Curve staking
Vertical curve length fitted to a fixed elevation at a fixed station
A sag curve must pass through an existing manhole rim at a set station and elevation. The required length comes out of a quadratic with two roots, only one of which puts the point on the curve.
Given
- An equal-tangent parabolic vertical curve is to be designed at a PVI at station 48+00.00, elevation 512.00 ft.
- Grade in g₁ = −2.00 percent; grade out g₂ = +3.00 percent.
- An existing manhole rim at station 50+00.00 must be met exactly. Its elevation is 519.25 ft.
- The PVI station, the PVI elevation and both grades are fixed by other constraints. Only the curve length may be varied.
- Elevations to 0.01 ft.
Required
- The curve length L that puts the finished profile through elevation 519.25 ft at station 50+00.00.
- The stations and elevations of the BVC and the EVC for that length.
- The low point, the K value and 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
Write the elevation at an arbitrary length
Measure the offset distance d from the PVI to the fixed point, positive ahead. For an equal-tangent curve of unknown length L, the BVC is L/2 back from the PVI, so the fixed point sits at x = d + L/2 from the BVC.
Substituting into the parabola and simplifying eliminates the BVC elevation entirely, leaving an expression in the PVI elevation, the entry grade, d and L alone. That is the form to solve.
elevation(x) = elev(BVC) + g₁ x + (A / 2L) x², with x measured from the BVC
elev(BVC) = elev(PVI) − g₁ L / 2 and x = d + L / 2
so elevation = elev(PVI) + g₁ d + (A / 2L)(d + L/2)²
A = g₂ − g₁ = 3.00 − (−2.00) = +5.00 percent = 0.0500
d = 50+00.00 − 48+00.00 = 200.00 ftReduce to a quadratic in L
Move everything that does not involve L to one side. What is left is the required parabolic offset at the fixed point, and it must be produced by a curve of the right length.
Multiplying through by 2L and collecting terms gives an ordinary quadratic. Note that the coefficient pattern is worth remembering: A/4 on L squared, A d minus twice the required offset on L, and A d squared as the constant.
required offset Δe = 519.25 − 512.00 − (−0.0200)(200.00) = 7.25 + 4.00 = 11.25 ft
(A / 2L)(d + L/2)² = Δe
A(d² + dL + L²/4) = 2 L Δe
(A/4) L² + (A d − 2Δe) L + A d² = 0
0.0125 L² + (10.00 − 22.50) L + 2000.00 = 0
0.0125 L² − 12.50 L + 2000.00 = 0
L² − 1000.00 L + 160,000 = 0Solve and reject the false root
The quadratic has two positive roots. That is not an accident of the numbers: the algebra was derived by assuming the fixed point lies on the parabola, and a second parabola can always be found whose extension passes through the same point even though the point is beyond its EVC.
The test is geometric, not algebraic. The fixed point lies on the curve only if d is less than L/2, that is only if L exceeds 400.00 ft. The short root fails that test.
Confirm the rejection numerically rather than by assertion. For L = 200 ft the EVC is at 49+00.00 at elevation 515.00 ft, and the forward tangent carries the profile to 518.00 ft at 50+00.00, not 519.25 ft. The short root satisfies the algebra of the extended parabola and not the profile.
L = [1000.00 ± √(1,000,000 − 640,000)] / 2 = [1000.00 ± 600.00] / 2
L = 800.00 ft or L = 200.00 ft
point is on the curve only if d < L/2, i.e. L > 400.00 ft
reject L = 200.00 ft; adopt L = 800.00 ftComplete the adopted curve
With L = 800.00 ft the curve runs from 44+00.00 to 52+00.00. The K value of 160 ft per percent is generous for a sag and comfortably satisfies ordinary headlight sight-distance requirements.
The low point falls at 47+20.00, upstation of the PVI, as it must when the entry grade is the flatter of the two.
BVC = 48+00.00 − 400.00 = 44+00.00, elevation = 512.00 + 0.0200 × 400.00 = 520.00 ft
EVC = 48+00.00 + 400.00 = 52+00.00, elevation = 512.00 + 0.0300 × 400.00 = 524.00 ft
K = 800.00 / 5.00 = 160.00 ft per percent
E = 0.0500 × 800.00 / 8 = 5.00 ft
x(low) = 800.00 × 2.00 / 5.00 = 320.00 ft, station 47+20.00
low point elevation = 520.00 − 0.0200 × 320.00 / 2 = 516.80 ftGrade sheet
Offsets are (A/2L)x² = 0.00003125x², measured up from the entry tangent produced from the BVC at 520.00 ft on a −2.00 percent grade.
The row at 50+00.00 is the one the whole problem was built around: it must read 519.25 ft, and it does.
| Station | Note | x from BVC (ft) | Tangent elevation (ft) | Offset (ft) | Curve elevation (ft) |
|---|---|---|---|---|---|
| 44+00.00 | BVC | 0.00 | 520.00 | 0.00 | 520.00 |
| 45+00.00 | 100.00 | 518.00 | 0.31 | 518.31 | |
| 46+00.00 | 200.00 | 516.00 | 1.25 | 517.25 | |
| 47+00.00 | 300.00 | 514.00 | 2.81 | 516.81 | |
| 47+20.00 | Low point | 320.00 | 513.60 | 3.20 | 516.80 |
| 48+00.00 | PVI station | 400.00 | 512.00 | 5.00 | 517.00 |
| 49+00.00 | 500.00 | 510.00 | 7.81 | 517.81 | |
| 50+00.00 | Manhole rim | 600.00 | 508.00 | 11.25 | 519.25 |
| 51+00.00 | 700.00 | 506.00 | 15.31 | 521.31 | |
| 52+00.00 | EVC | 800.00 | 504.00 | 20.00 | 524.00 |
Answer
- L = 800.00 ft. The other root of the quadratic, L = 200.00 ft, is rejected because station 50+00.00 would lie 100 ft beyond its EVC.
- BVC at 44+00.00, elevation 520.00 ft. EVC at 52+00.00, elevation 524.00 ft.
- Low point at 47+20.00, elevation 516.80 ft. K = 160.00 ft per percent, external offset E = 5.00 ft.
- The finished profile passes through 519.25 ft at station 50+00.00 exactly, as required.
Check
Substitute back into the parabola independently of the quadratic. At station 50+00.00, x = 600.00 ft, tangent elevation 520.00 − 0.0200(600.00) = 508.00 ft, offset (0.0500/1600)(600.00²) = 11.25 ft, curve elevation 519.25 ft. That is the required value to the hundredth.
Demonstrate that the short root fails. For L = 200.00 ft the EVC lies at 49+00.00 at elevation 515.00 ft, and the forward tangent gives 515.00 + 0.0300(100.00) = 518.00 ft at station 50+00.00, which is 1.25 ft below the manhole rim.
The offset at the PVI station must equal |A|L/8 = 5.00 ft, and the sheet shows 5.00 ft at x = 400.00 ft.
The rise from the low point to the EVC, 524.00 − 516.80 = 7.20 ft, must equal the average of 0 and +3.00 percent applied over 480.00 ft: 0.0150 × 480.00 = 7.20 ft.
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