---
title: "Minimum sight distance to avoid head-on collision"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-74bh-headon-sd"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-74bh-headon-sd
site: BCT Engineering
---

# Minimum sight distance to avoid head-on collision

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2074 Bhadra, 8 marks
- Video: [calculate minimum sight distance to avoid head on collision of 2 cars](https://www.youtube.com/watch?v=hFmJEtzpdqc) — Civil Engineering YouTube Channel 
- 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-74bh-headon-sd
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Calculate the minimum sight distance required to avoid a head-on collision of vehicles approaching from the opposite directions speed at 60 kmph. Use the total perception reaction time of 2.5 seconds, coefficient of friction 0.40 and brake efficiency of 50%. The section of the road under consideration has a grade of 10%.

## Short answer

Sum the stopping distances for uphill and downhill vehicles using effective friction 0.20 and 10% grade. Uphill SD = 88.9 m, downhill SD = 183.4 m. Minimum sight distance required is 272 m.

## Answer

For head-on approach, both drivers must stop in the total distance available, so the required sight distance is the **sum of the stopping distances of the two vehicles**. On a 10% grade one vehicle goes uphill and the other downhill.

#### Data
$V=60$ km/h each, $t=2.5$ s, $f=0.40$, brake efficiency 50%, so effective $f=0.40\times0.5=0.20$. Gradient $n=0.10$.

- Lag distance (each): $0.278Vt=0.278(60)(2.5)=41.7$ m

#### Vehicle going uphill (grade helps braking)

$$
SD_1=0.278Vt+\frac{V^2}{254(f+n)}=41.7+\frac{3600}{254(0.30)}=88.9\ \text{m}
$$

#### Vehicle going downhill (grade opposes braking)

$$
SD_2=0.278Vt+\frac{V^2}{254(f-n)}=41.7+\frac{3600}{254(0.10)}=183.4\ \text{m}
$$

#### Minimum sight distance

$$
SD=SD_1+SD_2=88.9+183.4=272.4\ \text{m}
$$

**Answer:** minimum sight distance to avoid head-on collision $\approx$ **272 m**. (On level ground it would be $2\times112.6=225.1$ m.)

## Check yourself

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

1. For head-on approach, what is the minimum sight distance?
   A. The sum of the stopping distances of both vehicles
   B. The stopping distance of one vehicle
   C. Half the stopping distance
   D. The overtaking sight distance
   Answer: A. Both drivers must stop in the available distance, so SD = SD1 + SD2. (Key point: Required distance)

2. How is brake efficiency of 50% used?
   A. f becomes 0.90
   B. The effective coefficient is f = 0.40 x 0.5 = 0.20
   C. f is unchanged at 0.40
   D. Distance is halved
   Answer: B. Reduced brake efficiency reduces usable friction to 0.20. (Key point: Effective friction)

3. How do the two vehicles' braking terms differ on a 10% grade?
   A. Both use f + n
   B. Both use f - n
   C. Uphill uses V^2/(254(f+n)); downhill uses V^2/(254(f-n))
   D. Both ignore the grade
   Answer: C. Uphill braking is helped by gravity (88.9 m); downhill is hindered (183.4 m). (Key point: Grade effect)

4. What is the total sight distance?
   A. About 225 m
   B. About 183 m
   C. About 112 m
   D. About 272 m (88.9 + 183.4)
   Answer: D. 88.9 + 183.4 = 272.4 m, larger than the 225 m for level ground. (Key point: Final value)

## 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
- [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
- [Posted speed limit at head-on crash](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79j-posted-speed) — Transportation Engineering I, chapter 3, asked 2×
- [Valley curve design, 100 kmph, 1 in 40](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-valley-100-312) — Transportation Engineering I, chapter 3, asked 2×
- [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
- [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

## Related in other subjects

- [Stopping distance and brake energy](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Abraking-distance-energy-numerical) — Automotive Technology, chapter 7
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- [Acceleration for 25 kmph schedule speed, 800 m stations](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aacceleration-25kmph-800m-stations) — Utilization of Electrical Energy, chapter 4
- [Time and space mean speed from 14 travel times](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Atime-space-mean-speed-14-vehicles) — Transportation Engineering II, chapter 1