Area & partitioning
Area of a right-of-way strip take along an alignment
A road widening takes a fifteen-foot strip along a frontage that runs partly on tangent and partly around a curve. The tangent part is a rectangle; the curved part is the difference of two circular sectors.
Given
- An existing public right of way is 30.00 ft wide measured from the alignment centerline. It is being widened to 45.00 ft from the same centerline, on the subject parcel's side only.
- The parcel frontage begins at the point of curvature station and runs back along the tangent 240.00 ft, and forward around the curve for the whole of the curve.
- The centerline curve has radius R = 800.00 ft and central angle Δ = 28°00′00″.
- The parcel lies on the outside of the curve, so its right-of-way line is at a radius greater than the centerline radius.
- The take is the 15.00 ft strip between the existing 30.00 ft line and the proposed 45.00 ft line, along the whole frontage.
- Distances in US survey feet; report the take to 1 ft² and 0.001 acre.
Required
- The take area along the tangent portion.
- The take area along the curved portion.
- The total take, in square feet and acres.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Set up the two radii and the two frontage lengths
The strip is bounded by two offset lines from the same alignment, so it has to be treated in two pieces: along the tangent the two lines are parallel straight lines, and around the curve they are concentric arcs.
Because the parcel is on the outside of the curve, the offsets add to the centerline radius. The existing right-of-way line lies at R + 30.00 = 830.00 ft and the proposed line at R + 45.00 = 845.00 ft. Had the parcel been on the inside, the radii would be 770.00 and 755.00 ft and the curved area would come out smaller, not larger — the same strip width covers less ground on the inside of a curve.
Work the central angle in radians once and reuse it. Every arc and sector below is that angle times a radius or a radius squared.
Δ = 28°00′00″ = 28 × π / 180 = 0.4886922 rad
existing R/W radius R₁ = 800.00 + 30.00 = 830.00 ft
proposed R/W radius R₂ = 800.00 + 45.00 = 845.00 ft
strip width w = R₂ − R₁ = 15.00 ftTangent portion
Along the tangent the two right-of-way lines are parallel, 15.00 ft apart, and the frontage is 240.00 ft. The take is a rectangle.
The tangent frontage length is the same on both offset lines, which is the property that fails as soon as the alignment curves — and the reason the two portions cannot be lumped together and multiplied by 15.00 ft.
A_tangent = w × L = 15.00 × 240.00 = 3600.00 ft²Curved portion as the difference of two sectors
The curved part of the strip is an annular sector: the region between two concentric arcs sharing the same central angle. Its area is the larger sector minus the smaller one, and since a sector is ½R²Δ the difference collects to Δ/2 times the difference of the squares of the radii.
This is exact — no approximation is involved, and no arc length needs computing to get it. Note that the strip is not simply 15.00 ft times the centerline arc: the ground the strip covers depends on where in the curve it sits, and on the outside of a curve there is more of it.
Compute the squares carefully. The difference of squares is a small number arising from two large ones, so carry the full products rather than rounding them.
A_curve = ½ Δ ( R₂² − R₁² )
R₂² = 845.00² = 714,025.00 ft²
R₁² = 830.00² = 688,900.00 ft²
R₂² − R₁² = 25,125.00 ft²
A_curve = 0.4886922 / 2 × 25,125.00 = 0.2443461 × 25,125.00 = 6139.20 ft²Total take and the frontage lengths for the description
Add the two portions and convert. A deed for the take needs the boundary lengths as well as the area, so record the arc lengths on both right-of-way lines: they are the calls that will appear along the front and back of the strip.
The two arcs differ by 7.33 ft over a 28° curve — the outer arc is longer, which is the geometric fact that makes the curved strip larger than 15.00 ft times the inner arc. Anyone who computes the curved take as width times the existing right-of-way arc will be short by 55 ft².
| Portion | Existing R/W frontage (ft) | Proposed R/W frontage (ft) | Take area (ft²) |
|---|---|---|---|
| Tangent | 240.00 | 240.00 | 3600.00 |
| Curve | 405.61 (arc at R = 830.00) | 412.94 (arc at R = 845.00) | 6139.20 |
| Total | 645.61 | 652.94 | 9739.20 |
Report the result
The total take is 9739 ft², which is 0.224 acre. State it to 0.001 acre on the plat and to the square foot in the legal description, and show the tangent and curve components separately so the figure can be audited without redoing the geometry.
One further note for the description: the centerline arc is 390.95 ft, which is neither of the two right-of-way frontage lengths. Calling the centerline arc as if it were the frontage is a common transcription error on a widening plan, and it understates the front boundary by nearly 15 ft here.
A_total = 3600.00 + 6139.20 = 9739.20 ft²
acres = 9739.20 / 43,560 = 0.223581 ac → 0.224 ac
centerline arc = R Δ = 800.00 × 0.4886922 = 390.95 ft (not a frontage length)Answer
- Tangent portion of the take: 3600 ft².
- Curved portion of the take: 6139 ft².
- Total take = 9739 ft² = 0.224 acre.
- Frontage along the existing 30.00 ft right-of-way line: 240.00 ft of tangent plus 405.61 ft of arc = 645.61 ft.
- Frontage along the proposed 45.00 ft right-of-way line: 240.00 ft of tangent plus 412.94 ft of arc = 652.94 ft.
Check
Recompute the curved take as mean arc times width. The mean radius of the strip is (830.00 + 845.00)/2 = 837.50 ft, and its arc is 837.50 × 0.4886922 = 409.28 ft. Multiplied by the 15.00 ft width that gives 6139.20 ft², matching the sector-difference result exactly.
The two methods agree exactly rather than approximately, and they must: ½Δ(R₂² − R₁²) factors to ½Δ(R₂ + R₁)(R₂ − R₁), which is the mean-radius arc times the width. Getting a difference between them means an arithmetic slip, not a modelling choice.
Test the temptation the problem is built around: 15.00 × 405.61 = 6084.15 ft², using the inner arc, which is 55 ft² short. Using the outer arc, 15.00 × 412.94 = 6194.10 ft², overshoots by 55 ft². The correct value sits exactly between them, which is what the mean-radius derivation predicts.
Rough total check: the whole frontage is about 649 ft on average and the strip is 15.00 ft wide, so 649 × 15 = 9735 ft² — within 5 ft² of the computed 9739 ft², as it should be for a strip this narrow relative to the radius.
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