Area & partitioning
The same parcel by double meridian distance
The 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.
Given
- The same five-corner parcel as the coordinate problem: 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 coordinate (shoelace) method gave 297,109.70 ft², or 6.821 acres.
- The reference meridian for the DMD sheet is to be taken through the most westerly corner, A.
Required
- The area by double meridian distance, worked as a full sheet.
- A demonstration that the DMD result and the shoelace result are the same number, not merely close.
- A second, independent accumulation (double parallel distance) as a cross-check.
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
Latitudes and departures from the coordinates
A DMD sheet works on the courses, not on the corners, so the first move is to difference consecutive coordinates. The latitude of a course is the change in northing and the departure is the change in easting; both are signed, and the signs carry the whole computation.
Because these latitudes and departures come from coordinates rather than from field observations, they close exactly — the columns must sum to 0.00. If they do not, the coordinate list is not a closed figure and nothing that follows is meaningful.
| Course | Latitude (ft) | Departure (ft) |
|---|---|---|
| A–B | +312.44 | +188.60 |
| B–C | −107.34 | +453.75 |
| C–D | −374.44 | +87.83 |
| D–E | −118.46 | −514.78 |
| E–A | +287.80 | −215.40 |
| Σ | 0.00 | 0.00 |
Accumulate the double meridian distances
The meridian distance of a course is the distance from the reference meridian to the midpoint of that course. Doubling it removes the halving and keeps the sheet in whole numbers, which is why the hand method was always worked doubled.
The recursion is short. The first course starts on the meridian, so its DMD is simply its own departure. Every course after that picks up the previous DMD, the previous departure and its own departure. Starting the sheet at the most westerly corner keeps every DMD positive, which is convenient but not required — the algebra works from any starting corner.
There is a built-in check at the bottom of the column: the DMD of the last course must equal the departure of that course with the sign reversed, because the traverse has returned to the meridian. Here the last DMD is 215.40 and the last departure is −215.40. If that identity fails, the DMD column has an arithmetic slip in it and there is no need to go any further.
DMD₁ = dep₁ = +188.60
DMD₂ = 188.60 + 188.60 + 453.75 = 830.95
DMD₃ = 830.95 + 453.75 + 87.83 = 1372.53
DMD₄ = 1372.53 + 87.83 − 514.78 = 945.58
DMD₅ = 945.58 − 514.78 − 215.40 = 215.40
check: DMD₅ = −dep₅ = −(−215.40) = 215.40 ✓Multiply DMD by latitude and sum
Each course contributes DMD × latitude to twice the area. The sign comes entirely from the latitude — the DMDs are all positive on this sheet — so the courses running south subtract and the courses running north add.
Note how lopsided the contributions are. Course C–D alone accounts for −513,930 of the total, because it is a long southerly course a long way from the reference meridian. That is not an error; it is the geometry, and it is why the DMD sheet is unforgiving of a sign slip in a single latitude.
| Course | Latitude (ft) | Departure (ft) | DMD (ft) | DMD × Lat (ft²) |
|---|---|---|---|---|
| A–B | +312.44 | +188.60 | 188.60 | +58,926.18 |
| B–C | −107.34 | +453.75 | 830.95 | −89,194.17 |
| C–D | −374.44 | +87.83 | 1372.53 | −513,930.13 |
| D–E | −118.46 | −514.78 | 945.58 | −112,013.41 |
| E–A | +287.80 | −215.40 | 215.40 | +61,992.12 |
| Σ = 2A | −594,219.41 |
Halve, convert, and compare with the shoelace answer
Twice the area is −594,219.41 ft², so the area is 297,109.70 ft². That is the shoelace result to the last hundredth of a square foot, because the two are the same algebra rearranged: substitute the DMD recursion into the sum and the cross-product form falls out.
The value of running both is not that they might differ numerically. It is that they fail differently. A corner keyed out of sequence gives a shoelace answer that is quietly wrong but plausible; on a DMD sheet the same mistake breaks the closing identity DMD_last = −dep_last and stops you before you convert anything.
A = | −594,219.41 | / 2 = 297,109.70 ft²
shoelace 2A = −594,219.41 ft² DMD 2A = −594,219.41 ft² difference 0.00
acres = 297,109.70 / 43,560 = 6.820700 acCross-check by double parallel distance
The mirror-image method accumulates a parallel distance from an east–west reference line and multiplies by the departure instead of the latitude. DPD₁ = lat₁, DPDᵢ = DPDᵢ₋₁ + latᵢ₋₁ + latᵢ, and 2A = Σ (DPDᵢ × depᵢ).
It uses the same input numbers in a completely different pairing, so an error in a single latitude or departure will not survive both sheets. The sign convention flips — a clockwise figure gives a positive DPD sum where it gave a negative DMD sum — so compare magnitudes, not signs.
| Course | Latitude (ft) | Departure (ft) | DPD (ft) | DPD × Dep (ft²) |
|---|---|---|---|---|
| A–B | +312.44 | +188.60 | 312.44 | +58,926.18 |
| B–C | −107.34 | +453.75 | 517.54 | +234,833.78 |
| C–D | −374.44 | +87.83 | 35.76 | +3,140.80 |
| D–E | −118.46 | −514.78 | −457.14 | +235,326.53 |
| E–A | +287.80 | −215.40 | −287.80 | +61,992.12 |
| Σ = 2A | +594,219.41 |
Answer
- Σ (DMD × Lat) = −594,219.41 ft², so 2A = −594,219.41 ft².
- Area = 297,110 ft² = 6.821 acres.
- The DMD result and the coordinate result agree to 0.00 ft² — the two methods are the same computation written two ways, and any disagreement between them would be an arithmetic fault, never a methodological one.
- The closing identity holds: DMD of the last course, 215.40 ft, equals the negative of that course's departure, −215.40 ft.
Check
The DPD accumulation gives Σ (DPD × Dep) = +594,219.41 ft², the same magnitude with the opposite sign, so the area is again 297,109.70 ft².
The sign reversal between the two sheets is expected, not a fault: swapping the roles of latitude and departure reflects the figure, which reverses the sense in which the boundary is traversed.
Three independent accumulations — shoelace, DMD and DPD — now agree on 297,109.70 ft². Each pairs the coordinates differently, so a single mistyped digit could not survive all three.
More area & partitioning
All in this category- Area of a closed parcel from coordinates by the shoelace ruleA 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.
- 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.