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

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

Surveying II · Chapter 6: Curves

- Asked in: 2078 Baisakh, 6 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-05up-35down-1266-at
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of 0.5% meets another grade of -3.5%. The elevation and chainage of IP are 1267 m and 780 m respectively. Field condition requires that the vertical curve should pass through a point of elevation 1266 m at a chainage 780 m. Compute a suitable equal tangent vertical curve including full stations elevation, take peg interval = 30 m.

## Short answer

Equal-tangent vertical curve length 200 m. Curve passes through elevation 1266 m at chainage 780 m. Highest point at chainage 705.00 m, elevation 1266.562 m. PVC at 680 m (1266.500 m), PVT at 880 m (1263.500 m).

## Answer

Data: $g_1 = +0.5\% = +0.005$, $g_2 = -3.5\% = -0.035$ (a summit curve), IP at chainage 780.00 m, RL 1267.00 m. The curve passes through RL 1266.00 m at chainage 780 m (below the IP, as it must for a summit curve).

#### Length of the curve

$e = 1267.00 - 1266.00 = 1.00$ m, and for an equal tangent parabolic curve $e = \dfrac{(g_1 - g_2)L}{8}$:

$$
L = \frac{8e}{g_1 - g_2} = \frac{8\times 1.00}{0.005 + 0.035} = 200.00\ \text{m}
$$

#### Elements

- PVC chainage $= 780 - 100 = 680.00$ m; PVT chainage $= 880.00$ m
- RL of PVC $= 1267.00 - 0.005\times 100 = 1266.500$ m; RL of PVT $= 1267.00 - 0.035\times 100 = 1263.500$ m
- $r = \dfrac{g_2 - g_1}{L} = -0.000200$ per m

#### Levels at full stations (30 m pegs)

$$
RL_x = RL_{PVC} + g_1x + \frac{r}{2}x^2
$$

| Chainage (m) | x from PVC (m) | Tangent level (m) | Offset $y=\frac{r}{2}x^2$ (m) | Curve level (m) |
|---|---|---|---|---|
| 680.00 | 0.00 | 1266.500 | 0.000 | 1266.500 |
| 690.00 | 10.00 | 1266.550 | -0.010 | 1266.540 |
| 720.00 | 40.00 | 1266.700 | -0.160 | 1266.540 |
| 750.00 | 70.00 | 1266.850 | -0.490 | 1266.360 |
| 780.00 | 100.00 | 1267.000 | -1.000 | 1266.000 |
| 810.00 | 130.00 | 1267.150 | -1.690 | 1265.460 |
| 840.00 | 160.00 | 1267.300 | -2.560 | 1264.740 |
| 870.00 | 190.00 | 1267.450 | -3.610 | 1263.840 |
| 880.00 | 200.00 | 1267.500 | -4.000 | 1263.500 |

#### Highest point

$$
x = \frac{-g_1}{r} = \frac{0.005}{0.000200} = 25.00\ \text{m}
$$

Chainage $= 680.00 + 25.00 = 705.00$ m and RL $= 1266.562$ m.

Check: level at chainage 780 m (x = 100 m) $= 1266.000$ m = 1266.000 m as required.

**Answer: L = 200 m; highest point at chainage 705.00 m, RL 1266.562 m.**

## Check yourself

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

1. Grades +0.5% then -3.5% with the curve passing 1 m below the IP. The curve is a
   A. summit curve, lying below the IP
   B. sag curve, lying above the IP
   C. sag curve, lying below the IP
   D. summit curve, lying above the IP
   Answer: A. The grade changes from rising to falling, so it is a summit curve and lies below the IP. (Key point: Type of curve)

2. What is L when e = 1.00 m and the grade change is 4%?
   A. 100 m
   B. L = 8e/(g1 - g2) = 8 x 1/0.04 = 200 m
   C. 400 m
   D. 50 m
   Answer: B. Using e = (g1 - g2)L/8 gives L = 200 m. (Key point: Length of curve)

3. How is RL of PVC found?
   A. RL of IP plus g1 x L/2 = 1267.5 m
   B. RL of IP minus g1 x L = 1266.0 m
   C. RL of IP minus g1 x L/2: 1267 - 0.005 x 100 = 1266.5 m
   D. RL of IP = 1267 m
   Answer: C. Going back from the IP along the first grade (rising), the PVC is lower by 0.005 x 100 m. (Key point: RL of PVC)

4. Where is the highest point of the curve?
   A. At the IP, chainage 780 m
   B. At x = 100 m from PVC
   C. At the PVT, chainage 880 m
   D. x = -g1/r = 0.005/0.0002 = 25 m from PVC, chainage 705 m
   Answer: D. The slope is zero where g1 + r x = 0, which gives x = 25 m, RL 1266.562 m. (Key point: Highest point)

## Close variants set in other exams

- [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 +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 +3.5% and -0.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-05-1266-780) — Surveying II, chapter 6, asked 8×
- [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 -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×
- [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 +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 +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×

## 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 -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
- [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 180 m, -3.5% then +4.5%, IP 2450](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-45up-ip-2450) — Surveying II, chapter 6, asked 2×

## 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