---
title: "Overtaking sight distance and zone numerical"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-62j-osd-numeric"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-62j-osd-numeric
site: BCT Engineering
---

# Overtaking sight distance and zone numerical

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2062 Jestha (old course), 10 marks
- Video: [overtaking sight distance with numericals](https://www.youtube.com/watch?v=dXd-TqPPrrk) — THE CIVIL BANDA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-62j-osd-numeric
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

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.

## 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).

## Answer

#### Data (IRC 66 / NRS 2070)
Overtaking vehicle speed $V=60$ km/h, overtaken vehicle speed $V_b=30$ km/h ($v_b=8.33$ m/s), acceleration $a=3.6$ km/h/s $=1.0$ m/s², reaction time $t=2$ s. The vehicle coming from the opposite direction is taken at the design speed $V=60$ km/h.

#### (i) Safe overtaking sight distance
OSD $=d_1+d_2+d_3$.

1. Distance travelled by the overtaking vehicle during the reaction time:

$$
d_1=0.278V_bt=0.278(30)(2)=16.68\ \text{m}
$$

2. Spacing between vehicles ($v_b$ in m/s): $s=0.7v_b+6=0.7(8.33)+6=11.83$ m. Time of overtaking:

$$
T=\sqrt{\frac{4s}{a}}=\sqrt{\frac{4(11.83)}{1.0}}=6.88\ \text{s}
$$

3. Distance covered by the overtaken vehicle during the manoeuvre, plus the spacings:

$$
d_2=0.278V_bT+2s=0.278(30)(6.88)+2(11.83)=81.05\ \text{m}
$$

4. Distance covered by the opposing vehicle:

$$
d_3=0.278VT=0.278(60)(6.88)=114.76\ \text{m}
$$


$$
OSD=16.68+81.05+114.76=212.49\ \text{m}
$$

#### (ii) Minimum length of overtaking zone
IRC recommends a minimum length of 3 × OSD (desirable 5 × OSD):

$$
L_{zone}=3\times212.5=637.5\ \text{m}
$$

**Answer:** OSD $\approx212.5$ m; minimum overtaking zone $\approx638$ m (desirable 1062 m).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What three distances make up OSD?
   A. d1 reaction distance, d2 overtaking distance, d3 opposing vehicle distance
   B. d1 braking, d2 lag, d3 spacing
   C. d1 + 2s only
   D. SSD x 2 x T
   Answer: A. OSD = d1 + d2 + d3. (Key point: OSD components)

2. Which formula gives T?
   A. T = s / a
   B. T = sqrt(4s/a) = 10.39 s with s = 27 m, a = 1 m/s^2
   C. T = sqrt(s/a)
   D. T = 4s/a
   Answer: B. Relative acceleration from rest across the spacing gives s = a T^2/4 style, T = sqrt(4s/a). (Key point: Overtaking time)

3. How is d3 found?
   A. 0.278 V T with V = 60 km/h = 173.34 m
   B. 0.278 v_b T
   C. 0.278 V t
   D. 2s
   Answer: A. The opposing vehicle travels at design speed during overtaking time T. (Key point: Opposing vehicle)

4. What is the minimum overtaking zone length?
   A. 3 x OSD, about 992 m
   B. OSD, 331 m
   C. 2 x OSD, 661 m
   D. 10 x OSD
   Answer: A. IRC recommends minimum 3 x OSD (desirable 5 x OSD). (Key point: Overtaking zone)

## Related in Transportation Engineering I

- [Overtaking sight distance and zone, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-76b-osd-60) — Transportation Engineering I, chapter 3
- [Vertical curve for overtaking sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81c-vc-overtaking) — Transportation Engineering I, chapter 3
- [Stopping sight distance at 50 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-66m-ssd-50) — Transportation Engineering I, chapter 3
- [Overtaking sight distance derivation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Aosd-derive) — Transportation Engineering I, chapter 3, asked 6×
- [Intersection sight distance numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81c-intersection-sd) — Transportation Engineering I, chapter 3