---
title: "Vertical curve -3.5% and -0.5%, 1268 m"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-neg35-neg05-1268-lowest"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-neg35-neg05-1268-lowest
site: BCT Engineering
---

# Vertical curve -3.5% and -0.5%, 1268 m

Surveying II · Chapter 6: Curves

- Asked in: 2076 Baisakh, 6 marks
- Video: [Design of Vertical Curves - Highway Geomteric Design](https://www.youtube.com/watch?v=W7ha0Jaoq9s) — Lecture Factory
- 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-neg35-neg05-1268-lowest
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of -3.5% meets another grade of -0.5%. The elevation of intersection is 1267 m & chainage is (1+780) km. Field coordinates require that the vertical curve should pass through a point of elevation 1268 m at chainage (1+780) km. Compute a suitable equal tangent vertical curve full station elevations including lowest point elevation. Use parabolic method.

## Short answer

Using mid-ordinate e = 1 m, L = 800×1/|−0.5−(−3.5)| = 266.67 m. Equal-tangent sag curve: BVC at 1646.67 m (1271.667 m), EVC at 1913.33 m (1266.333 m). No turning point; highest BVC, lowest EVC.

## Answer

Chainage of IP $= 1+780$ km $= 1780$ 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:

$$
e = 1268 - 1267 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}
$$

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

$$
L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|-0.5-(-3.5)|} = \frac{800}{3} = 266.67\ \text{m}
$$

#### Step 2: BVC, EVC and their levels
The curve is a sag (valley) curve with equal tangent lengths $L/2 = 133.33$ m.

$$
\begin{aligned}
\text{Ch. of BVC} &= 1780 - 133.33 = 1646.67\ \text{m}\\
\text{Ch. of EVC} &= 1780 + 133.33 = 1913.33\ \text{m}\\
\text{RL of BVC} &= 1267 - \frac{(-3.5)}{100}\times 133.33 = 1271.667\ \text{m}\\
\text{RL of EVC} &= 1267 + \frac{(-0.5)}{100}\times 133.33 = 1266.333\ \text{m}
\end{aligned}
$$

#### 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 $g_1 = -3.5\%$, $g_2 = -0.5\%$, $L = 266.67$ m, so the offset is $y = \frac{(3)x^2}{200\times 266.67}$ and pegs are taken at multiples of 30 m.

| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 1646.67 | 0 | 1271.667 | 0.000 | 1271.667 |
| 1650 | 3.33 | 1271.550 | +0.001 | 1271.551 |
| 1680 | 33.33 | 1270.500 | +0.062 | 1270.562 |
| 1710 | 63.33 | 1269.450 | +0.226 | 1269.676 |
| 1740 | 93.33 | 1268.400 | +0.490 | 1268.890 |
| 1770 | 123.33 | 1267.350 | +0.856 | 1268.206 |
| 1800 | 153.33 | 1266.300 | +1.322 | 1267.623 |
| 1830 | 183.33 | 1265.250 | +1.891 | 1267.141 |
| 1860 | 213.33 | 1264.200 | +2.560 | 1266.760 |
| 1890 | 243.33 | 1263.150 | +3.331 | 1266.481 |
| 1913.33 | 266.67 | 1262.333 | +4.000 | 1266.333 |

Check: the last curve RL (1266.333 m) equals the RL of EVC (1266.333 m), and the curve at the IP chainage lies 1 m above the IP, at RL 1268 m.

#### Step 4: Highest and lowest points
Both grades are falling, so the grade never becomes zero within the curve and there is no turning point; the highest level is at the BVC (RL 1271.667 m) and the lowest level is at the EVC (RL 1266.333 m).

**Answer:** $L = 266.67$ m, BVC at 1646.67 m (RL 1271.667 m), EVC at 1913.33 m (RL 1266.333 m); curve levels as tabulated.

## Check yourself

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

1. Grades -3.5% and -0.5%, e = 1 m. L =
   A. 400 m
   B. 800 m
   C. 133.33 m
   D. 266.67 m
   Answer: D. L = 800 x 1/3 = 266.67 m. (Key point: Length)

2. With IP at 1780 m the BVC is at
   A. 1646.67 m
   B. 1513.33 m
   C. 1913.33 m
   D. 1780 m
   Answer: A. 1780 - 133.33 = 1646.67 m. (Key point: BVC chainage)

3. The curve is a sag curve because
   A. L is large
   B. it lies above the IP (1268 vs 1267)
   C. it is below the IP
   D. the grades are negative
   Answer: B. Curve above the IP means a sag curve (grade increases from -3.5% to -0.5%). (Key point: Sag type)

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