Area & partitioning
Area of a parcel with one circular-arc boundary
A five-sided parcel whose southerly boundary is a circular arc. Compute the area of the chord polygon by coordinates, add the circular segment R²/2·(Δ − sin Δ), and check the segment as sector minus triangle.
Given
- Parcel: LOT 6, MILLRACE ADDITION. Corners on a local plane grid, US survey feet, northing first.
- A N 2500.00 E 3000.00
- B N 2500.00 E 3450.00
- C N 2160.00 E 3560.00
- D N 1849.64 E 3380.82
- E N 1980.00 E 3000.00
- The boundary from C to D is not a straight line. It is a circular arc of radius R = 500.00 ft and central angle Δ = 42°00′00″, concave toward the parcel, so the arc bulges outward beyond the chord C–D.
- The four remaining boundaries A–B, B–C, D–E and E–A are straight.
- The coordinates given for C and D are the ends of the arc, i.e. the ends of its long chord.
Required
- The area of the parcel including the ground between the chord C–D and the arc.
- The answer in square feet and acres, with the segment contribution stated separately.
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
Split the parcel into a polygon plus a segment
A coordinate area computation only knows about straight lines between the points it is given, so running the shoelace rule on A, B, C, D, E returns the area of the figure bounded by the chord C–D, not by the arc. The curved boundary is handled as a correction to that polygon.
The correction is the circular segment: the sliver between the chord and the arc. Because this arc is concave toward the parcel — its radius point lies inside the boundary — the arc bows away from the interior and the segment is land that belongs to the parcel. It is added. Reverse the curvature and the same sliver would be land the parcel does not own, and the segment would be subtracted. Get this sign from a sketch, every time; it is worth 0.18 acre here.
Splitting this way keeps the two parts honest. The polygon is exact from the coordinates, and the segment depends only on R and Δ, so an error in one cannot hide in the other.
A_parcel = A_polygon(A, B, C, D, E) + A_segment(R, Δ)
A_segment = R² / 2 · ( Δ − sin Δ ), Δ in radiansShoelace area of the chord polygon
Run the cross products around A–B–C–D–E with the closing course E–A. The chord C–D is treated as an ordinary straight course for this part of the work.
Two of the courses are cardinal — A–B runs due east and E–A runs due north — which makes their terms easy to verify by eye and gives some confidence in the rest of the column.
| Course | Eᵢ · Nᵢ₊₁ | Eᵢ₊₁ · Nᵢ | Difference |
|---|---|---|---|
| A–B | 3000.00 × 2500.00 = 7,500,000.00 | 3450.00 × 2500.00 = 8,625,000.00 | −1,125,000.00 |
| B–C | 3450.00 × 2160.00 = 7,452,000.00 | 3560.00 × 2500.00 = 8,900,000.00 | −1,448,000.00 |
| C–D | 3560.00 × 1849.64 = 6,584,718.40 | 3380.82 × 2160.00 = 7,302,571.20 | −717,852.80 |
| D–E | 3380.82 × 1980.00 = 6,694,023.60 | 3000.00 × 1849.64 = 5,548,920.00 | +1,145,103.60 |
| E–A | 3000.00 × 2500.00 = 7,500,000.00 | 3000.00 × 1980.00 = 5,940,000.00 | +1,560,000.00 |
| Σ = 2A | −585,749.20 |
Verify the curve data against the coordinates
Before using R and Δ, confirm they describe the same arc the coordinates imply. The long chord of a simple curve is 2R·sin(Δ/2); if that does not reproduce the C-to-D inverse, either the radius, the central angle or one of the two coordinates is wrong, and the segment correction would be wrong with it.
The chord computed from the curve data is 358.368 ft and the chord inversed from the coordinates is 358.370 ft. The 0.002 ft difference is the rounding of C and D to 0.01 ft, and it is far below anything that matters in a segment area.
The other elements are worth writing down because they are what a field crew will ask for: the arc length is what gets called on the plat, and the middle ordinate says how far the arc departs from the chord at its worst, which is a useful reality check on the size of the segment.
Δ = 42°00′00″ = 0.7330383 rad
long chord = 2R·sin(Δ/2) = 2 × 500.00 × sin 21°00′00″ = 1000.00 × 0.3583679 = 358.37 ft
chord C–D from coordinates = 358.37 ft, bearing S 29°59′57″ W ✓
arc length = R·Δ = 500.00 × 0.7330383 = 366.52 ft
middle ordinate = R(1 − cos(Δ/2)) = 500.00 × (1 − 0.9337865) = 33.21 ftArea of the circular segment
The segment formula subtracts the triangle formed by the two radii and the chord from the circular sector they bound. Written out that is R²Δ/2 − R²sin Δ/2, which collects to R²/2 · (Δ − sin Δ).
Δ must be in radians in both terms. Using degrees in the first term is the classic failure here and it inflates the segment by a factor of about 57, which is large enough to be obvious — unlike using the sine of Δ/2 by mistake, which is wrong by a plausible-looking amount.
Δ = 42° × π / 180 = 0.7330383 rad
sin Δ = sin 42° = 0.6691306
Δ − sin Δ = 0.7330383 − 0.6691306 = 0.0639077
R² / 2 = 500.00² / 2 = 125,000 ft²
A_segment = 125,000 × 0.0639077 = 7,988.46 ft²Combine and convert
Add the segment to the chord polygon and convert. The segment is 0.183 acre — 2.7 percent of the parcel — which is far more than any surveyor would accept losing to a shortcut, and it is exactly what gets lost if the arc is treated as a chord.
Report the two parts separately on the computation sheet. A reviewer can then confirm the polygon independently of the curve data, and can see at a glance whether the segment was added or subtracted.
A_polygon = | −585,749.20 | / 2 = 292,874.60 ft² = 6.723 ac
A_segment = 7,988.46 ft² = 0.183 ac
A_parcel = 292,874.60 + 7,988.46 = 300,863.06 ft²
acres = 300,863.06 / 43,560 = 6.906865 acAnswer
- Chord polygon A–B–C–D–E = 292,875 ft² = 6.723 acres.
- Circular segment on chord C–D = 7,988 ft² = 0.183 acre, added because the arc is concave toward the parcel.
- Total parcel area = 300,863 ft² = 6.907 acres.
- Supporting curve elements: long chord 358.37 ft, arc length 366.52 ft, middle ordinate 33.21 ft.
Check
Recompute the segment as sector minus triangle, without collecting the formula. Sector = ½R²Δ = 0.5 × 250,000 × 0.7330383 = 91,629.79 ft². Triangle on the two radii and the chord = ½R²·sin Δ = 0.5 × 250,000 × 0.6691306 = 83,641.33 ft².
91,629.79 − 83,641.33 = 7,988.46 ft², matching the collected formula exactly.
Order-of-magnitude check on the segment: a crude parabolic estimate of a segment is two thirds of chord × middle ordinate, or 2/3 × 358.37 × 33.21 = 7,935 ft². That is within 0.7 percent of the exact 7,988 ft², which is the right kind of agreement for a shallow segment — close, but not close enough to use.
The middle ordinate of 33.21 ft against a 358 ft chord makes a shallow arc, so a segment of about 2.7 percent of a 6.9 acre parcel is the right size. A segment coming out at 0.02 acre or 2 acres would mean Δ was used in the wrong units.
More area & partitioning
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- The same parcel by double meridian distanceThe HAWTHORN TRACT area recomputed on a DMD sheet — the form a plat's computation panel still shows — and set against the coordinate answer to prove the two methods are the same computation.
- Area from a metes-and-bounds descriptionA deed of five calls reduced to latitudes and departures, carried to coordinates, and closed out to an area — with the description's own closure tested before the area is trusted.