Area & partitioning
Area of a closed parcel from coordinates by the shoelace rule
A five-corner parcel is given as plane coordinates. Work the cross-product sum term by term to square feet and acres, and use the sign of the result to prove the corner list is in boundary order.
Given
- Parcel: HAWTHORN TRACT, five corners, on a local plane grid in US survey feet. Northing is listed first, as it is on the plat.
- A N 5000.00 E 5000.00
- B N 5312.44 E 5188.60
- C N 5205.10 E 5642.35
- D N 4830.66 E 5730.18
- E N 4712.20 E 5215.40
- The corners are listed in the order they are walked on the ground: A to B to C to D to E and back to A. No boundary is curved.
- Coordinates are carried to 0.01 ft; the area is to be reported to 1 ft² and 0.001 acre.
Required
- The enclosed area, in square feet and in acres.
- Whether the corner list runs clockwise or counter-clockwise on a north-up plan, taken from the sign of the computation rather than read off the plat.
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
What the shoelace sum is actually doing
The coordinate method computes twice the signed area of a closed polygon as a sum of cross products taken around the boundary. Each term is the area of a parallelogram spanned by the position vectors of two consecutive corners; summed around a closed figure, the parts outside the parcel cancel and what is left is exactly twice the enclosed area.
The sum is unsigned in magnitude but signed in direction. That sign is not decoration: it tells you which way the corner list runs, and a corner list that is out of boundary order produces a bow-tie whose area is meaningless rather than merely negative. Compute the sign, do not assume it.
The index i+1 wraps, so the last course runs from E back to A. Forgetting the closing course is the single most common way this computation goes wrong, and it does not announce itself — it just returns a smaller area.
2A = Σ ( Eᵢ · Nᵢ₊₁ − Eᵢ₊₁ · Nᵢ )
i runs 1 … n, and the index i+1 wraps: after E comes A
A = | 2A | / 2Cross products, one course at a time
Lay the five courses out in a column and evaluate the two products for each. Carry all the digits your calculator gives; the products run to eight figures and rounding them to the nearest foot at this stage throws away the third decimal of the acreage.
The differences are large numbers of opposite sign that nearly cancel. That near-cancellation is normal for coordinates carried on a grid with a large false origin, and it is the reason the products are kept to 0.01 ft² rather than being rounded early.
| Course | Eᵢ · Nᵢ₊₁ | Eᵢ₊₁ · Nᵢ | Difference |
|---|---|---|---|
| A–B | 5000.00 × 5312.44 = 26,562,200.00 | 5188.60 × 5000.00 = 25,943,000.00 | +619,200.00 |
| B–C | 5188.60 × 5205.10 = 27,007,181.86 | 5642.35 × 5312.44 = 29,974,645.83 | −2,967,463.97 |
| C–D | 5642.35 × 4830.66 = 27,256,274.45 | 5730.18 × 5205.10 = 29,826,159.92 | −2,569,885.47 |
| D–E | 5730.18 × 4712.20 = 27,001,754.20 | 5215.40 × 4830.66 = 25,193,824.16 | +1,807,930.03 |
| E–A | 5215.40 × 5000.00 = 26,077,000.00 | 5000.00 × 4712.20 = 23,561,000.00 | +2,516,000.00 |
| Σ | −594,219.41 |
Sum, halve, and read the sign
Adding the difference column gives twice the signed area. Halving it and taking the absolute value gives the area; keeping the sign gives the direction of travel.
The sum is negative. In this library's convention a positive double area means the vertices run counter-clockwise in the (easting, northing) plane, which is counter-clockwise on a north-up plan; a negative one means clockwise. So A–B–C–D–E is a clockwise list, which agrees with a rough sketch: the traverse leaves A heading north-east, swings east, then south, then west, then back north to A.
Σ = +619,200.00 − 2,967,463.97 − 2,569,885.47 + 1,807,930.03 + 2,516,000.00
2A = −594,219.41 ft²
A = | −594,219.41 | / 2 = 297,109.70 ft²Convert to acres and to square chains
One acre is 43,560 ft² exactly, and it is also 10 square chains exactly, since a chain is 66 ft and 66² × 10 = 43,560. Running both conversions costs nothing and gives a free check on the division: the square-chain figure must be exactly ten times the acreage.
The 43,560 relationship holds in whichever foot the coordinates are in, because both sides of the ratio are squared in the same foot. Which foot it is still matters when the answer has to be stated in metric — that is a separate problem, and it is where the 4 ppm area consequence of the two feet shows up.
acres = 297,109.70 / 43,560 = 6.820700 ac
square chains = 297,109.70 / 66² = 297,109.70 / 4356 = 68.2070 sq ch
68.2070 / 10 = 6.82070 ac ✓ the two conversions agreeConfirm the figure is a sensible parcel
Inverse each course before you trust the area. Course lengths in the 350–530 ft range on a 6.8 acre parcel are plausible; a length of a few feet or several thousand would say the coordinate list has a transposed digit or a corner out of sequence.
The latitudes and departures of a closed figure computed from coordinates must sum to exactly zero, because each coordinate is added once and subtracted once. That is a check on the arithmetic of the inverse, not on the field work.
| Course | Length (ft) | Bearing | Latitude (ft) | Departure (ft) |
|---|---|---|---|---|
| A–B | 364.95 | N 31°07′00″ E | +312.44 | +188.60 |
| B–C | 466.27 | S 76°41′26″ E | −107.34 | +453.75 |
| C–D | 384.60 | S 13°12′03″ E | −374.44 | +87.83 |
| D–E | 528.23 | S 77°02′27″ W | −118.46 | −514.78 |
| E–A | 359.48 | N 36°48′45″ W | +287.80 | −215.40 |
| Σ | 2103.53 | 0.00 | 0.00 |
Answer
- Twice the signed area is −594,219.41 ft².
- Area = 297,110 ft² of square US survey feet.
- Area = 6.821 acres (297,109.70 / 43,560 = 6.820700).
- The double area is negative, so the corner list A–B–C–D–E runs clockwise on a north-up plan. The list is in boundary order — no bow-tie.
- Closed perimeter 2103.54 ft (the five courses rounded to 0.01 ft sum to 2103.53 ft).
Check
Translate the whole parcel and recompute. The shoelace sum is invariant under a shift of origin, so subtract 4700.00 from every northing and 5000.00 from every easting to get A (300.00, 0.00), B (612.44, 188.60), C (505.10, 642.35), D (130.66, 730.18), E (12.20, 215.40).
The five differences become −56,580.00, −298,138.97, −284,884.47, −19,235.97 and +64,620.00, which sum to −594,219.41 ft² — identical to the original sum, on numbers small enough to check by hand.
That the two sums agree to the last cent of a square foot on completely different multiplicands is strong evidence the eight-figure products in the main table were entered correctly.
Second check: ΣLat = 0.00 and ΣDep = 0.00 in the course table, so the coordinate list really does close on itself.
More area & partitioning
All in this category- 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 of a parcel with one circular-arc boundaryA 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.
- 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.