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High or Low Point on a Vertical Curve

xt=−g1LAx_t = -\frac{g_1 L}{A}

Worked example: Sag low point 120 m past the BVC, −2 % and A = 5 % → 300 m curve — press Try an example to run it live, then adjust anything.

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High or Low Point on a Vertical Curve explained

g1AxtL

On a parabola the grade changes linearly from g₁ at the BVC to g₂ at the EVC, so it passes through zero at the fraction g₁/A of the way along — hence x = −g₁L/A. That point is the crest of a hill or the bottom of a sag, and it is where the drainage engineer needs to be standing: sag low points are where water collects and inlets go, crest high points are where the sightline breaks. If the result comes out negative or larger than L, the profile has no turning point inside the curve at all, meaning both grades run the same way and the road simply steepens or flattens.

A worked example: a road climbing at g₁ = +3 % meets a −2 % descent over a 500 ft crest, so A = −5 %. The high point sits x = −3 × 500/(−5) = 300 ft past the BVC — not at midcurve, which is the intuition to unlearn. The turning point is at midcurve only when the two grades are equal and opposite. Combine this with the elevation formula and you get the crest elevation itself, which is what determines whether a stop sign will be visible over the hill.

High or Low Point on a Vertical Curve formula

xt=−g1LAx_t = -\frac{g_1 L}{A}
Where
  • xtx_t= Distance from the BVC to the turning point (m)
  • g1g_1= Grade entering the curve (%)
  • LL= Total curve length (m)
  • AA= Algebraic grade change (%)

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