Why is the Braking Coefficient - Slip curve of this shape?

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The braking force coefficient initially increases and then decreases with wheel slip due to the conversion of momentum to heat, which affects friction and can lead to brake fade. As brakes heat up, their effectiveness diminishes, resulting in a softer brake pedal and potential issues like boiling brake fluid and tire overheating. The curve shape reflects the difference between static and kinetic friction, where maximum traction occurs before the tire begins to lock up. Once slipping starts, the contact patch and pavement move at different velocities, reducing traction. Understanding this interaction is crucial for grasping the braking dynamics involved.
Raj
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Hi all,

I want to know the why the Braking force coefficient is increasing first and then decreasing with respect to wheel slip. I attached the graph for your reference
 

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the short answer is HEAT – more specifically, friction. you are converting momentum of a vehicle at speed to heat when applying the brakes. Energy must go some where so we have mucho friction. this causes the brakes to fade ( reducing braking capability.) Brake fade means a softer brake pedal. We also can boil the brake fluid present in the disc brake caliper. Actually not boil it but any moisture ( water) present in the fluid turns to steam thus reducing the psi the caliper sees. We also have the tire heating up the closer we get to wheel lock. You are beginning to scrub the very small tire contact patch vs. tire rotating and presenting a new cooler tire contact patch to the pavement. Depending upon how abusive you want to get you can glaze the brake pads until they are totally useless, you can melt the piston caliper seals so fluid blows thru them and you can melt the disc brake rotors if they are thin enough.

so the curve you see is an indication of how effective the brakes are initially when cool and how the efficiency trails off when getting hot.
 
MIke, that graph has nothing to do with fade or even temperature.

It depicts brake force vs % tyre slip, and it reduces because the co-efficient of sliding friction is lower than the co-efficient of static friction. Essentially when you deploy more brake force than you have grip available, the tyre begins to lock up and when this happens the contact patch and the tarmac start moving at different velocities relative to each other. Thus it has less traction than when the tyre is not locked up and the contact patch and tarmac are at the same velocity. The bigger difference there is between the velocity of the contact patch and the tarmac, the less traction there is, which gives the shape of that graph.

Why does that happen? That's a lot more complex, you're talking about the complicated interaction between rubber and asphalt and all the forces at play.
 
Like Kozy said, it has to do with the basic understanding of friction.

If you try to push a box that rest on the ground, you will need a certain force to start moving it. Once this happened, you will need less force to keep it moving; It is the concept of static and kinetic friction.

When a tire is in pure rolling, there is no relative motion between the tire and the ground, so static friction applies. When it starts slipping, then a relative motion is set and kinetic friction began. In the real world, part of the tire's contact patch is rolling, part of it is slipping, thus the tire slip ratio.

As to why there is a difference between static and kinetic friction, the following video and this web page should help you understand:

 
ha...i learned something just now. I maintain there is impact on the mechanical portion of the brake system due to friction as well as tire to pavement contact

but

this is not a portion of the original post question and
apologize for getting off on the wrong track, fellows...
 
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