Curve staking
Compound curve: tangent distances and stationing from the main PI
Two arcs of different radius meet at a point of compound curvature. The vertex triangle gives the two tangent distances back to the main PI, and the alignment is stationed through both arcs.
Given
- A compound curve to the right made of two arcs meeting at a PCC.
- First arc: R₁ = 800.00 ft, Δ₁ = 18°30′00″.
- Second arc: R₂ = 1150.00 ft, Δ₂ = 26°10′00″.
- The total deflection between the back and forward tangents is Δ = Δ₁ + Δ₂ = 44°40′00″.
- Station of the main PI, where the back and forward tangents intersect, = 52+30.55.
- All distances are US survey feet.
Required
- The elements of each arc.
- The distance from the main PI back to the PC and ahead to the PT.
- Stations of the PC, the PCC and the PT.
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
Elements of the two arcs
Each arc is an ordinary circular curve and is solved on its own from its radius and its central angle. The two are joined at the PCC by a tangent common to both circles.
The two tangent distances T₁ and T₂ are measured from the PCC back to the first arc vertex and ahead to the second arc vertex. Their sum is the length of the common tangent between those two vertices, and it is the key dimension of the vertex triangle in the next step.
Arc 1: T₁ = 800.00 tan(9°15′00″) = 800.00 × 0.162860 = 130.29 ft
L₁ = 800.00 × 0.322886 = 258.31 ft, C₁ = 1600.00 × 0.160743 = 257.19 ft
Arc 2: T₂ = 1150.00 tan(13°05′00″) = 1150.00 × 0.232401 = 267.26 ft
L₂ = 1150.00 × 0.456694 = 525.20 ft, C₂ = 2300.00 × 0.226368 = 520.65 ft
Common tangent t = T₁ + T₂ = 130.29 + 267.26 = 397.55 ftThe vertex triangle
Call the first arc vertex V₁, the second arc vertex V₂, and the main intersection point PI. The three form a triangle whose side V₁V₂ is the common tangent t.
The exterior angle at V₁ is Δ₁ and the exterior angle at V₂ is Δ₂, so the interior angles of the triangle at those vertices are Δ₁ and Δ₂, and the angle at the PI is 180° − Δ. The sine rule then gives the two remaining sides directly.
Because sin(180° − Δ) = sin Δ, the denominator is simply sin of the total deflection.
sin Δ₁ = sin 18°30′00″ = 0.3173047
sin Δ₂ = sin 26°10′00″ = 0.4409838
sin Δ = sin 44°40′00″ = 0.7029811
PI to V₁ = t sin Δ₂ / sin Δ = 397.549 × 0.4409838 / 0.7029811 = 249.38 ft
PI to V₂ = t sin Δ₁ / sin Δ = 397.549 × 0.3173047 / 0.7029811 = 179.44 ftTotal tangent distances
The distance from the main PI back to the PC is the distance from the PI to V₁ plus the first arc tangent T₁, because the PC lies T₁ beyond V₁ on the back tangent. The same construction ahead gives the distance from the PI to the PT.
The two are not equal. A compound curve is asymmetric by design: that asymmetry is exactly what it is for, letting an alignment turn a corner where the two approach tangents are not the same length.
PI to PC = 249.38 + 130.29 = 379.67 ft
PI to PT = 179.44 + 267.26 = 446.70 ft
Difference: 446.70 − 379.67 = 67.03 ftStation the alignment
The PC is back-stationed from the PI by the total back tangent distance. From the PC, stationing runs along the first arc to the PCC and then along the second arc to the PT.
As always the arc lengths are what carry the chainage, never the tangent distances.
PC = 52+30.55 − 379.67 = 48+50.88
PCC = PC + L₁ = 48+50.88 + 258.31 = 51+09.19
PT = PCC + L₂ = 51+09.19 + 525.20 = 56+34.38
Total curve length = 258.31 + 525.20 = 783.51 ftSummary sheet
This is the block of data that goes on the plan. Both arcs are staked independently, each from its own beginning point: the first arc by deflections from the PC, the second by deflections from the PCC after the instrument is moved up and oriented on the common tangent.
The deflection at the PCC from the PC is Δ₁/2 = 9°15′00″, and the deflection at the PT from the PCC is Δ₂/2 = 13°05′00″. Neither arc knows anything about the other once the PCC is set.
| Quantity | Arc 1 | Arc 2 |
|---|---|---|
| Radius (ft) | 800.00 | 1150.00 |
| Central angle | 18°30′00″ | 26°10′00″ |
| Tangent from PCC (ft) | 130.29 | 267.26 |
| Arc length (ft) | 258.31 | 525.20 |
| Long chord (ft) | 257.19 | 520.65 |
| Beginning station | 48+50.88 (PC) | 51+09.19 (PCC) |
| Ending station | 51+09.19 (PCC) | 56+34.38 (PT) |
Answer
- Arc 1: T₁ = 130.29 ft, L₁ = 258.31 ft, C₁ = 257.19 ft. Arc 2: T₂ = 267.26 ft, L₂ = 525.20 ft, C₂ = 520.65 ft.
- Distance from the main PI back to the PC = 379.67 ft; ahead to the PT = 446.70 ft.
- PC = 48+50.88, PCC = 51+09.19, PT = 56+34.38.
- Total length of the compound curve = 783.51 ft.
Check
Close the figure by coordinates. Put the PC at the origin with the back tangent on azimuth 0°. The PCC is one long chord C₁ = 257.19 ft on azimuth Δ₁/2 = 9°15′00″, giving N 253.844, E 41.341. The PT is one long chord C₂ = 520.65 ft on azimuth Δ₁ + Δ₂/2 = 31°35′00″ from there, giving N 697.372, E 314.023.
Now find the main PI two independent ways. Running 379.67 ft ahead from the PC on azimuth 0° gives N 379.673, E 0.000. Running 446.70 ft back from the PT on the reverse of the forward tangent, azimuth 44°40′00″ + 180° = 224°40′00″, also gives N 379.673, E 0.000. The two agree, which proves both tangent distances and the vertex triangle at once.
Angle check: Δ₁ + Δ₂ = 18°30′00″ + 26°10′00″ = 44°40′00″ = Δ, so the two arcs together turn the alignment through exactly the deflection observed at the main PI.
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.