Camber elevations, straight and parabolic

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

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