A Five-Sided Boundary Traverse That 'Closes'
A five-sided closed traverse was run with a 2-second total station and fixed prisms to control a proposed replat of four lots in a Kansas town. Interior angles observed: A 98°12'10", B 112°44'30", C 105°03'50", D 121°58'00", E 102°01'48". Horizontal distances: AB 486.25 ft, BC 512.40 ft, CD 500.10 ft, DE 545.30 ft, EA 406.00 ft. After the angular misclosure is distributed equally, the latitudes and departures close with ΔN = +0.10 ft and ΔE = −0.098 ft. The firm's written standard for boundary surveys is a linear misclosure no worse than 1:10,000 and an angular misclosure no greater than 10″√n, where n is the number of angles. The client needs 0.05-ft stakeout of lot corners for fence construction, and the plat is due in nine days. The crew chief proposes to accept the traverse as-is because "it closes," skip the angular adjustment entirely, and apply only a compass-rule correction to the coordinates.
1. Compute the angular misclosure, compare it with the firm standard, and state how it must be handled before latitudes and departures are used.
Read the worked answer
Sum the observed angles. Seconds: 10 + 30 + 50 + 0 + 48 = 138″ = 2′18″. Minutes: 12 + 44 + 3 + 58 + 1 = 118′, plus the 2′ carried = 120′ = 2°00′00″. Degrees: 98 + 112 + 105 + 121 + 102 = 538°, plus 2° = 540°.
Theoretical sum. (n − 2) × 180° = 3 × 180° = 540°00′00″. Misclosure = +18″ (observed high).
Compare. Allowable = 10″√5 = 22.4″. The 18″ is inside the standard, so the angles are acceptable - but only 4.4″ inside it, which is poor for a 2″ instrument over five short backsights. That pattern usually means a systematic cause: forced-centering error, a misread or misbooked angle, one target set at the wrong height driving a vertical-axis effect, or a single angle turned only once.
Handling. Distribute the 18″ equally with opposite sign (3.6″ per angle, i.e., subtract 3.6″ from each) before computing latitudes and departures. Skipping this folds angular error into what the compass rule is supposed to fix, and it will put the correction in the wrong place. After distributing angles, recompute bearings and distances, then compute the linear misclosure - the given ΔN and ΔE must be understood as post-angular-adjustment values.
2. Compute the linear misclosure and relative precision and judge them against the standard.
Read the worked answer
Linear misclosure. √(0.10² + 0.098²) = √(0.010000 + 0.009604) = √0.019604 = 0.140 ft.
Perimeter. 486.25 + 512.40 + 500.10 + 545.30 + 406.00 = 2,450.05 ft.
Relative precision. 2,450.05 / 0.140 = 17,500, so about 1:17,500.
Judgment. It passes 1:10,000, with roughly a factor of 1.75 in reserve. Two cautions, though. First, the ratio is a whole-traverse statistic; it does not certify any single line, and the 406-ft leg EA is the one most likely to be weak. Second, passing a closure standard says the internal geometry is consistent - it says nothing about whether the control points are correctly tied to record monuments. A traverse can close beautifully around a wrongly identified corner.
Both tests are therefore met: angular 18″ < 22.4″, linear 1:17,500 < 1:10,000. Accept the traverse for computation, but note in the file that the angular residual is larger than a 2″ instrument should produce and that the closing ratio is not a positional-accuracy guarantee for the 0.05-ft stakeout.
3. Compute the compass-rule correction to line CD's latitude and departure, and evaluate the crew chief's plan and the 0.05-ft stakeout request.
Read the worked answer
Compass rule. Correction per line = −(total misclosure) × (line length / perimeter). With total latitude misclosure +0.10 ft and departure misclosure −0.098 ft:
- CD latitude correction = −0.10 × (500.10 / 2,450.05) = −0.0204 ft.
- CD departure correction = +0.098 × (500.10 / 2,450.05) = +0.0200 ft.
So CD's corrected latitude = computed latitude − 0.0204 ft and corrected departure = computed departure + 0.0200 ft.
The crew chief's plan is wrong on two counts. Applying only a compass rule without distributing the 18″ angular misclosure forces the coordinate correction to absorb angular error, which biases bearings instead of correcting them. And "it closes" is not the firm's standard; the standard is a stated ratio plus an angular tolerance.
The stakeout request. The 18″ angular residual, if left unadjusted, produces lateral displacement of about 500 × tan 18″ ≈ 0.044 ft across a 500-ft line - the same order as the 0.05-ft tolerance the fence depends on. Closure ratio alone does not buy that precision. Recommend turning a second set of angles (or reobserving the weakest legs) and, if the angles repeat, adjusting properly and reporting the residual uncertainty to the client rather than implying 0.05-ft certainty.