---
title: "Valley curve 1/35 and 1/45, 80 kmph"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-70bh-valley-1-35"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-70bh-valley-1-35
site: BCT Engineering
---

# Valley curve 1/35 and 1/45, 80 kmph

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2070 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-70bh-valley-1-35
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Design the length of valley curve with a descending grade of 1/35 and ascending grade of 1/45. The design speed is 80 kmph. Determine the RL of beginning, lowest and end point of curve if the RL of PVI is 212.36 m so as to fulfill both comfort condition and head light sight distance for night visibility. Also determine the apex distance and mid ordinate of the curve. Assume coefficient of friction = 0.35, Rate of change of centrifugal acceleration = 60 cm/sec³.

## Short answer

Valley curve with grades -1/35 and +1/45 at 80 km/h. Headlight criterion governs giving length 140 m. RL of BVC 214.36 m, lowest point 213.24 m, EVC 213.92 m. Apex distance and mid-ordinate are 0.89 m.

## Answer

A descending 1 in 35 grade meeting an ascending 1 in 45 grade gives a **valley curve**.

#### Data
$g_1=-\dfrac{1}{35}=-2.857\%$, $g_2=+\dfrac{1}{45}=2.222\%$, so $N=\dfrac{1}{35}+\dfrac{1}{45}=0.05079$. $V=80$ km/h, $f=0.35$, $t=2.5$ s, $C=0.6$ m/s³, headlight height 0.75 m, beam angle 1° (IRC/NRS).

#### Stopping sight distance

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

#### Length of valley curve
- Comfort: $L=2\sqrt{\dfrac{NV^3}{C}}=2\sqrt{\dfrac{0.05079(22.22)^3}{0.6}}=61.0$ m
- Headlight: $L=\dfrac{NS^2}{1.5+0.035S}=\dfrac{0.05079(127.6)^2}{1.5+0.035(127.6)}=138.6$ m ($L>S$, valid)

Adopt **L = 140 m** (headlight criterion governs and also satisfies comfort).

#### Levels (PVI RL = 212.36 m, half length = 70 m)

$$
RL_{BVC}=212.36+\frac{1}{35}(70)=214.360\ \text{m}
$$


$$
RL_{EVC}=212.36+\frac{1}{45}(70)=213.916\ \text{m}
$$

Lowest point at $x=\dfrac{g_1L}{N}=\dfrac{(140)/35}{0.05079}=78.75$ m from BVC:

$$
RL_{low}=214.360-\frac{78.75}{35}+\frac{0.05079(78.75)^2}{2(140)}=213.235\ \text{m}
$$

#### Apex distance and mid-ordinate
The vertical distance from the PVI (apex) to the curve, which also equals the mid-ordinate of the curve above/below its chord, is

$$
E=\frac{NL}{8}=\frac{0.05079(140)}{8}=0.889\ \text{m}
$$

(The curve lies 0.889 m above the PVI, at mid-length, level 213.249 m.)

**Answer:** L = 140 m; RL of BVC = 214.36 m; lowest point = 213.24 m; RL of EVC = 213.92 m; apex distance = mid-ordinate = 0.89 m.

## Check yourself

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

1. A descending 1/35 meeting ascending 1/45 gives?
   A. Valley curve, N = 1/35 + 1/45 = 0.0508
   B. Summit curve, N = 0.0508
   C. Valley curve, N = 1/35 - 1/45
   D. Summit, N = 0.0079
   Answer: A. Down-then-up is a valley. (Key point: Curve type and N)

2. Which criterion governs?
   A. Comfort, 61 m
   B. Headlight sight distance, 138.6 m (adopt 140 m), over comfort 61 m
   C. Both equal
   D. Neither, L = SSD
   Answer: B. The longest requirement governs. (Key point: Governing length)

3. How is the lowest point located?
   A. x = L/2
   B. x = g2 L/N
   C. x = g1 L/N = 78.75 m from BVC
   D. x = NL/8
   Answer: C. Slope zero. (Key point: Lowest point)

4. What is the apex distance (and mid-ordinate)?
   A. NL/2
   B. NL^2/8
   C. N/8L
   D. E = NL/8 = 0.889 m
   Answer: D. Curve midpoint lies 0.889 m from the PVI. (Key point: Apex distance)

## Related in Transportation Engineering I

- [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
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