SRT6_Roadster is absolutely correct-our cars do NOT have a mechanical LSD from the factory-the rear end is OPEN-When you have an OPEN rear end, and power is being supplied via the driveshaft, and only one wheel is spinning, if you STOP that wheel from spinning, the power is going to be transferred through the spider gear cluster (the little collection of gears INSIDE the ring/pinion that you would WELD UP if you wanted a 'RED-NECK' LOCKER) and spin the other side-Not to argue with your 22 years of racing and building cars, but Chrysler has been using the computer braking system in place of a mechanical LSD on a variety of cars...in fact the new Caliber SRT-4 uses that same technology to give it 'Limited Slip characteristics' without having a mechanical LSD-in its front wheel drive application...it is rare but not uncommon for an occasional 2 stripe burn out from a car with an open differential...if both surfaces are equally tractionless, then both tires will spin-until one gets less traction and spins faster (path of least resistance). Furthermore, when only one tire is spinning in an open rear end, the rotational speed of that tire is DOUBLED because the input driveshaft speed (RPMs) is not being split in half between both sides...that is why a one leg burnout will provide immediate smoke...the tire is spinning at TWICE the readout of your speedo...gotta keep that in mind when the teenager in the neighborhood does a 1 leg burnout around the corner!! Pretty good chance that spinning tire just approached 100mph under an extreme load!!

So back to the topic...if the electronic nannies were not applying the brakes to compensate for traction, and both wheels were applying power down to the ground MECHANICALLY, the car should be quicker because it will not have the parasitic drag of the brakes being applied when one tire looses traction...?