Curve staking
Sag vertical curve: low point, overhead clearance and the drainage check
A 500 ft sag curve is reduced to a grade sheet, the low point is located for drainage, clearance under an overhead structure is verified, and the flat zone around the low point is measured against a 100 ft criterion.
Given
- An equal-tangent parabolic sag curve carrying a curbed roadway.
- Grade in g₁ = −3.50 percent; grade out g₂ = +1.80 percent.
- Curve length L = 500.00 ft.
- PVI at station 61+00.00, elevation 428.60 ft.
- An existing structure crosses the roadway at station 60+50.00. The underside of its beams is at elevation 447.10 ft, and the required vertical clearance is 14.50 ft.
- Drainage policy: the profile grade must be steeper than 0.30 percent except over a length of at most 100 ft either side of the low point taken together.
- Elevations to 0.01 ft.
Required
- Stations and elevations of the BVC and the EVC, and the curve constants.
- Station and elevation of the low point.
- A grade sheet at full stations.
- Whether the clearance under the structure is met.
- The length of profile over which the grade is flatter than 0.30 percent, and whether it passes the drainage criterion.
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 and constants
The curve is centered on the PVI, so the BVC is 250.00 ft back and the EVC 250.00 ft ahead. The algebraic grade change is positive, which makes this a sag: the pavement lies above the extended entry grade everywhere past the BVC.
K, the length per percent of grade change, is the number that both the headlight sight-distance criterion and the drainage criterion are written against, so it is worth computing early.
BVC = 61+00.00 − 250.00 = 58+50.00, elevation = 428.60 + 0.0350 × 250.00 = 437.35 ft
EVC = 61+00.00 + 250.00 = 63+50.00, elevation = 428.60 + 0.0180 × 250.00 = 433.10 ft
A = +1.80 − (−3.50) = +5.30 percent (sag)
r = 5.30 / 500.00 = 0.0106 percent per foot = 1.06 percent per station
K = 500.00 / 5.30 = 94.34 ft per percent
External offset E = 0.0530 × 500.00 / 8 = 3.31 ftLocate the low point
The low point is where the grade passes through zero. Measured from the BVC that is at a distance L g₁ divided by the algebraic difference of the grades, using magnitudes for a sag.
This is the point a curb inlet has to be set at or near. Set the inlet at the PVI station instead and it sits 0.34 ft above the true low point, which on a curbed section is the difference between a puddle and a drain.
x(low) = L |g₁| / |A| = 500.00 × 3.50 / 5.30 = 330.19 ft from the BVC
station = 58+50.00 + 330.19 = 61+80.19
elevation = 437.35 − 0.0350 × 330.19 / 2 = 437.35 − 5.78 = 431.57 ftGrade sheet
Tangent elevations run from the BVC on the entry grade, 437.35 − 0.0350x. Offsets are (A/2L)x² = 0.0000530x², all positive because the curve is a sag.
The station of the structure, 60+50.00, is included because that elevation is needed for the clearance check.
| Station | Note | x from BVC (ft) | Tangent elevation (ft) | Offset (ft) | Curve elevation (ft) |
|---|---|---|---|---|---|
| 58+50.00 | BVC | 0.00 | 437.35 | 0.00 | 437.35 |
| 59+00.00 | 50.00 | 435.60 | 0.13 | 435.73 | |
| 60+00.00 | 150.00 | 432.10 | 1.19 | 433.29 | |
| 60+50.00 | Under structure | 200.00 | 430.35 | 2.12 | 432.47 |
| 61+00.00 | PVI station | 250.00 | 428.60 | 3.31 | 431.91 |
| 61+80.19 | Low point | 330.19 | 425.79 | 5.78 | 431.57 |
| 62+00.00 | 350.00 | 425.10 | 6.49 | 431.59 | |
| 63+00.00 | 450.00 | 421.60 | 10.73 | 432.33 | |
| 63+50.00 | EVC | 500.00 | 419.85 | 13.25 | 433.10 |
Clearance under the structure
Clearance is the beam soffit elevation less the profile grade elevation at the same station. The critical station is not necessarily where the beam is lowest and not necessarily where the profile is highest; on a sag it is wherever the two come closest, which here is at the crossing itself because the profile is still falling.
The margin is 0.13 ft. That is real but thin, and it assumes the structure elevation is exact and that the pavement will be built to grade. Any future overlay eats it.
profile elevation at 60+50.00 = 430.35 + 2.12 = 432.47 ft
clearance = 447.10 − 432.47 = 14.63 ft
required = 14.50 ft; 14.63 > 14.50, so the clearance is met by 0.13 ftDrainage check around the low point
The grade on a parabola is g₁ + r x, a straight line in x. Solve it for the two stations where the grade magnitude is 0.30 percent, one on each side of the low point, and the distance between them is the flat zone.
There is a shortcut worth knowing: the length over which the grade is flatter than p percent is 2pK. Here 2 × 0.30 × 94.34 = 56.60 ft, which is the answer without touching the stations at all. The criterion of 100 ft therefore translates into a maximum K of 167 for a 0.30 percent threshold.
grade(x) = −3.50 + 0.0106 x percent
grade = −0.30 percent at x = 3.20 / 0.0106 = 301.89 ft, station 61+51.89
grade = +0.30 percent at x = 3.80 / 0.0106 = 358.49 ft, station 62+08.49
flat zone = 358.49 − 301.89 = 56.60 ft
shortcut: 2 p K = 2 × 0.30 × 94.34 = 56.60 ft
56.60 ft ≤ 100 ft, so the drainage criterion is satisfiedAnswer
- BVC at 58+50.00, elevation 437.35 ft. EVC at 63+50.00, elevation 433.10 ft.
- A = +5.30 percent, K = 94.34 ft per percent, external offset E = 3.31 ft.
- Low point at station 61+80.19, elevation 431.57 ft.
- Clearance under the structure at 60+50.00 is 14.63 ft against a required 14.50 ft, so it passes with 0.13 ft to spare.
- The grade is flatter than 0.30 percent over 56.60 ft, from 61+51.89 to 62+08.49, which is within the 100 ft limit.
Check
The offset at the PVI station must equal the external offset: |A|L/8 = 0.0530 × 500.00 / 8 = 3.31 ft, and the sheet shows 3.31 ft at x = 250.00 ft.
The EVC elevation from the forward tangent instead of the parabola: 428.60 + 0.0180(250.00) = 433.10 ft, matching the last row.
The low point elevation two ways: 425.79 + 5.78 = 431.57 ft from the sheet, and 437.35 − 0.0350(330.19)/2 = 431.57 ft from the average-grade shortcut.
The flat zone two ways: 56.60 ft from the two solved stations, and 2pK = 56.60 ft from the K value. Agreement confirms both the K value and the grade equation.
Net fall from BVC to EVC: 437.35 − 433.10 = 4.25 ft. Running the two tangents out of the PVI independently gives 0.0350 × 250.00 = 8.75 ft up to the BVC and 0.0180 × 250.00 = 4.50 ft up to the EVC, and 8.75 − 4.50 = 4.25 ft.
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