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Traction-slip control: braking a low mu-wheel to support driving tor

  1. Mar 16, 2014 #1
    Traction-slip control: braking a low mu-wheel to "support" driving tor

    I've been doing some studying on traction-slip control(ASC). I've also been trying to draw comparisons between ASC with differential locking mechanisms on a μ-split situation. I don't quite understand how braking the μ-low wheel would still help maintain the max.torque on the μ-high wheel. Some simple explanation please.

    Say I'm on a μ-split road.

    Left wheel Right wheel

    Ice(μ-low) tarmac(μ-high)

    An open diff(without ASC) transfers 50/50 torque -> this would cause the μ-low wheel to spin -> limiting the torque on the high μ wheel -> vehicle in standstill.

    My book then says:
    "Traction control hence "brakes" the μ-low wheel to support the higher driving torque. By this measure the longitudinal slip of the μ-low wheel decreases with a nearly constant longitudinal force transmission,whereas the μ-high wheel transmits a higher traction force".

    This seems far different than the working of a differential lock.In a differential lock,I have all the power diverted to the μ-high wheel to push the vehicle forward. I can't see how I'm diverting power in case of a traction control.

    Note:I assume a scenario of traction control with a open differential here.
     
    Last edited: Mar 16, 2014
  2. jcsd
  3. Mar 17, 2014 #2

    jack action

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    Science Advisor
    Gold Member

    Whether a lock or open diff, both mechanisms split the POWER between the left and right wheels. The lock diff split the TORQUE while keeping the rpm the same on both wheels and the open diff split the RPM while keeping the same torque on each wheel.

    With an open diff, the wheel that is stationary still receive the torque input, it just doesn't rotate which, obviously, means it doesn't move. By locking the μ-low wheel, you force some rpm to go on the μ-high wheel. The torque split is still 50/50 but now the μ-low wheel torque is counteract by the brake system instead of being lost through friction (through slip) and wheel acceleration. The μ-high wheel still has its 50% torque, but now it can rotate such that the car will move instead of being a meaningless reaction point to the μ-low wheel rotation.

    Re-read this post:

    https://www.physicsforums.com/showpost.php?p=4689744&postcount=4
     
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