---
title: "Camber elevations, straight and parabolic"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-65c-camber-320"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-65c-camber-320
site: BCT Engineering
---

# Camber elevations, straight and parabolic

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2065 Chaitra (old course), 8 marks
- Video: [Camber | Numerical on Camber | Straight line Camber | Parabolic Camber | Jkssb JE  | SSC JE](https://www.youtube.com/watch?v=F91OX32TMjE) — Civil Onspot
- 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-65c-camber-320
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

The centre line of a two lane road has an elevation of 320.00 m. The camber of the pavement is 3.0% and cross-slope of shoulder is 5%. Calculate the elevation of pavement at centre of lane, edges of pavement and at road edge if i) Straight line camber is to be provided. ii) Parabolic camber is to be provided. Take width of lane 3.5 m and shoulder width as 1.5 m.

## Short answer

Straight camber: centre of lane 319.948 m, pavement edge 319.895 m, road edge 319.820 m. Parabolic camber: centre of lane 319.974 m, pavement edge 319.895 m, road edge 319.820 m.

## Answer

#### Data
Centre line RL = 320.0 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), shoulder width = 1.5 m with cross-slope 5%. Distances are measured from the centre line: centre of lane at $x=1.75$ m, pavement edge at $x=3.5$ m, road edge (shoulder edge) at $x=5$ m.

#### (i) Straight-line camber
Fall $=c\,x$.
- Centre of lane: $320.0-0.03(1.75)=\mathbf{319.948}$ m
- Edge of pavement: $320.0-0.03(3.5)=\mathbf{319.895}$ m
- Road edge (shoulder): $319.895-0.05(1.5)=\mathbf{319.820}$ 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 $=\mathbf{319.974}$ m
- Edge of pavement ($x=3.5$): $y=0.10500$ m, RL $=\mathbf{319.895}$ m
- Road edge: $319.895-0.05(1.5)=\mathbf{319.820}$ m

| Location | Straight camber (m) | Parabolic camber (m) |
|---|---|---|
| Centre line | 320.000 | 320.000 |
| Centre of lane | 319.948 | 319.974 |
| Edge of pavement | 319.895 | 319.895 |
| Road edge | 319.820 | 319.820 |

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 a two-lane road with 3.5 m lanes, where are the centre of lane, pavement edge and road edge from the centre line?
   A. 1.75 m, 3.5 m, 5.0 m
   B. 3.5 m, 7 m, 8.5 m
   C. 1.75 m, 3.5 m, 3.5 m
   D. 1.75 m, 7 m, 8.5 m
   Answer: A. Lane centre is half a lane, pavement edge is half the width, and the shoulder adds 1.5 m. (Key point: Distances)

2. How is RL found for straight-line camber?
   A. Centre RL minus 2c x^2/B
   B. Centre RL minus c x, e.g. 320 - 0.03(1.75) = 319.948
   C. Centre RL plus c x
   D. Centre RL minus c B
   Answer: B. The fall is linear with distance from the centre line. (Key point: Straight camber)

3. Which equation gives the fall for parabolic camber?
   A. y = c x
   B. y = c x^2 / 2B
   C. y = 2 c x^2 / B
   D. y = 2 c x / B
   Answer: C. The parabola has the same total fall at the edge but is flatter near the crown. (Key point: Parabolic camber)

4. How is the road edge RL found?
   A. Pavement edge RL minus 5% x 1.5 m
   B. Pavement edge RL minus 3% x 1.5 m
   C. Centre RL minus 5% x 5 m
   D. Pavement edge RL plus 0.075 m
   Answer: A. The shoulder falls at its own 5% cross-slope beyond the pavement edge. (Key point: Shoulder)

## Related in Transportation Engineering I

- [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
- [Camber elevations at 315.5 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-63k-camber-315) — 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
- [Raising outer edge for superelevation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-63k-super-raise) — Transportation Engineering I, chapter 3

## Related in other subjects

- [Road earthwork, four lane NRS-1070](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-g-four-lane) — Estimating and Costing, chapter 6