---
title: "Vertical curve 180 m, -3.5% then +4.5%, IP 2450"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-35down-45up-ip-2450"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-35down-45up-ip-2450
site: BCT Engineering
---

# Vertical curve 180 m, -3.5% then +4.5%, IP 2450

Surveying II · Chapter 6: Curves

- Asked in: 2066 Magh (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-35down-45up-ip-2450
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A down grade of 3.5% is followed by an upgrade of 4.5%. The reduced level and chainage of IP are 900.00 m and 2450.00 m respectively. A vertical parabolic curve 180 m long is to be introduced to connect the two grades. The pegs are to be fixed at 20 m intervals. Calculate the RLs of curve points including lowest point.

## Short answer

The 180 m parabolic sag curve has BVC at chainage 2360 m RL 903.150 m and EVC at 2540 m RL 904.050 m. The lowest point occurs at 78.75 m from BVC (chainage 2438.75 m) with RL 901.772 m. RLs at 20 m intervals are obtained by adding grade line elevation and offset.

## Answer

A parabolic vertical curve of the given length is set out by the tangent-offset method.

#### Step 1: Given data
$g_1 = -3.5\%$, $g_2 = 4.5\%$, $L = 180$ m, change of grade $= 8\%$. Mid-ordinate $e = \dfrac{L|g_2-g_1|}{800} = 1.8$ m.

#### Step 2: BVC, EVC and their levels
The curve is a sag (valley) curve with equal tangent lengths $L/2 = 90$ m.

$$
\begin{aligned}
\text{Ch. of BVC} &= 2450 - 90 = 2360\ \text{m}\\
\text{Ch. of EVC} &= 2450 + 90 = 2540\ \text{m}\\
\text{RL of BVC} &= 900 - \frac{(-3.5)}{100}\times 90 = 903.15\ \text{m}\\
\text{RL of EVC} &= 900 + \frac{(4.5)}{100}\times 90 = 904.05\ \text{m}
\end{aligned}
$$

#### Step 3: RL of curve points
Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$
y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}
$$

with $g_1 = -3.5\%$, $g_2 = 4.5\%$, $L = 180$ m, so the offset is $y = \frac{(8)x^2}{200\times 180}$ and pegs are taken at multiples of 20 m.

| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 2360 | 0 | 903.150 | 0.000 | 903.150 |
| 2380 | 20 | 902.450 | +0.089 | 902.539 |
| 2400 | 40 | 901.750 | +0.356 | 902.106 |
| 2420 | 60 | 901.050 | +0.800 | 901.850 |
| 2440 | 80 | 900.350 | +1.422 | 901.772 |
| 2460 | 100 | 899.650 | +2.222 | 901.872 |
| 2480 | 120 | 898.950 | +3.200 | 902.150 |
| 2500 | 140 | 898.250 | +4.356 | 902.606 |
| 2520 | 160 | 897.550 | +5.689 | 903.239 |
| 2540 | 180 | 896.850 | +7.200 | 904.050 |

Check: the last curve RL (904.05 m) equals the RL of EVC (904.05 m), and the curve at the IP chainage lies 1.8 m above the IP, at RL 901.8 m.

#### Step 4: Lowest point
At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 180}{8} = 78.75$ m from BVC.

- Chainage = 2360 + 78.75 = 2438.75 m
- RL = 903.15 + (-3.5/100)(78.75) + (8)(78.75)²/(200 × 180) = 901.772 m

**Answer:** $L = 180$ m, BVC at 2360 m (RL 903.15 m), EVC at 2540 m (RL 904.05 m); curve levels as tabulated.

## Check yourself

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

1. A 3.5% fall followed by 4.5% rise makes which curve?
   A. Transition
   B. Sag (valley)
   C. Compound
   D. Summit
   Answer: B. A falling grade followed by a rising grade makes a sag. (Key point: Curve type)

2. For IP at 2450 and L = 180 m, what are the BVC and EVC chainages?
   A. 2360 and 2540
   B. 2400 and 2500
   C. 2270 and 2630
   D. 2450 and 2630
   Answer: A. IP -/+ L/2 = 2450 -/+ 90. (Key point: BVC and EVC)

3. What is the mid-ordinate and where is the curve at the IP chainage?
   A. e = 1.8 m, curve 1.8 m above IP, RL 901.8 m
   B. e = 1.8 m, curve below IP
   C. e = 0.9 m
   D. e = 8 m
   Answer: A. e = 180 x 8/800 = 1.8 m, added for a sag. (Key point: Mid-ordinate)

4. Where is the lowest point of the curve?
   A. x = 90 m, chainage 2450 m
   B. x = 101.25 m
   C. x = 78.75 m from BVC, chainage 2438.75 m
   D. x = 100 m
   Answer: C. x = 3.5 x 180/8 = 78.75 m. (Key point: Lowest point)

## Close variants set in other exams

- [Vertical curve 180 m, -4.5% then +3.5%, IP 450](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-45down-35up-ip-450) — Surveying II, chapter 6, asked 2×

## Related in Surveying II

- [Vertical curve -2% and +3%, IP 2600, 0.5% per 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-2down-3up-ip-2600) — Surveying II, chapter 6
- [Vertical curve +0.5% and -0.7%, IP RL 350.750](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-pos05-neg07-ip-500) — Surveying II, chapter 6
- [Vertical curve +3.5% and -2.75%, 30 m pegs](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35-down-275-through) — Surveying II, chapter 6, asked 7×
- [Vertical curve -0.7% and +0.5%, IP RL 650](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg07-pos05-rl-ip) — Surveying II, chapter 6
- [Vertical curve -0.5% and +0.7%, IP RL 500](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg05-pos07-ip-1000) — Surveying II, chapter 6
- [Vertical curve -1% and +0.5% by parabolic method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-parabolic-neg1-pos05-ip) — Surveying II, chapter 6, asked 2×