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

Instructions:

The task was to solve some design math problems.

source..Content:

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:

Â

* 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=ÂµFz

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 formulaeâ€™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...

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