---
title: "Setback on 400 m curve, 1.95 m lane distance"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-68m-setback-1-95"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-68m-setback-1-95
site: BCT Engineering
---

# Setback on 400 m curve, 1.95 m lane distance

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2068 Magh (old course), 8 marks
- Video: [Setback Distances | Highway Engineering | Lec-5 Part-2 | GATE](https://www.youtube.com/watch?v=QtgXO8_dZ6g) — Unacademy GATE & ESE - Civil & ES
- 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-68m-setback-1-95
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

There is a horizontal curve with radius of 400 m and length of 200 m on the highway. Compute the set-back distance required from the centre line on the inner side of the curve so as to provide for (i) stopping sight distance of 90 m (ii) safe overtaking sight distance of 300 m. The distance between the centre line of the road and inner lane is 1.95 m.

## Short answer

For SSD = 90 m (< L), setback m = 4.49 m from road centre line. For OSD = 300 m (> L), setback = 26.87 m.

## Answer

#### Data
$R=400$ m (centre line), $L=200$ m, $n=1.95$ m (centre line of road to centre of inner lane). The sight line runs along the centre of the inner lane, radius $R-n=398.05$ m. The setback $m$ is measured from the road centre line.

#### (i) SSD = 90 m ($S<L$)

$$
\frac{\alpha}{2}=\frac{S}{2(R-n)}=\frac{90}{2(398.05)}=0.11305\ \text{rad}=6.477^\circ
$$


$$
m=R-(R-n)\cos\frac{\alpha}{2}=400-398.05\cos(6.477^\circ)=4.491\ \text{m}
$$

#### (ii) OSD = 300 m ($S>L$)

$$
\frac{\alpha}{2}=\frac{L}{2(R-n)}=\frac{200}{2(398.05)}=0.25122\ \text{rad}=14.394^\circ
$$


$$
m=R-(R-n)\cos\frac{\alpha}{2}+\frac{S-L}{2}\sin\frac{\alpha}{2}=14.445+50\sin(14.394^\circ)=26.875\ \text{m}
$$

**Answer:** setback from the centre line = **4.49 m** (SSD = 90 m) and **26.87 m** (OSD = 300 m).

## Check yourself

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

1. Which radius is used for the sight line and why?
   A. R = 400 m, the centre line
   B. R - n = 398.05 m, since sight runs along the centre of the inner lane
   C. R + n = 401.95 m
   D. R - 2n = 396.1 m
   Answer: B. The driver's line of sight follows the inner lane centre, n = 1.95 m inside the centre line. (Key point: Sight line radius)

2. For SSD = 90 m (S < L), how is the half angle found?
   A. alpha/2 = L / (2(R - n))
   B. alpha/2 = S / (R - n)
   C. alpha/2 = S / (2(R - n)) = 6.477°
   D. alpha/2 = S / (2R)
   Answer: C. The sight line is a chord of length S wholly on the curve, so the angle is based on S. (Key point: S < L case)

3. For OSD = 300 m (S > L), what extra term is added to the setback?
   A. (S + L)/2 x cos(alpha/2)
   B. None, the same formula as S < L
   C. (S - L) x tan(alpha)
   D. (S - L)/2 x sin(alpha/2), with alpha based on L
   Answer: D. The part of the sight line beyond the curve lies on the tangent, adding (S-L)/2 sin(alpha/2): 26.87 m. (Key point: S > L case)

## Close variants set in other exams

- [Setback on curve 400 m radius, 1.9 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78p-setback-400) — Transportation Engineering I, chapter 3, asked 2×

## Related in Transportation Engineering I

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