Traffic Engineering

Any vertical line drawn in the graph which cuts arrow lines give positions of vehicles in different regions at a particular time.
Slope of a line in a region represents speed of vehicles in that particular region.

From the Plot :
1. Distance between position of vehicles in region ' P ' is greater than that in region R \Rightarrow Density in region ' P ' is greater than that in region ' R '.
2. Slope of lines in region ' Q ' =0 \Rightarrow Speed of vehicles in region ' \mathrm{Q} ' is zero.
3. Slope of lines in region ' P ' is greater than in region ' R ' \Rightarrow speed of vehicles in region ' P ' is greater than in region ' R '.
4. Since vehicles in region ' Q ' are stationary, flow or volume in region ' Q ' is zero and hence lower than in region \mathrm{P} or region \mathrm{R}. And density in region ' \mathrm{Q} ' is higher than in region \mathrm{P} or region \mathrm{R}.
5. Boundary between region ' P and ' R ' represents 'forward moving slope wave' means slope is positive.
i.e. \quad slope =\frac{q_{R}-q_{P}}{K_{R}-K_{P}} \gt 0
q - Flow
\mathrm{K} - Density.
Since \mathrm{K}_{\mathrm{R}} \gt \mathrm{K}_{\mathrm{P}}
\Rightarrow \mathrm{q}_{\mathrm{R}} should be greater than \mathrm{q}_{\mathrm{P}}.
\Rightarrow Volume/flow in region ' R ' is higher than in Region 'P'.

For, Road 'A':
\begin{aligned} \mathrm{v}_{\mathrm{f}} & =\mathrm{u}_{\mathrm{A}}, \quad \mathrm{k}_{\mathrm{j}}=\mathrm{k}_{\mathrm{A}} \\ \mathrm{c}_{\mathrm{A}} & =\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{4} \\ \mathrm{c}_{\mathrm{A}} & =\text { Capacity of road, } \end{aligned}

For Road 'B':
\begin{aligned} & v_{f}=\text { Free flow speed }=u_{B} \\ & k_{j}=\text { Jam density }=k_{B} \end{aligned}

Capacity of Road 'B' \left(c_{B}\right)=\frac{u_{B} k_{B}}{4}
\therefore \quad \frac{\mathrm{C}_{\mathrm{A}}}{\mathrm{c}_{\mathrm{B}}}=\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{4\left(\frac{\mathrm{u}_{\mathrm{B}} \mathrm{k}_{\mathrm{B}}}{4}\right)}=\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{\mathrm{u}_{\mathrm{B}} \mathrm{k}_{\mathrm{B}}}

From figure
\begin{aligned} y_{N-S}&=\left [ \frac{360}{1800},\frac{396}{1800} \right ]_{max}\\ &=0.22\\ y_{E-W}&=\left [ \frac{504+504}{2 \times 18},\frac{440}{1800} ,\frac{460}{1800}\right ]_{max}\\ &=0.28\\ &\text{Optimum cycle length}\\ &=\frac{1.5L+5}{1-y}\\ &=\frac{1.5 \times 9+5}{1-(y_{N-S}+y_{E-W})}\\ &=\frac{1.5 \times 9+5}{1-(0.22+0.28)}\\ C_o&=37secs. \end{aligned}

\text{Speed of shock wave}=\frac{\text{Change in flow}}{\text{Change in density}}
\text{Flow}=\text{Speed} \times \text{Density}
\begin{aligned} Density&=\frac{Flow}{speed}\\ &=\frac{q_Q-q_P}{\frac{q_Q}{V_Q}-\frac{q_P}{V_P}}\\ &=\frac{1800-1200}{\frac{1800}{30}-\frac{1200}{60}}\\ &=\frac{600}{60-20}\\ &=15Kmph \end{aligned}

Given that,
Traffic density = 40 Veh/Km.
Time mean speed = 40 Kmph
Space mean speed = 30 kmph
Time Headway (sec) = ?
We know that traffic volume
= Density x Space mean speed = 40x30 =1200 Veh hr.
and we also know that
q=\frac{3600}{H_t}
where H_t- average headway (sc)
1200=\frac{3600}{H_t}
H_t=\frac{3600}{1200}=3 secs.