Vertical curve table, -4.5% and +3.5%, 200 m

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

Step 1: Given data

$g_1 = -4.5\%$, $g_2 = 3.5\%$, $L = 200$ m, change of grade $= 8\%$. Mid-ordinate $e = \dfrac{L|g_2-g_1|}{800} = 2$ m.

Step 2: BVC, EVC and their levels

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

$$\begin{aligned} \text{Ch. of BVC} &= 1500 - 100 = 1400\ \text{m}\\ \text{Ch. of EVC} &= 1500 + 100 = 1600\ \text{m}\\ \text{RL of BVC} &= 1000 - \frac{(-4.5)}{100}\times 100 = 1004.5\ \text{m}\\ \text{RL of EVC} &= 1000 + \frac{(3.5)}{100}\times 100 = 1003.5\ \text{m} \end{aligned}$$

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

Check: the last curve RL (1003.5 m) equals the RL of EVC (1003.5 m), and the curve at the IP chainage lies 2 m above the IP, at RL 1002 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{4.5\times 200}{8} = 112.5$ m from BVC.

• Chainage = 1400 + 112.5 = 1512.5 m
• RL = 1004.5 + (-4.5/100)(112.5) + (8)(112.5)²/(200 × 200) = 1001.969 m

Answer: $L = 200$ m, BVC at 1400 m (RL 1004.5 m), EVC at 1600 m (RL 1003.5 m); curve levels as tabulated.