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Vertical curve -3.5% and +4%, 250 m night

Short answer

Adopt valley curve length L = 330 m. BVC RL = 156.28 m, lowest point = 153.58 m (154 m from BVC), EVC = 157.10 m. At 75 m left of PVI: 154.05 m; 60 m right: 154.15 m.

A vertical curve connects a -3.5% grade with +4% grade on a highway at an elevation of [?]50.5 m. 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 point of vertical curve and at a distance of 75 m left and 60 m 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 0.6 m respectively.

Reading of the question: the first digit of the PVI elevation is unclear in the scan; I have taken the elevation as 150.5 m. The method is the same for any value (add or subtract the difference to every level).

A −3.5% grade meeting +4% is a valley curve; N=0.035+0.04=0.075N=0.035+0.04=0.075.

Length (night visibility, S=250S=250 m)

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, so the formula is valid. Adopt L = 330 m; half length = 165 m.

Levels at BVC and EVC (PVI = 150.50 m)

RLBVC=150.50+0.035(165)=156.275 m,RLEVC=150.50+0.04(165)=157.100 mRL_{BVC}=150.50+0.035(165)=156.275\ \text{m},\qquad RL_{EVC}=150.50+0.04(165)=157.100\ \text{m}

Lowest point

x=g1LN=0.035(330)0.075=154.00 m from BVCx=\frac{g_1L}{N}=\frac{0.035(330)}{0.075}=154.00\ \text{m from BVC}

For a valley curve with xx from BVC: y=RLBVC−0.035x+0.075x22(330)y=RL_{BVC}-0.035x+\dfrac{0.075x^2}{2(330)}.

RLlowest=156.275−0.035(154.00)+0.075(154.00)22(330)=153.580 mRL_{lowest}=156.275-0.035(154.00)+\frac{0.075(154.00)^2}{2(330)}=153.580\ \text{m}

Levels 75 m left and 60 m right of PVI

  • Left: x=165−75=90x=165-75=90 m, y=156.275−0.035(90)+0.075(90)22(330)=154.045y=156.275-0.035(90)+\dfrac{0.075(90)^2}{2(330)}=\mathbf{154.045} m
  • Right: x=165+60=225x=165+60=225 m, y=156.275−0.035(225)+0.075(225)22(330)=154.153y=156.275-0.035(225)+\dfrac{0.075(225)^2}{2(330)}=\mathbf{154.153} m

Answer: BVC 156.28 m; lowest point 153.58 m (at 154.0 m from BVC); EVC 157.10 m; 75 m left 154.05 m; 60 m right 154.15 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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