Curve staking
Curve stationing and the PT equals PC plus L trap
A 900 ft radius curve is stationed from its PI, with the 29.63 ft of chainage lost at the corner made explicit and every full station on the curve tabulated.
Given
- A simple circular curve with R = 900.00 ft and Δ = 41°18′00″.
- Station of the PI = 87+42.36.
- Alignment stationing runs ahead through the curve; there is no station equation on this alignment.
- All distances are US survey feet.
Required
- The stations of the PC and the PT.
- The chainage difference between the true PT and the incorrect PI plus T.
- The arc distance from the PC to each full station lying on the curve.
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
Curve elements
Only T and L are needed to station the curve, but computing the full set costs nothing and gives the later checks something to work with.
Half the central angle is 20°39′00″. Every element is 900.00 ft times a function of that half angle.
Δ/2 = 20°39′00″
T = 900.00 × tan(20°39′00″) = 900.00 × 0.376872 = 339.18 ft
L = 900.00 × 0.720821 = 648.74 ft
C = 1800.00 × sin(20°39′00″) = 1800.00 × 0.352658 = 634.79 ft
M = 900.00 × 0.064248 = 57.82 ft
E = 900.00 × 0.068659 = 61.79 ftBack-station the PC
The PC is on the back tangent, one tangent distance behind the PI, and stationing along the back tangent is continuous with the alignment. So the PC station is simply the PI station less the tangent distance.
Carry the unrounded tangent distance, 339.1844 ft, into the subtraction and round the resulting station once. Rounding T first and the station second can move the PC by a hundredth.
PC = PI − T
PC = 8742.36 − 339.1844 = 8403.1756 ft
PC = 84+03.18Forward-station the PT along the arc
This is where the problem is usually lost. The PT is one tangent distance ahead of the PI on the ground, but stationing does not run along the tangents once the curve exists. Chainage follows the alignment, and the alignment between the PC and the PT is the arc. So the PT station is the PC station plus the arc length.
The two tangents together are longer than the arc they replace, so a fixed amount of chainage disappears at the corner. That amount is 2T − L, and every station ahead of the PT is short by it relative to the naive tangent computation.
PT = PC + L = 8403.1756 + 648.7389 = 9051.9145 ft
PT = 90+51.91
Wrong answer, PI + T = 8742.36 + 339.18 = 9081.54 = 90+81.54
Chainage lost at the corner: 2T − L = 678.37 − 648.74 = 29.63 ftFull stations on the curve
The first full station ahead of the PC is 85+00.00 and the last one before the PT is 90+00.00, giving six full stations on the curve plus two odd sub-arcs at the ends.
Arc distance from the PC is the quantity every staking method needs, whether the curve is set by deflection angles, by offsets or by coordinates. The deflection column below is included because it is the cheapest possible independent check on the arc column.
| Station | Arc from PC (ft) | Deflection from back tangent | Chord from PC (ft) |
|---|---|---|---|
| 84+03.18 (PC) | 0.00 | 0°00′00″ | 0.00 |
| 85+00.00 | 96.82 | 3°04′55″ | 96.78 |
| 86+00.00 | 196.82 | 6°15′54″ | 196.43 |
| 87+00.00 | 296.82 | 9°26′54″ | 295.48 |
| 88+00.00 | 396.82 | 12°37′53″ | 393.62 |
| 89+00.00 | 496.82 | 15°48′52″ | 490.54 |
| 90+00.00 | 596.82 | 18°59′51″ | 585.95 |
| 90+51.91 (PT) | 648.74 | 20°39′00″ | 634.79 |
What the trap costs downstream
If the PT were stationed at 90+81.54 instead of 90+51.91, every station ahead of the curve would be 29.63 ft too large. A driveway staked at 95+00.00 would land 29.63 ft down the road, quantities computed by station would be wrong over the whole remaining alignment, and the error would not show up in any check confined to the curve itself.
The same arithmetic runs in reverse for a station equation. If a designer wants to hold the ahead stationing fixed, the plan records BACK 90+81.54 = AHEAD 90+51.91, a loss of 29.63 ft, and the alignment continues without a break in the numbering.
Answer
- PC = 84+03.18.
- PT = 90+51.91.
- The PT is 29.63 ft back from the incorrect PI plus T station of 90+81.54; that figure is 2T − L.
- Six full stations lie on the curve, 85+00.00 through 90+00.00, at arc distances 96.82, 196.82, 296.82, 396.82, 496.82 and 596.82 ft from the PC.
Check
Subtract the two computed stations: 90+51.91 − 84+03.18 = 648.73 ft, against an arc length of 648.74 ft. The 0.01 ft is the rounding of the two stations and confirms the arithmetic.
The deflection at the PT must be exactly Δ/2 = 20°39′00″, and the table's last row reads 20°39′00″.
The chord from the PC to the PT, 634.79 ft, equals the long chord C = 2R sin(Δ/2) = 634.79 ft computed in step 1 without reference to the stationing.
Independent check on the lost chainage: 2T − L = 2(339.18) − 648.74 = 29.62 ft from rounded values, 29.63 ft from unrounded ones.
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.
- Recover a curve radius from a measured long chord and middle ordinateAn 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.
- 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.