---
title: "Vertical curve +3.5% and -0.5%, 1266 m at 780"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-35down-05-1266-780"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-35down-05-1266-780
site: BCT Engineering
---

# Vertical curve +3.5% and -0.5%, 1266 m at 780

Surveying II · Chapter 6: Curves

- Asked in: 2074 Bhadra, 5 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-05-1266-780
- 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 and chainage of intersection pt are 1267.00 m and 780.00 m respectively. Field condition requires that vertical curve should pass through a point of elevation 1266.00 m at chainage 780.00 m. Compute a suitable equal tangent vertical curve and full station elevation including highest point. Take peg interval = 30 m.

## Short answer

An equal-tangent parabolic curve is used; mid-ordinate e = 1 m gives length L = 800×1/|−0.5−3.5| = 200 m. BVC at chainage 680 m (RL 1263.5 m), EVC at 880 m (RL 1266.5 m). Highest point at chainage 855 m, RL 1266.56 m.

## Answer

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 summit curve, so the curve lies below the IP:

$$
e = 1266 - 1267 = -1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}
$$

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

$$
L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|-0.5-(3.5)|} = \frac{800}{4} = 200\ \text{m}
$$

#### Step 2: BVC, EVC and their levels
The curve is a summit (crest) curve with equal tangent lengths $L/2 = 100$ m.

$$
\begin{aligned}
\text{Ch. of BVC} &= 780 - 100 = 680\ \text{m}\\
\text{Ch. of EVC} &= 780 + 100 = 880\ \text{m}\\
\text{RL of BVC} &= 1267 - \frac{(3.5)}{100}\times 100 = 1263.5\ \text{m}\\
\text{RL of EVC} &= 1267 + \frac{(-0.5)}{100}\times 100 = 1266.5\ \text{m}
\end{aligned}
$$

#### 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 $g_1 = 3.5\%$, $g_2 = -0.5\%$, $L = 200$ m, so the offset is $y = \frac{(-4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 680 | 0 | 1263.500 | 0.000 | 1263.500 |
| 690 | 10 | 1263.850 | -0.010 | 1263.840 |
| 720 | 40 | 1264.900 | -0.160 | 1264.740 |
| 750 | 70 | 1265.950 | -0.490 | 1265.460 |
| 780 | 100 | 1267.000 | -1.000 | 1266.000 |
| 810 | 130 | 1268.050 | -1.690 | 1266.360 |
| 840 | 160 | 1269.100 | -2.560 | 1266.540 |
| 870 | 190 | 1270.150 | -3.610 | 1266.540 |
| 880 | 200 | 1270.500 | -4.000 | 1266.500 |

Check: the last curve RL (1266.5 m) equals the RL of EVC (1266.5 m), and the curve at the IP chainage lies 1 m below the IP, at RL 1266 m.

#### Step 4: Highest point
At the highest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{-3.5\times 200}{-4} = 175$ m from BVC.

- Chainage = 680 + 175 = 855 m
- RL = 1263.5 + (3.5/100)(175) + (-4)(175)²/(200 × 200) = 1266.562 m

**Answer:** $L = 200$ m, BVC at 680 m (RL 1263.5 m), EVC at 880 m (RL 1266.5 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% then -0.5% with curve below the IP. This is a
   A. summit curve
   B. sag curve
   C. valley curve
   D. reverse curve
   Answer: A. A rising then falling grade is a summit curve, below the IP. (Key point: Curve type)

2. Using e = L|g2 - g1|/800 with e = 1 m, L is
   A. 400 m
   B. 200 m
   C. 100 m
   D. 50 m
   Answer: B. L = 800 x 1/4 = 200 m. (Key point: Length)

3. RL of EVC is found as
   A. 1267 + 0.5 = 1267.5 m
   B. 1263.5 m
   C. RL IP + g2 x L/2/100 = 1267 - 0.5 = 1266.5 m
   D. 1270.5 m
   Answer: C. EVC lies L/2 beyond the IP along the falling grade. (Key point: RL of EVC)

4. Highest point is at
   A. x = 100 m, chainage 780 m
   B. x = 200 m, chainage 880 m
   C. x = 125 m, chainage 805 m
   D. x = 175 m from BVC, chainage 855 m, RL 1266.562 m
   Answer: D. x = -g1 L/(g2 - g1) = 3.5 x 200/4 = 175 m. (Key point: Highest point)

## Close variants set in other exams

- [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 -2.5% and +3.5%, 1268.50 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-25down-35up-12685) — Surveying II, chapter 6, asked 9×
- [Vertical curve +0.5% and -3.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-05up-35down-1266-at) — Surveying II, chapter 6, asked 9×
- [Vertical curve 5% and 3%, 1474.50 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-5-3-ip-1475) — Surveying II, chapter 6, asked 7×
- [Vertical curve +1.25% and +4.75%, 1518.30 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-151760-2030) — Surveying II, chapter 6, asked 7×
- [Vertical curve (-)3% and (-)35%, IP 2477 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg3-neg35-ip-2477) — Surveying II, chapter 6, asked 6×
- [Vertical curve -3.5% and +0.5%, IP 1300 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-up-05-1300) — Surveying II, chapter 6, asked 5×

## Related in Surveying II

- [Vertical curve +1.25% and +4.75%, 1267.70 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-126770) — Surveying II, chapter 6, asked 3×
- [Vertical curve -3.5% and +5%, 1268 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-5up-1268-at) — Surveying II, chapter 6, asked 3×
- [Vertical curve, -3.5% and +0.5%, 3268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-parabolic-why-3267-1780) — Surveying II, chapter 6, asked 3×
- [Vertical curve -3.5% and -0.5%, 1268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-neg05-1268-lowest) — Surveying II, chapter 6

## Related in other subjects

- [Vertical curve +4% and -3%, through given point](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-vc-4-3) — Transportation Engineering I, chapter 3
- [Vertical curve chainage and elevations, R 8000 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-69bh-vc-chainage) — Transportation Engineering I, chapter 3
- [Vertical curve, -3% and 1.5% at chainage 2000](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79a-vc-2000) — Transportation Engineering I, chapter 3