---
title: "Vertical curve -3.25% and +3.75%, 225 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-73m-vc-225"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-73m-vc-225
site: BCT Engineering
---

# Vertical curve -3.25% and +3.75%, 225 m

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2073 Magh, 8 marks
- 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-73m-vc-225
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A vertical curve connects -3.25% grade with +3.75% grade. The curve should be designed at least to provide the visibility of the road surface to a distance of 225 m at night time. Calculate elevation of BVC, lowest point and EVC if the RL of the curve at 18 m distance from EVC is 125.32 m.

## Short answer

Valley curve length using standard headlight values (h=0.75 m, β=1°) is 380 m. BVC RL = 125.02 m, lowest point RL = 122.15 m at 176.4 m from BVC, EVC RL = 125.97 m.

## Answer

A −3.25% grade meeting +3.75% is a **valley curve**; $N=0.0325+0.0375=0.07$.

#### Length ($S=225$ m)
Headlight height and beam angle are not given, so the standard values $h_1=0.75$ m and $\alpha=1^\circ$ (IRC/NRS) are assumed:

$$
L=\frac{NS^2}{1.5+0.035S}=\frac{0.07(225)^2}{1.5+0.035(225)}=378.0\ \text{m}
$$

$L>S$, valid. Adopt **L = 380 m** (half = 190 m).

#### RL of EVC from the given point
The point is 18 m before EVC. Measured back from EVC along the +3.75% tangent, the curve lies above the tangent by $\dfrac{Nx^2}{2L}$:

$$
125.32=RL_{EVC}-0.0375(18)+\frac{0.07(18)^2}{2(380)}\Rightarrow RL_{EVC}=125.32+0.675-0.030=125.965\ \text{m}
$$

#### RL of PVI and BVC

$$
RL_{PVI}=RL_{EVC}-0.0375(190)=118.840\ \text{m},\qquad RL_{BVC}=RL_{PVI}+0.0325(190)=125.015\ \text{m}
$$

#### Lowest point

$$
x=\frac{g_1L}{N}=\frac{0.0325(380)}{0.07}=176.43\ \text{m from BVC}
$$


$$
RL_{low}=125.015-0.0325(176.43)+\frac{0.07(176.43)^2}{2(380)}=122.148\ \text{m}
$$

**Answer:** RL of BVC = 125.02 m; lowest point = 122.15 m (at 176.4 m from BVC); RL of EVC = 125.97 m.

## Check yourself

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

1. A -3.25% grade meeting +3.75% gives?
   A. Valley curve with N = 0.07
   B. Summit curve with N = 0.07
   C. Valley curve with N = 0.005
   D. Summit curve with N = 0.005
   Answer: A. N is the algebraic difference 0.0325 + 0.0375. (Key point: Valley curve type)

2. Which standard form is used for night visibility with assumed h=0.75 m and α=1°?
   A. L = NS^2/4.4
   B. L = NS^2/(1.5 + 0.035S) = 378 m
   C. L = NS^2/9.6
   D. L = NR
   Answer: B. Substituting gives 0.07 x 225^2 / 9.375 = 378 m, adopt 380 m. (Key point: Length formula)

3. How is RL_EVC found from the point 18 m before EVC with RL 125.32?
   A. RL_EVC = y - g2 x
   B. RL_EVC = y
   C. RL_EVC = y + g2 x - N x^2/(2L), because the curve is above the tangent in a valley measured back from EVC
   D. RL_EVC = y + L g2
   Answer: C. 125.32 + 0.675 - 0.030 = 125.965 m. (Key point: RL of EVC)

4. Where is the lowest point?
   A. x = L/2 = 190 m
   B. x = 18 m from EVC
   C. x = g2 L/N = 203.6 m
   D. x = g1 L/N = 176.4 m from BVC
   Answer: D. At lowest point slope is zero; x = 0.0325 x 380 / 0.07. (Key point: Lowest point)

## Related in Transportation Engineering I

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