Horizontal Curve Layout for a Cass County Road Realignment
Your firm is laying out a horizontal curve for a Cass County road realignment. The point of intersection (PI) is at station 214+37.50. The deflection angle is 47°18'20" to the right, and the design centerline radius is 1,200.00 ft. The incoming tangent bearing is N 12°45'00" E. A 60-ft-wide right-of-way is required. The county engineer proposes a culvert crossing at station 217+00.00 on the curve centerline, and a 200-ft-long box culvert will be set parallel to the chord from station 216+00 to 218+00. The contractor asks whether the curve fits between the existing fence lines, which are 250.0 ft apart and centered on the PI. You must compute the curve elements, check the culvert station, and determine whether the 200-ft chord stays inside the right-of-way.
1. Compute the tangent distance, curve length, external distance, middle ordinate, and the PC and PT stations. Show the formulas and stationing.
Read the worked answer
Given: Δ = 47°18'20" = 47.30556°, R = 1,200.00 ft, PI = 214+37.50.
Half angle = 23°39'10" = 23.65278°.
Tangent: T = R tan(Δ/2) = 1,200.00 × tan 23.65278° = 1,200.00 × 0.4379 = 525.5 ft.
Curve length: L = RΔ in radians = 1,200.00 × (47.30556° × π/180) = 1,200.00 × 0.82550 = 990.6 ft.
External: E = R[sec(Δ/2) − 1] = 1,200.00 × (1.0917 − 1) = 110.0 ft.
Middle ordinate: M = R[1 − cos(Δ/2)] = 1,200.00 × (1 − 0.9160) = 100.8 ft.
PC = PI − T = 214+37.50 − 5+25.5 = 209+12.0.
PT = PC + L = 209+12.0 + 9+90.6 = 219+02.6.
The arithmetic should be carried with more decimal places in the field, but the stationing is consistent to 0.1 ft.
2. Determine the deflection angle and chord bearing from the PC to the culvert at station 217+00.00.
Read the worked answer
First check that station 217+00 lies between PC = 209+12.0 and PT = 219+02.6. It does.
Arc distance from PC to the culvert: l = 217+00.0 − 209+12.0 = 787.98 ft.
Central angle for that arc: θ = l/R = 787.98/1,200.00 = 0.65665 rad = 37.623° = 37°37'23".
Deflection angle from the tangent at PC to the chord is half the central angle: δ = 37.623°/2 = 18.8115° = 18°48'41".
Chord length: c = 2R sin(θ/2) = 2,400 × sin 18.8115° = 2,400 × 0.3226 = 774.2 ft.
Incoming tangent azimuth = 12°45'00" = 12.7500°. Because the curve turns right, add the deflection: chord azimuth = 12.7500° + 18.8115° = 31.5615° = 31°33'41".
Chord bearing = N 31°33'41" E. The field check is that the chord bearing falls between the incoming tangent bearing and the outgoing tangent bearing, which it does.
3. Check whether the 200-ft chord from station 216+00 to 218+00 stays inside the 250-ft fence corridor centered on the PI. Show your reasoning.
Read the worked answer
Both stations are on the curve: PC = 209+12.0, PT = 219+02.6, so 216+00 and 218+00 are between them. The arc length between the two stations is 200.00 ft, giving a central angle of Δc = 200.00/1,200.00 = 0.166667 rad = 9.5493°.
The chord length is c = 2R sin(Δc/2) = 2,400 × sin 4.77465° = 2,400 × 0.08324 = 199.78 ft, essentially the specified 200-ft chord.
The mid-ordinate from the curve to the chord is M = R[1 − cos(Δc/2)] = 1,200 × (1 − 0.99653) = 4.17 ft. The chord lies inside the curve, toward the PI, by about 4.17 ft.
The curve's maximum distance from the PI is the external distance E = 110.0 ft, at the curve midpoint. The fence corridor is 250.0 ft wide, or 125.0 ft each side of the PI. Since 110.0 ft < 125.0 ft, the curve itself fits with about 15.0 ft to spare. The chord is 4.17 ft closer to the PI than the curve at midspan, so it also fits. The 60-ft right-of-way is centered on the centerline, not on the PI, so the controlling check for construction is the curve-to-fence clearance; on this geometry the layout is feasible, but the contractor should verify the actual fence locations in the field.