Curve staking
Fit a horizontal curve to a required external distance
An obstruction near the PI forces the curve to stay at least 65.00 ft out from the intersection point. The minimum radius is solved from the external distance, rounded to a design value, and stationed.
Given
- Two tangents intersect at a PI with a deflection angle Δ = 36°20′00″. Station of the PI = 96+14.50.
- A masonry headwall stands on the bisector of the interior angle at the PI. Its face is 65.00 ft from the PI, measured along the bisector toward the outside of the curve.
- The alignment must pass outside the headwall, so the midpoint of the curve must be at least 65.00 ft from the PI.
- Design policy is to adopt a radius that is a multiple of 25 ft.
- All distances are US survey feet.
Required
- The minimum radius that satisfies the external-distance constraint.
- The design radius adopted, and the external distance it actually gives.
- The complete curve elements and the stations of the PC and PT for the adopted radius.
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
The external distance is the controlling element
The external distance E is measured from the PI along the bisector of the interior angle to the midpoint of the curve. That is exactly the direction in which the headwall lies, so E is the element the obstruction constrains.
For a fixed Δ, E grows in direct proportion to R: E = R(sec(Δ/2) − 1). The bracketed factor depends only on Δ, so it can be computed once and reused for every trial radius.
Δ/2 = 36°20′00″ / 2 = 18°10′00″
cos(18°10′00″) = 0.950154
sec(Δ/2) = 1 / 0.950154 = 1.052461
sec(Δ/2) − 1 = 0.052461Solve for the minimum radius
Rearrange the external-distance relation for R and substitute the required clearance. Because E is proportional to R, a larger radius pushes the curve further from the PI and further from the headwall; the constraint is therefore a lower bound on R, not an upper one.
Getting that direction backwards is the common error in this problem. Sketching the PI, the bisector and the curve before computing settles it in ten seconds.
R = E / (sec(Δ/2) − 1)
R(min) = 65.00 / 0.052461 = 1239.01 ft
Round up to the next 25 ft multiple: R = 1250 ftElements of the adopted curve
With R = 1250 ft and Δ = 36°20′00″ every element scales directly off the trigonometric functions of the half angle. The external distance is checked first, since it is the reason the radius was chosen.
The middle ordinate is also worth recording. It is the offset from the long chord to the curve at its midpoint, and it is what a crew uses to spot the center of the curve with a tape when the PI is not occupiable.
| Element | Formula | Value (ft) |
|---|---|---|
| External E | 1250 × 0.052461 | 65.58 |
| Tangent T | 1250 × 0.328139 | 410.17 |
| Arc length L | 1250 × 0.634136 | 792.67 |
| Long chord C | 2500 × 0.311782 | 779.46 |
| Middle ordinate M | 1250 × 0.049846 | 62.31 |
Station the curve
The PC is one tangent distance back from the PI; the PT is the PC station plus the arc length, never the PI station plus the tangent distance.
Carry the unrounded tangent and arc into the station arithmetic and round once at the end. Rounding T to 410.17 before subtracting and then rounding again introduces a hundredth that will not reconcile when the next curve down the line is stationed.
PC = 96+14.50 − 410.17 = 92+04.33
PT = PC + L = 92+04.33 + 792.67 = 99+97.00
Chainage lost at the corner: 2T − L = 820.35 − 792.67 = 27.68 ftConfirm the clearance
The adopted curve gives E = 65.58 ft against a required 65.00 ft, so the midpoint of the curve clears the headwall face by 0.58 ft along the bisector. That is thin. If the headwall position carries any measurement uncertainty, or if the curve will be widened for superelevation, the next radius up should be examined.
For comparison, R = 1225 ft would give E = 64.27 ft, which fails the constraint. R = 1275 ft would give E = 66.89 ft and a 1.89 ft clearance, at the cost of 8.21 ft of extra tangent distance at each end.
Answer
- Minimum radius R(min) = 1239.01 ft.
- Adopted design radius R = 1250 ft, giving E = 65.58 ft, which clears the required 65.00 ft by 0.58 ft.
- T = 410.17 ft, L = 792.67 ft, C = 779.46 ft, M = 62.31 ft.
- PC = 92+04.33, PT = 99+97.00.
Check
Work the external distance backwards from the adopted radius: E = 1250 × (1/cos 18°10′00″ − 1) = 1250 × 0.052461 = 65.58 ft, which is greater than the required 65.00 ft.
The external and middle ordinate must satisfy E − M = E M / R for any circular curve. Here E − M = 65.58 − 62.31 = 3.27 ft, and E M / R = 65.58 × 62.31 / 1250 = 3.27 ft.
Independent station check: PT − PC = 99+97.00 − 92+04.33 = 792.67 ft, which is the arc length, as it must be.
More curve staking
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