W211 AMG Discuss the W211 AMG's such as the E55 and the E63
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Limited Slip Diff avail. from AMG

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Old 06-27-2003, 05:24 PM
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2002 ML55, 2006 C6 Z06
Limited Slip Diff avail. from AMG

Check this out, http://www.amg-owners-club.org/board...706e65e69cfecd.

But only in Germany???
Old 06-27-2003, 05:35 PM
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E55, F550, S600 Ducati 999
AMG "upgrades" your differential and transmission software at the factory, but the price is cheap compared to the tuners. Brabus and Vath both do LSD conversions for the car for 4990 euro and 4500 euro respectively. You also have the option of a ratio change as well. I'm sure MKB probably offers something, but they never answer sales enquiries. Their service is shocking, it is a surprise they manage to stay in business..
Old 06-28-2003, 01:29 AM
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E55
I am a little surprized that E 55 does not have limited slip differential. How then does traction control work?

When a car starts slipping the only thing that can happen in a sotck E 55 is differential application of brakes and reduction of engine torque. Is htis correct.

Would this imply that limited slip diff will enhance performance a lot?
Old 06-28-2003, 09:54 AM
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That's the theory. With an LSD you should be able to get the power down more effectively
Old 06-29-2003, 05:16 AM
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one of seven
question - what type of rear dif does the E55 have (ie - positraction, lim slip, one wheel drive)?
Old 06-30-2003, 01:43 AM
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E55, F550, S600 Ducati 999
The E55 has no mechanical limited slip capabilitiy at all.
Old 07-01-2003, 02:47 AM
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one of seven
so.......is it positraction (both rear wheels driven equally & locked together)? In other words, when you burn rubber from a standstill do you get 1 wheel smokin' or 2??

I'm ordering my 2004 E55 tomorrow, just curious.
Old 07-01-2003, 04:24 AM
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As far as I am aware, there is no positraction or limited slip capability at all. With no interference power will generally transmit to the wheel with least traction, the opposite of what you actually want. Having said that, the electronic traction control operates by braking the spinning wheel and reducing power if necessary.
Old 07-01-2003, 09:27 AM
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I am a bit confused by this. Please help me out. If I turn off the traction controll on my 211 E55 and land on the gas hard it will liquify both rear tires as long as I feel like wasting good new rubber and carve two nice black strips into the ground. If it does not have some kind of limited slip wouldn't it nove to one wheel. If the traction controll was involved wouldn't it brake the wheels down to a stop?

I know the TC is never really off and it can not defy gravity once the wheels get really spinning but it seems like sime type of limited slip is in play here.

Thanks!
Old 07-01-2003, 09:48 AM
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Originally posted by NewBlack55
If I turn off the traction controll on my 211 E55 and land on the gas hard it will liquify both rear tires as long as I feel like wasting good new rubber and carve two nice black strips into the ground.
Why don't you give it a try and see what happens? That way you can tell us. Oh and take some pics too if you're going to do that.
Old 07-01-2003, 11:11 AM
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E55
I am a bit confused now. In theory without LSD, if one wheel slips completely as on ice, the car will not travel. Under traction control the wheel will be braked. Whihc should mean the car will neither travel nor will the wheels travel. In other words we should have something like being in neutral.

Is this so.
Old 07-01-2003, 01:04 PM
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2003 E55,2005 Ferrari F430 Spider, 2005 Corvette 427 TT, 2005 Range Rover
If the spinning wheel is braked, the other wheel propels the car. Based on my experience with this car, on the street there is no need for any additional limited slip above whatever is already implemented. Even when getting the rear planted on corner exits, both wheels seem to break away pretty evenly.
Old 07-01-2003, 06:51 PM
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C63S past: E90 M3 6M, w211 E55
if one of the wheels is braked, that wheel is seeing some resistance, and for an open diff, the power will transfer to the wheel that isn't braked

going completely straight, there is no difference in torque between the two wheels, so both should light up just fine. When turning, there is still power going through both wheels. even if one is moving faster than the other, but they still both have resistance. In the case where there is no resistance to a wheel, then there's the problem

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