A highway vertical curve is a parabola, not a circle, and the reason is comfort rather than geometry. A parabola gives a constant rate of change of grade, which is what a driver feels as a steady push into or out of the seat. Everything about the computation follows from that one property, and it is why the elevation equation is quadratic and why the offsets from the tangent go up as the square of the distance.
The question that actually gets asked - on the exam, on the plan sheet and on site - is where the high or low point falls. On a crest it decides sight distance. In a sag it decides where water stands, which is to say where the inlet goes. Getting it wrong by fifty feet is the difference between a drain that works and a puddle.
What follows is the equation, the derived quantities that plan sheets label, a crest curve worked to a full grade-elevation table, and the specific case where the turning point does not exist at all.
The equation and everything derived from it
Measure x horizontally from the BVC, the beginning of the vertical curve, take the grades as decimals, and let L be the horizontal length of the curve. The elevation on the curve is the tangent grade line plus a parabolic offset.
elev(x) = elev_BVC + g₁·x + (g₂ − g₁)·x² ⁄ (2L)
A = g₂ − g₁ r = A ⁄ L E = |A|·L ⁄ 8 K = L ⁄ |A%|A is algebraic, which is the detail candidates lose marks on. A curve from +2.4% to −3.6% has A = −6.0, not −1.2 and not 6.0. Negative A is a crest, positive A is a sag, and the sign carries all the way through to the offsets, which are negative on a crest because the curve sags below its own tangent.
- K value
- Length of curve per percent of grade change, L ⁄ |A|. Sight-distance tables are indexed by K rather than by L because K captures the flatness of the curve independently of how much grade it has to absorb. Design charts give a minimum K; the curve length falls out as K × |A|.
- External offset E
- The vertical distance from the PVI to the curve, |A|·L ⁄ 8 with A as a decimal. It is the deepest the curve ever departs from the tangents, and it occurs at the PVI station, which is the midpoint of an equal-tangent curve.
A crest curve, worked
Take an 800 ft curve from +2.40% to −3.60%, with the PVI at station 32+00 and elevation 528.40. Because the curve is equal-tangent, the BVC and EVC are 400 ft either side of the PVI.
| Quantity | Value |
|---|---|
| Grade change A | −6.00% |
| Curve type | Crest |
| K value | 133.33 ft per percent |
| Rate of change r | −0.75% per station |
| External offset E | 6.00 ft |
| BVC | 28+00.00 at elevation 518.80 |
| EVC | 36+00.00 at elevation 514.00 |
The BVC elevation is the PVI elevation less the entering grade over half the length: 528.40 − 0.024 × 400 = 518.80. The EVC is the PVI elevation plus the exiting grade over the other half: 528.40 − 0.036 × 400 = 514.00. Both are on the tangents, because the curve meets its tangents at those two points.
The turning point
The high or low point is where the grade on the curve reaches zero. Differentiate the elevation equation, set the derivative to zero, and the offset from the BVC falls out.
x_t = −g₁ ⁄ (A ⁄ L) = L · g₁ ⁄ (g₁ − g₂)
= 800 × 0.024 ⁄ (0.024 + 0.036) = 320.00 ft from the BVCSubstituting x = 320 into the elevation equation gives 522.64. So the crest of this curve sits at station 31+20 at elevation 522.64, while the PVI sits at 32+00 at a tangent elevation of 528.40 - 5.76 ft higher and 80 ft further on. The turning point coincides with the PVI only when the two grades are equal and opposite.
| Station | x from BVC (ft) | Tangent elevation | Offset | Curve elevation |
|---|---|---|---|---|
| 28+00.00 (BVC) | 0 | 518.80 | 0.00 | 518.80 |
| 29+00.00 | 100 | 521.20 | −0.375 | 520.825 |
| 30+00.00 | 200 | 523.60 | −1.500 | 522.100 |
| 31+00.00 | 300 | 526.00 | −3.375 | 522.625 |
| 31+20.00 (high) | 320 | 526.48 | −3.840 | 522.640 |
| 32+00.00 (PVI) | 400 | 528.40 | −6.000 | 522.400 |
| 33+00.00 | 500 | 530.80 | −9.375 | 521.425 |
| 34+00.00 | 600 | 533.20 | −13.500 | 519.700 |
| 35+00.00 | 700 | 535.60 | −18.375 | 517.225 |
| 36+00.00 (EVC) | 800 | 538.00 | −24.000 | 514.000 |
The offsets are the check. They go as the square of the distance from the BVC: at 200 ft the offset is 1.500, at 400 ft it is four times that at 6.000, and at 800 ft it is sixteen times at 24.000. If your offsets do not scale as x², the arithmetic is wrong somewhere and no amount of checking the elevations will find it.
When there is no turning point
The formula x_t = L·g₁ ⁄ (g₁ − g₂) returns a number for any pair of grades, but that number only means something when it falls between 0 and L. It does so exactly when the two grades have opposite signs.
On a curve from +1.5% to +4.0% both grades are uphill throughout, so the curve never levels out. The algebraic A is +2.5, which technically makes it a sag curve, and the lowest point on it is the BVC - an endpoint, not a turning point. Substituting into the formula gives an x outside the curve, which is the arithmetic telling you the vertex of the parabola lies off the ends of the piece you have drawn.
- Opposite-sign grades: the turning point is inside the curve. Compute it.
- Same-sign grades: the extreme elevation is at the BVC or the EVC. Compare the two and report the endpoint.
- Equal and opposite grades: the turning point is at the PVI station, and only then.
- A zero exiting grade: the turning point is exactly at the EVC, where the curve becomes level.
For drainage the distinction is not academic. A sag curve between two downgrades has no low point on the curve at all, so the water keeps running and the inlet belongs somewhere else entirely. Checking the sign of the grades before reaching for the formula takes two seconds and prevents the most expensive kind of vertical curve mistake.