Question

A downgrade of 1 in 100 meets an upgrade of 1 in 125 on a road. If the rate of change of grade is 0.10 % per 30 m, the length (in m) of the vertical curve between the two grades is ______ (rounded off to the nearest integer).

Hint:

Total grade change = g2-g1=1.8%, divide by the given rate per unit length to get L.

Answer 1

Correct Answer: 540

A simpler way to see this problem is as a straight proportion: 0.10% of grade change happens every 30 m of curve length, so ask how many such 30 m chunks are needed to cover the total grade change of this curve.

1. Convert both grades to percentages: the downgrade 1 in 100 is a slope of $-1\%$ (falling), and the upgrade 1 in 125 is a slope of $+0.8\%$ (rising). Using the sign convention that rising grades are positive, $g_1 = -1\%$ and $g_2 = +0.8\%$.
2. Find the total swing in grade: going from $g_1$ to $g_2$, the grade swings by $g_2 - g_1 = 0.8 - (-1) = 1.8\%$. This is the total turning the curve must do, since the road goes from falling at 1% to rising at 0.8%.
3. Use the unitary method with the given rate: every 30 m of curve produces $0.10\%$ of that turning. So the number of 30 m chunks needed is $\dfrac{1.8\%}{0.10\%} = 18$ chunks.
4. Multiply chunks by chunk length: total length $L = 18 \times 30\text{ m} = 540\text{ m}$.

Let's summarize:

• Total grade change between the two grades is $1.8\%$.
• At $0.10\%$ change per 30 m, that needs 18 such intervals.

So the length of the vertical curve is $540$ m.