Vertical curves – maximum offset (sagitta) of vertex below PVI For a parabolic vertical curve of length L joining grades +g1% and -g2% (g1 > g2), what is the maximum vertical distance of the curve apex (vertex) from the point of intersection (PVI) of the two tangents?

Introduction / Context:
The vertex (apex) of a parabolic vertical curve is the point where the slope becomes zero. Its maximum offset below (for summit) or above (for sag) the PVI is a key geometric quantity used to compute levels and check clearance.

Given Data / Assumptions:

• Parabolic vertical curve of total length L.
• Approach grades are +g1% and -g2% with g1 > g2.
• Offsets are measured vertically from the PVI down to the curve (for a summit case).

Concept / Approach:
For a parabolic vertical curve with uniform rate of change of grade, the slope at a distance x from the beginning is: slope(x) = g1 - ((g1 + g2)/L) * x. The vertex occurs where slope(x) = 0. The offset from PVI at a general x is y(x) = ((g1 + g2) * x * (L - x)) / 200 when grades are in percent and L in meters.

Step-by-Step Solution:
Set slope(x) = 0 → g1 - ((g1 + g2)/L) * x = 0.Solve for x: x_v = (g1 * L) / (g1 + g2).Compute offset at x_v: y_max = ((g1 + g2) * x_v * (L - x_v)) / 200.Substitute x_v: y_max = (g1 * g2 * L^2) / (200 * (g1 + g2)).

Verification / Alternative check:
For symmetric grades (g1 = g2 = g), x_v = L/2 and y_max reduces to (g * L^2) / 800, which is the well-known mid-ordinate for a symmetric summit curve.

Why Other Options Are Wrong:

• ((g1 + g2) * L) / 200 and (g1 * g2 * L) / 200: dimensions wrong (length vs. length^2 factor missing).
• (g1 + g2) * L^2 / 800: valid only when g1 = g2; not correct for general unequal grades.

Common Pitfalls:
Using the symmetric formula for unequal grades, or forgetting to divide by 200 when grades are in percent.

Final Answer:
(g1 * g2 * L^2) / (200 * (g1 + g2))