Mathematical Explanation of ABS Brakes by Engr Shoaib Ahmed

1. Anti-Lock Breaking System (ABS): Mathematical
Explanation by Engr Shoaib Ahmed.
Above graph is of Experimental Behavior of Rubber Tyre against Road. Where X-axis is Slip Ratio, and Y-
axis is Coefficient of Friction.
So interpretation of above graph is that At 12% of Slip Ratio, Maximum Coefficient of Friction can be
achieved.
Where Slip Ratio= S
(Important Note: Whenever S=0, It means Tyre is at Pure rolling. And if S=1, it means Tyre is at Pure
Sliding Motion)
S = 1 - Rt ω/VCar
Where,
VCar = Linear Velocity of Car
ω = Angular Velocity of Tyre
Rt = Radius of Tyre

2. For Example:
A car traveling at 15 m/s, has a radius of tyre about 0.3m. What will be Angular Velocity of Tyre? And
What Angular velocity should be When ABS Break will be applied?
ω = V/r
ω = 15/0.3
ω = 50 rad/s. This will be speed of tyre while traveling car at 15 m/s.
Verifying:
S = 1 - Rt ω/VCar
S = 1 – (0.3)(50)/15
S = 1 – 15/15
S = 1 – 1
S = 0 (Means Tyre is in sliding motion.)
So we need to get S = 0.12, because when ABS Brakes are applied At 12% of slip ratio, Maximum Friction
is achieved.
So,
S = 1 - Rt ω/VCar
0.12 = 1 – (0.3) ω / 15
(0.3) ω / 15 = 1 – 0.12
ω = ((1 – 0.12)15)/0.3
ω = 44 Rad/s (So, this will be Angular Velocity of Tyre when ABS Brakes will be applied.)