Horizontal Curve Calculator
Professional horizontal (circular) curve design tool for highway and railway engineers. Calculate tangent length, curve length, external distance, mid-ordinate, degree of curve, superelevation and the full setting-out deflection table — following AASHTO, NHA Pakistan, IRC and BS standards.
Horizontal Curve Calculator
PC · PI · PT · Tangent · Curve Length · Degree of Curve · Superelevation · Setting-Out TableHow Horizontal Curve Calculation Works — Step by Step
A horizontal curve is a circular arc in plan that smoothly connects two straight road or railway alignments (tangents). It allows a gradual change of direction so that vehicles can negotiate the turn safely at the design speed. The geometry is fixed by just two inputs — the deflection angle between the tangents and the radius of the arc.
Establish the Deflection Angle and PI
The Point of Intersection (PI) is where the two tangent lines meet. The deflection angle Δ (also called the intersection angle I) is the angle through which the alignment turns. It is normally measured in the field with a total station.Choose Radius R or Degree of Curve D
The designer selects a radius R that satisfies the minimum-radius requirement for the design speed. Alternatively the curve is defined by its Degree of Curve D. Arc definition (metric): D = (arc × 180) / (π × R); for a 30 m arc, D = 1718.873 / R.Compute the Primary Curve Elements
Tangent length T = R·tan(Δ/2), curve length L = π·R·Δ/180, long chord LC = 2R·sin(Δ/2), external distance E = R(sec(Δ/2) − 1) and mid-ordinate M = R(1 − cos(Δ/2)).Locate PC and PT Stations
The Point of Curvature PC = PI − T (start of curve) and the Point of Tangency PT = PC + L (end of curve). Note PT is found by adding the curve length to PC — never by adding the tangent length twice.Check Superelevation and Minimum Radius
For the design speed, e + f = V²/(127R) (metric). The required superelevation must not exceed emax, and R must be ≥ the minimum radius Rmin = V²/[127(emax+f)].Generate the Setting-Out Deflection Table
For field layout, the curve is pegged at regular chord intervals from PC. The deflection angle for a chord of length c is δ = 1718.873·c / R minutes. Cumulative deflections at PT must equal exactly Δ/2 — the standard field check.∑ Core Horizontal Curve Formulas
Types of Horizontal Curves — Comparison
Highway and railway alignments use four families of horizontal curve depending on the site geometry and speed-transition requirements.
| Type | Description | Where Used | Key Note |
|---|---|---|---|
| Simple | Single circular arc of one radius joining two tangents | Most highway & road curves | Defined by Δ and R only |
| Compound | Two or more arcs of different radii curving the same way | Interchange ramps, hilly terrain | Radii ratio kept ≤ 1.5:1 (AASHTO) |
| Reverse | Two arcs curving in opposite directions, common tangent | Constrained alignments, channels | Needs straight between for superelevation reversal |
| Transition (Spiral) | Spiral easing radius from ∞ to R | High-speed roads, all railways | Length Ls from rate of change of radial accel. |
Minimum Radius & Superelevation — by Design Speed
Indicative minimum radii computed from Rmin = V²/[127(emax+f)] using emax = 0.07 (NHA) and AASHTO maximum side-friction factors. Always verify the exact value against the official AASHTO Green Book 2018 / NHA Geometric Design Manual for the project authority.
| Design Speed | Side Friction f | Rmin (e=0.07) | SSD | Typical Use |
|---|---|---|---|---|
| 50 km/h | 0.16 | ≈ 85 m | 65 m | Collector / urban |
| 60 km/h | 0.15 | ≈ 125 m | 85 m | Arterial road |
| 70 km/h | 0.14 | ≈ 185 m | 105 m | Secondary highway |
| 80 km/h | 0.14 | ≈ 240 m | 130 m | National highway |
| 90 km/h | 0.13 | ≈ 320 m | 170 m | NHA highway |
| 100 km/h | 0.12 | ≈ 415 m | 185 m | Motorway / expressway |
| 110 km/h | 0.11 | ≈ 530 m | 220 m | Motorway |
| 120 km/h | 0.09 | ≈ 710 m | 250 m | M-roads (M-1, M-2) |
Simple Horizontal Curve — NHA Highway
Problem: A National Highway alignment deflects through Δ = 36° at PI chainage 1+000.000 m. A radius of R = 300 m is proposed for a design speed of 100 km/h (emax = 0.07). Find all curve elements, PC/PT stations, and check the radius.
Given: Δ = 36° R = 300 m PI = 1+000.000 V = 100 km/h e_max = 0.07 Step 1 — Tangent Length: T = R·tan(Δ/2) = 300 · tan(18°) = 300 · 0.32492 = 97.476 m Step 2 — Curve Length: L = π·R·Δ/180 = π·300·36/180 = 188.496 m Step 3 — Long Chord, External, Mid-Ordinate: LC = 2·300·sin(18°) = 185.410 m E = 300·(sec18° − 1) = 300·(1.05146 − 1) = 15.439 m M = 300·(1 − cos18°) = 300·(1 − 0.95106) = 14.683 m Step 4 — Degree of Curve (30 m arc): D = 1718.873 / R = 1718.873 / 300 = 5.730° (5° 43' 48") Step 5 — Stations: PC = PI − T = 1000 − 97.476 = 0+902.524 PT = PC + L = 902.524 + 188.496 = 1+091.020 Step 6 — Radius Check (100 km/h, f = 0.12): R_min = V²/[127(e_max+f)] = 100²/[127·0.19] = 414.6 m R = 300 m < 414.6 m → INADEQUATE for 100 km/h Either raise R to ≥ 415 m, or reduce design speed to ~85 km/h. ANSWER: T = 97.476 m | L = 188.496 m | LC = 185.410 m E = 15.439 m | M = 14.683 m | D = 5.730° PC = 0+902.524 | PT = 1+091.020 Radius FAILS 100 km/h check — increase to 415 m.
Expert Design Tips for Horizontal Curves
Radius & Speed
- Never design below the minimum radius for the speed — round the chosen radius up to a practical value (e.g. 300, 350, 400 m)
- For ride comfort, NHA practice keeps mainline radii well above the absolute minimum
- On compound curves keep the radius ratio ≤ 1.5 : 1 between adjoining arcs
Superelevation
- Apply the full superelevation transition over the spiral or, for simple curves, over a runoff length before PC
- Limit e to emax (0.07 for NHA) — higher values cause slow-vehicle drift on ice/rain
- Provide a minimum 0.5% drainage grade across the superelevated section
Sight Distance
- Check lateral clearance: the setback m = R(1 − cos(28.6479·S/R)) must be free of obstructions (cut slopes, barriers, walls)
- Sharp curves often fail sight distance even when radius passes — clear the inside of the curve
Setting Out
- Always close the deflection table: cumulative δ at PT must equal Δ/2 to the second
- Peg at 10 m on open ground, 5 m on sharp curves; use sub-chords at PC and PT to hit round chainages
- Re-check PI, PC and PT with the total station before earthwork starts