Vertical curve +1.25% and +4.75%, 1518.30 m

Chainage of IP $= 2+030$ km $= 2030$ 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:

$$e = 1518.3 - 1517.6 = 0.7\ \text{m}\ \Rightarrow\ |e| = 0.7\ \text{m}$$

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

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.7}{|4.75-(1.25)|} = \frac{560}{3.5} = 160\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 80$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 2030 - 80 = 1950\ \text{m}\\ \text{Ch. of EVC} &= 2030 + 80 = 2110\ \text{m}\\ \text{RL of BVC} &= 1517.6 - \frac{(1.25)}{100}\times 80 = 1516.6\ \text{m}\\ \text{RL of EVC} &= 1517.6 + \frac{(4.75)}{100}\times 80 = 1521.4\ \text{m} \end{aligned}$$

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 $g_1 = 1.25\%$, $g_2 = 4.75\%$, $L = 160$ m, so the offset is $y = \frac{(3.5)x^2}{200\times 160}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1521.4 m) equals the RL of EVC (1521.4 m), and the curve at the IP chainage lies 0.7 m above the IP, at RL 1518.3 m.

Step 4: Highest and lowest points

Both grades are rising, so the grade never becomes zero within the curve and there is no turning point; the lowest level is at the BVC (RL 1516.6 m) and the highest level is at the EVC (RL 1521.4 m).

Answer: $L = 160$ m, BVC at 1950 m (RL 1516.6 m), EVC at 2110 m (RL 1521.4 m); curve levels as tabulated.