Curve staking
Stake a simple circular curve by deflection angles from the PC
A 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.
Given
- A simple circular curve to the right on a two-lane rural alignment. All distances are US survey feet.
- Radius R = 1150.00 ft, fixed by the design speed.
- Central angle Δ = 28°42′00″, taken as the deflection between the back and forward tangents.
- Station of the PI = 42+61.20.
- Stakes are wanted at every full station and at the PT, set from an instrument occupying the PC and backsighting the PI.
Required
- The curve elements T, L, C, M and E, and the arc-definition degree of curve.
- Stations of the PC and the PT.
- A deflection-angle table giving, for each staked point, the deflection from the back tangent, the chord from the PC, and the chord from the previous stake.
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
Solve the elements from R and Δ
A simple curve is fully determined by any two of its elements, and R with Δ is the pair a plan sheet almost always supplies. Every remaining element is the radius multiplied by a function of the half angle, so the whole computation reduces to evaluating four trigonometric functions of Δ/2 = 14°21′00″ and scaling each by 1150.00 ft.
Carry six decimals on the trigonometric values and round only the final lengths. Rounding the half angle or the trigonometric value first is the usual source of a foot-level discrepancy between two people working the same curve.
Δ/2 = 28°42′00″ / 2 = 14°21′00″
T = R tan(Δ/2) = 1150.00 × 0.255826 = 294.20 ft
L = R Δ(rad) = 1150.00 × 0.500909 = 576.05 ft
C = 2R sin(Δ/2) = 2300.00 × 0.247845 = 570.04 ft
M = R(1 − cos(Δ/2)) = 1150.00 × 0.031200 = 35.88 ft
E = R(sec(Δ/2) − 1) = 1150.00 × 0.032205 = 37.04 ft
D(arc) = 5729.5780 / 1150.00 = 4.98224° = 4°58′56″Station the PC and the PT
The PC lies one tangent distance back from the PI along the back tangent, so its station is the PI station less T. The PT does not lie one tangent distance ahead of the PI in stationing: chainage runs along the alignment, and the alignment now follows the arc rather than the two tangents. The PT station is therefore the PC station plus the arc length.
The chainage lost by rounding the corner is 2T − L = 588.40 − 576.05 = 12.35 ft, which is exactly the difference between PI + T and the true PT.
PC = PI − T = 42+61.20 − 294.20 = 39+67.00
PT = PC + L = 39+67.00 + 576.05 = 45+43.05
(PI + T = 45+55.40 would be wrong by 2T − L = 12.35 ft)Deflection angle per unit of arc
The deflection angle from the back tangent to a point on the curve is half the central angle that the arc to that point subtends. For an arc length l the central angle is l/R radians, so the deflection is l/(2R) radians. That constant, once computed, converts any arc distance straight into a plate reading.
Two derived rates are worth writing at the top of the field book: the deflection for a full 100 ft station, and the deflection per foot for the odd sub-arcs at each end.
δ = l / (2R) radians
per 100 ft: δ₁₀₀ = 100 / 2300.00 = 0.0434783 rad = 2°29′28″
per 1 ft: δ₁ = 1 / 2300.00 = 0.00043478 rad = 0°01′30″
chord to a point: c = 2R sin δBuild the staking table
The first sub-arc runs from the PC at 39+67.00 to station 40+00.00, a length of 33.00 ft; the last runs from 45+00.00 to the PT at 45+43.05, a length of 43.05 ft. Between them the arcs are full 100 ft stations.
Deflections accumulate: each row carries the total deflection from the back tangent, which is what the instrument reads without re-setting. The chord from the PC is the taped distance to that point if you can reach it directly; the chord from the previous stake is what the tape actually spans when the crew leapfrogs along the curve.
| Station | Arc from PC (ft) | Deflection from back tangent | Chord from PC (ft) | Chord from previous (ft) |
|---|---|---|---|---|
| 40+00.00 | 33.00 | 0°49′19″ | 33.00 | 33.00 |
| 41+00.00 | 133.00 | 3°18′48″ | 132.93 | 99.97 |
| 42+00.00 | 233.00 | 5°48′16″ | 232.60 | 99.97 |
| 43+00.00 | 333.00 | 8°17′44″ | 331.84 | 99.97 |
| 44+00.00 | 433.00 | 10°47′12″ | 430.45 | 99.97 |
| 45+00.00 | 533.00 | 13°16′40″ | 528.24 | 99.97 |
| 45+43.05 (PT) | 576.05 | 14°21′00″ | 570.04 | 43.04 |
Notes for the field
Set the plate to zero on the PI, turn the tabulated deflection, and tape the chord from the previous stake along the line of sight to intersect the new deflection ray. Because a chord is always shorter than the arc it subtends, the 100 ft stations are 99.97 ft apart on the tape, not 100.00 ft.
If the crew has to move up to an intermediate point because of a sight obstruction, the same table still works: occupy the point, backsight the PC with the plate set to the deflection already turned to that point, plunge, and the remaining tabulated values run on unchanged.
Answer
- T = 294.20 ft, L = 576.05 ft, C = 570.04 ft, M = 35.88 ft, E = 37.04 ft, D(arc) = 4°58′56″.
- PC = 39+67.00, PT = 45+43.05.
- Deflections from the PC: 0°49′19″ at 40+00, 3°18′48″ at 41+00, 5°48′16″ at 42+00, 8°17′44″ at 43+00, 10°47′12″ at 44+00, 13°16′40″ at 45+00, and 14°21′00″ at the PT.
- Chords from the previous stake: 33.00, 99.97, 99.97, 99.97, 99.97, 99.97 and 43.04 ft.
Check
The total deflection at the PT must equal Δ/2 exactly: 14°21′00″ = 28°42′00″ / 2. It does, which proves the whole accumulated column.
The chord from the PC to the PT in the last row, 570.04 ft, must equal the long chord C = 2R sin(Δ/2) = 570.04 ft computed independently in step 1.
The sub-chords sum to 33.00 + 5(99.97) + 43.04 = 575.89 ft, which is 0.16 ft short of the arc length 576.05 ft. A small positive shortfall is required; a negative one would mean an arc has been used where a chord belongs.
More curve staking
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