Vertical curve, -3.5% and +0.5%, 3268 m

Why a parabola?

A parabola is used for vertical curves because it gives a constant rate of change of grade along the curve, so the change in slope is uniform and the ride is smooth. Its equation is simple ($y = ax^2 + bx$), offsets from the tangents vary as the square of the distance and are very easy to compute, the curve is symmetrical about the IP for equal tangents, and a circle and a parabola are almost identical for the very flat curves used on roads.

Computation

Chainage of IP $= 1+780$ km $= 1780$ m.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

$$e = 3268 - 3267 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|0.5-(-3.5)|} = \frac{800}{4} = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

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

$$\begin{aligned} \text{Ch. of BVC} &= 1780 - 100 = 1680\ \text{m}\\ \text{Ch. of EVC} &= 1780 + 100 = 1880\ \text{m}\\ \text{RL of BVC} &= 3267 - \frac{(-3.5)}{100}\times 100 = 3270.5\ \text{m}\\ \text{RL of EVC} &= 3267 + \frac{(0.5)}{100}\times 100 = 3267.5\ \text{m} \end{aligned}$$

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 $g_1 = -3.5\%$, $g_2 = 0.5\%$, $L = 200$ m, so the offset is $y = \frac{(4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (3267.5 m) equals the RL of EVC (3267.5 m), and the curve at the IP chainage lies 1 m above the IP, at RL 3268 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 200}{4} = 175$ m from BVC.

• Chainage = 1680 + 175 = 1855 m
• RL = 3270.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 3267.438 m

Answer: $L = 200$ m, BVC at 1680 m (RL 3270.5 m), EVC at 1880 m (RL 3267.5 m); curve levels as tabulated.