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Define stopping site distance. Calculate the SSD required to avoid head on collision of two cars approaching from opposite direction at 65 and 55 kmph.

Assume a reaction time of 2.5 sec, coefficient of friction as 0.6 and brake efficiency of 50% in either case.

Mumbai university > civil engineering > sem 5 > Transport Engineering 1

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Given

$V_1 = 65\hspace{0.05cm} kmph\hspace{1.5cm}V_2 = 55\hspace{0.05cm}kmph\\ \hspace{0.25cm} = 18.05\hspace{0.05cm}m/s\hspace{2cm} = 15.28\hspace{0.05cm}m/s\\ \\ t = \textit{reaction time} = 2.5\hspace{0.05cm}sec\\ f = \textit{coefficient of friction} = 0.6\\ \textit{Brake efficiency} = 50\% \hspace{0.05cm}of\hspace{0.05cm}0.6\\ f = \frac{50}{100}\times0.6 = 0.3$

Stopping Sight Distance

$SSD = (\textit{Log Distance + Breaking Distance})_1 + (\textit{Lag Distance + Breaking Distance})_2\\ \hspace{0.25cm} = SD_1 + SD_2$

$\textit{For 1st car} = SD_1 = V_1t + \frac{V_1^2}{2gf}\\ \hspace{0.25cm} = 18.05\hspace{0.05cm}\times\hspace{0.05cm}2.5 + \frac{18.05^2}{2\hspace{0.05cm}\times\hspace{0.05cm}9.81\hspace{0.05cm}\times\hspace{0.05cm}0.3} = 100.48\hspace{0.05cm}m\\ \\ \textit{For 2nd car} = SD_2 = V_2t + \frac{V_2^2}{2gf}\\ \hspace{0.25cm} = 15.28\hspace{0.05cm}\times\hspace{0.05cm}2.5 + \frac{15.28^2}{2\hspace{0.05cm}\times\hspace{0.05cm}9.81\hspace{0.05cm}\times\hspace{0.05cm}0.3} = 77.87\hspace{0.05cm}m$

Sight Distance to avoid head on collision of the two approaching cars

$SSD = SD_1 + SD_2\\ \hspace{0.25cm} = 100.48 + 77.87\\ \hspace{0.25cm} = 178.35$