---
title: "Setback on four lane carriageway, 400 m radius"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-70bh-setback-four-lane"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-70bh-setback-four-lane
site: BCT Engineering
---

# Setback on four lane carriageway, 400 m radius

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2070 Bhadra, 8 marks
- Video: [Calculate Minimum Setback Distance for Building Safety on Curved Roads || Transportation Engineering](https://www.youtube.com/watch?v=CzXVg6BI-8I) — 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-70bh-setback-four-lane
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A four lane carriageway has a curve of 220 m length and 400 m radius. The safe stopping sight distance and overtaking sight distance are 152 m and 300 m respectively. Calculate the minimum set-back distance from the inner edge of the road to the edge of the obstruction to ensure safe visibility for the both cases of sight distances if the width of the pavement per lane is 3.75 m.

## Short answer

Four-lane curve radius 400 m, length 220 m, lane width 3.75 m. Setback from inner edge for SSD 152 m is 5.43 m; for OSD 300 m is 24.38 m.

## Answer

#### Assumptions
Four-lane undivided carriageway, lane width $w=3.75$ m (so pavement = 15 m), $R=400$ m to the road centre line, $L=220$ m. The driver keeps to the centre of the inner (kerb-side) lane, which is $n=3.75+1.875=5.625$ m from the centre line. Path radius $R_i=400-5.625=394.375$ m. The setback from the inner edge of the road is $m_c-\dfrac{w}{2}=m_c-1.875$ m.

#### Case 1: stopping sight distance S = 152 m ($S<L$)

$$
\frac{\alpha}{2}=\frac{S}{2R_i}=\frac{152}{2(394.375)}=0.19271\ \text{rad}
$$


$$
m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)=7.300\ \text{m}
$$

Setback from inner edge $=7.300-1.875=\mathbf{5.43}$ m.

#### Case 2: overtaking sight distance S = 300 m ($S>L$)

$$
\frac{\alpha}{2}=\frac{L}{2R_i}=\frac{220}{2(394.375)}=0.27892\ \text{rad}
$$


$$
m_c=R_i\left(1-\cos\frac{\alpha}{2}\right)+\frac{S-L}{2}\sin\frac{\alpha}{2}=15.242+40\sin(15.981^\circ)=26.254\ \text{m}
$$

Setback from inner edge $=26.254-1.875=\mathbf{24.38}$ m.

**Answer:** minimum setback from the inner edge = **5.43 m** (SSD) and **24.38 m** (OSD).

## Check yourself

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

1. Where is the driver path on a four-lane road with 3.75 m lanes?
   A. 5.625 m from the centre line, centre of the inner lane
   B. 1.875 m
   C. 3.75 m
   D. 7.5 m
   Answer: A. Ri = 400 - 5.625 = 394.375 m. (Key point: Path position)

2. How do you get the setback from the inner edge?
   A. m_c + 1.875
   B. m_c - w/2 = m_c - 1.875
   C. m_c - 3.75
   D. m_c
   Answer: B. SSD gives 7.300 - 1.875 = 5.43 m. (Key point: Setback conversion)

3. Why is no extension term used for SSD = 152 m?
   A. Because S > L
   B. Because friction is zero
   C. Because S < L = 220 m
   D. Because it is four lane
   Answer: C. Sight line lies on the curve. (Key point: Case SSD)

4. What is added for OSD = 300 m?
   A. (S-L) cos
   B. S/2
   C. None
   D. (S-L)/2 sin(α/2) = 40 sin(15.98°)
   Answer: D. m_c = 26.254, setback 24.38 m. (Key point: Case OSD)

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