Vertical curve -0.5% and +0.7%, IP RL 500

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 30 m, algebraic change in grade $= 0.7 - (-0.5) = 1.2\%$.

$$L = \frac{|g_2-g_1|}{r}\times 30 = \frac{1.2}{0.1}\times 30 = 360\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 180$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 1000 - 180 = 820\ \text{m}\\ \text{Ch. of EVC} &= 1000 + 180 = 1180\ \text{m}\\ \text{RL of BVC} &= 500 - \frac{(-0.5)}{100}\times 180 = 500.9\ \text{m}\\ \text{RL of EVC} &= 500 + \frac{(0.7)}{100}\times 180 = 501.26\ \text{m} \end{aligned}$$

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 $g_1 = -0.5\%$, $g_2 = 0.7\%$, $L = 360$ m, so the offset is $y = \frac{(1.2)x^2}{200\times 360}$ and pegs are taken at multiples of 20 m.

Check: the last curve RL (501.26 m) equals the RL of EVC (501.26 m), and the curve at the IP chainage lies 0.54 m above the IP, at RL 500.54 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{0.5\times 360}{1.2} = 150$ m from BVC.

• Chainage = 820 + 150 = 970 m
• RL = 500.9 + (-0.5/100)(150) + (1.2)(150)²/(200 × 360) = 500.525 m

Answer: $L = 360$ m, BVC at 820 m (RL 500.9 m), EVC at 1180 m (RL 501.26 m); curve levels as tabulated.