Area & partitioning
Parcel area from a compass-rule-adjusted traverse
Field bearings and distances taken all the way through: closure, compass-rule adjustment, coordinates, and the area those adjusted coordinates enclose — with the adjustment's effect on the acreage measured rather than assumed.
Given
- A five-course closed boundary traverse was run around a parcel with a total station. Observed courses, in order from corner A:
- A–B N 26°59′20″ E 423.12 ft
- B–C S 55°49′15″ E 519.74 ft
- C–D S 19°03′50″ W 431.68 ft
- D–E S 68°22′05″ W 379.80 ft
- E–A N 15°27′10″ W 480.37 ft
- Distances are horizontal, in US survey feet. Corner A is held at assumed coordinates N 2000.00, E 2000.00.
- Angles and distances were observed with comparable care, so the compass (Bowditch) rule is the appropriate adjustment.
- Report coordinates to 0.01 ft and the area to 1 ft² and 0.001 acre.
Required
- The closure in latitude and departure, the linear misclosure, and the relative precision.
- Compass-rule corrections and adjusted latitudes and departures.
- Adjusted coordinates for all five corners, and the enclosed area.
- How much the adjustment changed the area.
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
Reduce the courses to latitudes and departures
Convert each bearing to an azimuth clockwise from north, then take latitude = D·cos α and departure = D·sin α. Carry three decimals through the whole adjustment: the corrections are hundredths of a foot, and rounding the raw components to 0.01 ft would leave nothing to distribute.
The sums of the two columns are the closures. They are what the traverse failed to return to A by, resolved into a north–south and an east–west part.
| Course | Azimuth | Distance (ft) | Latitude (ft) | Departure (ft) |
|---|---|---|---|---|
| A–B | 26°59′20″ | 423.12 | +377.040 | +192.019 |
| B–C | 124°10′45″ | 519.74 | −291.981 | +429.973 |
| C–D | 199°03′50″ | 431.68 | −408.004 | −140.996 |
| D–E | 248°22′05″ | 379.80 | −140.011 | −353.051 |
| E–A | 344°32′50″ | 480.37 | +463.005 | −127.992 |
| Σ | 2234.71 | +0.049 | −0.047 |
Misclosure and precision
The linear misclosure is the resultant of the two closures. Compared with the perimeter it gives the relative precision, which is how a traverse specification is written and how the work is judged before anything is adjusted.
At 0.0676 ft in 2234.71 ft the traverse closes at 1:33,074. That is good work for a five-sided boundary traverse and comfortably inside a 1:10,000 requirement, so the misclosure is random error to be distributed rather than a blunder to be found.
The distinction matters. Adjusting a traverse that contains a blunder does not remove the blunder — it smears it over every course and pushes several corners off their true positions, while making the sheet look tidy. Test the precision first; adjust only if it passes.
ΣLat = +0.0488 ft ΣDep = −0.0467 ft (the three-decimal table columns show +0.049 and −0.047)
e = √( 0.0488² + 0.0467² ) = √( 0.0023814 + 0.0021809 ) = √0.0045623 = 0.0676 ft
relative precision = 2234.71 / 0.0676 = 1 : 33,074
closure direction: azimuth 136°13′14″ — the traverse fell north and west of ACompass-rule corrections
The compass rule distributes each closure in proportion to course length, on the assumption that angles and distances are of comparable quality — the usual case for a total-station traverse. The correction to a course's latitude is minus the latitude closure times that course's share of the perimeter, and likewise for departure.
The corrections are all of the same sign here because both closures are single-signed, and they are all about one hundredth of a foot: the misclosure spread over 2234.71 ft of perimeter simply does not amount to much per course. Carry them to four decimals — rounding a 0.009 ft correction to 0.01 ft is a 10 percent error in the correction itself.
Check the correction columns before applying them. They must sum to the negative of the closures, which is what makes the adjusted columns sum to zero.
| Course | D / ΣD | Corr. to lat (ft) | Corr. to dep (ft) | Adj. latitude (ft) | Adj. departure (ft) |
|---|---|---|---|---|---|
| A–B | 0.18934 | −0.0092 | +0.0089 | +377.031 | +192.028 |
| B–C | 0.23258 | −0.0113 | +0.0109 | −291.992 | +429.984 |
| C–D | 0.19317 | −0.0094 | +0.0090 | −408.014 | −140.987 |
| D–E | 0.16995 | −0.0083 | +0.0079 | −140.019 | −353.043 |
| E–A | 0.21496 | −0.0105 | +0.0100 | +462.994 | −127.982 |
| Σ | 1.00000 | −0.0487 | +0.0467 | 0.000 | 0.000 |
Adjusted coordinates
Run the adjusted latitudes and departures from A (2000.00, 2000.00), rounding each corner to 0.01 ft as it is set. The fifth course must return to A exactly, because the adjusted columns sum to zero — that is a check on the adjustment, not on the field work.
These are the coordinates that go on the plat and into the area computation. The unadjusted coordinates should be kept on the sheet but not published: they do not close, so any area taken from them depends on which corner you started at.
| Corner | Northing (ft) | Easting (ft) |
|---|---|---|
| A | 2000.00 | 2000.00 |
| B | 2377.03 | 2192.03 |
| C | 2085.04 | 2622.01 |
| D | 1677.02 | 2481.02 |
| E | 1537.01 | 2127.98 |
Area from the adjusted coordinates
Shoelace the five adjusted corners with the closing course E–A. Twice the area is −641,361.40 ft², so the parcel encloses 320,681 ft², or 7.362 acres. The negative sign says the corner list runs clockwise, which agrees with the traverse as observed.
This is the number the plat carries. It rests on adjusted coordinates, which rest on a traverse that closed at 1:33,074, and every one of those steps belongs on the computation sheet behind the acreage.
| Course | Difference Eᵢ·Nᵢ₊₁ − Eᵢ₊₁·Nᵢ |
|---|---|
| A–B | +370,000.00 |
| B–C | −1,662,126.20 |
| C–D | −775,862.73 |
| D–E | +244,687.53 |
| E–A | +1,181,940.00 |
| Σ = 2A | −641,361.40 |
What the adjustment was worth
Compute the area a second time from the raw, unadjusted coordinates: 320,663 ft², or 7.361 acres. The adjustment moved the acreage by 18 ft², which is 0.0004 acre, or 6 parts per hundred thousand.
That is a useful thing to know and an easy thing to misread. It does not mean the adjustment was pointless — the adjustment is what makes the coordinates mutually consistent, so that the corners inverse back to the record calls and a later retracement finds the boundary where the plat says it is. It means that on a well-closed traverse the area is remarkably insensitive to the adjustment, because the misclosure enters the area computation as a small perturbation of a nearly correct figure.
The practical consequence: a plat area is not evidence that a traverse closed well. A traverse closing at 1:3,000 would return an area within a few hundred square feet of this one, and would still be unfit to set boundary corners from.
adjusted area = 320,681 ft² = 7.362 ac
unadjusted area = 320,663 ft² = 7.361 ac
change = 18 ft² = 0.0004 ac = 0.006 %Answer
- ΣLat = +0.0488 ft, ΣDep = −0.0467 ft, linear misclosure e = 0.0676 ft, relative precision 1:33,074 on a perimeter of 2234.71 ft.
- Compass-rule corrections run −0.0083 to −0.0113 ft in latitude and +0.0079 to +0.0109 ft in departure, and the adjusted columns sum to 0.000.
- Adjusted coordinates: A (2000.00, 2000.00), B (2377.03, 2192.03), C (2085.04, 2622.01), D (1677.02, 2481.02), E (1537.01, 2127.98).
- 2A = −641,361.40 ft², so the parcel area is 320,681 ft² = 7.362 acres.
- Adjusting the traverse changed the computed area by only 18 ft², or 0.0004 acre.
Check
Recompute the area from the adjusted coordinates on a DMD sheet. DMDs accumulate 192.03, 814.04, 1103.03, 609.00, 127.98, and the last equals the negative of the last departure (−(−127.98) = 127.98) as required.
The DMD × latitude column is +72,401.07, −237,691.54, −450,058.30, −85,266.09, +59,253.46, summing to −641,361.40 ft². Halved that is 320,680.70 ft² — the shoelace answer to the hundredth of a square foot, from a completely different pairing of the same coordinates.
Check the adjustment arithmetic directly: the correction fractions must sum to 1.00000, and 0.18934 + 0.23258 + 0.19317 + 0.16995 + 0.21496 = 1.00000. The correction columns sum to −0.0487 and +0.0467 ft, cancelling the closures of +0.0488 and −0.0467 ft to within the rounding of the individual corrections.
Check that the adjusted traverse really returns to A: starting from E (1537.01, 2127.98) and applying the adjusted E–A components (+462.994, −127.982) gives (2000.00, 2000.00) to the hundredth.
Independent scale check on the area: the five courses average 447 ft and enclose a roughly pentagonal figure, so an area of the order of 300,000 ft² is expected. 320,681 ft² is in range.
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.
- 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.