Playboy. Just read.

And do it in a room without a computer.

I beg you.
 
Yeah that was real helpful. I can see how cynicism is a great diversion from actually engaging in discussion. I have been Googling it this entire time under different terms with no avail. Our front wheels see more weight on them, I am talking about normal driving not launching or racing.

I can apply many "high school physics" concepts that would imply the front wheels will get more traction. What I can't seem to find is people who disagree and provide useful information.
 
OK, you really need to stop posting (although this is now the all-time train-wreck thread, given the merge, so maybe you should keep going ... I dunno).

Read my two tech articles or anything else on Ferguson AWD. Read something basic on weight-transfer and learn how the tires with the most Normal force (i.e., weight) have the most grip, especially when the tires with more weight usually have less camber, too. Read a high-school physics text. Read last Sunday's New York Times. Read a Playboy. Just read.

And do it in a room without a computer.

I beg you.

He does, did you miss it? Modified Magazine duh!

Thanks for breaking it down, I've always heard we were a 50/50 split but never really got around to trying to understand it, but it makes perfect sense, assuming all 4 wheels have traction of course :thumb:
 
Yeah that was real helpful. I can see how cynicism is a great diversion from actually engaging in discussion. I have been Googling it this entire time under different terms with no avail. Our front wheels see more weight on them, I am talking about normal driving not launching or racing.

I can apply many "high school physics" concepts that would imply the front wheels will get more traction. What I can't seem to find is people who disagree and provide useful information.

It was helpful; I was going to say the same thing, but in different words maybe.

How can you define normal driving, when concerning tire spin? How are you going to have equal weight/traction at every wheel when spinning the tires? The car isn't even equally weighted sitting still. The closest you are going to come to a viable test would be putting the car on jackstands and driving it.
 
It was helpful, assuming you saw that part about being away from your computer.

But, OK, let's continue.

So, we're now talking about cruising on the highway. And we're now allowing the center's VC to play a role. Do I have the new conditions correct? Yes? Cool.

So please tell why the center's VC is doing anything at all. Why would the center's VC play any role in the torque distribution when you're just cruising along on the highway?
 
Thanks for breaking it down, I've always heard we were a 50/50 split but never really got around to trying to understand it, but it makes perfect sense, assuming all 4 wheels have traction of course :thumb:

The mistake that many people make is mixing up the action of the (usually) piggy-backed limited-slip devices with the action of the actual diffs. (Note how I wrote two, as-separate-as-possible tech articles on these two issues; there was a reason for this.) All stock DSMs have 50/50 centers. But they also have a limited-slip device piggybacked on it. So the actual, moment-to-moment distribution of torque can be almost anything.

Just keep "split" (which is a description of the diff) separate from "distribution" (which is the diffs plus the LSDs) and you'll be in good shape.
 
Take your transmission apart and count the teeth on your differential side gears. You'll find that the split is 50/50. The math does not lie.

Ps. If it was anything other than 50/50, the center VC would be acted against while driving in a straight line, and it would rob power and eventually cook I assume. Like having two different sized tires front to rear.
 
Doesn't the fact that it's a spider tell you that it's 50/50? I suppose that it's possible (although I can't even imagine it right now), but I have never heard of a spider-type diff that wasn't 50/50.
 
I didn't say cruising on the highway I said daily normal driving, which includes rain and turns and snow and all other variables that will account for traction loss, and I am saying that the rear wheels are more likely to break loose. I also said in that post I quoted for reference that when it is not engaged it is closer to true 50/50.
 
I didn't say cruising on the highway I said daily normal driving, which includes rain and turns and snow and all other variables that will account for traction loss, and I am saying that the rear wheels are more likely to break loose. I also said in that post I quoted for reference that when it is not engaged it is closer to true 50/50.

Although I'm no transmission expert that makes zero sense at all. And your physics is also extremely flawed... The wheel thats most likely to lose traction completetly depends on the condition, type of road, bank of the road, potholes, the way the car is facing(as far as weight distribution). It looks like you are completely wrong but you keep wanting to go until your somehow feel correct...
 
I didn't say cruising on the highway I said daily normal driving, which includes rain and turns and snow and all other variables that will account for traction loss, and I am saying that the rear wheels are more likely to break loose. I also said in that post I quoted for reference that when it is not engaged it is closer to true 50/50.

I was having quite a bit of fun with you, but now it's seriously worn thin. I actually understand what your trying to say and there's a germ of truth to it. So, rather than continue to be an a-hole, I wrote a little sticky for this forum that I hope works for you. If you have any questions after reading it, post them here, as I'm hoping that the sticky will be locked. If the sticky needs tweaking, you're the perfect person to help with that.

cheers

http://www.dsmtuners.com/forums/newbie-forum/410949-awd-dsm-torque-split-distribution.html

- Jtoby
 
I already read it and I understand where I went wrong in my logic (despite phunny's assumptions that I am incapable of saying I was wrong when I am, something he should work on). I learned something today and for that I am happy.
 
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