---
title: "Vertical curve with 300 m grade sections"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-65k-vc-300"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-65k-vc-300
site: BCT Engineering
---

# Vertical curve with 300 m grade sections

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2065 Kartik (old course), 8 marks
- Video: [Vertical Curve in Surveying/Highway Engineering | Formula Derivation + High & Low Points](https://www.youtube.com/watch?v=WQvvrnJNbdI) — CIVIL Solved
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-65k-vc-300
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A 300 m long ascending section of a double lane road with two way traffic road with 4% grade meets with a 300 m long descending section with 3% grade. Design the vertical curve to meet the visibility requirement. Design speed is 100 kmph. The braking efficiency is 90%. Calculate the formation levels of main points on curve at a distance of 50 m from PVI on either sides, at the highest point of the formation lines and at the beginning and end section of road, given the final formation level of the road at a distance of 25 m right from PVI as 120.105.

## Short answer

For summit curve N=0.07, effective f=0.315 gives SSD=194.5 m and L=601.7 m, adopt 600 m. PVI=125.266 m, BVC=113.266 m, EVC=116.266 m, highest point=120.124 m, 50 m before PVI=119.621 m, 50 m after PVI=120.121 m.

## Answer

An ascending 4% grade followed by a descending 3% grade gives a **summit curve**, $N=0.04+0.03=0.07$.

#### Stopping sight distance
$V=100$ km/h, $t=2.5$ s, coefficient of friction $f=0.35$ (assumed, IRC value for 100 km/h), so the effective $f=0.35\times0.90=0.315$:

$$
SSD=0.278(100)(2.5)+\frac{100^2}{254(0.315)}=194.5\ \text{m}
$$

#### Length of curve

$$
L=\frac{NS^2}{4.4}=\frac{0.07(194.5)^2}{4.4}=601.7\ \text{m}
$$

The grade sections are each 300 m long, so the curve can extend at most 300 m on each side of the PVI. Adopt **L = 600 m** (half = 300 m). The sight distance this gives, $S=\sqrt{4.4L/N}=194.2$ m, is practically the required SSD (194.5 m), so the requirement is met.

#### Formation level of the PVI (from the given point)
The point 25 m right of PVI is $300-25=275$ m from EVC. Offset below the tangent:

$$
y_o=\frac{N(275)^2}{2L}=\frac{0.07(275)^2}{2(600)}=4.411\ \text{m}
$$


$$
120.105=RL_{PVI}-0.03(25)-4.411\Rightarrow RL_{PVI}=125.266\ \text{m}
$$

#### Main levels

$$
RL_{BVC}=125.266-0.04(300)=113.266\ \text{m},\qquad RL_{EVC}=125.266-0.03(300)=116.266\ \text{m}
$$

- 50 m before PVI ($x=250$ m from BVC): offset $=\dfrac{0.07(250)^2}{1200}=3.646$ m; level $=125.266-0.04(50)-3.646=\mathbf{119.621}$ m
- 50 m after PVI: level $=125.266-0.03(50)-3.646=\mathbf{120.121}$ m
- Highest point: $x=\dfrac{g_1L}{N}=\dfrac{0.04(600)}{0.07}=342.86$ m from BVC

$$
RL_{max}=113.266+\frac{0.04^2(600)}{2(0.07)}=120.124\ \text{m}
$$

**Answer:** RL of PVI = 125.27 m; beginning (BVC) = 113.27 m; end (EVC) = 116.27 m; highest point = 120.12 m; 50 m before PVI = 119.62 m; 50 m after PVI = 120.12 m.

## Check yourself

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

1. How is the SSD friction value obtained?
   A. 0.35 unchanged
   B. 0.35 x 0.90 = 0.315 for 90% brake efficiency
   C. 0.35 / 0.90
   D. 0.35 + 0.9
   Answer: B. Brake efficiency scales the friction coefficient. (Key point: Effective friction)

2. Calculated L = 601.7 m. Why is L = 600 m adopted?
   A. 600 m is the minimum length for 100 km/h
   B. L must be a multiple of 100
   C. Each grade section is 300 m, so the curve can extend 300 m each side of the PVI
   D. Tangents are 600 m long
   Answer: C. Half-length cannot exceed the 300 m available; 600 m still gives S = 194.2 m, practically the SSD. (Key point: Length restriction)

3. The given point is 25 m right of the PVI. Which equation gives RL_PVI?
   A. 120.105 = RL_PVI - 0.03(25) - N(275)^2/(2L)
   B. 120.105 = RL_PVI + 0.03(25)
   C. 120.105 = RL_PVI - 0.04(25) - offset
   D. RL_PVI = 120.105 - NL/8
   Answer: A. The point is 275 m from EVC; its level is the tangent level minus the offset. (Key point: PVI level)

4. Where is the highest point from BVC?
   A. x = 300 m
   B. x = g1 L / N = 342.86 m
   C. x = g2 L / N = 257 m
   D. x = 250 m
   Answer: B. Zero grade occurs at x = g1 L / N. (Key point: Highest point)

## Related in Transportation Engineering I

- [Valley curve design, 100 kmph, 1 in 40](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-valley-100-312) — Transportation Engineering I, chapter 3, asked 2×
- [Vertical curve 35 m headlight and momentum grade](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-vc-headlight-35m) — Transportation Engineering I, chapter 3
- [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 for overtaking sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81c-vc-overtaking) — Transportation Engineering I, chapter 3
- [Vertical curve +2% and +5.5%, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79c-vc-2-55) — Transportation Engineering I, chapter 3
- [Vertical curve at station 6+525, night visibility](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-74bh-vc-6525) — Transportation Engineering I, chapter 3

## Related in other subjects

- [Night visibility, road marking, traffic islands](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Anight-visibility-road-marking-traffic-island) — Transportation Engineering II, chapter 1
- [Transition curve, road 8 m, R=300 m, 60°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-300m-60-70kmph) — Surveying II, chapter 6, asked 2×
- [Transition curve road 8 m, R=300 m, IP 3605](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-300m-3605-50kmh) — Surveying II, chapter 6, asked 3×
- [Road earthwork, upward grade 1:100](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-u-upward-1-100) — Estimating and Costing, chapter 6