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

Step 1: Write the two grades with sign convention.
Take the direction of travel as reference, and treat upgrades as positive and downgrades as negative. The downgrade of 1 in 100 means the road falls 1 unit for every 100 units travelled, so as a percentage
\[ g_1 = -\frac{1}{100} \times 100 = -1\% \]
The upgrade of 1 in 125 means the road rises 1 unit for every 125 units travelled, so
\[ g_2 = +\frac{1}{125} \times 100 = +0.8\% \]

Step 2: Find the total change in grade.
Since a downgrade meets an upgrade, this is a valley (sag) curve, and the total algebraic change in grade at the point where the curves join is
\[ N = g_2 - g_1 = 0.8 - (-1) = 1.8\% \]
This N is the total change in grade the vertical curve must smoothly absorb, spread out over its length.

Step 3: Use the given rate of change of grade to find the length.
The rate of change of grade r is defined as the change in grade per unit length of curve, that is r = N/L, so
\[ L = \frac{N}{r} \]
We are told the rate is 0.10% per 30 m, which as a rate per metre is
\[ r = \frac{0.10\%}{30 \text{ m}} \]

Step 4: Solve for L.
\[ L = \frac{N}{r} = \frac{1.8\%}{0.10\%/30\text{ m}} = 1.8 \times \frac{30}{0.10} \]
\[ L = 1.8 \times 300 = 540 \text{ m} \]

Final Answer:
The length of the vertical curve between the two grades is 540 m.
\[ \boxed{L = 540 \text{ m}} \]