Vertical curve +0.5% and -0.7%, IP RL 350.750

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 summit (crest) curve with equal tangent lengths $L/2 = 180$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 500 - 180 = 320\ \text{m}\\ \text{Ch. of EVC} &= 500 + 180 = 680\ \text{m}\\ \text{RL of BVC} &= 350.75 - \frac{(0.5)}{100}\times 180 = 349.85\ \text{m}\\ \text{RL of EVC} &= 350.75 + \frac{(-0.7)}{100}\times 180 = 349.49\ \text{m} \end{aligned}$$

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 30 m.

Check: the last curve RL (349.49 m) equals the RL of EVC (349.49 m), and the curve at the IP chainage lies 0.54 m below the IP, at RL 350.21 m.

Step 4: Highest point

At the highest 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 = 320 + 150 = 470 m
• RL = 349.85 + (0.5/100)(150) + (-1.2)(150)²/(200 × 360) = 350.225 m

Peg interval is not stated, so it is taken as 30 m (equal to the length over which the rate of change of grade is specified). The BVC and EVC are included.

Answer: $L = 360$ m, BVC at 320 m (RL 349.85 m), EVC at 680 m (RL 349.49 m); curve levels as tabulated.