Curves for roads and railways
Roads and railways cannot change direction abruptly, so straights are joined by curves. A simple circular curve has one radius. A compound curve has two or more arcs of different radii turning the same way; a reverse curve turns one way then the other. Between a straight and a circular curve, a transition curve, usually a clothoid or cubic spiral, gradually changes the curvature from zero to the circular value, so that superelevation and centrifugal force build up smoothly. Vertical curves, parabolic in shape, join changes of gradient.
A simple curve is defined by its radius R and the deflection angle Δ between the two straights, measured at the intersection point (PI). From these come the tangent length, curve length, external distance and mid-ordinate. Chainages of the tangent points fix where the curve starts and ends. Degree of curve is an older way of describing sharpness, defined by the angle subtended by a standard chord.
Setting out uses either linear methods, such as offsets from the tangent or from long chords, for short curves with good access, or angular methods. Rankine's method of tangential (deflection) angles sets out successive chord points with a theodolite at the tangent point and a tape. With a total station, coordinates of every curve point are computed and set out directly from any convenient station, which is now standard practice.
Tangent length: T = R tan(Δ/2)Curve length: L = π R Δ / 180External: E = R [sec(Δ/2) − 1]; mid-ordinate: M = R [1 − cos(Δ/2)]Rankine deflection angle for chord c: δ = 1 718.9 c / R minutes
Adding the tangent length to the PI chainage for the second tangent point; the second point comes from the first plus the curve length.
Problem. Two straights meet at chainage 1 250.00 m with Δ = 36°. A curve of R = 400 m is used. Find T, L, the chainages of both tangent points and the deflection angle for a 20 m chord.
- T = 400 × tan 18° = 129.97 m; chainage of first tangent point = 1 250.00 − 129.97 = 1 120.03 m.
- L = π × 400 × 36 / 180 = 251.33 m; second tangent point at 1 120.03 + 251.33 = 1 371.36 m.
- δ for 20 m = 1 718.9 × 20 / 400 = 85.9′ = 1° 25.9′.
Answer. T ≈ 130 m, L ≈ 251.3 m, tangent points at 1 120.03 m and 1 371.36 m
Quick quiz
1What is a transition curve for?
To change curvature, superelevation and lateral force gradually.
2Give the tangent length of a simple curve.
R tan(Δ/2).
3Which method sets out curves with deflection angles?
Rankine's method.
- IRC:73 (India): Geometric design standards for rural (non-urban) highways.
Clauses and documents to read
- IRC:73-1980 (geometric design standards for rural highways)
- IRC:38-1988 (design of horizontal curves for highways)
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.
- S. K. Duggal, Surveying (Vols 1 and 2), McGraw Hill Education (India), latest edition.
- B. C. Punmia, A. K. Jain, A. K. Jain, Surveying (Vols 1 to 3), Laxmi Publications, latest edition.
- C. D. Ghilani, P. R. Wolf, Elementary Surveying: An Introduction to Geomatics, Pearson, latest edition.
Free and official sources
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