10.2 Horizontal Curve Stationing and Layout

Stationing is chainage along the finished alignment centerline: 100 ft of centerline is one full station. Station 10+00 is 1,000 ft from 0+00; station 45+00 is 4,500.00 ft. Horizontal-curve layout on the Civil Engineering Surveying (CES) exam is almost always this sequence: given a PI station plus T and L (or enough data to compute them), find the PC and PT stations, then station a point at a given arc length and compute the deflection angle used to stake it. Domain III.E supplies the geometry; Domain V later occupies the same points with hubs and tacks.

Centerline stations from the PI

PC and PT from the PI

The PI is where the tangents meet. It is not on the curve and it is not a stationing checkpoint between PC and PT. Centerline stationing runs along the back tangent to the PC, then along the arc of length L, then along the forward tangent from the PT.

PC station = PI station − T

PT station = PC station + L

You subtract T because you occupy the back tangent from the PI back to the PC. You add L, not T, because the occupied centerline between PC and PT is the arc. A standing trap is writing PT = PI + T. That is a tangent distance from the PI to a point that is not on the curve; it is not a centerline station.

A useful check after you have PC and PT: PT − PI = L − T, which is not equal to T unless L = 2T (a special Δ, not the general case).

Worked stationing: PI 45+00, T = 250.00 ft, L = 480.00 ft

PI = 45+00.00 means 4,500.00 ft along the alignment.

PC = 4,500.00 − 250.00 = 4,250.00 = 42+50.00

PT = 4,250.00 + 480.00 = 4,730.00 = 47+30.00

Checks: T = 2+50.00 of station; L = 4+80.00 of station. PI − PC = 2+50.00, which equals T. PT − PC = 4+80.00, which equals L. PT − PI = 2+30.00, which is not T. That 230 ft equals L − T = 480 − 250. You can remember PT = PI + (L − T) as a check, but on exam day compute PC first, then add L.

Station of a point 120 ft along the curve

Arc length ℓ is measured from the PC along the curve, never from the PI.

Point station = PC + ℓ = 4,250.00 + 120.00 = 4,370.00 = 43+70.00

The point is still 360 ft of arc short of the PT (480 − 120). If the stem asks for a point "120 ft before the PT," that station is PT − 120 = 46+10.00, which is a different point (ℓ = 360 ft). Read whether 120 ft is from the PC, from the PT, or along a tangent.

Deflection angle and chord from the PC

Field layout from the PC uses the deflection angle δ from the forward tangent to the chord that reaches the new point. The central angle to the point is θ = ℓ / R radians = (ℓ / L) Δ. The deflection is half of that central angle (tangent-chord theorem):

δ = ℓ / (2R) radians

δ = (ℓ / L) × (Δ / 2) (δ and Δ in the same angular unit)

δ° = ℓ D / 200 with arc-definition D and ℓ in feet, because 100 ft of arc subtends D°, so 100 ft of deflection is D/2.

The chord from the PC to the point is c = 2 R sin(δ), with δ in the same mode as the sine function.

To apply those formulas to the PI 45+00 example you still need R or Δ. The stem gave T and L only, so recover them from T = R tan(Δ/2) and L = R Δ π / 180. Dividing gives

tan(Δ/2) / (Δ in radians) = T / L = 250 / 480 = 0.520833.

Solving yields Δ = 39.537° (39°32'13") and

R = T / tan(Δ/2) = 250.00 / tan(19.7685°) = 695.60 ft.

Check: L = 695.60 × 39.537 × π / 180 = 480.00 ft. Arc D = 5729.58 / 695.60 = 8.237°.

Now the 120-ft point:

• δ = 120 / (2 × 695.60) = 0.086257 rad = 4.942° = 4°56'32"
• Same result: δ = (120 / 480) × (39.537 / 2) = 0.25 × 19.7685 = 4.942°
• Same result again: δ° = 120 × 8.237 / 200 = 4.942°

Chord from PC: c = 2 × 695.60 × sin(4.942°) = 119.85 ft.

The chord is 0.15 ft shorter than the 120-ft arc—the usual LC < L relationship at a small central angle. Staking by deflection, you occupy the PC, backsight the PI (or another point on the back tangent), flip to the forward tangent, and turn 4°56'32" to sight the 119.85-ft chord.

Even-station stakes

Construction crews usually want 43+00, 44+00, 45+00, 46+00, 47+00—not an even 100-ft arc from the PC. The first even station after PC 42+50.00 is 43+00, so ℓ = 50.00 ft, δ = 50 × 8.237 / 200 = 2.059°, and the chord is 2 × 695.60 × sin(2.059°) = 49.99 ft. Each additional 100 ft of arc adds D/2 = 4.118° of deflection. At the PT, ℓ = L and δ = Δ/2 = 19.769°, matching the last row of the table.

Layout traps

• PT = PI + T as a station. Wrong. PT = PC + L.
• Measuring 120 ft from the PI. The PI is off the curve.
• Using δ = ℓ / R. That is the central angle θ, twice the deflection.
• Degree/radian mix in sin(δ) when converting the chord.
• Adding T along the curve. T is not centerline between PC and PT.
• Instrument at the PI to turn Δ/2 to the PC. You can occupy the PI to turn the intersection angle, but stationing and deflection layout for points on the arc start from the PC (or PT).

If the stem already gives R or Δ, skip the T–L recovery and go straight to δ = ℓ / (2R). The CES calculator rules still apply: two handhelds, no QWERTY, and you must know whether the trig is in degrees or radians before you hit sine.