---
title: "Camber elevations at 315.5 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-63k-camber-315"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-63k-camber-315
site: BCT Engineering
---

# Camber elevations at 315.5 m

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2063 Kartik (old course), 8 marks
- Video: [Camber in highway engineering, Shoulder in highway engineering | Lecture 5](https://www.youtube.com/watch?v=suUw8G59YBg) — Self study goals
- 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-63k-camber-315
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

The centre line of a double lane road has an elevation of 315.5 m as recorded from longitudinal profile. The camber is 3.0% and the lane width is 3.5 m. Find the elevation of the road at the edges of the pavement and the center of lane if (i) straight line camber is provided, (ii) parabolic camber is provided.

## Short answer

Straight camber: centre of lane 315.448 m, pavement edge 315.395 m. Parabolic: centre of lane 315.474 m, edge 315.395 m.

## Answer

#### Data
Centre line RL = 315.5 m, camber = 3.0% (1 in 33.3), two-lane carriageway, lane width = 3.5 m so pavement width $B=7$ m (half-width 3.5 m). Distances are measured from the centre line: centre of lane at $x=1.75$ m, pavement edge at $x=3.5$ m.

#### (i) Straight-line camber
Fall $=c\,x$.
- Centre of lane: $315.5-0.03(1.75)=\mathbf{315.448}$ m
- Edge of pavement: $315.5-0.03(3.5)=\mathbf{315.395}$ m

#### (ii) Parabolic camber
Parabola with crown at the centre line: $y=\dfrac{2c\,x^2}{B}$ (so the total fall at the edge, $x=B/2$, is the same, $cB/2=0.105$ m).
- Centre of lane ($x=1.75$): $y=\dfrac{2(0.03)(1.75)^2}{7}=0.02625$ m, RL $=\mathbf{315.474}$ m
- Edge of pavement ($x=3.5$): $y=0.10500$ m, RL $=\mathbf{315.395}$ m

| Location | Straight camber (m) | Parabolic camber (m) |
|---|---|---|
| Centre line | 315.500 | 315.500 |
| Centre of lane | 315.448 | 315.474 |
| Edge of pavement | 315.395 | 315.395 |

The parabolic camber keeps the middle of the pavement flatter (better riding) and is steeper near the edges, which helps drainage.

## Check yourself

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

1. For 3.5 m lanes, what x is used for lane centre and pavement edge?
   A. 3.5 m and 7 m
   B. 1.75 m and 3.5 m
   C. 1.75 m and 7 m
   D. 0.875 m and 1.75 m
   Answer: B. Distances are measured from the centre line: half a lane and half the width. (Key point: Distances)

2. RL at pavement edge with straight camber 3%?
   A. 315.5 - 0.03(3.5) = 315.395 m
   B. 315.5 - 0.03(7) = 315.29 m
   C. 315.5 + 0.105 = 315.605 m
   D. 315.5 - 0.0525 = 315.448 m
   Answer: A. Fall = c x = 0.105 m at the edge. (Key point: Straight camber)

3. RL at lane centre with parabolic camber?
   A. 315.5 - 2(0.03)(1.75)^2/7 = 315.474 m
   B. 315.448 m
   C. 315.395 m
   D. 315.526 m
   Answer: A. y = 2cx^2/B = 0.02625 m, less than the straight camber fall of 0.0525 m. (Key point: Parabolic camber)

## Related in Transportation Engineering I

- [Camber elevations, straight and parabolic](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65c-camber-320) — Transportation Engineering I, chapter 3
- [Camber elevations at 320.50 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-64p-camber-320-50) — Transportation Engineering I, chapter 3
- [Transition curve and edge elevations, hilly terrain](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79a-tc-super-elevations) — Transportation Engineering I, chapter 3
- [Setback on six lane road, 500 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65c-setback-500) — Transportation Engineering I, chapter 3