CogoKit

Traverse adjustment

Angular closure with a suspected blunder in one angle

A six-sided traverse misses its angular closure by 2′36″ against a 37 second tolerance. The size of the misclosure identifies a single blunder, which is traced in the field book and removed before any adjustment.

Analyze· about 30 minutes by hand· 6 steps

Given

  • A closed six-sided loop traverse A-B-C-D-E-F-A, interior angles observed to the right.
  • Observed angles as booked: A 84°38′15″, B 141°10′05″, C 136°26′11″, D 106°46′50″, E 118°49′40″, F 132°11′35″.
  • Distances: A-B 446.06, B-C 487.14, C-D 371.14, D-E 403.19, E-F 575.56, F-A 496.69 ft.
  • Azimuth of A-B fixed at 109°39′15″. Station A held at N 2000.00, E 3000.00.
  • Angles were turned in two sets with a 5 second instrument. The angular tolerance is 15 seconds times the square root of the number of angles.
  • Linear specification 1:10,000.

Required

  • The angular misclosure and whether it is within tolerance.
  • Identification of the blunder, and the corrected angle.
  • The adjusted angles, azimuths, coordinates and precision after the blunder is removed.
  • What the traverse would have produced if the misclosure had simply been distributed.

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

Angular closure as booked

Six interior angles must sum to (6 − 2) × 180° = 720°00′00″. The booked angles sum to 720°02′36″, a misclosure of 156 seconds.

The tolerance for six angles turned in two sets with a 5 second instrument is 15″√6 = 37 seconds. The observed misclosure is more than four times that.

This is the decision point. A misclosure inside tolerance is random error and is distributed. A misclosure several times tolerance is a blunder and must be found, not spread. Distributing 156 seconds over six angles would move every one of them by 26 seconds, corrupting five good angles to hide one bad one.

required sum = (6 − 2) × 180° = 720°00′00″ observed sum = 720°02′36″ misclosure = +156″ tolerance = 15″ × √6 = 36.7″, say 37″ 156″ / 37″ = 4.2 times tolerance: this is a blunder, not random error

Look for a blunder of the size of the misclosure

When a single blunder is present, the angular misclosure is the blunder, very nearly. So the first move is to look for a booked value that is wrong by about 2′30″ and to look in the places where 2′30″ errors come from: a misread minute, a transposed pair of digits, a mean of two sets computed from one set of the wrong pointing.

Reviewing the field book, station D was turned in two sets reading 106°44′18″ and 106°44′22″, mean 106°44′20″. The value carried forward to the reduction sheet was 106°46′50″. The minutes digit was transcribed as 46 instead of 44, a 2′30″ error given the accompanying seconds.

Correct the transcription and recompute the sum. The residual misclosure is then 6 seconds, comfortably inside the 37 second tolerance, which confirms the diagnosis: one blunder accounted for essentially the whole of the closure failure.

field book at D: sets read 106°44′18″ and 106°44′22″, mean 106°44′20″ value used on the reduction sheet: 106°46′50″ blunder = 106°46′50″ − 106°44′20″ = +2′30″ corrected sum = 720°02′36″ − 2′30″ = 720°00′06″ residual misclosure = +6″, which is well inside 37″ correction per angle = −6″ / 6 = −1″

Adjusted angles and azimuths

With the blunder removed, the remaining 6 seconds is ordinary random error and is spread equally, one second off each angle.

Azimuths are then carried from the fixed direction of A-B. The chain must return to 109°39′15″ after six stations.

Angle adjustment after removing the blunder, and the resulting azimuths
StationBooked angleCorrected for blunderAdjusted angleCourseAzimuth
A84°38′15″84°38′15″84°38′14″A-B109°39′15″
B141°10′05″141°10′05″141°10′04″B-C70°49′19″
C136°26′11″136°26′11″136°26′10″C-D27°15′29″
D106°46′50″106°44′20″106°44′19″D-E313°59′48″
E118°49′40″118°49′40″118°49′39″E-F252°49′27″
F132°11′35″132°11′35″132°11′34″F-A205°01′01″
Sum720°02′36″720°00′06″720°00′00″back to A-B109°39′15″

Latitudes, departures and precision

Resolve each course with the adjusted azimuths and the observed distances. The linear misclosure is 0.19 ft over 2779.78 ft, giving 1:14,598, which meets the 1:10,000 requirement.

The direction of the closure line, about 292°, does not lie along or across any single course, which is consistent with the remaining error being random.

e = √(0.0702² + 0.1770²) = 0.1904 ft precision = 2779.78 / 0.1904 = 14,598, that is 1:14,598
Latitudes and departures with the corrected angle set
CourseAzimuthDistance (ft)Latitude (ft)Departure (ft)
A-B109°39′15″446.06−150.029420.072
B-C70°49′19″487.14160.028460.105
C-D27°15′29″371.14329.926169.982
D-E313°59′48″403.19280.062−290.047
E-F252°49′27″575.56−169.966−549.892
F-A205°01′01″496.69−450.092−210.043
Sums2779.78−0.0700.177

Adjusted coordinates

Compass-rule corrections are applied in proportion to course length and the coordinates accumulate from station A.

These are the coordinates to publish. The field book entry, the transcription error and its correction all belong in the project record, because a later surveyor comparing the field book with the plat will otherwise find a 2′30″ discrepancy at station D and no explanation for it.

Compass-rule adjusted coordinates
StationNorthing (ft)Easting (ft)
A2000.003000.00
B1849.983420.04
C2010.023880.12
D2339.964050.08
E2620.033760.00
F2450.083210.08
A (closing)2000.003000.00

What spreading the blunder would have cost

Suppose the 156 second misclosure had simply been distributed, 26 seconds off each angle, with no field book review. Every azimuth after the first would shift by an accumulating multiple of 26 seconds, and the traverse would compute a linear misclosure of 0.30 ft over 2779.78 ft, or 1:9,180.

That is the trap. 1:9,180 fails 1:10,000 but only barely, and against a looser specification such as 1:5,000 it would have passed and been reported as acceptable. The angles at every station would have been wrong by up to 26 seconds, station D would have been wrong by 2′30″ less its share, and nothing in the linear closure would have said so.

The angular check is the one that catches this, which is why it is worked and compared to a tolerance before the distances are ever resolved.

misclosure spread without correcting the blunder: −26″ per angle resulting linear misclosure = 0.30 ft over 2779.78 ft resulting precision = 1:9,180

Answer

  • Angular misclosure as booked +156″ against a 37″ tolerance, a factor of 4.2, so a blunder is indicated.
  • The blunder is at station D: 106°46′50″ was booked where the field book mean is 106°44′20″, an error of +2′30″.
  • After correction the angular misclosure is +6″, adjusted by −1″ at each of the six angles.
  • Adjusted azimuths: A-B 109°39′15″, B-C 70°49′19″, C-D 27°15′29″, D-E 313°59′48″, E-F 252°49′27″, F-A 205°01′01″.
  • Linear misclosure 0.19 ft over 2779.78 ft, precision 1:14,598, which passes.
  • Adjusted coordinates: A 2000.00 / 3000.00; B 1849.98 / 3420.04; C 2010.02 / 3880.12; D 2339.96 / 4050.08; E 2620.03 / 3760.00; F 2450.08 / 3210.08.

Check

The blunder must account for the misclosure. Booked misclosure 156″ less the 150″ blunder leaves 6″, which is what the corrected sum shows and which is well inside the 37″ tolerance. A residual of the same order as the original misclosure would have meant a second blunder or a wrong diagnosis.

Run the azimuth chain all the way round with the adjusted angles: it returns to 109°39′15″, the fixed azimuth of A-B.

The adjusted latitude and departure columns sum to zero, and the traverse closes back on A at 2000.00 / 3000.00.

Comparison test: with the blunder left in and spread, the traverse computes 1:9,180; with it removed, 1:14,598. The improvement is real, and the fact that the uncorrected version was not catastrophically bad is the reason the angular tolerance test cannot be skipped.

More traverse adjustment

All in this category