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Vertical curve -3% and +4.5% at 320.8 m

Short answer

Valley curve length L = NS^2/(2(h+S tanα)) = 330 m. BVC RL = 325.75 m, lowest point = 323.77 m, EVC = 328.23 m. At 60 m left of PVI, RL = 323.85 m; 60 m right, RL = 324.75 m.

A vertical curve connects a -3% grade with +4.5% grade on a highway at an elevation of 320.8 m above the mean sea level. The curve should be designed at least to provide the visibility of the road surface to a distance of 250 m at night time. Calculate the elevation of beginning, lowest and end points of vertical curve and at a distance of 60 m left and right from the point of vertical intersection. The head light beam angle and height of the head light from the road surface is 1.5° and 0.60 m respectively.

A −3% grade meeting +4.5% is a valley curve; N=0.03+0.045=0.075N=0.03+0.045=0.075.

Length (S=250S=250 m, h=0.60h=0.60 m, α=1.5∘\alpha=1.5^\circ)

L=NS22(h+Stan⁡α)=0.075(250)22(0.6+250tan⁡1.5∘)=328.0 mL=\frac{NS^2}{2(h+S\tan\alpha)}=\frac{0.075(250)^2}{2(0.6+250\tan1.5^\circ)}=328.0\ \text{m}

L>SL>S, valid. Adopt L = 330 m; half length = 165 m.

Levels (PVI = 320.80 m)

RLBVC=320.80+0.03(165)=325.750 m,RLEVC=320.80+0.045(165)=328.225 mRL_{BVC}=320.80+0.03(165)=325.750\ \text{m},\qquad RL_{EVC}=320.80+0.045(165)=328.225\ \text{m}

Lowest point: x=g1LN=0.03(330)0.075=132.0x=\dfrac{g_1L}{N}=\dfrac{0.03(330)}{0.075}=132.0 m from BVC.

RLlow=325.750−0.03(132.0)+0.075(132.0)22(330)=323.770 mRL_{low}=325.750-0.03(132.0)+\frac{0.075(132.0)^2}{2(330)}=323.770\ \text{m}

At 60 m left and right of PVI

With y=RLBVC−0.03x+0.075x22(330)y=RL_{BVC}-0.03x+\dfrac{0.075x^2}{2(330)}:

  • 60 m left: x=165−60=105x=165-60=105 m, y=323.853y=\mathbf{323.853} m
  • 60 m right: x=165+60=225x=165+60=225 m, y=324.753y=\mathbf{324.753} m

Answer: BVC 325.75 m; lowest point 323.77 m; EVC 328.23 m; 60 m left 323.85 m; 60 m right 324.75 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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