Quote:
Originally Posted by The Cthulhu
don't argue that power doesn't get diverted, it's illogical and incorrect
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Explain that statement, please.
If you want, you can PM it to me instead of posting it (to avoid potential flames). Or you can post it at your own risk =P
Now, onto business.
In J Tylers thread, we were discussing how exactly the E-Diff functions in this car. Based on my experience: the car detects load bias in the differential (possibly measured at the transmission, since afaik there are no sensors on the differential) and applies the brakes to the wheel with significantly greater load bias up to a point at which it cannot compensate without doing more harm then good and the diff goes to full open action.
Some scenarios to clarify my hypothesis:
A) The car is stopped, tranny in gear, clutch disengaged. You dump the clutch. The power will go via the drivetrain to the wheels at which point one wheel will lose traction. The load bias is very high. The vehicle detects this and does not engage brakes. This is speaking from confirmed experience. When I have botched launches, the scenario I described is precisely what happened every time without fail or deviation.
B) Same launch situation but you do not dump the clutch, you let it out smoothly. The power find the wheels but one wheel does not lose traction (and when I say lose I mean completely lose... like smokey burnout lose... Im aware that you are always losing a littel bit of traction even when the car is driving perfectly straight and normal). The load bias is low (let's estimate 10%... so one side of the diff is seeing 10% greater load than the other... where the above example 'A' would be showing something like an 80% load bias). The vehicle detects this load bias is within the range neccesary to apply it's E-Diff magic (let's say that range is like 10-40% just for kicks) braking the wheel that has lost more traction than the other and reducing it's wheelspin, which in turn allows the wheel that has lower load to "catch up". Now keep in mind Im not talking about diverting power, Im simply talking about sapping power that's already traveling to the more loaded wheel.
C) Same launch, except you dont even launch. You just smoothly and slowly accelerate from the light. After all, you're just dropping your grandma off at home and don't need to give her a heart attack or something. Naturally, as stated above, there is always some loss of traction and some load bias. That's how tires work, by losing pieces of themselves. Otherwise we would get infinite mileage out of one set. Here the load bias is very low, percieveably non-existent to us lowly humans. Let's say its .5-1%. If the vehicle detects it at all (which it might not, naturally solving the problem Im about to mention) it identifies it as a "normal driving condition" and does not do any braking. The reason is that you would, obviously, prematurely wear your brakes if you were applying brakes (even just a little bit) for every moment you were driving.
That's my theory.
It's entirely conjecture, based only on my drag practice and my daily driving. I could be way off, but it all sounds pretty good to me and no ones called me crazy yet =P
So to give my opinion to your questions more specifically:
1)Yes, but it's the inside wheel, as proven
here.
2) In my experience, yes. My best launches have produced alternating dashed skid marks in the road where one wheel is spinning then stops and the other wheel spins and it goes back and forth. If you were to take these two dashed lines of skidmark and put them together they would precisely form one straight line. It's REALLY weird looking. I've never seen it before this car.
3) I would say... yes. A mechanical LSD would be doing this in infinitesimal time, essentially carying out both actions simulataneously. An E-Diff would only be doing this as fast as it's microprocessor can detect the change and react to it which might be fast but never as fast as 1/infinity.