---
title: "Vertical curve with intermediate sight distance"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-81a-vc-intermediate"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-81a-vc-intermediate
site: BCT Engineering
---

# Vertical curve with intermediate sight distance

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2081 Ashwin, 8 marks
- Video: [Sight Distance for Vertical Curves](https://www.youtube.com/watch?v=4bkRhnQ3dr0) — LDP CE
- 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-81a-vc-intermediate
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

An intermediate lane road is to be designed with intermediate sight distance of 255 m. The road is the two-way road having 5% gradient meets a 1.5% gradient at a chainage of (1 + 000) and at the reduced level of 500 m. If the design speed of the road is 80 km/hr, determine RL and chainage of the beginning of the curve, highest point and 50 m right of point of vertical intersection on the curve. Take frictional coefficient f = 0.35.

## Short answer

For an intermediate lane road with 80 km/h design speed, ISD 255 m, grades +5% and -1.5% meeting at chainage 1000 m (RL 500 m), the summit curve length is 440.3 m (using L = NS²/9.6 with N=0.065). BVC is at chainage 779.86 m (RL 488.99 m). Highest point is at chainage 1118.53 m (RL 497.46 m). Point 50 m right of PVI (chainage 1050 m) has RL 497.11 m.

## Answer

**Given:** ISD $= 255$ m, $V = 80$ km/h, $f = 0.35$ (the ISD is $2\times SSD$: $2\times127.6 = 255.2$ m, which is consistent with these values), PVI at chainage $1+000$, RL $500$ m, $g_1 = +5\%$.

**Reading:** a highest point exists only if the second grade falls, so the grades are $g_1 = +5\%$ and $g_2 = -1.5\%$. This is a **summit curve** with
$$
N = 0.05 + 0.015 = 0.065
$$

#### Length of the curve
Intermediate sight distance, eye height $H = 1.2$ m and object height $h = 1.2$ m (IRC): $(\sqrt{2H}+\sqrt{2h})^2 = 9.6$. Assuming $S < L$:

$$
L = \frac{N S^2}{9.6} = \frac{0.065\times255^2}{9.6} = 440.3\ \text{m}
$$
$S = 255 < 440.3$ so the assumption holds.

#### (a) Beginning of the curve (BVC)
$$
\begin{aligned}
\text{Chainage} &= 1000 - \frac{L}{2} = 1000 - 220.14 = 779.86\ \text{m}\ (0+779.86)\\
RL &= 500 - 0.05\times220.14 = 488.99\ \text{m}
\end{aligned}
$$

#### (b) Highest point
Distance from BVC:
$$
x = \frac{g_1 L}{N} = \frac{0.05\times440.27}{0.065} = 338.67\ \text{m}
$$
Chainage $= 779.86 + 338.67 = 1118.53$ m.

$$
RL = 488.99 + 0.05(338.67) - \frac{0.065\,(338.67)^2}{2\times440.27} = 488.99 + 16.93 - 8.47 = 497.46\ \text{m}
$$

#### (c) Point 50 m to the right of PVI
Chainage $= 1050$ m, and $x = 220.14 + 50 = 270.14$ m from BVC:
$$
RL = 488.99 + 0.05(270.14) - \frac{0.065\,(270.14)^2}{2\times440.27} = 488.99 + 13.51 - 5.39 = 497.11\ \text{m}
$$

**Answer:** BVC at chainage 779.86 m (RL 488.99 m); highest point at chainage 1118.53 m (RL 497.46 m); point 50 m right of PVI (chainage 1050 m) RL 497.11 m. End of curve at chainage 1220.14 m, RL 496.70 m.

## Check yourself

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

1. Why is the curve a summit curve with a highest point?
   A. Because grades +5% and -1.5% meet, so the curve has a downward second grade
   B. Because both grades are upward
   C. Because ISD is 255 m
   D. Because the speed is 80 km/h
   Answer: A. A highest point exists only if the grade changes from rising to falling; g1 = +5% and g2 = -1.5% give N = 0.065. (Key point: Type of curve)

2. What is L for ISD = 255 m with S < L, N = 0.065 and the IRC constant 9.6?
   A. About 220 m
   B. About 880 m
   C. About 440 m
   D. About 120 m
   Answer: C. L = N S^2/9.6 = 0.065 x 255^2/9.6 = 440.3 m, and S < L holds. (Key point: Length of summit curve)

3. BVC is half the curve length before the PVI at 1000 m. What are its chainage and RL?
   A. Chainage 779.86 m, RL 488.99 m
   B. Chainage 1220.14 m, RL 496.70 m
   C. Chainage 779.86 m, RL 511.01 m
   D. Chainage 559.7 m, RL 477.99 m
   Answer: A. BVC = 1000 - 220.14 = 779.86 m; RL = 500 - 0.05 x 220.14 = 488.99 m. (Key point: Chainage and RL of BVC)

4. How is the highest point located from BVC?
   A. x = g2 L / N
   B. x = L/2
   C. x = g1 L / N
   D. x = N L / g1
   Answer: C. Slope is zero where g1 - N x/L = 0, so x = g1 L/N = 0.05 x 440.27/0.065 = 338.67 m. (Key point: Highest point location)

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