Vertical curve -1% and +0.5%, IP 500 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 $r = 0.1\%$ per 20 m, algebraic change in grade $= 0.5 - (-1) = 1.5\%$.

$$L = \frac{|g_2-g_1|}{r}\times 20 = \frac{1.5}{0.1}\times 20 = 300\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 150$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 500 - 150 = 350\ \text{m}\\ \text{Ch. of EVC} &= 500 + 150 = 650\ \text{m}\\ \text{RL of BVC} &= 350 - \frac{(-1)}{100}\times 150 = 351.5\ \text{m}\\ \text{RL of EVC} &= 350 + \frac{(0.5)}{100}\times 150 = 350.75\ \text{m} \end{aligned}$$

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 $g_1 = -1\%$, $g_2 = 0.5\%$ and $L = 300$ m, the offset is $y = \frac{(1.5)x^2}{200\times 300}$. The pegs are taken at multiples of 30 m.

Check: the last curve RL (350.75 m) equals the RL of EVC (350.75 m), and at the IP chainage the curve lies 0.563 m above the IP, at RL 350.563 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{1\times 300}{1.5} = 200$ m from BVC.

• Chainage = 350 + 200 = 550 m
• RL = 351.5 + (-1/100)(200) + (1.5)(200)²/(200 × 300) = 350.5 m

The pegs are at multiples of 30 m; the BVC (350 m) and EVC (650 m) are added as the end points of the curve.

Answer: $L = 300$ m, BVC at 350 m (RL 351.5 m), EVC at 650 m (RL 350.75 m), lowest point at chainage 550 m (RL 350.5 m); curve levels as tabulated.