Curves, Super-elevation, Cant Deficiency & Transition Curves Notes

Last reviewed 30 Sept 2026 · 7 min read

Why curves need special treatment

On a curve a vehicle is subjected to a centrifugal force that pushes it outward. On a railway this must be balanced or the vehicle would ride hard against the outer rail, causing wear, discomfort and a risk of overturning or derailment. The remedy is to raise the outer rail above the inner rail — the cant or super-elevation — so that the resultant of weight and centrifugal force passes through the centre of the track. Curves are also joined to the straight by transition curves in which the curvature and the cant change gradually.

Degree and radius of curve

Indian Railways define the degree of curve as the angle subtended at the centre by a chord of 30 m:

D = \frac{1750}{R} (R in metres, D in degrees)

So a curve of radius 1750 m is a 1° curve, and a radius of 350 m a 5° curve. Sharper curves (bigger $D$ ) mean lower speed and more wear.

Equilibrium cant

The cant that exactly balances the centrifugal force at a speed $V$ (the equilibrium speed) is

C_{e q} = \frac{G V ^{2}}{127 R}

with $G$ = centre-to-centre distance between the rails (about 1750 mm for BG), $V$ in km/h and $R$ in metres; $C_{e q}$ in mm. For BG this is

C_{e q} \approx 13.78 \frac{V ^{2}}{R} mm

Since a track carries trains of different speeds (fast passenger, slow freight), the cant cannot suit all. It is fixed for the equilibrium speed, which is usually taken as a weighted average or determined by traffic pattern (in practice about the speed of the fastest train reduced by the cant deficiency).

Worked ExampleExample — equilibrium cant

A BG curve of radius 600 m is designed for an equilibrium speed of 80 km/h.

$C_{e q} = \frac{1750 \times 8 0 ^{2}}{127 \times 600} = \frac{1750 \times 6400}{76 200} = 147 mm$ .

This is less than the maximum permissible cant of 165 mm, so it can be provided.

Cant deficiency and cant excess

Cant deficiency ( $C_{d}$ ) — the amount by which the actual cant is less than the equilibrium cant needed at the speed of the fastest train: $C_{d} = C_{e q} (V_{ma x}) - C_{a}$ . The passengers feel a slight outward force, but the increase in outward thrust is accepted within limits.
Cant excess ( $C_{e}$ ) — the amount by which the actual cant exceeds the equilibrium cant for the slowest train (a slow train would tend to fall inward): $C_{e} = C_{a} - C_{e q} (V_{min})$ .

Limits on Indian Railways (BG)

Item	Limit
Maximum cant	About 165 mm (some routes up to 185 mm; check the current standard)
Maximum cant deficiency	About 75 mm (up to 100 mm on some routes)
Maximum cant excess	About 75 mm

(Values depend on the route category and speed; use the current Indian Railways Permanent Way Manual.)

Maximum permissible speed on a curve

The maximum speed for a given radius follows from the actual cant plus the permitted cant deficiency:

V_{ma x} = 0.27 (C_{a} + C_{d}) R (C in mm, R in m, V in km/h, BG)

A second limit comes from the sub-critical speed: the speed must not exceed the value that a transition curve of the available length can support.

Worked ExampleExample — maximum speed on a curve

$R$ = 600 m, actual cant $C_{a}$ = 100 mm, cant deficiency $C_{d}$ = 75 mm.

$V_{ma x} = 0.27 (100 + 75) \times 600 = 0.27 105000 = 0.27 \times 324 = 87.5 km/h$ .

With a cant of 140 mm the speed becomes $0.27 215 \times 600 = 0.27 \times 359 = 97 km/h$ — cant is increased only as far as the cant excess for the slowest train and the maximum cant limit allow.

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