Page 316 - Automotive Engineering Powertrain Chassis System and Vehicle Body
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Tyres and wheels      C HAPTER 10.1

             When braking on a bend, additional longitudinal  In the longitudinal direction the possible braking force
           forces F X,W,b occur on all wheels, and act against the  F X , W , b ¼ 3130 N is at a ¼ 0 and therefore (see Equation

           direction of travel. In this case Equation 10.1.18 also  10.1.5),
           applies.
             On standard vehicles and front-wheel drives, the   m X;W;max  ¼ F X;W;b =F Z;W ¼ 3130=2940ðN=NÞ
           front wheels take 70–80% of the braking force and the                       ¼ 1:06
           rear wheels only 20–30%. This means that the slip
           angles increase on both axles, but more at the front  and
           than the rear and the vehicle tends to understeer                       0:5      1
           (Fig. 10.1-41). If the wheels of an axle lock, the fric-  m X;W  ¼ 1:06 1    2  2
           tion becomes sliding friction and the vehicle pushes                   0:97
           with this pair of wheels towards the outside of the       ¼ 0:91
           bend.                                              The lateral forces that the tyre can absorb during braking
             Taking into consideration the maximum possible values  can also be calculated:
           in the longitudinal and lateral direction of the road –
           known respectively as m X,W,max and m X,W,min – the                        m X;W     1 2
                                                                                              2
           increasing force coefficient can be calculated:       m Y;W  ¼ m Y;W;max  1    m X;W;max   (10.1.19a)
                                            1
                                   m Y;W   2  2               m X,W ¼ 0.7 should be given. The lateral force coefficient
             m X;W  ¼ m X;W;max  1                (10.1.19)
                                  m Y;W;max                   (which can be used) is:

                                                                                  0:7  2  1 2
                                                                m    ¼ 0:97 1
           Consider as an example a braking process on a dry road at  Y;W         1:06
           100 km/h on a bend with R ¼ 156 m. Using Equation         ¼ 0:73
           10.1.9 the calculation gives m Y,W ¼ 0.5.

             Figure 10.1-48 shows a measurement on the tyre in  At S X,W,b ¼10% and a ¼ 4 the transferable lateral
           question where the greatest coefficient of friction in the  force is:
           lateral direction at F Z,W ¼ 2490 N, 3 w ¼ 10% and a ¼ 4    F Y;W ¼ m    F Z;W ¼ 0:73   2940
           (see Equation 10.1.11) amounts to                             Y;W
                                                                                   ¼ 2146 N
             m Y;W;max  ¼ F Y;W =F Z;W                        and the available braking force is:
                     ¼ 2850=2940 ðN=NÞ
                                                                F X;W;b  ¼ m X;W    F Z;M ¼ 0:7   2940
             m Y;W;max  ¼ 0:97                                                       ¼ 2058 N


























           Fig. 10.1-48 Tyre-tangential lateral force performance characteristics with slip angles and brake slip as parameters. The study was
           carried out on a 18565 R 14 86 S radial tyre loaded at 300 kg at p T ¼ 1.5 bar. The shape of the curves indicates that, with increasing
           longitudinal forces, those which can be absorbed laterally reduce. At 1.5 bar, the tyre carries a weight of 350 kg, i.e. it is only operating at
           86% capacity.


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