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Vertical curve at PVI 243.154 m

Short answer

L = 180 m; BVC = 245.85 m, EVC = 246.75 m; 30 m before PVI = 244.75 m, after = 245.05 m; lowest point at 77.1 m from BVC, RL = 244.70 m.

A descending section of a road with 3% grade meets an ascending section with 4% grade. Design the vertical curve. The stopping sight distance requirement is 120 m. Calculate the formation levels of main points on curve and at a distance of 30 m both sides from the point of vertical intersection (PVI). The reduce level of PVI is 243.154 m. Assume other data suitably.

A −3% grade meeting +4% is a valley curve; N=0.03+0.04=0.07N=0.03+0.04=0.07. PVI RL = 243.154 m.

Assumptions

SSD S=120S=120 m (given) corresponds to a design speed of about 77 km/h for t=2.5t=2.5 s and f=0.35f=0.35, so V=75V=75 km/h is taken for the comfort check. Headlight height 0.75 m, beam angle 1°, C=0.6C=0.6 m/s³.

Length of curve

  • Headlight criterion (L>SL>S): L=NS21.5+0.035S=0.07(120)21.5+0.035(120)=176.8L=\dfrac{NS^2}{1.5+0.035S}=\dfrac{0.07(120)^2}{1.5+0.035(120)}=176.8 m
  • Comfort criterion: L=2NV3C=20.07(20.83)30.6=65.0L=2\sqrt{\dfrac{NV^3}{C}}=2\sqrt{\dfrac{0.07(20.83)^3}{0.6}}=65.0 m

Adopt L = 180 m; half = 90 m.

Formation levels

RLBVC=243.154+0.03(90)=245.854 m,RLEVC=243.154+0.04(90)=246.754 mRL_{BVC}=243.154+0.03(90)=245.854\ \text{m},\qquad RL_{EVC}=243.154+0.04(90)=246.754\ \text{m}

With xx from BVC: y=RLBVC−0.03x+0.07x22(180)y=RL_{BVC}-0.03x+\dfrac{0.07x^2}{2(180)}.

  • 30 m before PVI (x=60x=60 m): y=244.754y=\mathbf{244.754} m
  • 30 m after PVI (x=120x=120 m): y=245.054y=\mathbf{245.054} m
  • Lowest point: x=0.03(180)0.07=77.14x=\dfrac{0.03(180)}{0.07}=77.14 m from BVC, y=244.697y=\mathbf{244.697} m

Answer: L = 180 m; BVC = 245.85 m; EVC = 246.75 m; 30 m before PVI = 244.75 m; 30 m after PVI = 245.05 m; lowest point = 244.70 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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