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.

7.2kviews

written 6.1 years ago by

teamques10 ★ 68k

• modified 5.3 years ago

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

ADD COMMENT EDIT

1 Answer

409views

written 6.1 years ago by

teamques10 ★ 68k

• modified 5.4 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

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

ADD COMMENT EDIT