---
title: "Valley curve design, 100 kmph, 1 in 40"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-valley-100-312"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-valley-100-312
site: BCT Engineering
---

# Valley curve design, 100 kmph, 1 in 40

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2076 Bhadra, 8 marks; 2075 Baisakh, 8 marks (asked 2 times)
- 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-valley-100-312
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Design the total length of the 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 100 kmph, so as to fulfil both comfort condition and head light sight distance for night driving. Locate the lowest point and the end of curve point too. Calculate their elevations if the elevation of the beginning of the curve (BVC) is 312.56 m above sea level. Assume other necessary data reasonably.

## Short answer

Total length is 250 m (comfort 91.3 m, headlight sight distance 246 m). Lowest point is at 107.1 m from BVC with elevation 311.22 m; end of curve at 250 m with elevation 313.60 m.

## Answer

**Given:** $V = 100$ km/h, $g_1 = -1/40 = -2.5\%$, $g_2 = +1/30 = +3.33\%$, BVC elevation $= 312.56$ m.

**Assumptions (IRC/NRS):** reaction time 2.5 s, $f = 0.35$, comfort rate $C = 0.6$ m/s³, headlight height 0.75 m, beam angle $1^\circ$.

$$
N = 0.025 + 0.03333 = 0.05833
$$
$v = 100/3.6 = 27.78$ m/s.

#### 1. Comfort condition
$$
L = 2\sqrt{\frac{N v^3}{C}} = 2\sqrt{\frac{0.05833\times 27.78^3}{0.6}} = 2\sqrt{2083.8} = 91.3\ \text{m}
$$

#### 2. Headlight sight distance
SSD:
$$
SSD = 0.278(100)(2.5) + \frac{100^2}{254\times0.35} = 69.5 + 112.5 = 182.0\ \text{m}
$$
Assuming $S < L$:
$$
L = \frac{N S^2}{2\,(h + S\tan\alpha)} = \frac{0.05833\,(182.0)^2}{2\,(0.75 + 182.0\tan1^\circ)} = \frac{1932.2}{2(3.927)} = 246.0\ \text{m}
$$
$S = 182 < 246$ so the assumption holds.

#### 3. Design length
Adopt the larger: $L = 246$ m, rounded to **$L = 250$ m**.

#### 4. Lowest point
Distance from BVC:
$$
x = \frac{g_1 L}{N} = \frac{0.025\times250}{0.05833} = 107.1\ \text{m}
$$
Elevation (curve equation $y = BVC + g_1x + \dfrac{N x^2}{2L}$ with $g_1 = -0.025$):
$$
\begin{aligned}
RL_{low} &= 312.56 - 0.025(107.14) + \frac{0.05833(107.14)^2}{2(250)} \\
&= 312.56 - 2.679 + 1.339 = 311.22\ \text{m}
\end{aligned}
$$

#### 5. End of curve (EVC)
$$
RL_{EVC} = 312.56 + \frac{(g_1+g_2)L}{2} = 312.56 + \frac{(-0.025+0.03333)(250)}{2} = 313.60\ \text{m}
$$
The EVC is at 250 m from the BVC.

**Answer:** $L = 250$ m (comfort 91.3 m, headlight 246.0 m); lowest point is 107.1 m from BVC at RL 311.22 m; end of curve at 250 m from BVC with RL 313.60 m.

## Check yourself

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

1. Comfort length for the valley curve is calculated as
   A. L = N S^2 / (2(h + S tan alpha))
   B. L = 2 sqrt(N v^3 / C) = 91.3 m
   C. L = 2 x SSD
   D. L = 250 m
   Answer: B. With N = 0.05833, v = 27.78 m/s and C = 0.6, L = 91.3 m. (Key point: Comfort length)

2. The headlight length (S = 182 m < L) is about
   A. 91 m
   B. 182 m
   C. 246 m
   D. 400 m
   Answer: C. L = N S^2/(2(0.75 + 182 tan 1 deg)) = 246 m. (Key point: Headlight length)

3. What length is adopted and why?
   A. 91 m, the comfort value
   B. 250 m, the larger value (headlight) rounded
   C. 182 m, the SSD
   D. 125 m, the average
   Answer: B. The larger of 91.3 and 246 m satisfies both criteria; rounded to 250 m. (Key point: Adopted length)

4. The lowest point lies from BVC at x = g1 L / N =
   A. 125 m
   B. 250 m
   C. 182 m
   D. 107.1 m
   Answer: D. x = 0.025 x 250/0.05833 = 107.1 m. (Key point: Lowest point)

## Related in Transportation Engineering I

- [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 design, 80 kmph, 1 in 40](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-76b-valley-80) — 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
- [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
- [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
- [Minimum sight distance to avoid head-on collision](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-74bh-headon-sd) — Transportation Engineering I, chapter 3

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