---
title: "Vertical curve chainage and elevations, R 8000 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-69bh-vc-chainage"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-69bh-vc-chainage
site: BCT Engineering
---

# Vertical curve chainage and elevations, R 8000 m

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2069 Bhadra, 8 marks
- Video: [Vertical Curve Numerical|Numerical from vertical Curve|calculate RL of various points of curve|](https://www.youtube.com/watch?v=7vZEF919Guk) — Hamro Engineering
- 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-69bh-vc-chainage
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

An ascending gradient of 3.75% meets with descending gradient of 3.25%. Calculate the chainage and elevation of beginning of the curve, end of the curve, highest point of the curve and 90 m left from the point of vertical intersection if the chainage and elevation of PVI are 1+225.00 and 875.62 m respectively. The radius of curve provided is 8000 m.

## Short answer

Length L = N×R = 0.07×8000 = 560 m. BVC chainage 0+945, RL 865.12 m. EVC chainage 1+505, RL 866.52 m. Highest point at 1+245, RL 870.75 m. At 90 m left of PVI, RL = 869.99 m.

## Answer

An ascending 3.75% grade followed by a descending 3.25% grade gives a **summit curve**; $N=0.0375+0.0325=0.07$.

#### Length and chainages

$$
L=NR=0.07(8000)=560\ \text{m},\qquad \frac{L}{2}=280\ \text{m}
$$

PVI chainage = 1+225.00 = 1225 m.
- Beginning of curve (BVC) = 1225 − 280 = **945 m (0+945)**
- End of curve (EVC) = 1225 + 280 = **1505 m (1+505)**

#### Levels of BVC and EVC

$$
RL_{BVC}=875.62-0.0375(280)=865.120\ \text{m},\qquad RL_{EVC}=875.62-0.0325(280)=866.520\ \text{m}
$$

#### Highest point

$$
x=g_1R=0.0375(8000)=300\ \text{m from BVC}
$$

Chainage $=945+300=\mathbf{1245}$ m (1+245).

$$
RL_{max}=RL_{BVC}+\frac{g_1^2R}{2}=865.120+\frac{0.0375^2(8000)}{2}=870.745\ \text{m}
$$

#### 90 m left of PVI
Chainage $=1225-90=1135$ m, distance from BVC $x=280-90=190$ m:

$$
y=RL_{BVC}+g_1x-\frac{x^2}{2R}=865.120+0.0375(190)-\frac{190^2}{2(8000)}=869.989\ \text{m}
$$

**Answer:** BVC at 0+945 (RL 865.12 m); EVC at 1+505 (RL 866.52 m); highest point at 1+245 (RL 870.75 m); 90 m left of PVI RL = 869.99 m.

## Check yourself

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

1. How is the curve length found for a given radius R of the vertical curve?
   A. L = R / N
   B. L = N R / 2
   C. L = N R = 560 m
   D. L = g1 R = 300 m
   Answer: C. For a parabola with radius R, L = N R with N = 0.07. (Key point: Curve length)

2. PVI is at 1+225. What are BVC and EVC chainages?
   A. BVC 0+945, EVC 1+505
   B. BVC 1+225, EVC 1+785
   C. BVC 0+665, EVC 1+785
   D. BVC 0+945, EVC 1+785
   Answer: A. BVC = PVI - L/2 = 1225 - 280 and EVC = PVI + L/2 = 1225 + 280. (Key point: BVC chainage)

3. Where is the highest point of the curve?
   A. x = g2 R = 260 m from BVC
   B. x = g1 R = 300 m from BVC (1+245)
   C. At the PVI, 1+225
   D. x = L = 560 m from BVC
   Answer: B. At the summit the grade is zero, so x = g1 R. (Key point: Highest point)

4. How is the RL 90 m left of the PVI found?
   A. Use x = 90 m from BVC
   B. Use x = 280 + 90 m from BVC
   C. RL_PVI - g1 x with no curve correction
   D. Use x = 280 - 90 = 190 m from BVC: RL = RL_BVC + g1 x - x^2/(2R)
   Answer: D. The point is 190 m from BVC and the curve falls below the tangent by x^2/(2R). (Key point: Level at 90 m left)

## Related in Transportation Engineering I

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