---
title: "Vertical curve -3% and +4.5% at 320.8 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-68m-vc-320"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-68m-vc-320
site: BCT Engineering
---

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

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2068 Magh (old course), 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-68m-vc-320
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

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.

## 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.

## Answer

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

#### Length ($S=250$ m, $h=0.60$ m, $\alpha=1.5^\circ$)

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

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

#### Levels (PVI = 320.80 m)

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

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

$$
RL_{low}=325.750-0.03(132.0)+\frac{0.075(132.0)^2}{2(330)}=323.770\ \text{m}
$$

#### At 60 m left and right of PVI
With $y=RL_{BVC}-0.03x+\dfrac{0.075x^2}{2(330)}$:
- 60 m left: $x=165-60=105$ m, $y=\mathbf{323.853}$ m
- 60 m right: $x=165+60=225$ m, $y=\mathbf{324.753}$ 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.

## Check yourself

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

1. A -3% grade meets +4.5%. Which curve and which governing sight condition?
   A. Valley curve, headlight sight distance at night
   B. Summit curve, SSD by day
   C. Valley curve, SSD by day
   D. Summit curve, headlight at night
   Answer: A. Descending then ascending makes a valley curve; at night the headlight beam limits visibility. (Key point: Curve type)

2. Which formula gives the valley curve length (L > S)?
   A. L = N S^2 / 4.4
   B. L = N S^2 / (2(h + S tan alpha))
   C. L = 2 sqrt(N V^3 / C)
   D. L = N S / (h + S tan alpha)
   Answer: B. The beam rises at angle alpha from height h, giving the headlight formula: 328 m, adopt 330 m. (Key point: Headlight length)

3. Where is the lowest point measured from BVC?
   A. x = L/2 = 165 m
   B. x = g2 L / N = 198 m
   C. x = g1 L / N = 132 m
   D. x = N L / 8
   Answer: C. The grade is zero when g1 - (N/L)x = 0, so x = g1 L / N. (Key point: Lowest point)

4. For a valley curve, how is RL of BVC found from the PVI?
   A. RL_PVI - 0.03(165)
   B. RL_PVI + g1 x L/2, i.e. 320.80 + 0.03(165)
   C. RL_PVI + 0.045(165)
   D. RL_PVI + N L / 8
   Answer: B. The BVC is on the descending tangent before the PVI, so it is higher than the PVI by g1 x L/2. (Key point: BVC level)

## Close variants set in other exams

- [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×

## 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 with intermediate sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81a-vc-intermediate) — 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

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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