---
title: "Actual grade on inner and outer lanes"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-64p-grade-inner-outer"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-64p-grade-inner-outer
site: BCT Engineering
---

# Actual grade on inner and outer lanes

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2064 Poush (old course), 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-64p-grade-inner-outer
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Determine the actual grade along the centre line of the inner most and outer most lanes of the road with 15 m wide carriageway along the circular curve of radius 75 m if the grade along the centre line of the road is 5% at that point. Do not consider the extra widening. The deflection angle at that point is 45°.

## Short answer

Inner lane grade = 5.41%, outer lane grade = 4.65%, using G_lane = G_c × R_c / R_lane with R_inner = 69.375 m, R_outer = 80.625 m.

## Answer

#### Principle
Over the same angle $\theta$ at the centre of the curve, every lane rises by the same height, but the length measured along the lane is $R_{lane}\,\theta$. The grade is therefore inversely proportional to the radius of the lane:

$$
G_{lane}=G_{c}\times\frac{R_c}{R_{lane}}
$$

#### Data
$R_c=75$ m (centre line), $G_c=5\%$, carriageway 15 m wide. Assuming four lanes of 3.75 m, the centre of the innermost and outermost lanes is $7.5-\dfrac{3.75}{2}=5.625$ m from the road centre line.
- $R_{inner}=75-5.625=69.375$ m
- $R_{outer}=75+5.625=80.625$ m

#### Grades

$$
G_{inner}=5\times\frac{75}{69.375}=5.405\%
$$


$$
G_{outer}=5\times\frac{75}{80.625}=4.651\%
$$

The deflection angle (45°) cancels out and is not needed.

**Answer:** grade on the innermost lane = **5.41%** (steeper) and on the outermost lane = **4.65%** (flatter), compared with 5% on the centre line. This is why grade compensation is applied on sharp hill-road bends.

## Check yourself

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

1. How does the grade along a lane relate to the centre line grade?
   A. G_lane = G_c x R_c / R_lane (inversely proportional to radius)
   B. G_lane = G_c x R_lane / R_c
   C. Equal to G_c
   D. G_lane = G_c + deflection angle
   Answer: A. The rise over the same central angle is equal, but the lane length is R_lane x theta. (Key point: Principle)

2. Assuming four 3.75 m lanes, what is the inner lane centre radius?
   A. 75 - 7.5 = 67.5 m
   B. 75 - 5.625 = 69.375 m
   C. 75 - 3.75 = 71.25 m
   D. 75 - 15 = 60 m
   Answer: B. Centre of innermost lane is 7.5 - 3.75/2 = 5.625 m from the centre line. (Key point: Lane radii)

3. Which lane is steeper, and by how much?
   A. Outer lane, 5.41%
   B. Both stay at 5%
   C. Inner lane, 5.41% (outer lane 4.65%)
   D. Inner lane, 4.65%
   Answer: C. The smaller radius gives a shorter path and hence steeper grade. (Key point: Result)

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