Curves and cant
A train on a curve needs an inward force to follow it. Without cant, all of that force would come from the flanges pressing on the outer rail, causing wear, noise and discomfort. Raising the outer rail above the inner one, called cant or superelevation, tilts the vehicle so that a component of its weight provides the inward force.
Cant can be exactly right for only one speed, the equilibrium speed. Faster trains feel an outward unbalanced acceleration; the amount of extra cant they would need is the cant deficiency. Slower trains, often heavy freight, lean inwards and load the inner rail; the excess of actual cant over their equilibrium cant is the cant excess. On mixed-traffic lines, cant is a compromise between fast passenger trains and slow freight, and both deficiency and excess are limited. Indian Railways' manual sets the maximum cant and permitted deficiency by route group.
Cant is run in and out gradually along transition curves, in which the curvature also grows from zero to its full value, so that the train's lateral acceleration and the rate of change of cant stay within comfort and safety limits. Lengths of transitions are set by rate of change of cant, rate of change of deficiency and the cant gradient.
Curves also need checks on gauge widening, check rails on sharp curves, and the clearance between adjacent tracks and structures, since a canted vehicle leans.
Set the equilibrium speed from the actual mix of traffic, not just the fastest train.
Equilibrium cant: e = G V² / (127 R), G = 1 750 mm for broad gaugeMaximum speed for actual cant e and deficiency C_d: V = √[127 R (e + C_d) / G]
Using the track gauge (1 676 mm) instead of G = 1 750 mm in the formula.
Problem. A broad-gauge curve of 800 m radius has an equilibrium speed of 80 km/h. Find the cant. Then, taking a permitted cant deficiency of 75 mm for illustration, find the maximum speed.
- e = 1 750 × 80² / (127 × 800) = 11 200 000 / 101 600 = 110.2 mm.
- Maximum speed: V = √[127 × 800 × (110.2 + 75) / 1 750] = √(127 × 800 × 185.2 / 1 750).
- V = √10 752 = 103.7 km/h.
- Check the result against the route's sectional speed and the turnouts and bridges on the curve.
Answer. Cant about 110 mm; maximum speed about 104 km/h
Quick quiz
1What is cant deficiency?
The extra cant a train faster than the equilibrium speed would need.
2Why are transition curves needed?
To introduce curvature and cant gradually, limiting the rate of change of lateral acceleration.
3Which train causes cant excess?
A slow train, typically freight, running below equilibrium speed.
- Indian Railways Permanent Way Manual (India): Standards and practice for track laying and maintenance.
- EN 13674 (series) (Europe): Railway rails.
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- Satish Chandra, M. M. Agarwal, Railway Engineering, Oxford University Press, latest edition.
- C. Esveld, Modern Railway Track, MRT-Productions, 2nd edition.
Free and official sources
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