Vertical curves — for an up-grade of +g1% meeting a down-grade of g2% (summit curve) with total curve length L (measured horizontally), what is the horizontal distance between the point of intersection (PVI) and the highest point on the parabolic vertical curve?

Introduction / Context:
On a summit vertical curve formed by +g1% (up) and −g2% (down), the highest point occurs where the slope (grade) becomes zero. Knowing its location relative to the point of vertical intersection (PVI) is important for sight distance checks and profile detailing.

Given Data / Assumptions:

• Parabolic vertical curve of total length L (PVC to PVT).
• Grades: +g1% approaching, −g2% leaving (g1, g2 > 0).
• Horizontal distances are used.

Concept / Approach:
Let x be the distance from the PVC to any point on the curve. For a simple parabola, the local grade at distance x is: G(x) = g1 − ((g1 + g2) / L) * x. At the highest point, G(x) = 0, so x_HP = g1 * L / (g1 + g2). The PVI lies midway in chainage between PVC and PVT, i.e., at L/2 from PVC. Therefore, the offset distance from PVI to the highest point equals |x_HP − L/2| = L * |g1 − g2| / (2 * (g1 + g2)). For a summit where g1 and g2 are positive magnitudes, we customarily write the algebraic expression as L * (g1 − g2) / (2 * (g1 + g2)), taking direction according to whether g1 > g2 or vice versa.

Step-by-Step Solution:

Verification / Alternative check:
For a symmetric curve (g1 = g2), the formula gives zero, i.e., the highest point is exactly at the PVI projection, which is correct.

Why Other Options Are Wrong:

• Option (b) inverts the numerator/denominator and does not reduce to zero for g1 = g2.
• Options (c) and (d) give the distance from PVC to the highest point, not from PVI to the highest point.

Common Pitfalls:
Confusing the distance from PVC with the offset from PVI; always check the reference point requested.

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