Research And Describe Car Braking System Modeling & Control (Math Problem Sample)

CAR BRAKING SYSTEM MODELLING & CONTROL
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* Simulink model
Display1sGain K+_Input signal (ft)
* Response to step input.
Generally,
so
We now note several features about this equation, namely
Thus we can write the general form of the unit step response as:
 
* Limitations of K
K is subject to changes in the spring stiffness.
The friction also affects the value of K
QUESTION TWO
* The wheel is like a torus; a hollow cylinder
* Jw dwdt=FxRw-Tb Taking the right hand side, we know:
J w=Tb-Ta rolling resistance neglected
But Ta=FxRw and Fx=µFz
Combining these equations
Jw =Tb- FxRw
Differentiating with respect to t we get:
Jw dwdt=FxRw-Tb
C. We know that Fzf=m*gLr*+axhLr+Lf
And Fzr=m*gLf*+axhLr+Lf
But Lr is the distance from the centre of gravity of the vehicle to the centre of the rear wheels
And Lf the distance from the centre of gravity of the vehicle to the centre of the front wheels
But Lf= Lr therefore:
Fzr=Fzf
The stopping distance of a vehicle can be calculated using a number of formulae’s dependent on the nature of the braking system. Fd=1/2mv2 mad=1/2mv2 d=v2/2a
158115029845D.
Using the above model of a car breaking system
For the stopping distance dx= u2/2a
Fd=1/2mv2 mad=1/2mv2 d=v2/2a
For a car stopping from 100km/ hr the estimated stopping distance =( 100* 1000)/ 3600= 36.67 m/s
The brake servo has turned out to be more regular in cars as circle brakes have supplanted drum brakes as the standard setup in vehicles. Circle brakes make it vital for cars to have control brakes to evacuate a dominant part of the force that a driver needs to apply to stop the car.
Inside the brake servo system, a vacuum increases the force that is applied by the driver on...