Brake performance of the vehicle can be seen by braking efficiency. Drivers are generally asked to stop their vehicle,between certain limit of stopping distance from particular maximum velocity to zero velocity. Stopping distance also depends on the Response time i.e time taken from actuation of brake pedal to full development of braking force.
μ*mg = md(ideal)
The braking efficiency indicates the extent to which the vehicle utilizes the available coefficient of road adhesion for braking. Thus, when d/g <μ, hence ηb < 1.0, the deceleration is less than the maximum achievable, resulting in an unnecessarily long stopping distance.
M= γb*W/g (1)
M is the mass of vehicle.
γb=> is an equivalent mass factor taking into account the mass moments of inertia of the rotating components involved during braking.
a*ds = v*dv (2)
a = ( Fb + ΣR)/M (3)
ΣR = Ra + fr*W*Cosθ +(-) W*Sinθ
(+) When car is going up hill .
(-) When car is going downhill.
Ra = [Cd * A*V^2]*ρ/2
Ra- aerodynamic force
Cd- drag coefficient
A- projected area
ρ - density
Cd*A*ρ/2 = Ca (As all are constants)
Ra = Ca * V^2
ΣR = Ca * V^2 + fr*W*Cosθ +(-) W*Sinθ (4)
Using equations (1),(2),(3) & (4) we get:-
∫ ds =γb*W/g ∫ (Vdv)/(Fb + fr*W*Cosθ +(-) W*Sinθ + Ca*V^2)
(Integrate this from V2 to V1)
V2- final velocity
V1- initial velocity
S=[γb*W/2*g*Ca]* ln [ ( Fb +fr*W*Cosθ +(-) W*Sinθ +Ca*V1^2)/(Fb +fr*W*Cosθ +(-) W*Sinθ +Ca*V2^2) ]
For stopping distance , V2=0 we get:-
S=[γb*W/2*g*Ca]* ln [ 1+ (Ca*V1^2)/(Fb + fr*W*Cosθ +(-) W*Sinθ) ]
For minimum value of stopping distance:-
- ηb = 100% - γb=1
In this case, the braking torque generated by the brakes has already overcome the inertia of the rotating parts connected with the wheels; the maximum brakmg forces developed at the tire ground contact are retarding only the transnational inertia. The mass factor γb , is therefore one. - Fb=μ*W
S=[W/2*g*Ca]* ln [ 1+ (Ca*V1^2)/( μ*W+ fr*W*Cosθ +(-) W*Sinθ) ]
Above is the value of minimum stopping distance.
If braking efficiency (ηb) is less then 100% then,
Stopping distance is given by,
S=[W/2*g*Ca]* ln [ 1+ (Ca*V1^2)/(ηb*μ*W+ fr*W*Cosθ +(-) W*Sinθ) ]
Note :- It should be pointed out that, in practice, there is a time lag between the application of brakes and the full development of the braking force. This time lag depends on the response of the brake system.
Additional stopping distance Sa, may be calculated from:-
Where, td - response time V1 - intial velocity
Final stopping distance will be the sum of Sa & S .