2 Forward Vehicle Dynamics

1 2 Forward Vehicle DynamicsStraight motion of an ideal rigid Vehicle is the subject of this ignore air friction and examine the load variation under the tires todetermine the Vehicle s limits of acceleration, road grade, and Parked Car on a Level RoadWhen a car is parked on level pavement, the normal force,Fz, under eachof the front and rear wheels,Fz1,Fz2,areFz1=12mga2l( )Fz2=12mga1l( )where,a1is the distance of the car s mass center,C,fromthefrontaxle,a2is the distance ofCfrom the rear axle, andlis the wheel +a2( )a1a22Fz22Fz1xzCmgFIGURE A parked car on level Forward Vehicle a longitudinally symmetrical car as shown in Figure can be modeled as a two-axel Vehicle .

2 A symmetric two-axel Vehicle isequivalent to a rigid beam having two supports. The vertical force underthe front and rear wheels can be determined using planar static ( )XMy=0( )Applying the equilibrium equations2Fz1+2Fz2 mg=0( ) 2Fz1a1+2Fz2a2=0( )provide the reaction forces under the front and rear +a2=12mga2l( )Fz2=12mga1a1+a2=12mga1l( )Example 39 Reaction forces under ,C,is78 cmbehind the frontwheel axis, and it has a235 cmwheel m( )l= m( )m= 890 kg( )The force under each front wheel isFz1=12mga2l=12890 ( )and the force under each rear wheel isFz2=12mga1l=12890 1449 N.( )2. Forward Vehicle Dynamics41 Example 40 Mass center ( ) and ( ) can be rearranged to calculate the position ofmass ( )a2=2lmgFz1( )Reaction forces under the front and rear wheels of a horizontally parkedcar, with a wheel basel= m,are:Fz1= 2000 N( )Fz2= 1800 N( )Therefore, the longitudinal position of the car s mass center is ata1=2lmgFz2= (2000 + 1800) 1800 = m( )a2=2lmgFz1= (2000 + 1800) 2000 = m.

3 ( )Example 41 Longitudinal mass center position of mass centerCcan be determined experimentally. Todetermine the longitudinal position ofC, we should measure the total weightof the car as well as the force under the front or the rear wheels. Figure a situation in which we measure the force under the front the force under the front wheels is2Fz1, the position of themass center is calculated by static equilibrium conditionsXFz=0( )XMy=0.( )Applying the equilibrium equations2Fz1+2Fz2 mg=0( ) 2Fz1a1+2Fz2a2=0( )422. Forward Vehicle Dynamicsl2Fz22Fz1xCmga1a2 FIGURE Measuring the force under the front the longitudinal position ofCand the reaction forces under the (mg 2Fz1)( )Fz2=12(mg 2Fz1)( )Example 42 Lateral mass center cars are approximately symmetrical about the longitudinal centerplane passing the middle of the wheels, and therefore, the lateral position ofthe mass centerCis close to the center plane.

4 However, the lateral positionofCmay be calculated by weighing one side of the 43 Height mass center determine the height of mass centerC, we should measure the forceunder the front or rear wheels while the car is on an inclined surface. Ex-perimentally, we use a device such as is shown in Figure The car isparked on a level surface such that the front wheels are on a scale jack. Thefront wheels will be locked and anchored to the jack, while the rear wheelswill be left free to turn. The jack lifts the front wheels and the requiredvertical force applied by the jacks is measured by a load that we have the longitudinal position ofCand the jack is liftedsuch that the car makes an angle with the horizontal plane.

5 The slopeangle is measurable using level meters. Assuming the force under thefront wheels is2Fz1, the height of the mass center can be calculated by2. Forward Vehicle Dynamics43xzmga2a12Fz2h H2Fz1 Ccosh ()sinhR ()sinhR FIGURE Measuring the force under the wheels tofind the height of the equilibrium conditionsXFZ=0( )XMy=0.( )Applying the equilibrium equations2Fz1+2Fz2 mg=0( ) 2Fz1(a1cos (h R)sin )+2Fz2(a2cos +(h R)sin )=0( )provides the vertical position ofCand the reaction forces under the Fz1( )h=Fz1(Rsin +a1cos )+Fz2(Rsin a2cos )mgsin =R+a1Fz1 a2Fz2mgcot =R+ 2Fz1mgl a2 cot ( )442. Forward Vehicle DynamicsA car with the following specificationsm= 2000 kg2Fz1= 18000 N =30deg rad( )a1= 110 cml= 230 cmR=30cmhas aCat ( )There are three assumptions in this calculation:1 thetiresareassumedto be rigid disks with radiusR,2 fluid shift, such as fuel, coolant, and oil,are ignored, and3 suspension deflection generates the maximum effect on height determi-nation error.

6 To eliminate the suspension deflection,weshouldlockthesuspension, usually by replacing the shock absorbers with rigid rods to keepthe Vehicle at its ride 44 Different front and rear on the application, it is sometimes necessary to use differenttype of tires and wheels for front and rear axles. When the longitudinalposition ofCfor a symmetric Vehicle is determined, we canfind the heightofCby measuring the load on only one axle. As an example, consider themotorcycle in Figure It has different front and rear the load under the rear wheel of the motorcycleFzis heighthofCcan be found by taking a moment of the forces about thetireprint of the front (a1+a2)mg a1cos sin 1Ha1+a2 +Rf+Rr2( )Example 45 Statically Vehicle with more than three wheels is statically indeterminate.

7 Todetermine the vertical force under each tire, we need to know the mechanicalproperties and conditions of the tires, such as the value of deflection at thecenter of the tire, and its vertical Parked Car on an Inclined RoadWhen a car is parked on an inclined pavement as shown in Figure , thenormal force,Fz, under each of the front and rear wheels,Fz1,Fz2,is:2. Forward Vehicle Dynamics45 CmgFz2Rf hFz1 RrHa2a1 FIGURE A motorcycle with different front and rear +12mghlsin ( )Fz2=12mga1lcos 12mghlsin ( )l=a1+a2where, is the angle of the road with the horizon. The horizon is perpen-dicular to the gravitational the car shown in Figure Let us assume the parkingbrake forces are applied on only the rear tires.

8 It means the front tires arefree to spin. Applying the planar static equilibrium equationsXFx=0( )XFz=0( )XMy=0( )shows that2Fx2 mgsin =0( )2Fz1+2Fz2 mgcos =0( ) 2Fz1a1+2Fz2a2 2Fx2h=0.( )462. Forward Vehicle Dynamics2Fx22Fz1xzCmga2a12Fz2ha FIGURE A parked car on inclined equations provide the brake force and reaction forces under the frontand rear 12mghlsin ( )Fz2=12mga1lcos +12mghlsin ( )Fx2=12mgsin ( )Example 46 Increasing the inclination =0, Equations ( ) and ( ) reduce to ( ) and ( ). Byincreasing the inclination angle, the normal force under the front tires ofa parked car decreases and the normal force and braking force under therear tires increase.

9 The limit for increasing is where the weight vectormggoes through the contact point of the rear tire with the ground. Such anangle is called atilting 47 Maximum inclination required braking forceFx2increases by the inclination angle. Be-causeFx2is equal to the friction force between the tire and pavement, itsmaximum depends on the tire and pavement conditions. There is a specificangle Mat which the braking forceFx2will saturate and cannot increaseany more. At this maximum angle, the braking force is proportional to thenormal forceFz2Fx2= x2Fz2( )2. Forward Vehicle Dynamics47where, the coefficient x2is thex-direction friction coefficient for the rearwheel.

10 At = M, the equilibrium equations will reduce to2 x2Fz2 mgsin M=0( )2Fz1+2Fz2 mgcos M=0( )2Fz1a1 2Fz2a2+2 x2Fz2h=0.( )These equations provideFz1=12mga2lcos M 12mghlsin M( )Fz2=12mga1lcos M+12mghlsin M( )tan M=a1 x2l x2h( )showing that there is a relation between the friction coefficient x2,max-imum inclination M, and the geometrical position of the mass angle Mincreases by a car having the specifications x2=1a1= 110 cm( )l= 230 cmh=35cmthetiltingangleis M rad deg.( )Example 48 Front wheel the front wheels are the only braking wheelsFx2=0andFx16= this case, the equilibrium equations will be2Fx1 mgsin =0( )2Fz1+2Fz2 mgcos =0( ) 2Fz1a1+2Fz2a2 2Fx1h=0.