Curve staking
Recover a curve radius from a measured long chord and middle ordinate
An existing curve with no plan record is measured with a tape: the long chord between two identifiable points and the middle ordinate at its center. Radius and central angle follow in closed form.
Given
- An existing roadway curve is being retraced. No plan or record of the alignment survives.
- Two points on the face of curb, both on the curve, are recovered. The straight tape distance between them is 682.34 ft.
- At the midpoint of that chord the perpendicular offset from the chord to the curb face measures 50.22 ft.
- The two recovered points are the PC and the PT, so the taped chord is the long chord of the whole curve.
- All distances are US survey feet.
Required
- The radius R of the existing curve.
- The central angle Δ.
- The tangent distance, arc length and external distance, so the curve can be re-established from the tangents.
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
Why two elements are enough
A circular curve has one shape parameter and one size parameter. Fix any two independent elements and the curve is fully determined. Here the two measured quantities are the long chord C and the middle ordinate M, neither of which is R or Δ, so the solution starts from geometry rather than from a plan table.
The relationship is exact and needs no iteration. Draw the radius to the midpoint of the arc: it bisects the long chord perpendicularly and passes through the middle ordinate. That gives a right triangle with legs C/2 and R − M and hypotenuse R.
(C/2)² + (R − M)² = R²
C²/4 + R² − 2RM + M² = R²
C²/4 + M² = 2RM
R = C²/(8M) + M/2Compute the radius
Substitute the two taped values. Keep the square of the chord to the full four decimals it carries; dropping to a rounded value here moves the radius by more than a tenth of a foot.
The result, 1183.98 ft, is not a round design radius. That is normal on an older alignment where the curve was laid out from a degree of curve or was field-fitted, and it is exactly why the retracement has to be measured rather than assumed.
C² = 682.34² = 465,587.8756 ft²
8M = 8 × 50.22 = 401.76 ft
C²/(8M) = 465,587.8756 / 401.76 = 1158.8707 ft
M/2 = 50.22 / 2 = 25.11 ft
R = 1158.8707 + 25.11 = 1183.98 ftCompute the central angle
With R known, the half angle comes straight out of the long chord: half the chord is R sin(Δ/2). Take the arcsine and double it.
Report the angle to the nearest second. On a curve this size one second of Δ is worth about 0.006 ft of arc length, so a second is the right place to stop.
sin(Δ/2) = C / (2R) = 682.34 / 2367.96 = 0.288155
Δ/2 = arcsin(0.288155) = 16°44′51″
Δ = 33°29′42″Fill out the remaining elements
With R and Δ in hand every other element follows in the usual way. These are the numbers needed to re-establish the curve from the reconstructed tangents rather than from the two recovered curb points.
The arc length is the figure that matters most for re-stationing, because it is the distance the alignment actually runs between the PC and the PT.
| Element | Formula | Value (ft) |
|---|---|---|
| Tangent distance T | R tan(Δ/2) = 1183.98 × 0.300919 | 356.28 |
| Arc length L | R Δ(rad) = 1183.98 × 0.584599 | 692.15 |
| Long chord C | 2R sin(Δ/2) = 2367.96 × 0.288155 | 682.34 |
| Middle ordinate M | R(1 − cos(Δ/2)) = 1183.98 × 0.042416 | 50.22 |
| External E | R(sec(Δ/2) − 1) = 1183.98 × 0.044295 | 52.44 |
Answer
- R = 1183.98 ft.
- Δ = 33°29′42″.
- T = 356.28 ft, L = 692.15 ft, E = 52.44 ft.
- The arc is 692.15 ft long between the two recovered curb points, 9.81 ft longer than the taped chord.
Check
Recompute the two measured quantities from the answers and see whether the tape is reproduced. C = 2R sin(Δ/2) = 2367.96 × 0.288155 = 682.34 ft, matching the taped chord to 0.01 ft.
M = R(1 − cos(Δ/2)) = 1183.98 × 0.042416 = 50.22 ft, matching the taped offset to 0.01 ft.
A second, cruder check that catches gross errors: for a flat curve M is close to C²/(8R) = 465,587.88 / 9471.85 = 49.16 ft. The true value is a little larger, 50.22 ft, as it must be, because the exact relation adds M/2 to the radius denominator.
More curve staking
All in this category- Stake a simple circular curve by deflection angles from the PCA 1150.00 ft radius curve with a 28°42′00″ central angle is solved for every element, stationed through the PI, and taken to a full deflection-angle and chord table for staking from the PC.
- Fit a horizontal curve to a required external distanceAn 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.
- Degree of curve on plan data: arc definition against chord definitionA plan labels a curve D = 3°30′00″ with Δ = 24°15′00″ but does not say which definition of degree of curve applies. Both radii are computed and the two sets of elements compared.