---
title: "Vertical curve radius 5000 m, BVC and highest point"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-73bh-vc-5000"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-73bh-vc-5000
site: BCT Engineering
---

# Vertical curve radius 5000 m, BVC and highest point

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2073 Bhadra, 8 marks
- Video: [Solved Example of Vertical Curve (Crest Curve): Complete calculations of vertical parabolic curve](https://www.youtube.com/watch?v=wX8KNsVjoT0) — CIVIL Solved
- 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-73bh-vc-5000
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

An ascending gradient of 2.75% meets with descending gradient of 2.25%. The radius of curve is 5000 m. If the reduced level of the curve at a distance of 60 m from BVC is 312.12 m, find the reduced level of BVC, EVC and highest point of the curve.

## Short answer

Summit curve length L = N R = 0.05 × 5000 = 250 m. BVC RL = 310.83 m, EVC RL = 311.46 m. Highest point is at 137.5 m from BVC with RL = 312.72 m.

## Answer

Ascending 2.75% meeting descending 2.25% gives a **summit curve**. $N=0.0275+0.0225=0.05$.

#### Length of curve

$$
L=NR=0.05(5000)=250\ \text{m}
$$

#### RL of BVC
Parabolic summit curve, $x$ measured from BVC: $y=RL_{BVC}+g_1x-\dfrac{x^2}{2R}$. At $x=60$ m, $y=312.12$:

$$
312.12=RL_{BVC}+0.0275(60)-\frac{60^2}{2(5000)}=RL_{BVC}+1.65-0.36
$$


$$
RL_{BVC}=310.830\ \text{m}
$$

#### RL of EVC

$$
RL_{EVC}=RL_{BVC}+g_1L-\frac{L^2}{2R}=310.830+0.0275(250)-\frac{250^2}{10000}=311.455\ \text{m}
$$

Check through the PVI: $RL_{PVI}=310.830+0.0275(125)=314.268$ m, and $RL_{PVI}-0.0225(125)=311.455$ m. It agrees.

#### Highest point
The highest point is where the slope is zero: $g_1-\dfrac{x}{R}=0$.

$$
x=g_1R=0.0275(5000)=137.5\ \text{m from BVC}
$$


$$
RL_{max}=RL_{BVC}+\frac{g_1^2R}{2}=310.830+\frac{0.0275^2(5000)}{2}=312.721\ \text{m}
$$

**Answer:** L = 250 m; RL of BVC = 310.83 m; RL of EVC = 311.46 m; highest point at 137.5 m from BVC with RL = 312.72 m.

## Check yourself

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

1. How is the length found?
   A. L = NR = 0.05 x 5000 = 250 m
   B. L = N/R
   C. L = R/N = 100000 m
   D. L = 2NR
   Answer: A. N = 0.0275 + 0.0225 = 0.05. (Key point: Curve length)

2. How is RL of BVC obtained from the given point at x = 60 m?
   A. RL_BVC = 312.12 + 60 x g1
   B. Substitute into y = RL_BVC + g1 x - x^2/(2R) and solve for RL_BVC
   C. RL_BVC = 312.12 since it is near BVC
   D. Use RL_BVC = y + x^2/R only
   Answer: B. 312.12 = RL_BVC + 1.65 - 0.36, so RL_BVC = 310.83 m. (Key point: RL of BVC)

3. Where does the highest point lie on a summit curve?
   A. At the PVI
   B. At x = L/2 = 125 m always
   C. Where slope is zero: x = g1 R = 137.5 m from BVC
   D. At x = g2 R
   Answer: C. Slope g1 - x/R = 0 gives x = 0.0275 x 5000 = 137.5 m. (Key point: Highest point)

4. What gives the maximum RL?
   A. RL_BVC + g1 L = 317.7 m
   B. RL_PVI
   C. RL_BVC - g1^2 R/2
   D. RL_BVC + g1^2 R / 2 = 312.72 m
   Answer: D. 310.83 + 0.0275^2 x 5000 / 2 = 312.72 m. (Key point: RL at highest point)

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