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Transition curve length and shift, 7.2 m pavement

Short answer

Transition length governed by centrifugal acceleration (35.1 m) and superelevation (30.2 m); adopt 40 m. Shift = Ls^2/(24R) = 0.303 m.

Calculate the length of the transition curve and the required shift, if the design speed is 60 kmph, the radius of the circular curve is 220 m. And allowable rate of change of centrifugal acceleration is 60 cm/sec³. Allowable rate of change of super elevation is 1 in 120. The pavement width including extra widening is 7.2 m.

Assumptions

Pavement width (with extra widening) W=7.2W=7.2 m, rotated about the centre line (rotation is not stated). e=V2225R=0.073e=\dfrac{V^2}{225R}=0.073, limited to the maximum e=0.07e=0.07.

Length of transition curve

  1. Rate of change of centrifugal acceleration (C=0.6C=0.6 m/s³, V=16.67V=16.67 m/s):
Ls=V3CR=16.6730.6(220)=35.07 mL_s=\frac{V^3}{CR}=\frac{16.67^3}{0.6(220)}=35.07\ \text{m}
  1. Rate of introduction of superelevation (1 in 120, centre-line rotation):
Ls=N e W2=120(0.07)(7.2)2=30.24 mL_s=\frac{N\,e\,W}{2}=\frac{120(0.07)(7.2)}{2}=30.24\ \text{m}

The greater value governs, so adopt L_s = 40 m.

Shift

s=Ls224R=40224(220)=0.303 ms=\frac{L_s^2}{24R}=\frac{40^2}{24(220)}=0.303\ \text{m}

Answer: L_s = 40 m (calculated 35.1 m); shift = 0.30 m. (If the pavement were rotated about the inner edge, Ls=NeW=120(0.07)(7.2)=60.5L_s=NeW=120(0.07)(7.2)=60.5 m.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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