---
title: "Vertical curve elevations at 185 m from PVI"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-72m-vc-185"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-72m-vc-185
site: BCT Engineering
---

# Vertical curve elevations at 185 m from PVI

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2072 Magh, 8 marks
- Video: [Stationing and Elevation of PVI, PVT and Lowest Point of Sag Vertical Curve|Sag Curve Fundamentals](https://www.youtube.com/watch?v=FGDcYFVzCpU) — Master Civil Engineering
- 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-72m-vc-185
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A ascending gradient 3% meets the 2.5% descending gradient. Calculate the elevation of the road at a distance of every 185 m on either side of the point of vertical intersection (PVI) if the elevation of PVI is 500.00 m. Assume the radius of vertical curve as 8500 m.

## Short answer

Summit curve with grades +3% and -2.5%, radius 8500 m, PVI elevation 500 m. Curve length 467.5 m. Using parabolic formula, elevation 185 m before PVI is 494.31 m and 185 m after PVI is 495.24 m.

## Answer

A +3% grade followed by a −2.5% grade is a **summit curve**. $N=0.03+0.025=0.055$, $R=8500$ m, PVI RL = 500.00 m.

#### Length of curve

$$
L=NR=0.055(8500)=467.5\ \text{m},\qquad \frac{L}{2}=233.75\ \text{m}
$$

Since 185 m < 233.75 m, both required points lie on the curve.

#### Levels of BVC and EVC

$$
RL_{BVC}=500.00-0.03(233.75)=492.9875\ \text{m},\qquad RL_{EVC}=500.00-0.025(233.75)=494.1563\ \text{m}
$$

#### Elevation 185 m before PVI (on the BVC side)
Distance from BVC: $x=233.75-185=48.75$ m.

$$
y=RL_{BVC}+g_1x-\frac{x^2}{2R}=492.9875+0.03(48.75)-\frac{48.75^2}{2(8500)}=494.310\ \text{m}
$$

#### Elevation 185 m after PVI (on the EVC side)
Distance from EVC: $x'=48.75$ m (measured back from EVC along the falling 2.5% grade):

$$
y=RL_{EVC}+g_2x'-\frac{x'^2}{2R}=494.1563+0.025(48.75)-\frac{48.75^2}{2(8500)}=495.235\ \text{m}
$$

(Check: tangent level 185 m before PVI is $500-0.03(185)=494.450$ m; the curve is below it by $x^2/2R=0.140$ m, giving 494.310 m.)

**Answer:** elevation 185 m before PVI = **494.31 m**; elevation 185 m after PVI = **495.24 m**.

## Check yourself

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

1. How is L found with R = 8500 m?
   A. L = NR = 0.055 x 8500 = 467.5 m
   B. L = N/R
   C. L = 8500/2
   D. L = 0.055 x 185
   Answer: A. N = 0.03 + 0.025. (Key point: Curve length)

2. Why are both points 185 m from PVI on the curve?
   A. Because 185 m equals L/2
   B. Because 185 m is less than L/2 = 233.75 m
   C. Because they are beyond the EVC
   D. Because R is greater than 185
   Answer: B. Points within L/2 of the PVI lie on the curve. (Key point: Points on the curve)

3. What is the x used for the point 185 m before PVI?
   A. x = 185 m from BVC
   B. x = 233.75 + 185
   C. x = 233.75 - 185 = 48.75 m from BVC
   D. x = 185/2
   Answer: C. BVC is 233.75 m before PVI. (Key point: Before-PVI point)

4. Why is x^2/(2R) subtracted?
   A. It lies above the tangent
   B. It is the superelevation
   C. It is the vertical clearance
   D. A summit curve lies below the tangent by x^2/(2R)
   Answer: D. Tangent level 494.45 minus 0.14 = 494.31 m. (Key point: Parabola correction)

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