Actual grade on inner and outer lanes

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.