Where a road's profile changes grade, the two tangents are joined by a parabola — not a circle. The reason is worth one sentence: on a parabola the grade changes at a constant rate, so a vehicle's vertical acceleration is constant, and constant is what a passenger's stomach and a headlight beam both want.

The letters, once, in words. $g_1$ is the grade entering the curve and $g_2$ the grade leaving it, both in percent and both signed — uphill positive, downhill negative. $A = g_2 - g_1$ is the algebraic grade change, also in percent, and its sign says which way the road bends: negative is a crest, positive a sag. $L$ is the whole curve length from BVC to EVC, where BVC is the beginning of vertical curve and EVC the end. $x$ is the distance measured forward from the BVC, and $E_{BVC}$ is the elevation there.

Length from K: $L = K A$ — L equals K A. $K$ is a design standard read off a table at the design speed, in metres of curve per percent of grade change. It is a rate, so it multiplies. A bigger kink needs a longer curve, and K says how much longer.

Elevation anywhere: $E = E_{BVC} + \dfrac{g_1 x}{100} + \dfrac{A x^{2}}{200 L}$. Three terms, three plain ideas: where you started, how far the straight entering tangent would have carried you, and how far the parabola has pulled away from that tangent by station $x$. The house trap lives in the divisions. Grades arrive as percents, so the first term divides by 100; the second divides by 200, which is that same 100 with the 2 that falls out of integrating a linear rate. Leave either out and the answer is off by hundreds of metres.

Turning point: $x_t = -\dfrac{g_1 L}{A}$, the distance past the BVC where the profile grade reaches zero — the crest of a hill or the low point of a sag. And here is the nugget: it is not at midcurve unless the two grades happen to be equal and opposite. Set a catch basin at midcurve on a curve joining −6% to +2% and it will sit three-quarters of the way along, watching the water pond somewhere else.