The consolidated data collected from speed and delay studies by floating car method on a stretch of urban road of length 3.5 km, running North-South are given below.

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written 6.4 years ago by

teamques10 ★ 69k

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Determine the average value of (i) volume (ii) journey speed (iii) running speed of the traffic stream along each direction

Trip no	Direction of trip	Journey time, min-sec	Total stopped delay, min-sec	No of vehicles overtaking	No of vehicles overtaken	No of vehicles from opposite direction
1	N-S	6-32	1-40	4	7	268
2	S-N	7-14	1-50	5	3	186
3	N-S	6-50	1-30	5	3	280
4	S-N	7-40	2-00	2	1	200
5	N-S	6-10	1-10	3	5	250
6	S-N	8-00	2-22	2	2	170
7	N-S	6-28	1-40	2	5	290
8	S-N	7-30	1-40	3	2	160

transportation engineering

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written 6.4 years ago by

teamques10 ★ 69k

• modified 5.7 years ago

The mean value of journey time, stopped delay, no of vehicles overtaking,overtaken and in opposite direction for North-South and South-North directions obtained are given below.

Direction	Journey time, min-sec	Total stopped delay, min-sec	No of vehicles overtaking	No of vehicles overtaken	No of vehicles from opposite direction
N-S	6-32	1-40	4	7	268
	6-50	1-30	5	3	280
	6-10	1-10	3	5	250
	6-28	1-40	2	5	290
Total:	26-00	6-00	14	20	1088
Mean:	6-30	1-30	3.5	5.0	272
S-N	7-14	1-50	5	3	186
	7-40	2-00	2	1	200
	8-00	2-22	2	2	170
	7-30	1-40	3	2	160
Total:	30-24	7-12	12	8	716
Mean:	7-36	1-40	3.0	2.0	179

(a) North-South direction

$n_y = \textit{average no. of vehicles overtaking minus overtaken} = 3.5 - 5.0 = -1.5$

$n_a$ = average no. of vehicles during trips in opposite direction ( fro S-N trips) $= 179$

$t_w = \textit{average journey time with the stream} , q = 6\hspace{0.05cm}min\hspace{0.05cm}30\hspace{0.05cm}sec = 6.5\hspace{0.05cm}min$

$t_a = \textit{average journey time during trips against the stream} = 7\hspace{0.05cm}min\hspace{0.05cm}36\hspace{0.05cm}sec = 7.6\hspace{0.05cm}min$

$\textit{Average volume}\hspace{0.05cm}(q) = \frac{n_a + n_y}{t_a + t_w} = \frac{179 - 1.5}{7.6 + 6.5} = 12.59\hspace{0.05cm}veh/min$

$\textit{Average journey speed} = \frac{3.5}{6.62}\hspace{0.05cm}km/min = \frac{3.5\hspace{0.05cm}\times\hspace{0.05cm}60}{6.62} = 31.7\hspace{0.05cm}kmph$

$\textit{Average stopped delay} = 1.5\hspace{0.05cm}min$

$\textit{Average running time} = \textit{Average journey time - Average stopped delay}$

$\hspace{3cm} = 6.62 - 1.50 = 5.12\hspace{0.05cm}min$ $\textit{Average running speed} = \frac{3.5\hspace{0.05cm}\times\hspace{0.05cm}60}{5.12} = 41.0\hspace{0.05cm}kmph$

(b) South-North direction

$n_y = 3.0 - 2.0 = 1.0\\ t_w = 7.6\hspace{0.05cm}min\\ t_a = 6.5\hspace{0.05cm}min\\ n_a = (\textit{for N-S strips}) =272\\ q = \frac{272 + 1.0}{6.5 + 7.6} = 19.36\hspace{0.05cm}veh/min\\ \overline{t} = 7.6 - \frac{1.0}{19.36} = 7.55\hspace{0.05cm}mm\\ \textit{Journey speed} = \frac{3.5\hspace{0.05cm}\times\hspace{0.05cm}60}{7.55} = 27.8\hspace{0.05cm}kmph\\ \textit{Average stopped delay} = 1.8\hspace{0.05cm}min\\ \textit{Average running time} = 7.55 - 1.80 = 5.75\hspace{0.05cm}min\\ \textit{Average running speed} = \frac{3.5\hspace{0.05cm}\times\hspace{0.05cm}60}{5.75} = 36.5\hspace{0.05cm}kmph$

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