Page - 25 - in Integration of Advanced Driver Assistance Systems on Full-Vehicle Level - Parametrization of an Adaptive Cruise Control System Based on Test Drives

2.4. ACC Controller where j describes the index for one front (f) or one rear (r) wheel. The forcewjFx is the contact force between tyre and road in theWj-point. The climbing resistance is given byFβj=mj g cosβ. The angular momentum for both wheels is given by Ij cjω˙y= cjTy− cjTyr−wjFx rj, (2.16) where Ij is the moment of inertia of each wheel, and cjTy equals the drive and brake torque. cjTyr describes the rolling resistance of the tyre and depends approximately linearly on the wheel load Fzj at lower speeds and the distance betweenWj and the pointof loadofwjFz, thedistancefrj. Hence, the formula for the rolling resistance reads cjTyr=frj wjFz rj. The moment of inertia at wheel j reads Ij= Iwj+ i 2 fj ( Ig 2 + i 2 g Ie 2 ) , where Iwj, Ig and Ie describe the moment of inertia of the wheel, the gear box and the engine, respectively. The moment of inertia depends quadratically on the transmission ratios of the final drive ifj and of the gear box ig. The model in eqs. (2.12) and (2.14) to (2.16) has five Degrees of Freedom (DOFs), vvx, cvxf, cvxr, cωyf and cωyr. Excluding elasto kinematics, the longitudinal velocity of the vehicle body and wheels has to be the same, vvx = cfvx = crvx. Using this relation, eqs. (2.12) and (2.15) can be simplified to (mB+2mf+2mr)︸ ︷︷ ︸ m vv˙x= 2wfFx+2wrFx−(FβB+2Fβf+2Fβr)︸ ︷︷ ︸ Fβ=mgcosβ −Fa. (2.17) Excluding longitudinal slip cjvx = cjωy rj, due to low accelerations and a high friction potential of the road tyre contact, a combination of eqs. (2.16) and (2.17) results in the simplified longitudinal vehicle model with one DOF (vvx), which reads( m+2 If r2f +2 Ir r2r ) ︸ ︷︷ ︸ m∗ vv˙x= 2 cfTy− cfTyr rf +2 crTy− crTyr rr −Fβ−Fa. (2.18) Here,m∗ is the so-called generalized vehicle mass. The torque cjTy consists of the sum of the drive torqueTdj and the brake torqueTbj. The drive torque at one wheel reads Tdj= 1 2 ηj ifj ig Te(ωe,ud)bdj. (2.19) The efficiency of the drive train is described by ηj. Depending on the engine throttle ud and its speed ωe, the engine generates the torque Te. The factor bdj describes the percentageof the torque that is sent to theaxle j (e.g. a front-drivenvehiclehaspdf = 1 andpdr= 0). Thedynamicsof the combustionengine canbedescribedwithafirst order lag, [Ise02]. Thus the differential equation for the engine torque reads τe T˙e+Te=Te,min(ωe)+(Te,max(ωe)−Te,min(ωe))ud, (2.20) where the time constant of the lag equals τe = 0.5s. The parameters Te,max(ωe) and Te,min(ωe) describe the maximum and minimum torque that can be generated by the 25