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Transition curve 220 m radius, 65 kmph

Short answer

Superelevation restricted to 0.07. Length by rate of centrifugal acceleration = 46.8 m, by superelevation introduction = 39.4 m, IRC formula = 51.9 m. Greatest value adopted as transition length = 55 m.

Calculate the length of transition curve using the following data: i) Design speed = 65 kmph ii) Radius of circular curve = 220 m iii) Allowable rate of introduction of super elevation (Pavement rotated about centre line) = 1 in 150 iv) Pavement width including extra widening = 7.5 m

Superelevation

e=V2225R=652225(220)=0.0854e=\dfrac{V^2}{225R}=\dfrac{65^2}{225(220)}=0.0854. This exceeds 0.07, the maximum for plain/rolling terrain (IRC/NRS), so e = 0.07. Friction required =V2127R−e=0.081<0.15=\dfrac{V^2}{127R}-e=0.081<0.15, so it is safe.

Length of transition curve (the greatest of three)

  1. Rate of change of centrifugal acceleration: C=8075+65=0.571C=\dfrac{80}{75+65}=0.571 m/s³, V=18.06V=18.06 m/s
Ls=V3CR=18.0630.571(220)=46.8 mL_s=\frac{V^3}{CR}=\frac{18.06^3}{0.571(220)}=46.8\ \text{m}
  1. Rate of introduction of superelevation (1 in 150, rotation about centre line, so each edge moves eW/2eW/2 from the centre line):
Ls=N e W2=150(0.07)(7.5)2=39.4 mL_s=\frac{N\,e\,W}{2}=\frac{150(0.07)(7.5)}{2}=39.4\ \text{m}
  1. IRC empirical formula (plain and rolling terrain):
Ls=2.7V2R=2.7(65)2220=51.9 mL_s=\frac{2.7V^2}{R}=\frac{2.7(65)^2}{220}=51.9\ \text{m}

Answer: the greatest value governs, so adopt L_s = 55 m (calculated 51.9 m). Shift s=Ls224R=0.57s=\dfrac{L_s^2}{24R}=0.57 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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