CivilSolve

Transportation & Surveying

Horizontal Circular Curve Geometry

Complete simple circular curve geometry from the central angle Δ (decimal degrees or DMS) and either the radius R or the arc-definition degree of curve D: tangent length T, curve (arc) length L, long chord LC, external distance E, middle ordinate M, degree of curve, and PC/PT stationing from a PI station. Use for: 'a horizontal curve has R = 500 ft and Δ = 30°...', finding curve stationing (PC/PT), degree-of-curve conversions. Not for: spiral transition curves, superelevation design, vertical curves.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
delta dmsCentral angle in degrees-minutes-seconds, e.g. 55°30′ = {d: 55, m: 30}. Alternative to deltaDeg.
pi station

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

US curve: R = 500 ft, Δ = 30°, PI at 12+50

Given

delta deg
30
radius
500 ft
pi station
12+50

Assumptions

  • Simple circular curve: Δ is the deflection angle between the tangents, equal to the central angle subtended by the arc.
  • Degree of curve uses the ARC definition (angle subtending 100 ft of arc, D = 5729.58/R with R in feet) — the standard for highway work; the railroad chord definition differs.
  • Stationing is US survey stationing in feet ("12+50" = 1250 ft), increasing in the direction of travel.

Solution steps

  1. Central angle

    Express Δ in radians for the arc-length formula.

    Δrad = Δ·π/180

    Δ = 30 deg = 0.523599 rad

  2. Tangent length

    Distance from PC (or PT) to the PI along the tangent.

    T = R·tan(Δ/2)

    T = 500 ft × 0.267949 = 133.975 ft

    = 134 ft

  3. Curve (arc) length

    Length along the arc from PC to PT — this, not the chord, carries the stationing.

    L = R·Δ with Δ in radians (equivalently L = 100·Δ/D in ft)

    L = 500 ft × 0.523599 = 261.799 ft

    = 261.8 ft

  4. Long chord

    Straight-line distance from PC to PT.

    LC = 2·R·sin(Δ/2)

    LC = 2 × 500 ft × 0.258819 = 258.819 ft

    = 258.8 ft

  5. External distance

    From the PI to the midpoint of the curve, along the bisector.

    E = R·(1/cos(Δ/2) − 1)

    E = 500 ft × (1/0.965926 − 1) = 17.6381 ft

    = 17.64 ft

  6. Middle ordinate

    From the midpoint of the long chord to the midpoint of the curve.

    M = R·(1 − cos(Δ/2))

    M = 500 ft × (1 − 0.965926) = 17.0371 ft

    = 17.04 ft

  7. Degree of curve

    Arc-definition sharpness: the central angle subtending 100 ft of arc.

    D = 5729.58/R(ft), degrees per 100 ft of arc

    D = 5729.58 / 500 ft = 11.459 degrees per station

    = 11.46

  8. Curve stationing

    The PC sits a tangent length BEFORE the PI; the PT is then reached along the ARC. Classic trap: PT = PC + L, NOT PI + T — station distance accumulates along the curve, so adding T at the PI overshoots.

    PC = PI − T; PT = PC + L (along the arc, never PI + T)

    PI at 1250 ft; PC = 1250 ft − 133.975 ft = 1116.03 ft; PT = 1116.03 ft + 261.799 ft = 1377.82 ft

    = 1377.82 ft

Results

Radius R

500ft

Degree of curve D (arc definition, degrees per 100 ft)

11.46

Tangent length T

134ft

Curve (arc) length L

261.8ft

Long chord LC

258.8ft

External distance E

17.64ft

Middle ordinate M

17.04ft

PI station 12+50.00

1250ft

PC station 11+16.03 (point of curvature)

1116.03ft

PT station 13+77.82 (point of tangency)

1377.82ft

Where this answer was checked
source
Circular-curve closed forms (NCEES FE Reference Handbook style)T = R·tan(Δ/2), L = R·Δrad, LC = 2R·sin(Δ/2), E, M, PC/PT stationing
verified by
hand-recomputed
derivation
tan 15° = 0.267949 → T = 500·0.267949 = 133.97 ft. Δrad = 30·π/180 = 0.523599 → L = 500·0.523599 = 261.80 ft. sin 15° = 0.258819 → LC = 1000·0.258819 = 258.82 ft. cos 15° = 0.965926 → E = 500·(1/0.965926 − 1) = 500·0.035276 = 17.638 ft; M = 500·(1 − 0.965926) = 17.037 ft. D = 5729.578/500 = 11.459°. PC = 1250 − 133.97 = 1116.03 ft (11+16.03); PT = 1116.03 + 261.80 = 1377.82 ft (13+77.82) — NOT PI + T = 1383.97.

Example 2

SI curve: R = 300 m, Δ = 40°

Given

delta deg
40
radius
300 m

Assumptions

  • Simple circular curve: Δ is the deflection angle between the tangents, equal to the central angle subtended by the arc.
  • Degree of curve uses the ARC definition (angle subtending 100 ft of arc, D = 5729.58/R with R in feet) — the standard for highway work; the railroad chord definition differs.

Solution steps

  1. Central angle

    Express Δ in radians for the arc-length formula.

    Δrad = Δ·π/180

    Δ = 40 deg = 0.698132 rad

  2. Tangent length

    Distance from PC (or PT) to the PI along the tangent.

    T = R·tan(Δ/2)

    T = 300 m × 0.36397 = 109.191 m

    = 109.2 m

  3. Curve (arc) length

    Length along the arc from PC to PT — this, not the chord, carries the stationing.

    L = R·Δ with Δ in radians (equivalently L = 100·Δ/D in ft)

    L = 300 m × 0.698132 = 209.44 m

    = 209.4 m

  4. Long chord

    Straight-line distance from PC to PT.

    LC = 2·R·sin(Δ/2)

    LC = 2 × 300 m × 0.34202 = 205.212 m

    = 205.2 m

  5. External distance

    From the PI to the midpoint of the curve, along the bisector.

    E = R·(1/cos(Δ/2) − 1)

    E = 300 m × (1/0.939693 − 1) = 19.2533 m

    = 19.25 m

  6. Middle ordinate

    From the midpoint of the long chord to the midpoint of the curve.

    M = R·(1 − cos(Δ/2))

    M = 300 m × (1 − 0.939693) = 18.0922 m

    = 18.09 m

  7. Degree of curve

    Arc-definition sharpness: the central angle subtending 100 ft of arc.

    D = 5729.58/R(ft), degrees per 100 ft of arc

    D = 5729.58 / 984.252 ft = 5.8213 degrees per station

    = 5.821

Results

Radius R

300m

Degree of curve D (arc definition, degrees per 100 ft)

5.821

Tangent length T

109.2m

Curve (arc) length L

209.4m

Long chord LC

205.2m

External distance E

19.25m

Middle ordinate M

18.09m

Where this answer was checked
source
Circular-curve closed formsT, L, LC, E, M for a metric curve
verified by
hand-recomputed
derivation
tan 20° = 0.363970 → T = 300·0.363970 = 109.19 m. Δrad = 40·π/180 = 0.698132 → L = 300·0.698132 = 209.44 m. sin 20° = 0.342020 → LC = 600·0.342020 = 205.21 m. cos 20° = 0.939693 → E = 300·(1/0.939693 − 1) = 300·0.064178 = 19.253 m; M = 300·(1 − 0.939693) = 18.092 m.

Example 3

degree of curve given: D = 4°, Δ = 55°30′ (DMS)

Given

delta dms
d 55m 30
degree of curve deg
4

Assumptions

  • Simple circular curve: Δ is the deflection angle between the tangents, equal to the central angle subtended by the arc.
  • Degree of curve uses the ARC definition (angle subtending 100 ft of arc, D = 5729.58/R with R in feet) — the standard for highway work; the railroad chord definition differs.

Solution steps

  1. Central angle

    Convert Δ from degrees-minutes-seconds to decimal degrees, then to radians for the arc-length formula.

    Δ = d + m/60 + s/3600, then Δrad = Δ·π/180

    Δ = 55° 30′ 0″ = 55.5 deg = 0.968658 rad

  2. Radius from degree of curve

    With the arc definition, D degrees of central angle subtend exactly 100 ft of arc, which fixes the radius.

    R = 18000/(π·D) = 5729.58/D, R in ft for D per 100 ft of arc

    R = 5729.58 / 4 = 1432.39 ft

    = 1432.39 ft

  3. Tangent length

    Distance from PC (or PT) to the PI along the tangent.

    T = R·tan(Δ/2)

    T = 436.594 m × 0.526125 = 229.703 m

    = 229.7 m

  4. Curve (arc) length

    Length along the arc from PC to PT — this, not the chord, carries the stationing.

    L = R·Δ with Δ in radians (equivalently L = 100·Δ/D in ft)

    L = 436.594 m × 0.968658 = 422.91 m

    = 422.9 m

  5. Long chord

    Straight-line distance from PC to PT.

    LC = 2·R·sin(Δ/2)

    LC = 2 × 436.594 m × 0.465615 = 406.569 m

    = 406.6 m

  6. External distance

    From the PI to the midpoint of the curve, along the bisector.

    E = R·(1/cos(Δ/2) − 1)

    E = 436.594 m × (1/0.884988 − 1) = 56.7394 m

    = 56.74 m

  7. Middle ordinate

    From the midpoint of the long chord to the midpoint of the curve.

    M = R·(1 − cos(Δ/2))

    M = 436.594 m × (1 − 0.884988) = 50.2137 m

    = 50.21 m

Results

Radius R

436.6m

Degree of curve D (arc definition, degrees per 100 ft)

4

Tangent length T

229.7m

Curve (arc) length L

422.9m

Long chord LC

406.6m

External distance E

56.74m

Middle ordinate M

50.21m

Where this answer was checked
source
Arc-definition degree of curve + circular-curve closed formsR = 5729.58/D, then T/L/LC; L cross-checked by the independent L = 100·Δ/D route
verified by
hand-recomputed
derivation
Δ = 55 + 30/60 = 55.5°. R = 5729.578/4 = 1432.39 ft. tan 27.75° = 0.526125 → T = 1432.39·0.526125 = 753.62 ft. L = R·Δrad = 1432.39·0.968658 = 1387.50 ft; independent route L = 100·Δ/D = 100·55.5/4 = 1387.50 ft (exact agreement). sin 27.75° = 0.465615 → LC = 2864.79·0.465615 = 1333.89 ft. cos 27.75° = 0.884988 → E = 1432.39·(1/0.884988 − 1) = 186.15 ft; M = 1432.39·(1 − 0.884988) = 164.74 ft.