Vertical curve -2% and +3%, IP 2600, 0.5% per 20 m

A parabolic vertical curve with a constant rate of change of grade is set out using the tangent-offset method.

Step 1: Length of the curve

Rate of change of grade $r = 0.5\%$ per 20 m, algebraic change in grade $= 3 - (-2) = 5\%$.

$$L = \frac{|g_2-g_1|}{r}\times 20 = \frac{5}{0.5}\times 20 = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 2600 - 100 = 2500\ \text{m}\\ \text{Ch. of EVC} &= 2600 + 100 = 2700\ \text{m}\\ \text{RL of BVC} &= 1200 - \frac{(-2)}{100}\times 100 = 1202\ \text{m}\\ \text{RL of EVC} &= 1200 + \frac{(3)}{100}\times 100 = 1203\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = -2\%$, $g_2 = 3\%$, $L = 200$ m, so the offset is $y = \frac{(5)x^2}{200\times 200}$ and pegs are taken at multiples of 20 m.

Check: the last curve RL (1203 m) equals the RL of EVC (1203 m), and the curve at the IP chainage lies 1.25 m above the IP, at RL 1201.25 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{2\times 200}{5} = 80$ m from BVC.

• Chainage = 2500 + 80 = 2580 m
• RL = 1202 + (-2/100)(80) + (5)(80)²/(200 × 200) = 1201.2 m

Answer: $L = 200$ m, BVC at 2500 m (RL 1202 m), EVC at 2700 m (RL 1203 m); curve levels as tabulated.