---
title: "Sight distance with obstruction on curve"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-78b-sd-obstruction"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-78b-sd-obstruction
site: BCT Engineering
---

# Sight distance with obstruction on curve

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2078 Baisakh, 8 marks
- Video: [Horizontal Curve Sight Distance Calculations](https://www.youtube.com/watch?v=TW87e_i3y2E) — Lorraine Cahill
- 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-78b-sd-obstruction
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A two-lane two-way highway has a curve 500 m long and radius of 300 m. The curve has an obstruction to sight of 10 m from its inner edge. Determine the sight distance that can be provided on the curve section. Check if this sight distance meets the overtaking sight distance requirement for a speed of 50 kmph. Assume the coefficient of longitudinal friction of 0.35 and overtaking acceleration of 3.6 km/hr/sec.

## Short answer

Sight distance provided is 168 m. Required overtaking sight distance at 50 km/h is about 210 m, so the curve does not meet the requirement.

## Answer

#### Sight distance available
Assumptions: two-lane road, 7.0 m wide (lane 3.5 m); $R=300$ m refers to the road centre line; the driver is at the centre of the inner lane, so the obstruction is $m=10+1.75=11.75$ m from the driver's path, and the path radius is $R_i=300-1.75=298.25$ m.

Assuming $S<L$ (500 m):

$$
\cos\theta=1-\frac{m}{R_i}=1-\frac{11.75}{298.25}=0.9606\Rightarrow\theta=16.14^\circ=0.2816\ \text{rad}
$$

$$
S=2R_i\theta=2(298.25)(0.2816)=168.0\ \text{m}
$$

$S<L$, so the assumption is valid. Sight distance available = **168 m**.

#### Required overtaking sight distance at V = 50 km/h (IRC)
Speed of the overtaken vehicle $v_b=V-16=34$ km/h $=9.44$ m/s; reaction time $t=2$ s; $a=3.6$ km/h/s $=1.0$ m/s²; speed of the overtaking and opposing vehicles $v=50/3.6=13.89$ m/s.

- $d_1=v_bt=9.44\times2=18.9$ m
- Spacing $s=0.7v_b+6=0.7(9.44)+6=12.6$ m ($v_b$ in m/s); overtaking time $T=\sqrt{4s/a}=\sqrt{4(12.61)/1.0}=7.10$ s
- $d_2=v_bT+2s=9.44(7.10)+2(12.61)=92.3$ m
- $d_3=vT=13.89(7.10)=98.6$ m (opposing vehicle at design speed)

$$
OSD=d_1+d_2+d_3=18.9+92.3+98.6=209.8\ \text{m}
$$

#### Check
Available (168 m) < required OSD (about 210 m), so the curve **does not** meet the overtaking sight distance requirement. Overtaking must be prohibited on this curve (centre line marking and signs), or the obstruction must be cleared. The available distance does exceed the SSD at 50 km/h (62.9 m, $f=0.35$), so stopping is safe.

## Check yourself

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

1. How is the available sight distance on the curve found?
   A. From m = Ri(1 - cosθ) with the obstruction clearance from the driver's path, then S = 2Riθ
   B. S = L, the curve length of 500 m
   C. S = sqrt(8 R m) using R = 300 on the edge
   D. S = V x t
   Answer: A. With m = 11.75 m and Ri = 298.25 m, θ = 16.14° and S = 168 m (< L, so valid). (Key point: Sight distance available)

2. What speed is used for the overtaken vehicle in the OSD calculation at V = 50 km/h?
   A. vb = V = 50 km/h
   B. vb = V - 16 = 34 km/h
   C. vb = V/2 = 25 km/h
   D. vb = V + 16 = 66 km/h
   Answer: B. IRC takes the overtaken vehicle 16 km/h slower than the overtaking one. (Key point: Overtaking speed of vehicle)

3. What are the three parts of OSD?
   A. Lag distance, braking distance and spacing
   B. d1 + d2 only, ignoring the oncoming vehicle
   C. d1 reaction distance, d2 distance during overtaking, d3 distance covered by the opposing vehicle
   D. Reaction, braking and clearance distances
   Answer: C. OSD = d1 + d2 + d3 = 18.9 + 162.8 + 151.8 = 333.5 m. (Key point: OSD components)

4. What is concluded when available 168 m is compared with the 333 m OSD?
   A. Overtaking is safe because 168 m exceeds SSD so it also exceeds OSD
   B. The curve is fine if the speed is raised
   C. The curve must be lengthened to 500 m
   D. Overtaking is unsafe on the curve, so it must be prohibited (centre line and signs) or the obstruction cleared
   Answer: D. 168 m is below the required OSD though above SSD (62.9 m), so stopping is safe but overtaking is not. (Key point: Conclusion)

## Related in Transportation Engineering I

- [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
- [Setback on six lane highway, 550 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-72a-setback-350) — Transportation Engineering I, chapter 3
- [Setback on six lane highway, 450 m curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-73m-setback-six-lane) — Transportation Engineering I, chapter 3
- [Setback on four lane carriageway, 400 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70bh-setback-four-lane) — Transportation Engineering I, chapter 3
- [Setback on 400 m curve, 1.95 m lane distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-68m-setback-1-95) — Transportation Engineering I, chapter 3, asked 2×
- [Vertical curve with intermediate sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81a-vc-intermediate) — Transportation Engineering I, chapter 3

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