Curve staking
Superelevation runoff and tangent runout layout for a highway curve
A 6 percent superelevation transition is laid out around a 1000 ft radius curve: runoff and runout lengths from the relative gradient, the five key stations, and edge-of-pavement elevations for the stakes.
Given
- A two-lane highway curve to the right, R = 1000.00 ft, Δ = 30°00′00″, PI at station 76+80.00.
- Travelled way 24.00 ft wide, two 12.00 ft lanes, rotated about the centerline.
- Normal crown 2.0 percent each way from the centerline.
- Design superelevation e = 6.0 percent.
- Maximum relative gradient between the centerline profile and the pavement edge = 0.50 percent for this design speed.
- Agency practice places two thirds of the superelevation runoff on the tangent ahead of the PC and one third on the curve.
- Centerline profile: a constant +1.20 percent grade, elevation 612.40 ft at station 74+00.00.
- All distances in US survey feet; elevations to 0.01 ft.
Required
- Stations of the PC and the PT.
- The superelevation runoff length and the tangent runout length.
- The five key transition stations at the approach end, with the cross slope at each.
- Centerline and edge-of-pavement elevations at those 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
Station the curve
The horizontal curve is solved and stationed first, because every transition station is referenced to the PC and the PT.
The transition is symmetric about the curve, so the approach end is computed in full and the departure end mirrored.
Δ/2 = 15°00′00″
T = 1000.00 × tan(15°00′00″) = 1000.00 × 0.267949 = 267.95 ft
L = 1000.00 × 0.523599 = 523.60 ft
PC = 76+80.00 − 267.95 = 74+12.05
PT = PC + L = 74+12.05 + 523.60 = 79+35.65Runoff and runout lengths
The superelevation runoff is the length over which the outside lane is rotated from level to full superelevation. It is governed by the relative gradient, the difference in longitudinal grade between the centerline and the pavement edge, which is capped so that the edge profile does not look or drive as a kink.
For rotation about the centerline the edge rises by the lane width times the cross slope, so the runoff length is that rise divided by the allowable relative gradient.
The tangent runout is the length before the runoff over which the outside lane is brought from normal crown to level. It uses the same relative gradient, so it is the runoff length scaled by the ratio of normal crown to design superelevation.
This alignment has no spiral. Where a spiral is used, the runoff is placed along the spiral, from the TS to the SC, so that full superelevation is reached exactly where the radius reaches its minimum, and the two thirds and one third split below is not needed.
edge rise at full superelevation = w e = 12.00 × 0.060 = 0.72 ft
L(runoff) = w e / relative gradient = 0.72 / 0.0050 = 144.00 ft
L(runout) = L(runoff) × (normal crown / e) = 144.00 × 2.0 / 6.0 = 48.00 ft
runoff on tangent = (2/3)(144.00) = 96.00 ft
runoff on curve = (1/3)(144.00) = 48.00 ftThe five key stations at the approach end
Work outward from the PC. Full superelevation is reached one third of the runoff past the PC. The runoff begins two thirds of the runoff before the PC, and at that point the outside lane is level. The tangent runout begins one runout length before that, where the section is at normal crown.
The reverse crown station, where the whole travelled way is a plane at 2.0 percent, falls one third of the way through the runoff, because the outside lane has to climb from 0 to 2.0 out of a total 6.0.
full superelevation begins = PC + 48.00 = 74+60.05
PC = 74+12.05, outside lane at 4.0 percent (two thirds through the runoff)
reverse crown = PC − 96.00 + 48.00 = 73+64.05, outside lane at +2.0 percent
crown removed, runoff begins = PC − 96.00 = 73+16.05, outside lane at 0.0 percent
normal crown, runout begins = 73+16.05 − 48.00 = 72+68.05, outside lane at −2.0 percentEdge elevations for the stakes
Centerline elevations come off the profile grade: 612.40 ft at 74+00.00 on a +1.20 percent grade. Edge elevations are the centerline elevation plus the lane width times the cross slope, signed.
The outside edge is the one that moves. The inside edge stays at −2.0 percent until the reverse crown station, after which the whole travelled way rotates as a plane and the inside edge falls away at the same rate the outside rises.
| Station | Condition | Outside slope | Centerline elevation (ft) | Outside edge (ft) | Inside edge (ft) |
|---|---|---|---|---|---|
| 72+68.05 | Normal crown, runout begins | −2.0% | 610.82 | 610.58 | 610.58 |
| 73+16.05 | Crown removed, runoff begins | 0.0% | 611.39 | 611.39 | 611.15 |
| 73+64.05 | Reverse crown, section is a plane | +2.0% | 611.97 | 612.21 | 611.73 |
| 74+12.05 | PC | +4.0% | 612.54 | 613.02 | 612.06 |
| 74+60.05 | Full superelevation | +6.0% | 613.12 | 613.84 | 612.40 |
The departure end
The layout mirrors about the curve. Full superelevation is held from 74+60.05 all the way to a point one third of a runoff before the PT, then the section is unwound over the same lengths.
Between 74+60.05 and 78+87.65, a length of 427.60 ft, the pavement is at a constant 6.0 percent, which is the only part of the alignment where the design superelevation is fully developed.
full superelevation ends = PT − 48.00 = 78+87.65
PT = 79+35.65, outside lane at 4.0 percent
runoff ends, crown removed = PT + 96.00 = 80+31.65
normal crown restored = 80+31.65 + 48.00 = 80+79.65Answer
- PC = 74+12.05, PT = 79+35.65.
- Superelevation runoff L(runoff) = 144.00 ft; tangent runout L(runout) = 48.00 ft; 96.00 ft of runoff on the tangent and 48.00 ft on the curve.
- Approach transition stations: normal crown 72+68.05, crown removed 73+16.05, reverse crown 73+64.05, PC 74+12.05 at 4.0 percent, full superelevation 74+60.05.
- Departure transition stations: full superelevation ends 78+87.65, PT 79+35.65 at 4.0 percent, crown removed 80+31.65, normal crown restored 80+79.65.
- Outside edge elevations across the approach transition: 610.58, 611.39, 612.21, 613.02 and 613.84 ft.
Check
The relative gradient must come back out of the finished layout. The outside edge rises from 611.39 ft at 73+16.05 to 613.84 ft at 74+60.05, a rise of 2.45 ft in 144.00 ft, which is 1.70 percent. Over the same length the centerline rises 613.12 − 611.39 = 1.73 ft, or 1.20 percent. The difference is 0.50 percent, the allowable relative gradient.
The same test across the runout: the outside edge rises 611.39 − 610.58 = 0.81 ft in 48.00 ft, or 1.69 percent, against a centerline gain of 611.39 − 610.82 = 0.57 ft, or 1.19 percent. The difference is again 0.50 percent.
Edge offset at full superelevation must equal w e: 613.84 − 613.12 = 0.72 ft = 12.00 × 0.060.
The transition lengths must sum consistently: 96.00 ft on tangent plus 48.00 ft on curve equals the 144.00 ft runoff, and the cross slope at the PC is therefore 6.0 × 96.00/144.00 = 4.0 percent, which is the value tabulated.
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