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Tools · Highways and surveying

Road curves

Guided step by step: set out a circular curve by Rankine’s deflection angles, work out superelevation, extra widening and transition length as per IRC, and design summit and valley curves with levels at every peg.

Guided · step 1 of 8

The data

Two straights meet at the point of intersection PI. The angle one straight turns through to become the other is the deflection angle Δ. The designer chooses the radius R; pegs are set at round chainages (every 20 or 30 m) along the curve.

Δ = 36° 00′ 00″, R = 300 m, chainage of PI = 1+000.000, peg interval = 30 m

Curve

Plan of the curve

PIΔ = 36° 00′ 00″PCPTT = 97.48R = 300 m · sketch to scale
T
97.476m
L
188.496m
Ch(PC) → Ch(PT)
902.52 → 1091.02m
E, M
15.439, 14.683m
Full solution, all steps at once
  1. The data

    Δ = 36° 00′ 00″, R = 300 m, chainage of PI = 1+000.000, peg interval = 30 m

  2. Tangent length TT = R·tan(Δ2)

    T = 300 × tan(18° 00′ 00″) = 300 × 0.32492 = 97.476 m

  3. Length of the curve LL = πRΔ180°

    L = π × 300 × 36180 = 188.496 m

  4. Chainages of PC and PTCh(PC) = Ch(PI) − TCh(PT) = Ch(PC) + L

    Ch(PC) = 1000 − 97.476 = 902.524 m (0+902.524)

    Ch(PT) = 902.524 + 188.496 = 1091.02 m (1+091.020)

  5. Apex distance, mid-ordinate and long chordE = R·(sec(Δ2) − 1)M = R·(1 − cos(Δ2))LC = 2R·sin(Δ2)

    E = 300 × (1.05146 − 1) = 15.439 m

    M = 300 × (1 − 0.95106) = 14.683 m

    LC = 2 × 300 × 0.30902 = 185.41 m

  6. Chords: first and last sub-chords

    First sub-chord = 930 − 902.524 = 27.476 m

    Full chords: 5 × 30 m

    Last sub-chord = 1091.02 − 1080 = 11.02 m

    Check: 27.476 + 5 × 30 + 11.02 = 188.496 m = L

  7. Rankine’s deflection anglesδ = 1718.9·cR minutes

    δ₁ = 1718.9 × 27.476300 = 157.43′ = 2° 37′ 26″

    δ for a full chord = 1718.9 × 30300 = 171.89′

    δₙ = 1718.9 × 11.02300 = 63.14′

  8. Total deflection angles and the checkΔₙ = Σδ = Δ2

    Δ2 = 18° 00′ 00″; Σδ = 18° 00′ 01″ ✓

    Full setting-out table on the right.