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Overtaking sight distance and zone numerical

Short answer

Using d1=0.278Vb t, s=0.7v_b+6, T=sqrt(4s/a), OSD=d1+d2+d3 = 212.5 m. Minimum overtaking zone = 3×OSD = 638 m (desirable 1062 m).

The speeds of overtaking and overtaken vehicles are 60 kmph and 30 kmph respectively on a two-way traffic road. If the acceleration of the overtaking vehicle is 3.6 kmph per second. i) Calculate the safe overtaking sight distance. ii) Determine the minimum length of the overtaking zone.

Data (IRC 66 / NRS 2070)

Overtaking vehicle speed V=60V=60 km/h, overtaken vehicle speed Vb=30V_b=30 km/h (vb=8.33v_b=8.33 m/s), acceleration a=3.6a=3.6 km/h/s =1.0=1.0 m/s², reaction time t=2t=2 s. The vehicle coming from the opposite direction is taken at the design speed V=60V=60 km/h.

(i) Safe overtaking sight distance

OSD =d1+d2+d3=d_1+d_2+d_3.

  1. Distance travelled by the overtaking vehicle during the reaction time:
d1=0.278Vbt=0.278(30)(2)=16.68 md_1=0.278V_bt=0.278(30)(2)=16.68\ \text{m}
  1. Spacing between vehicles (vbv_b in m/s): s=0.7vb+6=0.7(8.33)+6=11.83s=0.7v_b+6=0.7(8.33)+6=11.83 m. Time of overtaking:
T=4sa=4(11.83)1.0=6.88 sT=\sqrt{\frac{4s}{a}}=\sqrt{\frac{4(11.83)}{1.0}}=6.88\ \text{s}
  1. Distance covered by the overtaken vehicle during the manoeuvre, plus the spacings:
d2=0.278VbT+2s=0.278(30)(6.88)+2(11.83)=81.05 md_2=0.278V_bT+2s=0.278(30)(6.88)+2(11.83)=81.05\ \text{m}
  1. Distance covered by the opposing vehicle:
d3=0.278VT=0.278(60)(6.88)=114.76 md_3=0.278VT=0.278(60)(6.88)=114.76\ \text{m} OSD=16.68+81.05+114.76=212.49 mOSD=16.68+81.05+114.76=212.49\ \text{m}

(ii) Minimum length of overtaking zone

IRC recommends a minimum length of 3 × OSD (desirable 5 × OSD):

Lzone=3×212.5=637.5 mL_{zone}=3\times212.5=637.5\ \text{m}

Answer: OSD ≈212.5\approx212.5 m; minimum overtaking zone ≈638\approx638 m (desirable 1062 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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