Engineering, 05.10.2020 01:01 kaitlyn114433
g With aerodynamic resistance considered (using Eq. 2.42), SS= bbWW 2ggKKaa (1+ KKaaVV1 2 WW(bbµµ+ff±sinθθgg) ) = • For the same conditions but with bb ′ =0.85, Eq. 2.42 gives SS= bbWW 2ggKKaa (1+ KKaaVV1 2 WW(bbµµ+ff±sinθθgg) ) = • Now applying Eq. 2.47 (or Eq. 2.46 since GG=0) for the practical stopping distance, dd = VV1 2 2gg( aa gg±GG) = • In the first case, the difference is ; • In the case of 85% braking efficiency, the difference is .
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g With aerodynamic resistance considered (using Eq. 2.42), SS= bbWW 2ggKKaa (1+ KKaaVV1 2 WW(bbµµ+f...
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