Design the full superelevation transition for highway curves — superelevation rate, runoff length, tangent runout, total transition development and edge-of-pavement profile — using the AASHTO maximum-relative-gradient method, with NHA Pakistan, IRC and BS limits.
| Station | Distance (m) | Outer-Lane Cross-Slope | Region |
|---|
Superelevation is the banking of the carriageway toward the inside of a horizontal curve so that a component of gravity helps balance centrifugal force. But the curve cannot jump from a flat cross-slope to full banking instantly — it is developed gradually over a transition made of the tangent runout and the superelevation runoff. This calculator sizes that whole transition.
Every superelevation transition moves the pavement through these stages in order.
| Stage | What Happens | Length | End Condition |
|---|---|---|---|
| Normal Crown | Both lanes drain away from centerline at the normal cross-slope | — | Outer lane at −eNC |
| Tangent Runout | Outer lane rotates up from −eNC to level (0%) | Lt | Pavement is a single flat plane |
| Superelevation Runoff | Whole section rotates from 0% to full ed | Lr | Full superelevation ed reached |
| Full Superelevation | Constant banking maintained around the curve | (curve) | Held until the exit transition |
The maximum relative gradient between the pavement edge and the axis of rotation. Lower values at higher speed produce longer, smoother runoffs. From AASHTO Green Book 2018, Table 3-15.
| Design Speed | Max Relative Gradient Δ | Typical Runoff (e=0.06, 2 lanes) | Use |
|---|---|---|---|
| 50 km/h | 0.65% | ≈ 50 m | Collector |
| 60 km/h | 0.60% | ≈ 54 m | Arterial |
| 70 km/h | 0.55% | ≈ 59 m | Secondary highway |
| 80 km/h | 0.50% | ≈ 65 m | National highway |
| 90 km/h | 0.47% | ≈ 69 m | NHA highway |
| 100 km/h | 0.44% | ≈ 74 m | Motorway |
| 110 km/h | 0.41% | ≈ 79 m | Motorway |
| 120 km/h | 0.38% | ≈ 85 m | M-roads (M-1, M-2) |
Problem: A 2-lane (each direction) motorway curve has R = 400 m, design speed 100 km/h, emax = 0.07, lane width 3.6 m, normal crown 2%. Rotation about centerline, 0.67 of runoff on tangent. Find e, runoff, tangent runout and total transition.
Given:
R = 400 m V = 100 km/h e_max = 0.07 w = 3.6 m
n1 = 2 e_NC = 0.02 Δ = 0.44% p = 0.67 f = 0.12
Step 1 — Design Superelevation Rate:
e = V²/(127R) − f = 100²/(127·400) − 0.12
= 10000/50800 − 0.12 = 0.1969 − 0.12 = 0.0769
e exceeds e_max 0.07 → adopt e_d = 0.07 (curve near minimum radius)
Step 2 — Multilane Adjustment Factor:
b_w = (1 + 0.5(n1−1))/n1 = (1 + 0.5)/2 = 0.75
Step 3 — Superelevation Runoff:
L_r = (w·n1·e_d/Δ)·b_w
= (3.6·2·0.07 / 0.0044)·0.75
= (0.504 / 0.0044)·0.75 = 114.55·0.75 = 85.9 m
Step 4 — Tangent Runout:
L_t = (e_NC/e_d)·L_r = (0.02/0.07)·85.9 = 24.5 m
Step 5 — Total Transition:
L_total = L_t + L_r = 24.5 + 85.9 = 110.4 m
Step 6 — Position Relative to PC:
Runoff on tangent = 0.67·85.9 = 57.6 m (before PC)
Runoff on curve = 0.33·85.9 = 28.3 m (after PC)
ANSWER: e_d = 0.07 | L_r = 85.9 m | L_t = 24.5 m
Total transition = 110.4 m
Place 57.6 m of runoff before PC, 28.3 m beyond PC.
