With S = 250 m, h = 0.6, α = 2°, L is?

L = NS^2/(2(h + S tanα)) = 251 m, adopted 255 m

How are BVC and EVC chainages found?

PVI chainage minus and plus L/2: 6397.5 and 6652.5

PVI minus and plus L: 6270 and 6780

Where is the lowest point and why?

At x = g1 L/N = 102 m from BVC (chainage 6499.5), where slope is zero

2074 Bhadra · 8 marks

Vertical curve at station 6+525, night visibility

Short answer

Valley curve length from night visibility (headlight height 0.6 m, beam angle 2°) is 255 m. BVC RL = 415.43 m, lowest point RL = 413.90 m at 102 m from BVC, EVC RL = 417.34 m. At 50 m left of PVI RL = 413.98 m, 50 m right RL = 414.73 m.

A vertical curve connects a -3.0% grade with +4.5% grade on a rural highway at station 6+525 and elevation 411.6 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. Locate the starting, lowest, and end point of vertical curve. Calculate the elevation of road at all these points along the curve and at a distance of 50 m left and right from the point of vertical intersection. Assume the head light beam angle and heights of the head light from the road surface for the design vehicle are 2° and 0.6 m respectively.

A −3.0% grade meeting +4.5% is a valley curve; $N=0.03+0.045=0.075$ . PVI at station 6+525 ( $=6525$ m), RL 411.60 m.

Length (night visibility $S=250$ m, $h=0.6$ m, $\alpha=2^\circ$ )

L=\frac{NS^2}{2(h+S\tan\alpha)}=\frac{0.075(250)^2}{2(0.6+250\tan2^\circ)}=251.2\ \text{m}

$L>S$ , valid. Adopt L = 255 m; half length = 127.5 m.

Chainages

Starting point (BVC) = 6525 − 127.5 = 6397.5 m (station 6+397.50)
End point (EVC) = 6525 + 127.5 = 6652.5 m (station 6+652.50)
Lowest point: $x=\dfrac{g_1L}{N}=\dfrac{0.03(255)}{0.075}=102.0$ m from BVC, i.e. chainage 6499.5 m.

Elevations ( $y=RL_{BVC}-0.03x+\dfrac{0.075x^2}{2L}$ )

RL_{BVC}=411.60+0.03(127.5)=415.425\ \text{m},\quad RL_{EVC}=411.60+0.045(127.5)=417.338\ \text{m}

RL_{lowest}=415.425-0.03(102.0)+\frac{0.075(102.0)^2}{2(255)}=413.895\ \text{m}

50 m left of PVI ( $x=77.5$ m): $y=\mathbf{413.983}$ m
50 m right of PVI ( $x=177.5$ m): $y=\mathbf{414.733}$ m

Answer: BVC 415.43 m, lowest point 413.90 m, EVC 417.34 m; 50 m left 413.98 m, 50 m right 414.73 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Transportation Engineering I chapter 3 questions →

Vertical curve at station 6+525, night visibility

Length (night visibility S=250S=250S=250 m, h=0.6h=0.6h=0.6 m, α=2∘\alpha=2^\circα=2∘)

Chainages

Elevations (y=RLBVC−0.03x+0.075x22Ly=RL_{BVC}-0.03x+\dfrac{0.075x^2}{2L}y=RLBVC​−0.03x+2L0.075x2​)

Related questions

Length (night visibility $S=250$ m, $h=0.6$ m, $\alpha=2^\circ$ )

Elevations ( $y=RL_{BVC}-0.03x+\dfrac{0.075x^2}{2L}$ )