---
title: "Vertical curve -2.5% and +3.5%, 1268.50 m"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-25down-35up-12685"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-25down-35up-12685
site: BCT Engineering
---

# Vertical curve -2.5% and +3.5%, 1268.50 m

Surveying II · Chapter 6: Curves

- Asked in: 2078 Poush
- Video: [Highway engineering vertical valley curve numerical, transportation engineering numerical](https://www.youtube.com/watch?v=XDNka2YRB6k) — CTEVT PROBLEM
- 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-25down-35up-12685
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of (-) 2.5% 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 1268.50 m at a chainage 780.0 m. Compute a suitable equal tangent vertical curve including full stations elevation, take peg interval = 30 m.

## Short answer

The 200 m equal-tangent parabolic sag curve has its lowest point at chainage 763.33 m, RL 1268.458 m. Elevations at 30 m stations from PVC (RL 1269.500 m) are computed; the curve passes through IP at chainage 780 m at RL 1268.500 m.

## Answer

Data: $g_1 = -2.5\% = -0.025$, $g_2 = +3.5\% = +0.035$ (a valley or sag curve), IP at chainage 780.00 m, RL 1267.00 m. The curve passes through RL 1268.50 m at chainage 780 m (above the IP, as it must for a sag curve).

#### Length of the curve

$e = 1268.50 - 1267.00 = 1.50$ m, and for an equal tangent parabolic curve $e = \dfrac{(g_2 - g_1)L}{8}$:

$$
L = \frac{8e}{g_2 - g_1} = \frac{8\times 1.50}{0.035 + 0.025} = 200.00\ \text{m}
$$

#### Elements

- PVC chainage $= 780 - 100 = 680.00$ m; PVT chainage $= 880.00$ m
- RL of PVC $= 1267.00 + 0.025\times 100 = 1269.500$ m; RL of PVT $= 1267.00 + 0.035\times 100 = 1270.500$ m
- $r = \dfrac{g_2 - g_1}{L} = 0.000300$ 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 | 1269.500 | 0.000 | 1269.500 |
| 690.00 | 10.00 | 1269.250 | +0.015 | 1269.265 |
| 720.00 | 40.00 | 1268.500 | +0.240 | 1268.740 |
| 750.00 | 70.00 | 1267.750 | +0.735 | 1268.485 |
| 780.00 | 100.00 | 1267.000 | +1.500 | 1268.500 |
| 810.00 | 130.00 | 1266.250 | +2.535 | 1268.785 |
| 840.00 | 160.00 | 1265.500 | +3.840 | 1269.340 |
| 870.00 | 190.00 | 1264.750 | +5.415 | 1270.165 |
| 880.00 | 200.00 | 1264.500 | +6.000 | 1270.500 |

#### Lowest point

$$
x = \frac{-g_1}{r} = \frac{0.025}{0.000300} = 83.33\ \text{m}
$$

Chainage $= 680.00 + 83.33 = 763.33$ m and RL $= 1268.458$ m.

Check: level at chainage 780 m (x = 100 m) $= 1268.500$ m = 1268.500 m as required.

**Answer: L = 200 m; lowest point at chainage 763.33 m, RL 1268.458 m.**

## Check yourself

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

1. Grades -2.5% then +3.5% meet at the IP. What type of vertical curve is this?
   A. A sag (valley) curve, so the curve lies above the IP
   B. A summit curve, so the curve lies below the IP
   C. A sag curve, so the curve lies below the IP
   D. A summit curve, so the curve lies above the IP
   Answer: A. The grade changes from falling to rising, so it is a sag curve and the curve passes above the IP, which is why RL 1268.50 m is above 1267 m. (Key point: Type of curve)

2. What does the 1.50 m (1268.50 - 1267.00) represent?
   A. The rise of the PVT above the PVC
   B. The mid-ordinate e, the vertical gap between the IP and the curve at the IP chainage
   C. The tangent offset at the PVT
   D. The difference of the two grades in metres
   Answer: B. Since the curve passes through the given level at the IP chainage, that gap is e = (g2 - g1)L/8. (Key point: Mid-ordinate e)

3. Which relation gives the curve length here, and what is L?
   A. L = 8e/(g2 + g1) = 8 x 1.5/0.01 = 1200 m
   B. L = 4e/(g2 - g1) = 100 m
   C. L = 8e/(g2 - g1) = 8 x 1.5/0.06 = 200 m
   D. L = e/(g2 - g1) = 25 m
   Answer: C. For an equal tangent parabola e = (g2 - g1)L/8, so L = 8e/(g2 - g1) = 200 m. (Key point: Length of curve)

4. How is the lowest point located from PVC?
   A. At the IP, chainage 780 m
   B. At the mid-point, x = L/2 = 100 m
   C. x = g2/r = 116.67 m
   D. Set grade zero: x = -g1/r = 0.025/0.0003 = 83.33 m, so chainage 763.33 m
   Answer: D. The lowest point is where the slope is zero: g1 + r x = 0, giving x = 83.33 m from PVC at chainage 763.33 m. (Key point: Lowest point)

## Close variants set in other exams

- [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×
- [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
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- [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

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- [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