---
title: "Setback on curve 400 m radius, 1.9 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-78p-setback-400"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-78p-setback-400
site: BCT Engineering
---

# Setback on curve 400 m radius, 1.9 m

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2078 Poush, 4+4 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-78p-setback-400
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

There is a horizontal highway curve of radius 400 m and length 200 m on the highway. Compute the setback distance required from the center line of the pavement on the inner side of the curve so as to provide for
a) Stopping sight distance of 90 m.
b) Safe overtaking distance of 300 m.
The distance between center line of road and inner lane is 1.9 m.

## Short answer

For inner lane radius 398.1 m: setback from centre line for SSD=90 m (S<L) is4.44 m; for OSD=300 m (S>L) setback is26.82 m.

## Answer

#### Data
$R=400$ m (centre line), $L=200$ m, $n=1.9$ m (centre line of road to centre of inner lane). The sight line is taken along the centre of the inner lane, so its radius is $R-n=398.1$ m. The setback $m$ is measured from the centre line of the pavement to the obstruction line.

#### (a) Stopping sight distance S = 90 m ($S<L$)
Half the angle subtended by the sight distance:

$$
\frac{\alpha}{2}=\frac{S}{2(R-n)}=\frac{90}{2(398.1)}=0.11304\ \text{rad}=6.477^\circ
$$


$$
m=R-(R-n)\cos\frac{\alpha}{2}=400-398.1\cos(6.477^\circ)=4.441\ \text{m}
$$

#### (b) Overtaking sight distance S = 300 m ($S>L$)
For $S>L$ the angle is based on the curve length:

$$
\frac{\alpha}{2}=\frac{L}{2(R-n)}=\frac{200}{2(398.1)}=0.25119\ \text{rad}=14.392^\circ
$$


$$
m=R-(R-n)\cos\frac{\alpha}{2}+\frac{S-L}{2}\sin\frac{\alpha}{2}=14.394+50\sin(14.392^\circ)=26.822\ \text{m}
$$

**Answer:** setback from the centre line = **4.44 m** for SSD = 90 m and **26.82 m** for OSD = 300 m.

## Check yourself

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

1. What radius is used in the setback formula for the sight line here?
   A. R - n = 398.1 m, because the sight line runs along the centre of the inner lane
   B. R = 400 m, the centre line radius only
   C. R + n = 401.9 m
   D. R - 2n = 396.2 m
   Answer: A. The inner lane centre is n = 1.9 m inside the centre line, so its radius is 398.1 m. (Key point: Sight line radius)

2. For SSD = 90 m on a 200 m curve, which formula applies?
   A. m includes the extra term (S-L)/2 sin(α/2)
   B. m = R - (R-n)cos(α/2) with α/2 = S/(2(R-n))
   C. m = S^2/(8R) only using centre line radius with no lane offset
   D. α/2 = L/(2(R-n))
   Answer: B. Since S < L the whole sight line lies on the curve, so no straight extension term is needed; m = 4.44 m. (Key point: Case S<L)

3. What changes when OSD = 300 m exceeds L = 200 m?
   A. The angle is based on S and the formula is unchanged
   B. The setback is simply doubled
   C. The angle is based on L, and a term (S-L)/2 x sin(α/2) is added for the straight portions of the sight line
   D. The radius is replaced with R + n
   Answer: C. The sight line extends beyond the curve along the tangents, adding 50 sin(14.39°) to the setback, giving 26.82 m. (Key point: Case S>L)

4. Why is the setback for OSD (26.8 m) so much larger than for SSD (4.4 m)?
   A. OSD uses a larger friction coefficient
   B. OSD is measured from the outer edge
   C. A smaller radius is used for overtaking
   D. OSD is a much longer sight line, and the chord of a longer sight line lies farther from the curve
   Answer: D. Setback grows with sight distance, roughly with the square of S, so 300 m needs far more clearance than 90 m. (Key point: Result comparison)

## Close variants set in other exams

- [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×

## Related in Transportation Engineering I

- [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 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 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
- [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 rural highway, 80 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-71m-setback-80) — Transportation Engineering I, chapter 3
- [Setback with overtaking, deflection 70 degrees](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81a-setback-overtaking) — Transportation Engineering I, chapter 3

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