---
title: "Overtaking sight distance and zone, 60 kmph"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-76b-osd-60"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-76b-osd-60
site: BCT Engineering
---

# Overtaking sight distance and zone, 60 kmph

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2076 Baisakh, 8 marks
- Video: [Calculating Safe Overtaking Sight Distance & Overtaking Zone Length | Highway Design Explained](https://www.youtube.com/watch?v=_4vtpCon7ug) — OPengineer
- 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-76b-osd-60
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

The speed of overtaking vehicle is 60 kmph on a two way traffic road. If the acceleration of overtaking vehicle is 0.79 m/sec²,
a) Calculate safe overtaking sight distance
b) Mention the minimum length of overtaking zone
c) Draw a neat sketch of the overtaking zone and show the position of sign posts.

## Short answer

OSD ≈302 m. Minimum length of overtaking zone is 3 × OSD = 906 m.

## Answer

#### Data (IRC 66 / NRS 2070)
Speed of overtaking vehicle $V=60$ km/h, acceleration $a=0.79$ m/s², reaction time $t=2$ s. The speed of the overtaken vehicle is not given, so $V_b=V-16=44$ km/h (IRC), i.e. $v_b=44/3.6=12.22$ m/s. The opposing vehicle is taken at the design speed $V=60$ km/h.

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

- Distance in reaction time: $d_1=0.278V_bt=0.278(44)(2)=24.5$ m
- Minimum spacing ($v_b$ in m/s): $s=0.7v_b+6=0.7(12.22)+6=14.56$ m
- Overtaking time $T=\sqrt{\dfrac{4s}{a}}=\sqrt{\dfrac{4(14.56)}{0.79}}=8.58$ s
- Distance covered during overtaking: $d_2=0.278V_bT+2s=0.278(44)(8.58)+2(14.56)=134.1$ m
- Distance covered by the opposing vehicle: $d_3=0.278VT=0.278(60)(8.58)=143.2$ m


$$
OSD=24.5+134.1+143.2=301.8\ \text{m}
$$

**OSD is about 302 m.**

#### (b) Minimum length of overtaking zone
IRC recommends a minimum length of **3 × OSD** = 3 × 302 = **906 m** (desirable 5 × OSD = 1510 m).

#### (c) Sketch of overtaking zone
```text
 Direction of travel  ---->
 [Sign: Overtaking zone]       [Sign: End of overtaking zone]
        |<----- Overtaking zone (min 3 x OSD) ----->|
 =======|===========================================|=======
 ------- - - - - - - - - - - - - - - - - - - - - - - -------
 =======|===========================================|=======
 continuous    broken centre line inside zone     continuous
 centre line                                      centre line
 (A "No overtaking" sign is placed beyond the end of the zone.
  The opposite direction has its own zone and signs.)
```
Signs are placed at the start and end of the zone (for each direction of travel). The centre line is a broken line inside the zone and a continuous line outside it.

## Check yourself

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

1. With vb not given, what is assumed?
   A. vb = V - 16 = 44 km/h
   B. vb = V = 60 km/h
   C. vb = 0
   D. vb = V/2 = 30 km/h
   Answer: A. IRC suggests vb = V - 16 when not stated. (Key point: Speed of overtaken vehicle)

2. How is overtaking time T obtained?
   A. T = s/vb
   B. T = sqrt(4s/a), with s = 0.7vb + 6
   C. T = 2s/a
   D. T = sqrt(s/a)
   Answer: B. s = 36.8 m, so T = sqrt(4 x 36.8/0.79) = 13.65 s. (Key point: Overtaking time)

3. Which expression correctly describes d3?
   A. Distance during reaction: 0.278 vb t
   B. Distance of the overtaking vehicle: 0.278 vb T + 2s
   C. Distance travelled by the oncoming vehicle at design speed during overtaking time: 0.278VT
   D. Braking distance V^2/254f
   Answer: C. d3 = 0.278 x 60 x 13.65 = 227.7 m; OSD = 24.5 + 240.6 + 227.7 = 492.7 m. (Key point: OSD components)

4. What minimum length of overtaking zone is recommended?
   A. 1 x OSD
   B. OSD/3
   C. 10 x OSD
   D. 3 x OSD (desirable 5 x OSD)
   Answer: D. IRC recommends a minimum of 3 OSD, i.e. about 1478 m, with 5 OSD desirable. (Key point: Overtaking zone length)

## Related in Transportation Engineering I

- [Overtaking sight distance and zone numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-62j-osd-numeric) — 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
- [Minimum sight distance to avoid head-on collision](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-74bh-headon-sd) — 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×
- [Sight distance with obstruction on curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78b-sd-obstruction) — Transportation Engineering I, chapter 3

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