Derive RL on parabolic vertical curve

A parabolic vertical curve is used because the rate of change of grade is constant along the curve. This gives smooth riding and simple computation.

Derivation

Take the origin at BVC, $x$ horizontal along the chord distance, $y$ the level. The rate of change of grade is constant, so with grades in % (rise per 100 m):

$$\frac{d^2y}{dx^2} = \frac{g_2-g_1}{100\,L}$$

Integrate once. At $x=0$ the slope is the first grade, $dy/dx = g_1/100$:

$$\frac{dy}{dx} = \frac{g_1}{100} + \frac{(g_2-g_1)\,x}{100L}$$

Check: at $x = L$, $dy/dx = g_2/100$, as required. Integrate again, with $y = \text{RL of BVC}$ at $x=0$:

$$y_P = \frac{(g_2-g_1)\,x^2}{200L} + \frac{g_1 x}{100} + \text{RL of BVC}$$

The first term is the offset of the curve from the tangent, and $g_1x/100 + \text{RL}_{BVC}$ is the level on the first tangent.

Highest or lowest point

At the highest (summit) or lowest (sag) point the grade is zero, $dy/dx = 0$:

$$\frac{g_1}{100} + \frac{(g_2-g_1)\,x}{100L} = 0 \quad\Rightarrow\quad x = \frac{g_1 L}{g_1-g_2}$$

Substituting this $x$ in the level equation gives the RL of that point:

$$\text{RL}_{max/min} = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L} = \text{RL}_{BVC} + \frac{g_1 x}{200}$$

If $g_1 > g_2$ (summit curve) this is the highest point; if $g_1 < g_2$ (sag curve) it is the lowest point. A turning point exists on the curve only if $0 < x < L$, i.e. $g_1$ and $g_2$ have opposite signs.