---
title: "Vertical curve -3.5% and +0.5%, IP 1300 m"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-neg35-up-05-1300"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-neg35-up-05-1300
site: BCT Engineering
---

# Vertical curve -3.5% and +0.5%, IP 1300 m

Surveying II · Chapter 6: Curves

- Asked in: 2077 Chaitra, 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-neg35-up-05-1300
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of (-)3.5% meets with another grade of 0.5%. The elevation and chainage of point of intersection are 1300 m and 2600 m respectively. Compute the suitable equal tangent vertical curve for full stations elevation assuming that vertical curve should pass through a point of elevation 2601 m at a chainage 1300 m. Take peg interval = 30 m.

## Short answer

Equal-tangent vertical curve length 200 m. BVC at chainage 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m). Lowest point at chainage 2675 m, elevation 1300.438 m. Curve passes through 1301 m at the IP chainage.

## Answer

In the question, the elevation and chainage of the given curve point appear to be interchanged (a chainage of 1300 m would be far away from the IP). The data are therefore taken as: IP at chainage 2600 m with RL 1300 m, and the curve passes through RL 1301 m at the IP chainage (2600 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:

$$
e = 1301 - 1300 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}
$$

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 sag (valley) curve with equal tangent lengths $L/2 = 100$ m.

$$
\begin{aligned}
\text{Ch. of BVC} &= 2600 - 100 = 2500\ \text{m}\\
\text{Ch. of EVC} &= 2600 + 100 = 2700\ \text{m}\\
\text{RL of BVC} &= 1300 - \frac{(-3.5)}{100}\times 100 = 1303.5\ \text{m}\\
\text{RL of EVC} &= 1300 + \frac{(0.5)}{100}\times 100 = 1300.5\ \text{m}
\end{aligned}
$$

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

| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 2500 | 0 | 1303.500 | 0.000 | 1303.500 |
| 2520 | 20 | 1302.800 | +0.040 | 1302.840 |
| 2550 | 50 | 1301.750 | +0.250 | 1302.000 |
| 2580 | 80 | 1300.700 | +0.640 | 1301.340 |
| 2610 | 110 | 1299.650 | +1.210 | 1300.860 |
| 2640 | 140 | 1298.600 | +1.960 | 1300.560 |
| 2670 | 170 | 1297.550 | +2.890 | 1300.440 |
| 2700 | 200 | 1296.500 | +4.000 | 1300.500 |

Check: the last curve RL (1300.5 m) equals the RL of EVC (1300.5 m), and at the IP chainage the curve lies 1 m above the IP, at RL 1301 m.

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

- Chainage = 2500 + 175 = 2675 m
- RL = 1303.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 1300.438 m

**Answer:** $L = 200$ m, BVC at 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m), lowest point at chainage 2675 m (RL 1300.438 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% with curve 1 m above the IP gives
   A. a summit curve with e = 1 m
   B. a sag curve with e = 1 m
   C. a sag curve with e = 4 m
   D. a summit curve with e = 4 m
   Answer: B. A falling then rising grade is a sag curve, and the curve is above the IP by e = 1 m. (Key point: Curve type and e)

2. What is L using e = L|g2 - g1|/800 (grades in %)?
   A. 100 m
   B. 400 m
   C. 200 m
   D. 800 m
   Answer: C. L = 800 x 1/4 = 200 m. (Key point: Length)

3. Where is the lowest point measured from BVC?
   A. x = 100 m, chainage 2600 m
   B. x = 125 m, chainage 2625 m
   C. x = 200 m, chainage 2700 m
   D. x = -g1 L/(g2 - g1) = 175 m, chainage 2675 m
   Answer: D. Setting the grade to zero gives x = 3.5 x 200/4 = 175 m. (Key point: Lowest 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 +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 +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×

## Related in Surveying II

- [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×
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