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 can be calculated as, l = 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 $p_{SW}$. Then using equation,$p_{ES}$ = $\frac{450+550+700+520}{200+450+550+700+520+300}$= $\frac{2220}{2720}$=0.816
  $p_{WN}$ = $\frac{370+550+500+520}{350+370+550+500+520+420}$= $\frac{1740}{2510}$=0.69
  $p_{NE}$ = $\frac{420+500+700+300}{520+400+420+500+700+300}$= $\frac{1920}{2840}$=0.676
  $p_{SW}$ = $\frac{450+300+370+420}{550+450+400+370+420+350}$= $\frac{1540}{2540}$=0.630
  
• Thus the proportion of weaving traffic to non-weaving traffic is highest in the East-South direction.
• Therefore, the capacity of the rotary will be the capacity of this weaving section. From equation, $Q_{ES}= \frac{280\times13.5[1+\frac{10}{13.5}][1-\frac{0.816}{3}]}{1+\frac{13.5}{54}}$ = 380.56veh/hr.