---
title: "Vertical curve +1.25% and +4.75%, 1518.30 m"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-125-475-151760-2030"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-125-475-151760-2030
site: BCT Engineering
---

# Vertical curve +1.25% and +4.75%, 1518.30 m

Surveying II · Chapter 6: Curves

- Asked in: 2080 Chaitra, 6 marks
- Video: [Stations and Elevations of PVC, PVT and High point of Vertical Curve|Vertical Curve Fundamentals](https://www.youtube.com/watch?v=KLN9Oe0EJc4) — Master Civil Engineering
- 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-125-475-151760-2030
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of 1.25% meets another grade of 4.75%. The elevation and chainage of intersection point are 1517.60 and 2+030 km. Field condition requires that the vertical curve should pass through a point of elevation 1518.30 m at chainage (2+030) km. Compute a suitable equal tangent vertical curve and full station elevations. Use parabolic equation. Take peg interval 30 m.

## Short answer

Mid-ordinate e = 0.7 m gives L = 160 m. BVC at 1950 m (RL 1516.6 m), EVC at 2110 m (RL 1521.4 m). Curve elevations from parabolic equation; at IP chainage 2030 m, RL = 1518.3 m as required.

## Answer

Chainage of IP $= 2+030$ km $= 2030$ m.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

#### Step 1: Length of the curve
The curve is a sag curve, so the curve lies above the IP:

$$
e = 1518.3 - 1517.6 = 0.7\ \text{m}\ \Rightarrow\ |e| = 0.7\ \text{m}
$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$
L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.7}{|4.75-(1.25)|} = \frac{560}{3.5} = 160\ \text{m}
$$

#### Step 2: BVC, EVC and their levels
The curve is a sag (valley) curve with equal tangent lengths $L/2 = 80$ m.

$$
\begin{aligned}
\text{Ch. of BVC} &= 2030 - 80 = 1950\ \text{m}\\
\text{Ch. of EVC} &= 2030 + 80 = 2110\ \text{m}\\
\text{RL of BVC} &= 1517.6 - \frac{(1.25)}{100}\times 80 = 1516.6\ \text{m}\\
\text{RL of EVC} &= 1517.6 + \frac{(4.75)}{100}\times 80 = 1521.4\ \text{m}
\end{aligned}
$$

#### 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 $g_1 = 1.25\%$, $g_2 = 4.75\%$, $L = 160$ m, so the offset is $y = \frac{(3.5)x^2}{200\times 160}$ and pegs are taken at multiples of 30 m.

| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 1950 | 0 | 1516.600 | 0.000 | 1516.600 |
| 1980 | 30 | 1516.975 | +0.098 | 1517.073 |
| 2010 | 60 | 1517.350 | +0.394 | 1517.744 |
| 2040 | 90 | 1517.725 | +0.886 | 1518.611 |
| 2070 | 120 | 1518.100 | +1.575 | 1519.675 |
| 2100 | 150 | 1518.475 | +2.461 | 1520.936 |
| 2110 | 160 | 1518.600 | +2.800 | 1521.400 |

Check: the last curve RL (1521.4 m) equals the RL of EVC (1521.4 m), and the curve at the IP chainage lies 0.7 m above the IP, at RL 1518.3 m.

#### Step 4: Highest and lowest points
Both grades are rising, so the grade never becomes zero within the curve and there is no turning point; the lowest level is at the BVC (RL 1516.6 m) and the highest level is at the EVC (RL 1521.4 m).

**Answer:** $L = 160$ m, BVC at 1950 m (RL 1516.6 m), EVC at 2110 m (RL 1521.4 m); curve levels as tabulated.

## Check yourself

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

1. With e = 0.7 m and change of grade 3.5%, L = 800e/|g2 - g1| =
   A. 280 m
   B. 80 m
   C. 320 m
   D. 160 m
   Answer: D. 800 x 0.7/3.5 = 160 m. (Key point: Length)

2. With IP at 2030 m, L = 160 m, BVC and EVC are
   A. 1950 m and 2110 m
   B. 1870 m and 2190 m
   C. 2030 m and 2190 m
   D. 1950 m and 2030 m
   Answer: A. Half-lengths of 80 m on each side. (Key point: BVC and EVC)

3. RL of BVC with IP at 1517.6 and g1 = +1.25% is
   A. 1515.6 m
   B. 1516.6 m
   C. 1517.6 m
   D. 1518.6 m
   Answer: B. BVC is 80 m before the IP on a rising grade, so RL = 1517.6 - 0.0125 x 80 = 1516.6 m. (Key point: RL of BVC)

4. Why is there no highest or lowest turning point within the curve?
   A. The offset is zero
   B. The curve is straight
   C. Both grades are rising so the slope never becomes zero
   D. L is too short
   Answer: C. Grade changes from +1.25% to +4.75% and stays positive, so extremes lie at the ends: lowest at BVC and highest at EVC. (Key point: Turning point)

## Close variants set in other exams

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