Vertical curve -0.7% and +0.5%, IP RL 650

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 $= 0.5 - (-0.7) = 1.2\%$.

$$L = \frac{|g_2-g_1|}{r}\times 20 = \frac{1.2}{0.1}\times 20 = 240\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 120$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 1000 - 120 = 880\ \text{m}\\ \text{Ch. of EVC} &= 1000 + 120 = 1120\ \text{m}\\ \text{RL of BVC} &= 650 - \frac{(-0.7)}{100}\times 120 = 650.84\ \text{m}\\ \text{RL of EVC} &= 650 + \frac{(0.5)}{100}\times 120 = 650.6\ \text{m} \end{aligned}$$

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 $g_1 = -0.7\%$, $g_2 = 0.5\%$, $L = 240$ m, so the offset is $y = \frac{(1.2)x^2}{200\times 240}$ and pegs are taken at multiples of 20 m.

Check: the last curve RL (650.6 m) equals the RL of EVC (650.6 m), and the curve at the IP chainage lies 0.36 m above the IP, at RL 650.36 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{0.7\times 240}{1.2} = 140$ m from BVC.

• Chainage = 880 + 140 = 1020 m
• RL = 650.84 + (-0.7/100)(140) + (1.2)(140)²/(200 × 240) = 650.35 m

Answer: $L = 240$ m, BVC at 880 m (RL 650.84 m), EVC at 1120 m (RL 650.6 m); curve levels as tabulated.