---
title: "Vertical curve -3.5% and +4%, 250 m night"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-78b-vc-night"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-78b-vc-night
site: BCT Engineering
---

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

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2078 Baisakh, 8 marks
- Video: [Vertical Alignment - Highway Geometric Design - Transportation Engineering 1](https://www.youtube.com/watch?v=DRF2BxYzeso) — Ekeeda
- 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-78b-vc-night
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

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.

## 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.

## Answer

**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**; $N=0.035+0.04=0.075$.

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

$$
L=\frac{NS^2}{2(h+S\tan\alpha)}=\frac{0.075(250)^2}{2(0.6+250\tan1.5^\circ)}=328.0\ \text{m}
$$

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

#### Levels at BVC and EVC (PVI = 150.50 m)

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

#### Lowest point

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

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

$$
RL_{lowest}=156.275-0.035(154.00)+\frac{0.075(154.00)^2}{2(330)}=153.580\ \text{m}
$$

#### Levels 75 m left and 60 m right of PVI
- Left: $x=165-75=90$ m, $y=156.275-0.035(90)+\dfrac{0.075(90)^2}{2(330)}=\mathbf{154.045}$ m
- Right: $x=165+60=225$ m, $y=156.275-0.035(225)+\dfrac{0.075(225)^2}{2(330)}=\mathbf{154.153}$ 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.

## Check yourself

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

1. A -3.5% grade meeting +4% is which type of curve and N?
   A. Valley curve, N = 0.075
   B. Summit curve, N = 0.075
   C. Valley curve, N = 0.005
   D. Summit curve, N = 0.005
   Answer: A. Down then up is a valley; N = 0.035 + 0.04. (Key point: Curve type)

2. Which formula gives the valley curve length (L>S) for headlight sight distance?
   A. L = N S^2 / (2(sqrt(H)+sqrt(h))^2)
   B. L = N S^2 / (2(h + S tanα))
   C. L = 2 sqrt(N V^3 / C)
   D. L = N R
   Answer: B. The headlight beam of angle α at height h must hit the road surface at S; L = 328 m, adopt 330 m. (Key point: Night visibility length)

3. How are RL_BVC and RL_EVC found from the PVI?
   A. RL_BVC = RL_PVI - g1(L/2) for both ends
   B. RL_BVC = RL_PVI + L/2
   C. RL_BVC = RL_PVI + g1(L/2) and RL_EVC = RL_PVI + g2(L/2), since a valley curve sits above the tangent points' PVI
   D. RL_BVC = RL_EVC always
   Answer: C. The BVC lies on the falling grade, so it is higher than PVI by 0.035(165); EVC is higher by 0.04(165). (Key point: BVC and EVC levels)

4. Where is the lowest point of the valley curve measured from BVC?
   A. x = L/2 = 165 m
   B. x = g2 L / N = 176 m
   C. x = N L / 8
   D. x = g1 L / N = 154 m
   Answer: D. At lowest point slope is zero: -g1 + N x/L = 0, so x = 0.035(330)/0.075 = 154 m. (Key point: Lowest point)

## Close variants set in other exams

- [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×

## Related in Transportation Engineering I

- [Vertical curve at station 6+525, night visibility](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-74bh-vc-6525) — Transportation Engineering I, chapter 3
- [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 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
- [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 +4% and -3%, through given point](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-vc-4-3) — 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

## Related in other subjects

- [At-grade intersection requirements and night visibility](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aintersection-requirements-night-visibility-lighting) — Transportation Engineering II, chapter 1
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- [Vertical curve +3.5% and -2.75%, 30 m pegs](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35-down-275-through) — Surveying II, chapter 6, asked 7×
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