---
title: "Summit curve +4% and -4%, limited to 800 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-71m-summit-800"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-71m-summit-800
site: BCT Engineering
---

# Summit curve +4% and -4%, limited to 800 m

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2071 Magh, 8 marks
- Video: [Highway Engineering / Summit (Curve Crest) Curve Sub-Engineer SSC-JE MPPSC-AE 📚](https://www.youtube.com/watch?v=J6Vpd2cU1n4) — Civil Site Plus
- 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-71m-summit-800
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A vertical summit curve is to be designed when two grades +4% and -4% meet on a highway. The stopping sight distance and overtaking sight distance required are 150 m and 500 m respectively. But due to site conditions, the length of curve has to be restricted to a maximum value of 800 m if possible. Calculate the length of curve needed to fulfil the requirements of (a) stopping sight distance (b) overtaking sight distance or at least intermediate sight distance. Discuss the results.

## Short answer

Summit curve with grades +4% and -4% requires 409 m for SSD (150 m), 750 m for ISD (300 m), and 2083 m for OSD (500 m). With maximum length 800 m, SSD and ISD are satisfied but OSD cannot be provided; overtaking must be prohibited.

## Answer

$g_1=+4\%,\ g_2=-4\%$, so $N=0.08$. This is a summit curve (IRC/NRS 2070: eye height 1.2 m; object height 0.15 m for SSD and 1.2 m for OSD).

#### (a) Stopping sight distance S = 150 m
Assuming $L>S$:

$$
L=\frac{NS^2}{4.4}=\frac{0.08(150)^2}{4.4}=409.1\ \text{m}
$$

$L>S$, so valid. **L ≈ 410 m**, which is less than 800 m, so it is possible.

#### (b) Overtaking sight distance S = 500 m
For $h_1=h_2=1.2$ m:

$$
L=\frac{NS^2}{9.6}=\frac{0.08(500)^2}{9.6}=2083.3\ \text{m}
$$

This exceeds the 800 m limit. With L = 800 m the sight distance actually available is

$$
S=\sqrt{\frac{9.6L}{N}}=\sqrt{\frac{9.6(800)}{0.08}}=309.8\ \text{m}
$$

which is much less than the 500 m OSD.

#### Intermediate sight distance
ISD $=2\times SSD=300$ m:

$$
L=\frac{0.08(300)^2}{9.6}=750\ \text{m}
$$

This is within 800 m, so ISD can be provided.

#### Discussion
| Criterion | Required length | Within 800 m? |
|---|---|---|
| SSD 150 m | 409 m | Yes |
| ISD 300 m | 750 m | Yes |
| OSD 500 m | 2083 m | No |

**Answer:** adopt L = 800 m (or at least 750 m). It satisfies SSD (210 m available) and ISD (310 m), but full OSD cannot be provided within 800 m. Overtaking must be prohibited on this summit (no-overtaking signs and continuous centre line).

## Check yourself

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

1. Which formula gives L for SSD of 150 m?
   A. L = NS^2/4.4 = 409 m
   B. L = NS^2/9.6 = 187 m
   C. L = 2S - 4.4/N
   D. L = NR
   Answer: A. Eye 1.2 m and object 0.15 m give the 4.4 constant. (Key point: SSD length)

2. Why is L for OSD much longer (2083 m)?
   A. It uses a higher friction value
   B. OSD uses object height 1.2 m, giving constant 9.6 and a much larger S
   C. It uses the 4.4 constant
   D. It uses the valley formula
   Answer: B. L = 0.08 x 500^2 / 9.6. (Key point: OSD length)

3. What sight distance does 800 m give for OSD criteria?
   A. 500 m
   B. 150 m
   C. S = sqrt(9.6L/N) = 310 m
   D. 800 m
   Answer: C. Much less than 500 m. (Key point: Sight distance with 800 m)

4. What does intermediate sight distance equal and what length does it need?
   A. ISD = SSD, L = 409 m
   B. ISD = OSD, L = 2083 m
   C. ISD = 3 x SSD, L = 1688 m
   D. ISD = 2 x SSD = 300 m, L = 750 m
   Answer: D. Within the 800 m limit. (Key point: ISD)

## Close variants set in other exams

- [Summit curve limited to 750 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-75bh-vc-summit-750) — Transportation Engineering I, chapter 3, asked 3×

## Related in Transportation Engineering I

- [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 with intermediate sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81a-vc-intermediate) — Transportation Engineering I, chapter 3
- [Overtaking sight distance and zone numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-62j-osd-numeric) — Transportation Engineering I, chapter 3
- [Summit curve +1.5% and -2%](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-68bh-summit) — Transportation Engineering I, chapter 3
- [Sight distance with obstruction on curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78b-sd-obstruction) — Transportation Engineering I, chapter 3
- [Overtaking sight distance and zone, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-76b-osd-60) — Transportation Engineering I, chapter 3

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- [Vertical curve -1% and +0.5%, IP 500 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-falling-1-rising-05) — Surveying II, chapter 6, asked 2×
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