---
title: "Vertical curve at station 6+525, night visibility"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-74bh-vc-6525"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-74bh-vc-6525
site: BCT Engineering
---

# Vertical curve at station 6+525, night visibility

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2074 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-74bh-vc-6525
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

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.

## 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.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A -3% grade meeting +4.5% is?
   A. Valley curve, N = 0.075
   B. Summit curve, N = 0.075
   C. Valley curve, N = 0.015
   D. Summit curve, N = 0.015
   Answer: A. Down then up gives a valley; N = 0.03 + 0.045. (Key point: Curve type)

2. With S = 250 m, h = 0.6, α = 2°, L is?
   A. L = NS^2/4.4 = 1065 m
   B. L = NS^2/(2(h + S tanα)) = 251 m, adopted 255 m
   C. L = NR
   D. L = 2S = 500 m
   Answer: B. The headlight criterion applies with L > S, so it is valid. (Key point: Length)

3. How are BVC and EVC chainages found?
   A. PVI minus and plus L: 6270 and 6780
   B. PVI plus L/2 for both
   C. PVI chainage minus and plus L/2: 6397.5 and 6652.5
   D. Both equal the PVI chainage
   Answer: C. L/2 = 127.5 m on either side of 6525. (Key point: Chainages)

4. Where is the lowest point and why?
   A. At x = L/2 = 127.5 m
   B. At the BVC
   C. At x = g2 L/N = 153 m
   D. At x = g1 L/N = 102 m from BVC (chainage 6499.5), where slope is zero
   Answer: D. Set slope -g1 + Nx/L = 0, so x = 0.03(255)/0.075 = 102 m. (Key point: Lowest point)

## Related in Transportation Engineering I

- [Vertical curve -3.5% and +4%, 250 m night](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78b-vc-night) — Transportation Engineering I, chapter 3, asked 2×
- [Vertical curve -3% and +4.5% at 320.8 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-68m-vc-320) — Transportation Engineering I, chapter 3, asked 2×
- [Vertical curve -3.25% and +3.75%, 225 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-73m-vc-225) — Transportation Engineering I, chapter 3
- [Vertical curve 35 m headlight and momentum grade](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-vc-headlight-35m) — Transportation Engineering I, chapter 3
- [Vertical curve for overtaking sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81c-vc-overtaking) — Transportation Engineering I, chapter 3
- [Vertical curve with 300 m grade sections](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65k-vc-300) — Transportation Engineering I, chapter 3

## Related in other subjects

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- [Night visibility, road marking, traffic islands](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Anight-visibility-road-marking-traffic-island) — Transportation Engineering II, chapter 1