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Rotary capacity with weaving section numerical

Short answer

Use the IRC rotary capacity formula with weaving width 12 m, average entry width 8.5 m, weaving length 60 m, and weaving proportion 0.7. The capacity is 3667 PCU/h.

Calculate the capacity of rotary with entry and exit width of 8m, the width of non-weaving section is 9m, width of rotary is 12m, length of weaving section is 60m. The ratio of weaving to total traffic in the weaving section is 0.7.

Capacity of the weaving section of a rotary (IRC:65, based on the TRL/Wardrop formula):

Qp=280 w(1+ew)(1−p3)1+wl PCU/hQ_p = \frac{280\,w\left(1+\dfrac{e}{w}\right)\left(1-\dfrac{p}{3}\right)}{1+\dfrac{w}{l}}\ \text{PCU/h}

Data: weaving width w=12w = 12 m (width of the rotary), entry width e1=8e_1 = 8 m, non-weaving width e2=9e_2 = 9 m, so e=(e1+e2)/2=8.5e = (e_1+e_2)/2 = 8.5 m; l=60l = 60 m; p=0.7p = 0.7.

Check of limits: e/w=0.708e/w = 0.708 (0.4 to 1.0 OK), w/l=0.20w/l = 0.20 (0.12 to 0.4 OK), p=0.7p = 0.7 (0.4 to 1.0 OK).

Qp=280(12)(1+8.512)(1−0.73)1+1260=3360(1.7083)(0.7667)1.2=3667 PCU/h\begin{aligned} Q_p &= \frac{280(12)\left(1+\dfrac{8.5}{12}\right)\left(1-\dfrac{0.7}{3}\right)}{1+\dfrac{12}{60}}\\ &= \frac{3360(1.7083)(0.7667)}{1.2} = 3667\ \text{PCU/h} \end{aligned}

Answer: capacity of the rotary = 3667 PCU/h.

✓ Answer checked by Claude Opus 5.5 (high effort)

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