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.