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Composite curve, 65 kmph, centrifugal ratio 1/5

Short answer

For design speed 65 km/h and centrifugal ratio 1/5, circular curve radius is 166.3 m. Transition curve length by rate of change of centrifugal acceleration is 61.9 m. Total composite curve length with 48° deflection is 201.3 m.

A vehicle moving in a horizontal curve at a design speed of 65 kmph, develops a centrifugal ratio of 1/5. The deflection angle at curve is 48°. Calculate: a) radius of circular curve b) length of transition curve by rate of change of centrifugal acceleration criteria c) total length of composite curve

(a) Radius of circular curve

Centrifugal ratio =V2127R=15=\dfrac{V^2}{127R}=\dfrac{1}{5}:

R=5V2127=5(65)2127=166.34 mR=\frac{5V^2}{127}=\frac{5(65)^2}{127}=166.34\ \text{m}

(b) Length of transition curve (rate of change of centrifugal acceleration)

IRC: C=8075+V=80140=0.571C=\dfrac{80}{75+V}=\dfrac{80}{140}=0.571 m/s³ (within the permitted range 0.5 to 0.8). V=65/3.6=18.06V=65/3.6=18.06 m/s.

Ls=V3CR=18.0630.571(166.34)=61.93 mL_s=\frac{V^3}{CR}=\frac{18.06^3}{0.571(166.34)}=61.93\ \text{m}

(c) Total length of composite curve

  • Spiral angle θs=Ls2R=0.1861\theta_s=\dfrac{L_s}{2R}=0.1861 rad =10.665∘=10.665^\circ
  • Circular arc angle Δc=48∘−2θs=26.669∘\Delta_c=48^\circ-2\theta_s=26.669^\circ
  • Circular arc length Lc=RΔc=166.34(0.4655)=77.42L_c=R\Delta_c=166.34(0.4655)=77.42 m
L=Lc+2Ls=77.42+2(61.93)=201.28 mL=L_c+2L_s=77.42+2(61.93)=201.28\ \text{m}

Answer: R=166.3R=166.3 m; Ls=61.9L_s=61.9 m; total composite curve length =201.3=201.3 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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