Vertical curve 180 m, -3.5% then +4.5%, IP 2450

A parabolic vertical curve of the given length is set out by the tangent-offset method.

Step 1: Given data

$g_1 = -3.5\%$, $g_2 = 4.5\%$, $L = 180$ m, change of grade $= 8\%$. Mid-ordinate $e = \dfrac{L|g_2-g_1|}{800} = 1.8$ m.

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 90$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 2450 - 90 = 2360\ \text{m}\\ \text{Ch. of EVC} &= 2450 + 90 = 2540\ \text{m}\\ \text{RL of BVC} &= 900 - \frac{(-3.5)}{100}\times 90 = 903.15\ \text{m}\\ \text{RL of EVC} &= 900 + \frac{(4.5)}{100}\times 90 = 904.05\ \text{m} \end{aligned}$$

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 $g_1 = -3.5\%$, $g_2 = 4.5\%$, $L = 180$ m, so the offset is $y = \frac{(8)x^2}{200\times 180}$ and pegs are taken at multiples of 20 m.

Check: the last curve RL (904.05 m) equals the RL of EVC (904.05 m), and the curve at the IP chainage lies 1.8 m above the IP, at RL 901.8 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 180}{8} = 78.75$ m from BVC.

• Chainage = 2360 + 78.75 = 2438.75 m
• RL = 903.15 + (-3.5/100)(78.75) + (8)(78.75)²/(200 × 180) = 901.772 m

Answer: $L = 180$ m, BVC at 2360 m (RL 903.15 m), EVC at 2540 m (RL 904.05 m); curve levels as tabulated.