---
title: "Percent grade from stopping distance difference"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-71bh-grade-ssd"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-71bh-grade-ssd
site: BCT Engineering
---

# Percent grade from stopping distance difference

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2071 Bhadra, 8 marks
- Video: [Calculate Braking Distance and Percentage Grade for Vehicles on  an 80 km/h Grade Hill 🚗⛰️](https://www.youtube.com/watch?v=oumsf7Cy_7M) — OPengineer
- 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-71bh-grade-ssd
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

The driver of a vehicle travelling at 65 kmph down a grade required 12 m more stopping sight distance to stop than the driver travelling at same speed up the same grade. If the coefficient of friction between tire and pavement is 0.38. Determine the percent grade and stopping sight distance up the grade.

## Short answer

Given 12 m extra stopping distance down grade at 65 km/h with friction 0.38, the grade is 5.11%. Stopping sight distance up grade is 83.8 m and down grade is 95.8 m.

## Answer

#### Approach
The extra stopping distance down the grade comes only from the braking distance, because the lag distance is the same:

$$
SSD_{down}-SSD_{up}=\frac{V^2}{254}\left[\frac{1}{f-n}-\frac{1}{f+n}\right]=12\ \text{m}
$$

Here $V=65$ km/h, $f=0.38$ and $n$ = grade as a ratio. $\dfrac{V^2}{254}=\dfrac{65^2}{254}=16.634$.

#### Solving for the grade

$$
16.634\left[\frac{2n}{f^2-n^2}\right]=12\ \Rightarrow\ \frac{2n}{0.1444-n^2}=0.7214
$$


$$
0.7214\,n^2+2n-0.10417=0\ \Rightarrow\ n=0.0511
$$

**Grade $\approx5.11\%$.**

#### Stopping sight distance up the grade ($t=2.5$ s)

$$
SSD_{up}=0.278Vt+\frac{V^2}{254(f+n)}=0.278(65)(2.5)+\frac{65^2}{254(0.4311)}=45.18+38.58=83.8\ \text{m}
$$

Check: $SSD_{down}=45.18+\dfrac{65^2}{254(0.3289)}=95.8$ m; difference $=12.00$ m, which matches 12 m.

**Answer:** grade = 5.1%; stopping sight distance up the grade = 83.8 m (down the grade 95.8 m).

## Check yourself

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

1. Why does the 12 m difference depend only on braking distance?
   A. The lag distance 0.278Vt is the same for both drivers
   B. Because reaction time differs
   C. Because speeds differ
   D. Because friction differs
   Answer: A. Only the braking term includes the grade. (Key point: Why only braking differs)

2. Which expression equals 12 m?
   A. V^2/254 [1/(f+n) - 1/(f-n)]
   B. V^2/254 [1/(f-n) - 1/(f+n)]
   C. V^2/(127f)
   D. V^2/254 [1/f]
   Answer: B. Down minus up. (Key point: Difference expression)

3. Which simplification results?
   A. 2f/(f^2 - n^2)
   B. 2n/(f + n)
   C. 1/(f-n) - 1/(f+n) = 2n/(f^2 - n^2)
   D. n/(f^2 - n^2)
   Answer: C. Leads to a quadratic in n. (Key point: Simplification)

4. What grade and SSD up the grade result?
   A. About 3% and 112 m
   B. About 10% and 60 m
   C. About 5.1% and 95.8 m
   D. About 5.1% and 83.8 m
   Answer: D. 95.8 m is the downgrade SSD; up-grade is 83.8 m. (Key point: Result)

## Related in Transportation Engineering I

- [Stopping sight distance at 50 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-66m-ssd-50) — 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
- [Stopping sight distance and derivation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Assd-factors-derive) — Transportation Engineering I, chapter 3, asked 6×
- [Factors affecting stopping sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Assd-factors-only) — Transportation Engineering I, chapter 3, asked 3×
- [Stopping distance and stopping sight distance](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Astopping-distance-relation) — Transportation Engineering I, chapter 3, asked 3×
- [Minimum sight distance to avoid head-on collision](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-74bh-headon-sd) — Transportation Engineering I, chapter 3

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