---
title: "Vertical curve +3.5% and -2.75%, 30 m pegs"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-35-down-275-through"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-35-down-275-through
site: BCT Engineering
---

# Vertical curve +3.5% and -2.75%, 30 m pegs

Surveying II · Chapter 6: Curves

- Asked in: 2078 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-35-down-275-through
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of 3.5% meets another grade of (-)2.75%. The elevation and chainage of IP are 1470.00 m and 2800.00 m respectively. Field condition requires that the vertical curve should pass through a point of elevation 1468.75 m at a chainage of 2800.00 m. Compute a suitable equal tangent vertical curve of full stations elevation including highest point also. Take peg interval = 30 m.

## Short answer

The equal-tangent parabolic vertical curve length is 160 m. The highest point occurs at chainage 2809.60 m, elevation 1468.768 m. Elevations at 30 m stations from PVC (RL 1467.200 m) are computed with r = –0.0003906, satisfying the required IP elevation.

## Answer

Data: $g_1 = +3.5\% = +0.035$, $g_2 = -2.75\% = -0.0275$ (summit curve), IP at chainage 2800.00 m, RL 1470.00 m. The curve must pass through RL 1468.75 m at chainage 2800.00 m, i.e. at the IP vertical.

#### Length of the curve

The offset from the IP to the curve at mid-point is $e = 1470.00 - 1468.75 = 1.25$ m. For an equal-tangent parabolic curve

$$
e = \frac{(g_1 - g_2)L}{8} \quad\Rightarrow\quad L = \frac{8e}{g_1 - g_2} = \frac{8\times 1.25}{0.035 + 0.0275} = 160.00\ \text{m}
$$

Each tangent length $= L/2 = 80.00$ m.

#### Elements

- PVC chainage $= 2800 - 80 = 2720.00$ m; PVT chainage $= 2880.00$ m
- RL of PVC $= 1470.00 - 0.035\times 80 = 1467.200$ m; RL of PVT $= 1470.00 - 0.0275\times 80 = 1467.800$ m
- $r = \dfrac{g_2 - g_1}{L} = -0.0003906$ per m

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

$$
RL_x = RL_{PVC} + g_1 x + \frac{r}{2}x^2
$$

| Chainage (m) | x from PVC (m) | Tangent level (m) | Offset $y=\frac{r}{2}x^2$ (m) | Curve level (m) |
|---|---|---|---|---|
| 2720.00 | 0.00 | 1467.200 | 0.000 | 1467.200 |
| 2730.00 | 10.00 | 1467.550 | -0.020 | 1467.530 |
| 2760.00 | 40.00 | 1468.600 | -0.312 | 1468.288 |
| 2790.00 | 70.00 | 1469.650 | -0.957 | 1468.693 |
| 2820.00 | 100.00 | 1470.700 | -1.953 | 1468.747 |
| 2850.00 | 130.00 | 1471.750 | -3.301 | 1468.449 |
| 2880.00 | 160.00 | 1472.800 | -5.000 | 1467.800 |

#### Highest point

$$
x = \frac{-g_1}{r} = \frac{0.035}{0.0003906} = 89.60\ \text{m}
$$

Chainage $= 2720.00 + 89.60 = 2809.60$ m, and RL $= 1467.200 + 0.035(89.60) + \frac{-0.0003906}{2}(89.60)^2 = 1468.768$ m.

Check: level at the IP chainage (x = 80 m) $= 1468.750$ m = 1468.750 m as required.

**Answer: L = 160 m; highest point at chainage 2809.60 m, RL 1468.768 m.**

## Check yourself

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

1. The curve passes through 1468.75 m at the IP chainage where IP RL is 1470.00 m. What does this give?
   A. The grade difference
   B. The offset e = 1.25 m between IP and curve
   C. The PVC level
   D. The length of the tangent
   Answer: B. At the IP chainage (curve mid-point) the vertical offset is e = 1470.00 - 1468.75 = 1.25 m. (Key point: Mid-curve offset from given data)

2. Which equation gives L?
   A. L = 8e/(g1 - g2) = 8 × 1.25/0.0625 = 160 m
   B. L = e/(g1 - g2) = 20 m
   C. L = 8e/(g1 + g2) = 1000 m
   D. L = 4e/(g1 - g2) = 80 m
   Answer: A. From e = (g1 - g2)L/8 with g1 - g2 = 0.035 + 0.0275 = 0.0625, L = 160 m. (Key point: Length of curve)

3. What are the PVC chainage and RL?
   A. 2880.00 and 1467.800
   B. 2720.00 and 1467.200
   C. 2720.00 and 1472.800
   D. 2800.00 and 1470.000
   Answer: B. PVC is L/2 = 80 m before the IP: chainage 2720, and RL = 1470 - 0.035 × 80 = 1467.200 m. (Key point: PVC chainage and level)

4. How far from the PVC is the highest point and what is its chainage?
   A. x = 80 m, chainage 2800.00
   B. x = 100 m, chainage 2820.00
   C. x = 160 m, chainage 2880.00
   D. x = -g1/r = 89.60 m, chainage 2809.60
   Answer: D. The grade is zero where g1 + r·x = 0, so x = 0.035/0.0003906 = 89.60 m, chainage 2809.60, RL 1468.768 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 +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 -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×

## Related in Surveying II

- [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 +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 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×
- [Highway curve through M 16 m from IP](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahighway-curve-pqr-140-point-m) — 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
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- [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