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.