Find tangent length, arc length, chord, offsets, and PC/PT stations for a simple circular curve from its central angle and radius or degree of curve.
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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
Central angle
Express Δ in radians for the arc-length formula.
Δrad = Δ·π/180
Δ = 30 deg = 0.523599 rad
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
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
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
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
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
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
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
Central angle
Express Δ in radians for the arc-length formula.
Δrad = Δ·π/180
Δ = 40 deg = 0.698132 rad
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
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
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
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
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
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 forms — T, 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
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
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
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
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
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
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
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 forms — R = 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.