Curve staking
Equal-tangent crest vertical curve: grade elevations and the high point
A 600 ft crest curve joining a +3.20 percent grade to a −2.40 percent grade is reduced to a grade sheet at full stations, with the high point located and the K value reported.
Given
- An equal-tangent parabolic vertical curve on a highway profile.
- Grade in g₁ = +3.20 percent; grade out g₂ = −2.40 percent.
- Curve length L = 600.00 ft, measured horizontally.
- PVI at station 34+50.00, elevation 852.40 ft.
- Grade elevations are wanted at every full station, plus the BVC, EVC and the high point.
- Elevations to 0.01 ft; stations in 100 ft US customary form.
Required
- Stations and elevations of the BVC and the EVC.
- The algebraic grade change, the rate of change of grade, the K value and the external offset.
- Station and elevation of the high point.
- A grade sheet at full stations.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Ends of the curve
An equal-tangent curve is centered on the PVI, so the BVC lies half a length back and the EVC half a length ahead. Their elevations are found by running the two grades from the PVI over that half length.
Grades are given in percent because plans label them that way, but every elevation computation uses them as decimals: +3.20 percent is 0.0320 ft of rise per foot of station.
BVC station = 34+50.00 − 300.00 = 31+50.00
BVC elevation = 852.40 − 0.0320 × 300.00 = 852.40 − 9.60 = 842.80 ft
EVC station = 34+50.00 + 300.00 = 37+50.00
EVC elevation = 852.40 + (−0.0240) × 300.00 = 852.40 − 7.20 = 845.20 ftCurve constants
The algebraic grade change A is the exit grade minus the entry grade, carrying signs. A negative A is a crest, a positive A is a sag, and the sign is not optional bookkeeping: it is the sign of every tangent offset on the curve.
The rate of change of grade r is A divided by the length. Expressed per station it is the number a profile sheet prints. K, the length per percent of grade change, is the value sight-distance tables are indexed on.
A = g₂ − g₁ = −2.40 − (+3.20) = −5.60 percent (crest, since A is negative)
r = A / L = −5.60 / 600.00 = −0.009333 percent per foot = −0.9333 percent per station
K = L / |A| = 600.00 / 5.60 = 107.14 ft per percent
External offset E = |A| L / 8 = 0.0560 × 600.00 / 8 = 4.20 ftLocate the high point
On the parabola the grade changes linearly from g₁ to g₂. The high point is where the grade passes through zero, which happens at the distance from the BVC where the accumulated grade change equals the entry grade.
A turning point exists inside the curve only when the two grades have opposite signs, which they do here. If both grades ran the same way, the highest point would be at one of the ends and this computation would produce a meaningless value outside the curve.
The elevation at the turning point has an unusually tidy form: because the grade falls linearly from g₁ to zero over that distance, the average grade over it is g₁/2, so the rise is g₁ x / 2.
x(high) = L g₁ / (g₁ − g₂) = 600.00 × 3.20 / 5.60 = 342.86 ft from the BVC
station = 31+50.00 + 342.86 = 34+92.86
elevation = 842.80 + 0.0320 × 342.86 / 2 = 842.80 + 5.49 = 848.29 ftGrade sheet
The tangent-offset form is the one to compute by hand. Run the back tangent grade out from the BVC to get a tangent elevation, then apply the offset (A/2L)x², where x is measured from the BVC. On this curve the offset coefficient is −0.056/1200 = −0.00004667 per foot squared.
Every offset is negative because the curve is a crest: the road surface lies below the extended entry grade everywhere past the BVC.
| Station | Note | x from BVC (ft) | Tangent elevation (ft) | Offset (ft) | Curve elevation (ft) |
|---|---|---|---|---|---|
| 31+50.00 | BVC | 0.00 | 842.80 | 0.00 | 842.80 |
| 32+00.00 | 50.00 | 844.40 | −0.12 | 844.28 | |
| 33+00.00 | 150.00 | 847.60 | −1.05 | 846.55 | |
| 34+00.00 | 250.00 | 850.80 | −2.92 | 847.88 | |
| 34+50.00 | PVI station | 300.00 | 852.40 | −4.20 | 848.20 |
| 34+92.86 | High point | 342.86 | 853.77 | −5.49 | 848.29 |
| 35+00.00 | 350.00 | 854.00 | −5.72 | 848.28 | |
| 36+00.00 | 450.00 | 857.20 | −9.45 | 847.75 | |
| 37+00.00 | 550.00 | 860.40 | −14.12 | 846.28 | |
| 37+50.00 | EVC | 600.00 | 862.00 | −16.80 | 845.20 |
Reading the sheet
The curve rises 5.49 ft from the BVC to the high point over 342.86 ft, then falls 3.09 ft to the EVC over the remaining 257.14 ft. Total rise from BVC to EVC is 2.40 ft, which is what the two tangent grades demand.
Note how flat the profile is around the summit: between 34+00 and 35+00 the elevation changes by only 0.40 ft. That flatness is what a K value of 107 describes, and it is also why the high point station matters much less than the high point elevation for clearance questions and much more for drainage ones.
Answer
- BVC at 31+50.00, elevation 842.80 ft. EVC at 37+50.00, elevation 845.20 ft.
- A = −5.60 percent, r = −0.9333 percent per station, K = 107.14 ft per percent, external offset E = 4.20 ft.
- High point at station 34+92.86, elevation 848.29 ft.
- Grade elevations at full stations: 844.28 at 32+00, 846.55 at 33+00, 847.88 at 34+00, 848.28 at 35+00, 847.75 at 36+00, 846.28 at 37+00.
Check
The offset at the PVI station must equal the external offset computed independently: (A/2L)(L/2)² = |A|L/8 = 4.20 ft. The table shows −4.20 ft at x = 300.00 ft.
Compute the EVC elevation a second way, from the forward tangent instead of the parabola: 852.40 + (−0.0240)(300.00) = 845.20 ft, which is the last row of the sheet.
The high point elevation by the average-grade shortcut, 842.80 + 0.0320(342.86)/2 = 848.29 ft, matches the value obtained from the full tangent-plus-offset computation, 853.77 − 5.49 = 848.28 ft, to a hundredth.
The grade at any station is g₁ + r x. At the high point that gives 3.20 + (−0.009333)(342.86) = 0.00 percent, confirming the turning point location.
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