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Mathematics

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Answer

64.01 s\text{64.01 s}

Step-by-step solution

5 steps

  1. Calculate minimum green time for pedestrians (GminG_{min}).

    For Road A:

    GminA=21 m1.2 m/s+7 s=17.5 s+7 s=24.5 sG_{minA} = \frac{21 \text{ m}}{1.2 \text{ m/s}} + 7 \text{ s} = 17.5 \text{ s} + 7 \text{ s} = 24.5 \text{ s}

    For Road B:

    GminB=14 m1.2 m/s+7 s=11.67 s+7 s=18.67 sG_{minB} = \frac{14 \text{ m}}{1.2 \text{ m/s}} + 7 \text{ s} = 11.67 \text{ s} + 7 \text{ s} = 18.67 \text{ s}

  2. Calculate critical flow ratios (yiy_i) and total critical flow ratio (YY).

    yA=400 V/hr1800 V/hr/lane=0.2222y_A = \frac{400 \text{ V/hr}}{1800 \text{ V/hr/lane}} = 0.2222 yB=200 V/hr1800 V/hr/lane=0.1111y_B = \frac{200 \text{ V/hr}}{1800 \text{ V/hr/lane}} = 0.1111 Y=yA+yB=0.2222+0.1111=0.3333Y = y_A + y_B = 0.2222 + 0.1111 = 0.3333

  3. Calculate optimal cycle length using Webster's formula (C0C_0).

    Assuming total lost time L=2×4 s=8 sL = 2 \times 4 \text{ s} = 8 \text{ s}.

    C0=1.5L+51−Y=1.5(8 s)+51−0.3333=12 s+5 s0.6667=17 s0.6667≈25.5 sC_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(8 \text{ s}) + 5}{1 - 0.3333} = \frac{12 \text{ s} + 5 \text{ s}}{0.6667} = \frac{17 \text{ s}}{0.6667} \approx 25.5 \text{ s}

  4. Determine the minimum cycle length required to satisfy pedestrian green time (CminC_{min}).

    For Road A:

    C−L≥YyAGminA=0.33330.2222×24.5 s=1.5×24.5 s=36.75 sC - L \ge \frac{Y}{y_A} G_{minA} = \frac{0.3333}{0.2222} \times 24.5 \text{ s} = 1.5 \times 24.5 \text{ s} = 36.75 \text{ s}

    For Road B:

    C−L≥YyBGminB=0.33330.1111×18.67 s=3×18.67 s=56.01 sC - L \ge \frac{Y}{y_B} G_{minB} = \frac{0.3333}{0.1111} \times 18.67 \text{ s} = 3 \times 18.67 \text{ s} = 56.01 \text{ s}

    The maximum required value for (C−L)(C-L) is 56.01 s56.01 \text{ s}.

    Cmin=L+56.01 s=8 s+56.01 s=64.01 sC_{min} = L + 56.01 \text{ s} = 8 \text{ s} + 56.01 \text{ s} = 64.01 \text{ s}

  5. Select the design cycle length.

    Compare C0=25.5 sC_0 = 25.5 \text{ s} and Cmin=64.01 sC_{min} = 64.01 \text{ s}.
    The design cycle length is the greater of the two.
    Design Cycle Length = 64.01 s\boxed{\text{64.01 s}}

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