Converting two-way slope to one-way slope

When the inner edge cannot be lowered (because of a drainage problem), the pavement is rotated about the inner edge and the outer edge is raised. The two-way (cambered) section with slope $n$ on each side is changed to a one-way section of slope $e$ over the transition length $L$ in three stages.

Notation

$W$ = pavement width, $n$ = camber, $e$ = full superelevation ($e>n$), $L$ = length of transition. The inner edge level is held fixed. The outer edge must rise by $eW$ relative to the inner edge over $L$ (at a uniform rate $eW/L$).

Stages (cross-sections)

 A (normal)      B (outer flat)  C (one-way n)   D (one-way e)
   /\            ____/            ___.-'          ___.---'
  /  \          /                ./              ./   
 inner fixed at all sections --> ^ (same level)

• Section A (start): normal crown, both halves slope at $n$.
• Stage 1 (A to B): the outer half is rotated about the crown until it is level (slope $0$). The outer edge rises $nW/2$, so B is at a distance $$x_B = \frac{nW/2}{eW/L} = \frac{nL}{2e}$$
• Stage 2 (B to C): the outer half is rotated further until the whole pavement has a uniform one-way slope $n$. The outer edge has now risen $nW$ in all: $$x_C = \frac{nW}{eW/L} = \frac{nL}{e}$$
• Stage 3 (C to D): the whole pavement is rotated about the inner edge, increasing the one-way slope from $n$ to $e$ at the end of the transition, $x_D = L$. The outer edge has risen $eW$.

Level of the outer edge

At a distance $x$ from the start, the outer edge is $\dfrac{eW}{L}x$ above its original level (relative to the inner edge). At D, the outer edge is $eW$ above the inner edge, and the centre line has risen by $\dfrac{(e-n)W}{2}$ above its original level.

Any extra widening is provided over the same length $L$.