Highway Geometric Design: Sight Distance and Curves
Stopping and overtaking sight distance, superelevation and friction, extra widening, transition curve length, set-back distance, and summit and valley vertical curves as per IRC.
Geometric design fixes the visible shape of a road: how far drivers can see, how curves are banked, and how grades change. Indian practice follows IRC guidelines.
Stopping sight distance (SSD)
The distance to perceive, react and stop:
SSD = v t + v²/(2 g f) (v in m/s)
- Reaction time t = 2.5 s (IRC); longitudinal friction f = 0.35–0.40.
- On a gradient n (as a fraction): v²/(2g(f ± n)), + going uphill.
- Intermediate sight distance = 2 × SSD.
- On two-way single-lane roads, SSD is doubled.
Overtaking sight distance (OSD)
OSD = d₁ + d₂ + d₃ = v_b t + (v_b T + 2s) + v T, with s = 0.7v_b + 6 m and T = √(4s/a). It applies on two-lane roads; overtaking zones are marked where OSD is available.
Horizontal curves
Superelevation: e + f = v²/(g R).
- IRC design: e = (0.75v)²/(gR) (superelevation for 75% of the design speed, no friction), limited to e_max = 0.07 (plain and rolling terrain), and f = 0.15.
- Ruling minimum radius: R = v²/(g(e + f)).
- Extra widening = mechanical (n l²/(2R)) + psychological (V/(9.5√R)), with l = wheelbase, n = number of lanes, V in km/h.
- Set-back distance to keep sight distance on a curve: m = R − (R − d) cos(α/2), with α/2 = S/(2(R − d)) for S < curve length.
Transition curves
Spiral (clothoid) curves introduce curvature and superelevation gradually. Length from the rate of change of centrifugal acceleration:
L = v³/(C R), with C = 80/(75 + V) (V in km/h), limited to 0.5–0.8 m/s³.
Also checked against the rate of introducing superelevation and IRC empirical minimums.
Vertical curves
- Deviation angle N = |n₁ − n₂| (difference of gradients).
- Valley (sag) curves: designed for headlight sight distance and comfort; cubic parabola. For L > S: L = N S²/(1.5 + 0.035 S).
- Summit curves: designed for sight distance, square parabola. For L > SSD: L = N S²/4.4 (driver eye height 1.2 m, object height 0.15 m). For OSD (both heights 1.2 m): L = N S²/9.6.
Worked example
Design speed 80 km/h (22.22 m/s), f = 0.35, t = 2.5 s, level road.
- SSD = 22.22 × 2.5 + 22.22²/(2 × 9.81 × 0.35) = 55.6 + 71.9 = 127.5 m.
Practice
For a design speed of 65 km/h on a curve of radius 200 m, what superelevation does the IRC formula e = (0.75v)²/(gR) give? (v in m/s, g = 9.81; give 3 decimal places.)
Show answer and explanation
Answer: 0.093v = 18.06 m/s; e = (13.54)²/1962 = 0.093. This exceeds 0.07, so e is limited to 0.07 and friction is checked.
A vehicle travels at 20 m/s with reaction time 2.5 s and f = 0.4 (g = 9.81) on level ground. What is the SSD in m? (Give 1 decimal place.)
Show answer and explanation
Answer: 101SSD = 50 + 400/7.848 = 50 + 50.97 = 101.0 m.
Two gradients of +3% and −2% meet at a summit. What is the deviation angle N?
Show answer and explanation
Answer: 0.05N = |n₁ − n₂| = 0.03 + 0.02 = 0.05.
A summit curve must give an SSD of 120 m with N = 0.05 and L > SSD. What length (m) is needed? (Give 1 decimal place.)
Show answer and explanation
Answer: 163.6L = NS²/4.4 = 0.05 × 14,400/4.4 = 163.6 m (which is more than 120 m, as assumed).