CogoKit

Traverse adjustment

A traverse that fails its precision specification, and what to do about it

A four-sided traverse closes at 1:2,538 against a 1:10,000 requirement. The direction of the closure line points straight at the course carrying the blunder, which is remeasured and the traverse recomputed.

Apply· about 28 minutes by hand· 5 steps

Given

  • A closed-loop boundary traverse P-Q-R-S-P. Angles have already been adjusted and converted to bearings.
  • P-Q N 78°15′00″ E 512.98 ft
  • Q-R S 15°40′00″ E 399.82 ft
  • R-S S 77°50′00″ W 532.83 ft
  • S-P N 12°55′00″ W 402.84 ft
  • Station P is held at N 3000.00, E 4000.00.
  • The contract requires 1:10,000 linear precision.
  • The site is on a hillside; courses Q-R and R-S were measured with a hand-held tape over broken ground.

Required

  • The linear misclosure and the precision as computed.
  • Whether the traverse meets its specification.
  • The most likely source of the error, identified from the closure itself.
  • The recomputed traverse after the suspect course is remeasured at 531.96 ft.

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 as observed

Convert the bearings to azimuths and resolve each course. Nothing in this step is different from a traverse that closes well; the arithmetic is the same and the failure only appears in the sums.

The departure column sum is nearly five times the latitude column sum, which is the first clue: an error that lives almost entirely in the east-west direction.

Observed traverse
CourseBearingAzimuthDistance (ft)Latitude (ft)Departure (ft)
P-QN 78°15′00″ E78°15′00″512.98104.464502.231
Q-RS 15°40′00″ E164°20′00″399.82−384.966107.968
R-SS 77°50′00″ W257°50′00″532.83−112.297−520.862
S-PN 12°55′00″ W347°05′00″402.84392.647−90.048
Sums1848.47−0.153−0.712

Misclosure and the verdict

The linear misclosure is the resultant of the two sums, and the precision is the ratio of the perimeter to it, rounded down.

At 1:2,538 the traverse fails the 1:10,000 requirement by a factor of four. The allowable misclosure at 1:10,000 would be 1848.47 / 10,000 = 0.18 ft; the traverse produced 0.73 ft. This is not a case for adjustment. A misclosure four times the tolerance is a blunder, and adjusting a blunder simply smears it over every station.

ΣLat = −0.1527 ft, ΣDep = −0.7120 ft e = √(0.1527² + 0.7120²) = √(0.023317 + 0.506944) = √0.530261 = 0.7282 ft precision = 1848.47 / 0.7282 = 2538, that is 1:2,538 allowable at 1:10,000 = 1848.47 / 10,000 = 0.18 ft 0.73 ft observed against 0.18 ft allowable: reject

Read the direction of the closure line

The closure line has a direction as well as a length, and that direction is diagnostic. A blunder in a single distance leaves a closure line parallel to that course, because the traverse simply overshoots or undershoots along it. A blunder in a single angle leaves a closure line perpendicular to the line joining the station with the bad angle to the far end of the traverse, which is a different and usually less obliging geometry.

Compute the azimuth of the closure line as the direction the corrections must push, that is the direction from the computed closing position back to the true start.

It comes out within four minutes of the back azimuth of course R-S. That is not a coincidence at this precision: R-S is one of the two courses taped over broken ground, and it is the longest course in the traverse.

closure azimuth = atan2(−ΣDep, −ΣLat) = atan2(0.7120, 0.1527) = 77°53′38″ back azimuth of R-S = 257°50′00″ − 180° = 77°50′00″ difference = 3′38″ magnitude of the closure, 0.73 ft, is the approximate size of the suspected blunder

Remeasure and recompute

The crew returns and remeasures R-S with a total station: 531.96 ft, which is 0.87 ft shorter than the taped value. The original tape reading was long, consistent with a tape held over a sag or a missed break in the slope on a hillside course.

Nothing else changes. Substitute the new distance and recompute the latitude and departure of that one course.

R-S remeasured = 531.96 ft (taped 532.83 ft, difference 0.87 ft) latitude = 531.96 cos(257°50′00″) = −112.114 ft departure = 531.96 sin(257°50′00″) = −520.011 ft
Recomputed traverse with the remeasured course
CourseAzimuthDistance (ft)Latitude (ft)Departure (ft)
P-Q78°15′00″512.98104.464502.231
Q-R164°20′00″399.82−384.966107.968
R-S257°50′00″531.96−112.114−520.011
S-P347°05′00″402.84392.647−90.048
Sums1847.600.0310.139

Adjust and report

The recomputed misclosure is 0.14 ft over 1847.60 ft, or 1:13,024, which meets the specification. Only now is the compass rule appropriate: the residual is small, unbiased in direction and consistent with random error in the remaining measurements.

The final report should say what happened. A traverse that failed, was diagnosed, remeasured and recomputed is stronger evidence of good work than one that closed the first time, provided the failure and its resolution are on the record.

e = √(0.0306² + 0.1385²) = 0.1419 ft precision = 1847.60 / 0.1419 = 13,024, that is 1:13,024, which meets 1:10,000
Compass-rule adjustment and final coordinates
CourseLat correction (ft)Dep correction (ft)To stationNorthing (ft)Easting (ft)
P-Q−0.009−0.038Q3104.464502.19
Q-R−0.007−0.030R2719.484610.13
R-S−0.009−0.040S2607.364090.08
S-P−0.007−0.030P3000.004000.00

Answer

  • As observed: misclosure 0.73 ft over a 1848.47 ft perimeter, precision 1:2,538. The traverse fails the 1:10,000 specification, which allows only 0.18 ft.
  • The closure line bears 77°53′38″, within 3′38″ of the back azimuth of course R-S, which identifies R-S as carrying a distance blunder of roughly 0.7 to 0.9 ft.
  • R-S remeasured 531.96 ft, 0.87 ft shorter than taped.
  • Recomputed: misclosure 0.14 ft over 1847.60 ft, precision 1:13,024, which passes.
  • Final coordinates: P 3000.00 / 4000.00; Q 3104.46 / 4502.19; R 2719.48 / 4610.13; S 2607.36 / 4090.08.

Check

The blunder and the closure must be of the same size and along the same line. The remeasurement differed by 0.87 ft; the original misclosure was 0.73 ft on a bearing within four minutes of that course. The residual 0.14 ft is the ordinary random error that was present all along.

Verify the diagnosis arithmetically: subtracting 0.87 ft of length from course R-S changes its latitude by 0.87 cos(257°50′00″) = −0.18 ft and its departure by 0.87 sin(257°50′00″) = −0.85 ft. Adding those to the original sums, −0.153 + 0.184 = 0.031 and −0.712 + 0.851 = 0.139, reproduces the recomputed sums exactly.

The adjusted latitude and departure columns must sum to zero, and the traverse must close back on P at 3000.00 / 4000.00. It does.

What not to conclude: a good closure never proves the absence of a blunder. Two compensating errors close perfectly. The direction test above is what turned a failing number into a located fault.

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