Transportation Learning HubTransportation Learning Hub · PE Civil Transportation · Area 5 of 10
Vertical Design — PE Civil Transportation
Vertical design is the road in profile: parabolic curves joining grade to grade, elevations at any point along them, and curve lengths long enough that a driver can always see far enough to stop.
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Vertical alignment geometrics
An equal-tangent vertical curve is a parabola from the PVC (point of vertical curvature) to the PVT, with the PVI — the intersection of the two grade lines — at mid-length. Grades are signed (upgrade positive) and enter the elevation equation as decimals:
\[y(x) = y_{PVC} + g_{1} x + (g_{2} - g_{1}) \frac{x^{2}}{2L} \qquad x_{hp} = \frac{g_{1} L}{(g_{1} - g_{2})} \qquad K = \frac{L}{A}\]
| y(x) | elevation at distance x (ft) from the PVC |
|---|---|
| g1, g2 | entering and exiting grades, signed decimals (+2% = +0.02) |
| xhp | distance from the PVC to the high point (crest) or low point (sag) — real only if it lands between 0 and L |
| A | absolute algebraic grade difference, in percent |
| K | rate of vertical curvature: horizontal distance for a 1% change of grade, ft/% |
Vertical clearance (subtopic A) is the same geometry read at a structure: the curve equation gives the road surface elevation, and the clearance is the gap between that surface and the underside of whatever crosses it (or the top of whatever passes beneath) — take current minimum values from the current design references, never from memory.
Worked example Elevation on a crest curve, and its K value
Given: a crest curve of length \(L = 400\) ft joins \(g_{1} = +2\%\) to \(g_{2} = -3\%\). The PVC elevation is 100.00 ft. Find the elevation 100 ft from the PVC, the high-point location, and the curve’s K value.
Solution:
- Work in decimals: \(g_{1} = +0.02\), \(g_{2} = -0.03\), so \((g_{2} - g_{1}) = -0.05\).
- Elevation at \(x = 100\): \(y = 100.00 + (0.02)(100) + \dfrac{(-0.05)}{2 \times 400}(100)^{2} = 100.00 + 2.00 - 0.625 = 101.38\) ft.
- High point: \(x_{hp} = \dfrac{0.02 \times 400}{0.02 - (-0.03)} = \dfrac{8}{0.05} = 160\) ft from the PVC — inside the curve, so it is real.
- K value: \(A = |2 - (-3)| = 5\%\), so \(K = L/A = 400/5 = 80\) ft per percent of grade change.
Answer: elevation at \(x = 100\) ft is 101.38 ft; high point 160 ft from the PVC; \(K = 80\) ft/%.
Crest curves — sight distance over the top
A crest curve hides the road beyond it. Design therefore checks the line of sight over the crest between the driver’s eye and an object on the far side: the curve must be long enough that this sightline clears the pavement everywhere and delivers at least the stopping sight distance for the design speed. Where passing is to be permitted on a two-lane road, passing sight distance — much longer than stopping sight distance, because the manoeuvre occupies the opposing lane — can govern instead, and many crests are deliberately designed to stopping sight distance with passing prohibited. When you solve for a crest length, the working splits into two cases depending on whether the sight distance is shorter or longer than the curve itself; compute with one case, then check the case assumption against your answer and switch if it fails.
Sag curves — headlight sight distance
A sag curve hides nothing in daylight; at night the limit is the headlight beam, which points ahead and slightly down and lights only a finite length of the dip. Sag length is set so the headlight sight distance at least matches the stopping sight distance — a completely different control from the crest’s eye-over-the-top geometry, and the two must never be swapped. Two secondary checks ride along: rider comfort through the bottom of the dip, and drainage — a sag low point is where water collects, so the low point’s location (from the \(x_{hp}\) formula) matters for inlet placement as well as for the profile.