Module 6 : Traffic Intersection Control
Lecture 30 : Uncontrolled Intersection
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Numerical example

Figure 1: Three legged intersection
\begin{figure} \centerline{\epsfig{file=qfUnCtrlThreeLegInter.eps,width=8cm}} \end{figure}
For a three legged intersection given in figure 1 determine the control delay and level of service for movement 7. The total volume of both pedestrian and vehicular traffic at each movement is given in the figure itself. Following data is also given:
  • The speed of the pedestrians is 1.2m/s
  • All flows contains 10% trucks
  • The percentage of the grade is 0.00
  • Ignore moments coming from south bound
  • The analysis period is 15 min. (T=0.25)

Solution:

  1. Compute the critical gap and follow up time:
    1. Critical gap $ t_{cx} = t_{cb} + t_{cHV} PHV + t_{cG} G – t_{cT} – t_{LT}$. From table. [*] and table. [*] we have $ t_{cb}$ = 7.1 s , $ t_{cG}$ = 0.2, $ t_{cT}$ = 0.0, $ t_{LT}$ = 0.0. Then $ t_{cx}$ at movement 7 computed as: $ t_{c7}$ = 7.1 + 1.0 $ \times$ 0.1+0.2 $ \times$ 0.0 - 0.0 - 0.0 = 6.50 sec
    2. To compute the Follow up time: From table. [*] and table. [*] we have $ t_{fb}$ = 3.5 s , $ t_{fHV}$ = 0.9. Then $ t_{fx}$ at movement 7 computed as: $ t_{fx} = t_{fb} + t_{fHV} P_{HV}$ $ t_{f7}$ = 3.5 + 0.9 $ \times$ 0.1 = 3.59 sec.
  2. Compute the conflicting flow rate:
    $\displaystyle V_{c7}$ $\displaystyle =$ $\displaystyle 2V_4+V_5+V_{13}+V_2+0.5V_3+V_{15}$  
      $\displaystyle =$ $\displaystyle 40+400+15+200+0.5 \times 30+30$  
      $\displaystyle =$ $\displaystyle 700~{\mathrm{conflicts/hr}}$  

  3. Determining potential capacity:
    $\displaystyle C_{px}$ $\displaystyle =$ $\displaystyle v_{cx} \frac{e^{-(v_{cx} t_{cx}/3600)}}{1- e^{-(v_{cx} t_{fx}/3600)}}$  
      $\displaystyle =$ $\displaystyle 700 \frac{e^{-(700 \times 6.5/3600)}}{1-e^{-(700 \times 3.59/3600)}}$  
    $\displaystyle C_{p7}$ $\displaystyle =$ $\displaystyle 394~vph.$  

  4. Determine the impudence effect of the movement capacity for movement 7: From the given figure movement 7 is impeded by vehicular movement 4 and 1 and pedestrian 13 and 15.
    1. Pedestrian impedance probability computed as:
      $\displaystyle P_{pi}$ $\displaystyle =$ $\displaystyle 1- \frac{v_j \times \left[ \frac{w}{Sp} \right]}{3600}$  
      $\displaystyle P_p13$ $\displaystyle =$ $\displaystyle 1- \frac{15 \times \left[ \frac{6}{1.2} \right]}{3600}= 1-0.0417 = 0.958$  
      $\displaystyle P_p15$ $\displaystyle =$ $\displaystyle 1- \frac{30 \times \left[ \frac{4.5}{1.2} \right]}{3600}= 1-0.03125 =0.969.$  

    2. Vehicular impedance probabilities are:
      $\displaystyle P_{vi}$ $\displaystyle =$ $\displaystyle 1- \frac{v_i}{C_{mi}}$  
      $\displaystyle P_{v4}$ $\displaystyle =$ $\displaystyle 1-20/394 = 0.949$  

    3. Once the pedestrian and vehicular impedance is determined, the moment capacity is computed as:
      $\displaystyle C_{mx}$ $\displaystyle =$ $\displaystyle C_{px}PP_{vi} \times P_{pj}$  
      $\displaystyle C_{m7}$ $\displaystyle =$ $\displaystyle 394 \times (0.949)( 0.969)(0.958) = 347~vph.$  

  5. Delay computation: The delay is Calculated by using the formula
    $\displaystyle d_7$ $\displaystyle =$ $\displaystyle \frac{3600}{C_{mx}} + 900T\left[\left(\frac{V_x}{C_{mx}}-1\right)... ...x}{C_{mx}}-1)^2+\frac{\frac{3600}{C_{mx}}\frac{V_x}{C_{mx}}}{450T}} \,\right]+5$  
      $\displaystyle =$ $\displaystyle \frac{3600}{347} + 900 \times 0.25 \left[\left(\frac{75}{347}-1\r... ...}{347}-1)^2+\frac{\frac{3600}{347}\frac{75}{347}}{450 \times 0.25}} \,\right]+5$  
      $\displaystyle =$ $\displaystyle 18.213~sec/veh$  

    The delay of movement 7 is 18.213 sec/veh.
  6. Determine the level of service: From the computed delay (18.213 se) in step 5 the level of service is LOS C obtained from HCM table.