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

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Traction-slip control (ASC) helps maintain driving torque on a high-friction wheel by braking a low-friction wheel in a μ-split scenario. In a situation where one wheel is on ice and the other on tarmac, an open differential would typically cause the low-friction wheel to spin, limiting torque on the high-friction wheel and resulting in vehicle standstill. ASC counteracts this by applying brakes to the low-friction wheel, reducing its slip and allowing the high-friction wheel to transmit greater traction force. Unlike a differential lock, which directs all power to the high-friction wheel, ASC maintains a torque split while enabling the high-friction wheel to rotate and propel the vehicle forward. This mechanism effectively prevents power loss through slip, allowing for improved vehicle movement in low-traction conditions.
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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.
 
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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
 
My idea is that I want to use immerse Whitetail Antlers in a fishtank to measure their volumetric displacement (the Boone and Crockett system is the current record measurement standard to place in a juxtaposition with) I would use some sight glass plumbed into the side of the tank to get the change in height so that I can multiply by the tank cross-section. Simple Idea. But... Is there a simple mechanical way to amplify the height in the sight glass to increase measurement precision...

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