CogoKit

Traverse adjustment

Omitted measurement: the missing bearing and distance of one side

Four sides of a five-sided parcel are recovered but the fifth runs through a swamp. Both its bearing and its length are computed from the condition that the traverse must close.

Apply· about 22 minutes by hand· 5 steps

Given

  • A five-sided parcel A-B-C-D-E-A. Four sides were measured; the fifth, E-A, crosses standing water and could not be measured.
  • A-B N 62°15′00″ E 425.60 ft
  • B-C S 27°40′00″ E 316.85 ft
  • C-D S 55°10′00″ W 388.20 ft
  • D-E N 78°30′00″ W 254.75 ft
  • E-A bearing and distance both unknown.
  • The four measured courses are taken as correct; the whole of any error will fall into the computed course.
  • Distances to 0.01 ft, bearings to 1 second.

Required

  • The latitude and departure of the missing course.
  • Its bearing and its length.
  • The coordinates of every corner on an assumed grid with A at N 1000.00, E 1000.00.

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

The closure condition is the equation

A closed traverse must return to its starting point, so the sum of all latitudes is zero and the sum of all departures is zero. With four courses known and one unknown, those two conditions are two equations in the two unknowns of the missing course.

Because the two unknowns happen to be exactly the latitude and departure of one course, the equations are already solved: the missing latitude is the negative of the sum of the known latitudes, and likewise for departure.

Note what this costs. Every error in the four measured courses lands, undiminished, in the computed one. There is no closure left over to check it, so an omitted-measurement computation carries no internal quality control at all. It is a last resort, not a shortcut.

ΣLat(all) = 0 → lat(E-A) = −ΣLat(measured) ΣDep(all) = 0 → dep(E-A) = −ΣDep(measured)

Latitudes and departures of the measured courses

Convert each bearing to an azimuth, then resolve. Keeping three decimals through this table matters: the missing course is the difference of large numbers, so it inherits every rounding decision made here.

The two sums are what the missing course has to undo.

The four measured courses
CourseBearingAzimuthDistance (ft)Latitude (ft)Departure (ft)
A-BN 62°15′00″ E62°15′00″425.60198.166376.651
B-CS 27°40′00″ E152°20′00″316.85−280.623147.122
C-DS 55°10′00″ W235°10′00″388.20−221.736−318.641
D-EN 78°30′00″ W281°30′00″254.7550.789−249.636
Sums1385.40−253.405−44.504

Solve the missing course

Change the sign of each sum to get the components of the missing course. Both come out positive, so the course runs north and east: it must, since the four measured courses have carried the traverse south and west of where it started.

Length is the resultant of the two components; direction is their arctangent, taken with the easting first in the surveying convention so that the answer is an azimuth clockwise from north.

lat(E-A) = +253.405 ft, dep(E-A) = +44.504 ft length = √(253.405² + 44.504²) = √(64,214.1 + 1980.6) = √66,194.7 = 257.28 ft azimuth = atan2(44.504, 253.405) = 9°57′40″ bearing = N 9°57′40″ E

Coordinates

With the missing course solved, the traverse is complete and coordinates follow by ordinary accumulation from the held station.

The computed course closes the figure by construction, so the coordinates of A recovered at the end of the circuit are exactly the coordinates of A held at the start. That is a consequence of the method, not a check on it.

Coordinates, A held at N 1000.00 E 1000.00
StationNorthing (ft)Easting (ft)
A1000.001000.00
B1198.171376.65
C917.541523.77
D695.811205.13
E746.60955.50
A (closing)1000.001000.00

What to do about the missing check

Because the computation is forced to close, the only way to test it is from outside. Three practical options: measure the missing course by some indirect method such as a distance and angle from an offset point and compare; measure a diagonal across the parcel and compare it with the inverse from the computed coordinates; or occupy a fifth station on high ground and observe directions to two of the corners.

A diagonal is usually cheapest. Here the inverse from B to E is a natural choice: from the computed coordinates ΔN = −451.570 ft and ΔE = −421.155 ft, giving √(203,916 + 177,372) = 617.48 ft on a bearing of S 42°58′30″ W. A tape across that line either supports the computation or does not.

Record on the plat that the course E-A is computed rather than measured. A later surveyor who assumes it was taped will draw the wrong conclusions from any discrepancy.

Answer

  • Missing course latitude +253.405 ft, departure +44.504 ft.
  • E-A bears N 9°57′40″ E and is 257.28 ft long.
  • Coordinates with A held at 1000.00 / 1000.00: B 1198.17 / 1376.65; C 917.54 / 1523.77; D 695.81 / 1205.13; E 746.60 / 955.50.
  • The computed course carries the whole of any error in the four measured courses, so it must be labelled as computed on the plat.

Check

Resolve the computed course back into components and confirm it reverses the measured sums: 257.28 cos(9°57′40″) = 253.40 ft and 257.28 sin(9°57′40″) = 44.50 ft, which are the negatives of −253.405 and −44.504.

Compute the coordinates of E by accumulation, then inverse from E back to A: ΔN = 1000.00 − 746.60 = 253.40, ΔE = 1000.00 − 955.50 = 44.50, giving 257.28 ft at 9°57′40″. This confirms the accumulation arithmetic even though it cannot confirm the field work.

External check available in the field: the diagonal B to E computes as 617.48 ft on S 42°58′30″ W. Taping it is the only real test of the answer.

Sanity check on the direction: the four measured courses net 253.41 ft south and 44.50 ft west of the start, so the closing course must run north and slightly east, at a bearing well within the first quadrant. It does.

More traverse adjustment

All in this category