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Angle misclosure, closed link traverse AT1-ET2

Short answer

Angle misclosure 40'' (computed ET2 105°35'28''). Correction +8'' per angle, giving adjusted bearings: T1A 339°09'21'', AB 81°26'47'', BC 168°15'08'', CD 77°32'09'', DE 354°08'22'', ET2 105°36'08''.

Compute the angle misclosure and adjusted bearings in the following closed link traverse. Adjusted bearing AT1 = 329°09′21″ [?]. Adjusted bearing ET2 = 105°36′08″. Angles (angles to the right): A = 282°17′18″, B = 266°48′13″, C = 89°16′53″, D = 96°36′05″, E = 291°27′38″.

Reading of the data. The first bearing is printed as 329°09′21″ [?]. If 329°09′21″ is taken as the bearing of the line from T1 to A, the computed bearing of ET2 comes out as 95°35′28″, a misclosure of about 10°00′40″, which is impossible for angles measured to the second. The value consistent with the other data is 339°09′21″ for the line from T1 to A (its reverse bearing, from A to T1, is 159°09′21″). That value is used below. The angles are to the right, so the bearing of each forward line = bearing of the previous line + 180° + angle at the station.

Rule (angles to the right): FB of next line=FB of previous line+180∘+angle to the right\text{FB of next line} = \text{FB of previous line} + 180^\circ + \text{angle to the right} (add or subtract 360∘360^\circ to bring the result between 0∘0^\circ and 360∘360^\circ).

Step 1: Bearings computed from the observed angles

LineWorkingComputed FB
T1-Agiven339°09′21″
A-B339°09′21″ + 180° + 282°17′18″ − 720°81°26′39″
B-C81°26′39″ + 180° + 266°48′13″ − 360°168°14′52″
C-D168°14′52″ + 180° + 89°16′53″ − 360°77°31′45″
D-E77°31′45″ + 180° + 96°36′05″354°07′50″
E-T2354°07′50″ + 180° + 291°27′38″ − 720°105°35′28″

Computed FB of E-T2 = 105°35′28″; given (known) FB of E-T2 = 105°36′08″.

Step 2: Angular misclosure

Misclosure=given−computed=0∘00′40′′\text{Misclosure} = \text{given} - \text{computed} = 0^\circ 00'40''

The computed bearing is 40″ too small. This is only 8″ per angle for 5 angles, which is acceptable for a 20″ theodolite (the permissible value CnC\sqrt{n} with C=20′′C = 20'' is about 45′′45''). The error is distributed equally among the 5 angles.

Step 3: Adjustment

Total correction = +0°00′40″ (added to the computed bearing of the last line). It is distributed equally over the 5 observed angles: +0°00′40″ / 5 = +0°00′08″ per angle. The correction to a line is +0°00′08″ multiplied by the number of angles used to reach it.

LineComputed FBCorrectionCorrected FBCorrected BB
T1-A339°09′21″0339°09′21″159°09′21″
A-B81°26′39″+0°00′08″81°26′47″261°26′47″
B-C168°14′52″+0°00′16″168°15′08″348°15′08″
C-D77°31′45″+0°00′24″77°32′09″257°32′09″
D-E354°07′50″+0°00′32″354°08′22″174°08′22″
E-T2105°35′28″+0°00′40″105°36′08″285°36′08″

Check: the corrected FB of the last line equals the given value 105°36′08″.

Answer: angle misclosure = 40″ (computed ET2 = 105°35′28″, adjusted ET2 = 105°36′08″). Adjustment = +8″ per angle (bearing corrections +8″, +16″, +24″, +32″, +40″). Adjusted bearings: T1A 339°09′21″, AB 81°26′47″, BC 168°15′08″, CD 77°32′09″, DE 354°08′22″, ET2 105°36′08″.

Note: the adjusted angles are the observed angles plus 8″ each; the correction is added to each angle to the right because the computed bearing is too small.

✓ Answer checked by Claude Opus 5.5 (high effort)

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