CogoKit

Traverse adjustment

Closed traverse with the area required as well as the closure

A six-sided parcel traverse is closed and adjusted, then its area is computed twice, once by the coordinate method and once by double meridian distances, so the two can be made to agree.

Apply· about 34 minutes by hand· 5 steps

Given

  • A closed six-sided boundary traverse, stations 1 through 6, run clockwise on the ground.
  • 1-2 S 68°37′45″ E 494.02 ft
  • 2-3 N 55°22′35″ E 510.46 ft
  • 3-4 N 12°22′50″ W 419.90 ft
  • 4-5 N 71°08′50″ W 433.26 ft
  • 5-6 S 57°59′40″ W 471.70 ft
  • 6-1 S 2°47′35″ E 410.49 ft
  • Bearings are already adjusted for angular closure. Station 1 is held at N 4200.00, E 7100.00.
  • Specification 1:10,000. Area to be reported to 0.001 acre.
  • Distances and coordinates to 0.01 ft; latitudes and departures carried to 0.001 ft.

Required

  • The linear misclosure and precision.
  • Compass-rule adjusted coordinates for all six corners.
  • The area by the coordinate method and by double meridian distances.
  • The area in 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

Latitudes, departures, misclosure

Convert the bearings to azimuths and resolve. The sums are the closures in latitude and departure.

The misclosure of 0.16 ft over 2739.83 ft gives 1:16,970, which passes the 1:10,000 requirement, so the traverse can be adjusted and its area reported.

e = √(0.1462² + 0.0685²) = √(0.021374 + 0.004692) = 0.1614 ft precision = 2739.83 / 0.1614 = 16,970, that is 1:16,970
Observed traverse
CourseBearingAzimuthDistance (ft)Latitude (ft)Departure (ft)
1-2S 68°37′45″ E111°22′15″494.02−180.022460.052
2-3N 55°22′35″ E55°22′35″510.46290.035420.059
3-4N 12°22′50″ W347°37′10″419.90410.135−90.028
4-5N 71°08′50″ W288°51′10″433.26140.003−410.016
5-6S 57°59′40″ W237°59′40″471.70−250.002−400.000
6-1S 2°47′35″ E177°12′25″410.49−410.00220.003
Sums2739.830.1460.069

Compass-rule adjustment and coordinates

Corrections are proportional to course length: the negative of each closure times the course's share of the perimeter.

Area must always be computed from adjusted coordinates. An unadjusted traverse does not enclose a figure at all, and running the shoelace formula on it produces a number whose error is unbounded, not merely small.

Adjusted latitudes and departures, and coordinates
CourseLat correction (ft)Dep correction (ft)Adjusted latitude (ft)Adjusted departure (ft)To stationNorthing (ft)Easting (ft)
1-2−0.026−0.012−180.049460.04024019.957560.04
2-3−0.027−0.013290.007420.04634309.967980.09
3-4−0.022−0.011410.113−90.03944720.077890.05
4-5−0.023−0.011139.979−410.02754860.057480.02
5-6−0.025−0.012−250.027−400.01264610.027080.01
6-1−0.022−0.010−410.02419.99214200.007100.00

Area by coordinates

The coordinate, or shoelace, method computes twice the signed area as the sum around the polygon of each easting times the next northing, less each next easting times the current northing.

The sign tells you which way the vertices run. Take the absolute value and halve it for the area. A figure whose vertices are not in boundary order produces a bow tie and an area that is meaningless rather than merely negative, so the sign is worth looking at every time.

2A = Σ (Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ) 2A = 1,072,725.59 ft² A = 536,362.79 ft² A = 536,362.79 / 43,560 = 12.313 acres

Area by double meridian distances

The DMD method is the one a hand computation sheet prints, and it is what the coordinate method reduces to when the origin is placed on the first station. The double meridian distance of the first course is its departure; each subsequent DMD is the previous DMD plus the previous departure plus the current departure.

Multiply each DMD by its own latitude and sum. The total is twice the area, with a sign that again indicates the direction of travel.

The two methods are algebraically the same computation. A disagreement between them is never a rounding artefact of any size worth arguing about; it means a coordinate list is out of order or a departure has been carried with the wrong sign.

Double meridian distance sheet
CourseLatitude (ft)Departure (ft)DMD (ft)DMD × latitude (ft²)
1-2−180.049460.040460.040−82,829.48
2-3290.007420.0461340.125388,646.18
3-4410.113−90.0391670.132684,942.72
4-5139.979−410.0271170.066163,785.25
5-6−250.027−400.012360.027−90,016.45
6-1−410.02419.992−19.9928197.36
Sum0.0000.0001,072,725.59

Report the area

Both methods give twice the area as 1,072,725.59 square feet, so the area is 536,362.79 square feet, which is 12.313 acres.

Think about how many digits that deserves. The traverse closed at 1:16,970. A rough guide is that the relative error in the area is of the order of the relative error in the closure, so 12.313 acres carries an uncertainty around 0.001 acre. Reporting 12.3132 acres claims a precision the traverse does not support.

A = 536,362.79 ft² 1 acre = 43,560 ft² exactly A = 12.313 acres relative closure 1 : 16,970 implies an area uncertainty of roughly 0.001 acre

Answer

  • Latitude closure +0.146 ft, departure closure +0.069 ft, linear misclosure 0.16 ft over 2739.83 ft.
  • Precision 1:16,970, which meets the 1:10,000 specification.
  • Adjusted coordinates: 1 at 4200.00 / 7100.00; 2 at 4019.95 / 7560.04; 3 at 4309.96 / 7980.09; 4 at 4720.07 / 7890.05; 5 at 4860.05 / 7480.02; 6 at 4610.02 / 7080.01.
  • Area = 536,362.79 square feet by both the coordinate method and the DMD method.
  • Area = 12.313 acres.

Check

The two independent area methods must agree exactly, and they do: 2A = 1,072,725.59 square feet from the shoelace sum and 1,072,725.59 from the DMD column.

The DMD sheet contains its own check. The last DMD plus the last departure must come to zero, so that the recursion regenerates the first DMD as the first departure: −19.992 + 19.992 = 0.000, and 0.000 + 460.040 = 460.040, which is the departure of course 1-2 and the DMD in the first row.

The adjusted latitude and departure columns sum to 0.000, and the traverse closes back on station 1 at 4200.00 / 7100.00.

Order-of-magnitude check on the area. The adjusted coordinates span about 900 ft east to west and 660 ft north to south, so the parcel cannot exceed 900 × 660 / 43,560 = 13.6 acres, and with two corners cut off something in the 11 to 13 acre range is expected. 12.313 acres sits there, which is exactly the check that catches a factor-of-two error from forgetting to halve the double area.

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