---
title: "Valley curve design, 80 kmph, 1 in 40"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-76b-valley-80"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-76b-valley-80
site: BCT Engineering
---

# Valley curve design, 80 kmph, 1 in 40

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2076 Baisakh, 8 marks
- Video: [Design a Valley Curve for Comfort and Night Visibility at 100 km/h || Transportation Engineering-II](https://www.youtube.com/watch?v=ojz2f9IFT3o) — OPengineer
- 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-76b-valley-80
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Design a total length of valley curve at the junction of the descending gradient of 1 in 40 and an ascending gradient of 1 in 30 if the design speed is 80 kmph. Locate the lowest point and end points and calculate their elevations if the elevation of beginning of the curve is 2.12.86 [?] m above sea level. Take co-efficient of friction as 0.35.

## Short answer

Valley curve length governed by headlight sight distance = 160 m. Lowest point at 68.6 m from BVC with RL 212.00 m; BVC RL 212.86 m, EVC RL 213.53 m.

## Answer

**Reading:** the RL of the beginning of the curve is printed as "2.12.86" in the scan; I have read it as **212.86 m**. For any other value, simply add the difference to every level below.

A descending 1 in 40 grade meeting an ascending 1 in 30 grade forms a **valley curve**.

#### Data
$g_1=-\dfrac{1}{40}=-2.5\%$, $g_2=+\dfrac{1}{30}=+3.33\%$, $N=\dfrac{1}{40}+\dfrac{1}{30}=0.05833$. $V=80$ km/h, $f=0.35$, $t=2.5$ s, $C=0.6$ m/s³, headlight height 0.75 m, beam angle 1° (IRC/NRS).

#### Length of curve

$$
SSD=0.278(80)(2.5)+\frac{80^2}{254(0.35)}=127.6\ \text{m}
$$

- Comfort: $L=2\sqrt{\dfrac{NV^3}{C}}=65.3$ m
- Headlight: $L=\dfrac{NS^2}{1.5+0.035S}=\dfrac{0.05833(127.6)^2}{1.5+0.035(127.6)}=159.2$ m ($L>S$, valid)

Adopt **L = 160 m** (half = 80 m).

#### Levels

$$
RL_{BVC}=212.86\ \text{m}
$$


$$
RL_{PVI}=212.86-\frac{1}{40}(80)=210.860\ \text{m}
$$


$$
RL_{EVC}=210.860+\frac{1}{30}(80)=213.527\ \text{m}
$$

#### Lowest point

$$
x=\frac{g_1L}{N}=\frac{160/40}{0.05833}=68.57\ \text{m from BVC}
$$


$$
RL_{low}=212.86-\frac{68.57}{40}+\frac{0.05833(68.57)^2}{2(160)}=212.003\ \text{m}
$$

**Answer:** L = 160 m; beginning RL = 212.86 m; lowest point RL = 212.00 m at 68.6 m from BVC; end RL = 213.53 m.

## Check yourself

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

1. A descending 1 in 40 grade meeting an ascending 1 in 30 grade forms what, and what is N?
   A. A summit curve with N = 0.0083
   B. A valley curve with N = 0.0583
   C. A valley curve with N = 0.0083
   D. A summit curve with N = 0.0583
   Answer: B. Falling then rising grade gives a valley curve; N = 1/40 + 1/30 = 0.05833 (algebraic difference). (Key point: Valley curve identification)

2. Why is L = 160 m adopted instead of the comfort length 65.3 m?
   A. Comfort length is always ignored
   B. The headlight sight distance length (159.2 m) is the larger and governs
   C. Both lengths are added together
   D. 160 m is the stopping sight distance
   Answer: B. The longest length from comfort and headlight criteria governs; headlight gives 159.2 m, rounded to 160 m. (Key point: Governing length criterion)

3. How is SSD computed here?
   A. V t + V^2/(2 g f) with V in km/h
   B. V^2/(254 f) only
   C. 0.278 V t + V^2/(254 f) = 127.6 m
   D. 0.278 V t only
   Answer: C. SSD = lag distance (0.278 V t) plus braking distance V^2/(254 f), giving 127.6 m. (Key point: Stopping sight distance)

4. How is the lowest point of the valley curve located from BVC?
   A. x = L/2
   B. x = g1 L / N
   C. x = N L / g1
   D. x = g2 L / N^2
   Answer: B. At the lowest point the slope is zero, so x = g1 L/N = (160/40)/0.05833 = 68.57 m from BVC. (Key point: Lowest point location)

## Related in Transportation Engineering I

- [Valley curve design, 100 kmph, 1 in 40](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-valley-100-312) — Transportation Engineering I, chapter 3, asked 2×
- [Valley curve 2.5% and 3.5%, 80 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-63k-valley-80) — Transportation Engineering I, chapter 3
- [Valley curve 1/35 and 1/45, 80 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70bh-valley-1-35) — Transportation Engineering I, chapter 3
- [Valley curve 1 in 30 and 1 in 25](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-67m-valley-1-30) — 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 +2% and +5.5%, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79c-vc-2-55) — Transportation Engineering I, chapter 3

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