Module 5: Traffic Engineering

Lecture 40 Traffic rotaries

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Solution

• The traffic from the four approaches negotiating through the roundabout is illustrated in figure 1.

  Figure 1: Traffic negotiating a rotary

  
• Weaving width is calculated as, w = $[\frac{e_1+e_2}{2}]+3.5$ = 13.5 m
• Weaving length, l is calculated as = 4$\times$w = 54 m
• The proportion of weaving traffic to the non-weaving traffic in all the four approaches is found out first.
• It is clear from equation,that the highest proportion of weaving traffic to non-weaving traffic will give the minimum capacity. Let the proportion of weaving traffic to the non-weaving traffic in West-North direction be denoted as $p_{WN}$, in North-East direction as $p_{NE}$, in the East-South direction as $p_{ES}$, and finally in the South-West direction as $p_{SW}$.
• The weaving traffic movements in the East-South direction is shown in figure 2. Then using equation,$p_{ES}$ = $\frac{510+650+500+600}{510+650+500+600+250+375}$= $\frac{2260}{2885}$=0.783
  $p_{WN}$ = $\frac{505+510+350+600}{505+510+350+600+400+370}$= $\frac{1965}{2735}$=0.718
  $p_{NE}$ = $\frac{650+375+505+370}{650+375+505+370+510+408}$= $\frac{1900}{2818}$=0.674
  $p_{SW}$ = $\frac{350+370+500+375}{350+370+500+375+420+600}$= $\frac{1595}{2615}$=0.6099
  
• Thus the proportion of weaving traffic to non-weaving traffic is highest in the East-South direction.

  Figure 2: Traffic weaving in East-South direction

  
• Therefore, the capacity of the rotary will be capacity of this weaving section. From equation,
  

  $$Q_{ES}= \frac{280\times 13.5[1+\frac{10}{13.5}][1-\frac{0.783}{3}]}{1+\frac{13.5}{54}} = 2161.164 veh/hr.$$ (1)