---
title: "Vertical curve +2% and +5.5%, 60 kmph"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-79c-vc-2-55"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-79c-vc-2-55
site: BCT Engineering
---

# Vertical curve +2% and +5.5%, 60 kmph

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2079 Chaitra, 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-79c-vc-2-55
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A vertical curve connects an ascending gradient of +2% grade with another ascending gradient of +5.5% grade on a highway of design speed 60 kmph at an elevation of 1411.6 m. Design the length of the vertical curve and determine the elevations at the beginning and end point of the curve. Use coefficient of longitudinal friction of 0.35.

## Short answer

SSD=82.2 m; headlight criterion with L<S gives L=40 m. BVC elevation =1411.20 m, EVC elevation =1412.70 m.

## Answer

The grade rises from +2% to the steeper +5.5%, so the curve is a **valley (sag) curve**. $N=g_2-g_1=0.055-0.02=0.035$.

#### Stopping sight distance ($V=60$ km/h, $f=0.35$, $t=2.5$ s)

$$
SSD=0.278(60)(2.5)+\frac{60^2}{254(0.35)}=82.2\ \text{m}
$$

#### Length of curve (headlight beam, $h_1=0.75$ m, $\alpha=1^\circ$)
- Trial with $L>S$: $L=\dfrac{NS^2}{1.5+0.035S}=54.0$ m, which is less than $S=82.2$ m, so this case does not apply.
- Case $L<S$:

$$
L=2S-\frac{1.5+0.035S}{N}=2(82.2)-\frac{1.5+0.035(82.2)}{0.035}=39.3\ \text{m}
$$

- Comfort ($C=0.6$ m/s³): $L=2\sqrt{NV^3/C}=32.9$ m.

Adopt **L = 40 m** (greater of the two).

#### Elevations
PVI elevation = 1411.60 m, half length = 20 m.

$$
RL_{BVC}=1411.60-0.02(20)=1411.20\ \text{m}\qquad RL_{EVC}=1411.60+0.055(20)=1412.70\ \text{m}
$$

**Answer:** L = 40 m; elevation of beginning point = 1411.20 m; elevation of end point = 1412.70 m.

## Check yourself

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

1. Grade rising from +2% to +5.5% is which curve, and N?
   A. Summit, N = 0.075
   B. Valley, N = 0.035
   C. Valley, N = 0.075
   D. Summit, N = 0.035
   Answer: B. The upgrade becomes steeper so it is a valley curve with N = 0.055 - 0.02 = 0.035. (Key point: Type of curve)

2. Trial L > S gave 54.0 m, less than S = 82.2 m. What follows?
   A. Accept 54 m
   B. Use the case L < S: L = 2S - (1.5 + 0.035S)/N = 39.3 m
   C. Use comfort only
   D. Use S = L
   Answer: B. The result contradicts the L > S assumption, so the L < S formula is used. (Key point: S versus L)

3. Headlight gave 39.3 m, comfort 32.9 m. Adopted L is
   A. 32.9 m
   B. 40 m, the greater rounded
   C. 82.2 m
   D. 36 m
   Answer: B. The greater value governs, rounded to 40 m. (Key point: Adopted length)

4. With PVI at 1411.6 m and L = 40 m, the BVC and EVC elevations are
   A. 1411.20 m and 1412.70 m
   B. 1412.00 m and 1411.10 m
   C. 1411.20 m and 1411.60 m
   D. 1410.40 m and 1412.70 m
   Answer: A. BVC = 1411.6 - 0.02 x 20 = 1411.20; EVC = 1411.6 + 0.055 x 20 = 1412.70 m. (Key point: Elevation of BVC and EVC)

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