Research And Describe Car Braking System Modeling & Control - 275 Words

Research And Describe Car Braking System Modeling & Control (Math Problem Sample) Content: CAR BRAKING SYSTEM MODELLING CONTROLName:Institutional Affiliation: * Simulink modelDisplay1sGain K+_Input signal (ft) * Response to step input.Generally,soWe now note several features about this equation, namelyThus we can write the general form of the unit step response as: * Limitations of KK is subject to changes in the spring stiffness.The friction also affects the value of KQUESTION 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 neglectedBut Ta=FxRw and Fx=FzCombining these equationsJw =Tb- FxRwDifferentiating with respect to t we get:Jw dwdt=FxRw-TbC. We know that Fzf=m*gLr*+axhLr+LfAnd Fzr=m*gLf*+axhLr+LfBut Lr is the distance from the centre of gravity of the vehicle to the centre of the rear wheelsAnd Lf the distance from the centre of gravity of the vehicle to the centre of the front wheelsBut Lf= Lr therefore:Fzr=FzfThe stopping distance of a vehicle can be calculated using a number of formulaes dependent on the nature of the braking system. Fd=1/2mv2 mad=1/2mv2 d=v2/2a158115029845D.Using the above model of a car breaking systemFor the stopping distance dx= u2/2aFd=1/2mv2 mad=1/2mv2 d=v2/2aFor a car stopping from 100km/ hr the estimated stopping distance =( 100* 1000)/ 3600= 36.67 m/sThe 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...