Skip to main content

Design controls for valley curves

Short answer

Valley curve design controls are comfort (rate of change of centrifugal acceleration ≤ 0.6 m/s³), headlight sight distance at night, overhead clearance (min 5.0 m per IRC), appearance and drainage, and stopping sight distance. Length is calculated from comfort (L = 2√(Nv³/C)), headlight sight distance (different formulas for S<L and S>L), and overhead clearance (L = NS²/[8(Hc - (h1+h2)/2)] for S<L). The maximum of these lengths is adopted.

What are the basic design controls for valley curves? How is the length of valley curve calculated from different criteria?

A valley (sag) curve is a vertical curve with the centre of curvature above the road, formed where a downgrade meets an upgrade (or a lesser downgrade). A parabola is generally used.

Basic design controls

  1. Comfort: the centrifugal acceleration due to the vertical curvature adds to the weight; its rate of change must be limited (0.6 m/s³) to avoid jerk.
  2. Headlight sight distance at night: the beam must light up the road ahead for at least the stopping distance (in day time there is no obstruction to vision).
  3. Overhead clearance under bridges and structures (flyovers) - minimum 5.0 m headroom (as per IRC) with the sight line below the structure.
  4. Appearance and drainage: the curve should look smooth, and the lowest point needs special drainage (inlets and drains), the gradient near it should not be flat.
  5. Safety and stopping distance: SSD must be available.

Length of valley curve

NN = algebraic difference of grades (ratio), vv = speed in m/s.

1. Comfort criterion (C=0.6C = 0.6 m/s³):

L=2N v3CL = 2\sqrt{\frac{N\,v^3}{C}}

2. Headlight sight distance criterion (h=0.75h = 0.75 m, beam angle α=1∘\alpha = 1^\circ):

  • If S<LS < L:  L=NS22 (h+Stan⁡α)\ L = \dfrac{N S^2}{2\,(h + S\tan\alpha)}
  • If S>LS > L:  L=2S−2 (h+Stan⁡α)N\ L = 2S - \dfrac{2\,(h + S\tan\alpha)}{N}

3. Overhead clearance criterion (under a structure with clearance HcH_c, driver's eye height h1h_1 and object height h2h_2), for S<LS<L: L=NS28 (Hc−h1+h22)L = \dfrac{N S^2}{8\,\left(H_c - \tfrac{h_1+h_2}{2}\right)}.

The longest of these values is adopted for design.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Transportation Engineering I chapter 3 questions →