written 5.4 years ago by
teamques10
★ 64k
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modified 4.8 years ago
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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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teamques10
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