Traversing and Traverse Computations: Interior and Deflection Angle Traverses

Overview of Traversing and Traverse Computations

• Traverse Classification and Applications:

  • Traversing is a fundamental field procedure in plane surveying used to establish horizontal control networks.

  • Two primary traverse methods based on angular measurement types are:

  • Interior Angle Traverse: Applied primarily in boundary and land surveying of closed polygonal tracts.

  • Deflection Angle Traverse: Employed extensively in route surveys for continuous linear projects including highways, roads, railroads, pipelines, transmission lines, canals, and to a lesser extent in land boundary surveys and control networks for hydrographic and topographic surveys.

Interior Angle Traverse Principles and Procedures

• Definition of Interior Angles:

  • Interior angles are the horizontal angles formed between adjacent sides inside a closed geometric polygon.

  • In a closed traverse with stations $A, B, C, D, E, F$, the interior angles correspond to the vertex angles at each successive station: $\beta_a, \beta_b, \beta_c, \beta_d, \beta_e, \beta_f$.

• Direction of Measurement:

  • Interior angles may be measured in either a clockwise or counterclockwise direction.

  • The standard surveying practice is to measure all interior angles clockwise.

  • Systematic clockwise progression reduces common errors in instrument reading, field plotting, and recording:

  • Interior angle at station $A$ is turned clockwise from backsight $B$ to foresight $F$.

  • Interior angle at station $B$ is turned clockwise from backsight $C$ to foresight $A$.

  • Interior angle at station $C$ is turned clockwise from backsight $D$ to foresight $B$.

  • Interior angle at station $D$ is turned clockwise from backsight $E$ to foresight $C$.

  • Interior angle at station $E$ is turned clockwise from backsight $F$ to foresight $D$.

  • Interior angle at station $F$ is turned clockwise from backsight $A$ to foresight $E$.

• Instrumental Field Operation Procedure:

  • The transit or theodolite is set up and precisely leveled directly over the traverse station.

  • The horizontal circle vernier or digital readout is indexed to zero ($0^\circ 00'$).

  • A backsight is taken to the station on the left.

  • The upper motion clamp is loosened, and the telescope is rotated clockwise until the station on the right is sighted and bisected.

  • The horizontal circle reading is observed and recorded as the interior angle.

  • Repetition Method for Precision:

  • To achieve the desired level of precision, interior angles must be measured with sufficient repetitions.

  • For any specified number of repetitions, exactly half of the measurements are made with the telescope in the direct (normal) position, and the remaining half are made with the telescope in the reversed (plunged/inverted) position.

  • This procedure cancels out instrumental errors including collimation error, horizontal axis tilt, and index error.

• Geometric Conditions and Checks:

  • For any closed polygon having $n$ sides (or vertices), the theoretical sum of all interior angles is governed by the geometric relationship:     $\sum \text{Interior Angles} = (n - 2) \times 180^\circ$

  • This geometric condition provides an absolute mathematical check on the field measurements and serves as the baseline for distributing angular errors.

  • Angular Error of Closure:

  • The observed interior angles are summed ($\sum \text{Observed}$).

  • The angular closure error ($\text{Error}$) is determined by comparing the observed sum with $(n - 2) \times 180^\circ$:       $\text{Error} = (n - 2) \times 180^\circ - \sum \text{Observed}$

  • When an angular error exists, it is adjusted by distributing the total error equally to each station, assuming equal precision across all instrument setups:       $\text{Correction per station} = \frac{\text{Error}}{n}$

  • The adjusted angles are summed to confirm that their total equals exactly $(n - 2) \times 180^\circ$.

  • Horizon Closure:

  • For high-precision surveys, an additional check is performed at each station by measuring both the interior angle and the exterior angle to close the horizon ($\text{Interior Angle} + \text{Exterior Angle} = 360^\circ$).

• Linear Measurements and Bearing Computation:

  • Distances between traverse stations ($d_1, d_2, d_3, \dots$) are measured using a steel tape or Electronic Distance Measurement (EDM) equipment.

  • The precision of linear measurements must match the precision of the angular measurements.

  • Distances may be determined concurrently with angular measurements or as an independent operation before or after angle measurements.

  • Course bearings or azimuths are calculated sequentially by sketching the known reference meridian and angular turns at each vertex.

Interior Angle Field Notes and Data Tabulation

• Field Note Header Specifications (Sample Exercise):

  • Project: Interior Angle Traverse

  • Date: November 3, 1986

  • Time / Temperature: 9–12 AM / $28^\circ\text{C}$

  • Weather: Cloudy Bright

  • Location: PMA Parade Ground

  • Lab Exercise: No. 15

  • Field Party: Group No. 4 / CE 221

• Tabulation of Observed Data and Angular Measurements:

  • Station $A$:

  • Backsight to $E$, Foresight to $B$

  • Distance $AB = 139.11\,m$

  • Observed Interior Angle (Single): $166^\circ 53'$

  • Observed Interior Angle (Double): $333^\circ 45'$

  • Mean Observed Interior Angle: $166^\circ 52'$

  • Observed Bearing: $\text{S } 89^\circ 00' \text{ E}$

  • Station $B$:

  • Backsight to $A$, Foresight to $C$

  • Distance $BC = 50.03\,m$

  • Observed Interior Angle (Single): $103^\circ 16'$

  • Observed Interior Angle (Double): $205^\circ 30'$

  • Mean Observed Interior Angle: $103^\circ 15'$

  • Observed Bearing: $\text{S } 12^\circ 00' \text{ E}$

  • Station $C$:

  • Backsight to $B$, Foresight to $D$

  • Distance $CD = 210.46\,m$

  • Observed Interior Angle (Single): $89^\circ 29'$

  • Observed Interior Angle (Double): $178^\circ 56'$

  • Mean Observed Interior Angle: $89^\circ 28'$

  • Observed Bearing: $\text{S } 78^\circ 30' \text{ W}$

  • Station $D$:

  • Backsight to $C$, Foresight to $E$

  • Distance $DE = 80.23\,m$

  • Observed Interior Angle (Single): $90^\circ 05'$

  • Observed Interior Angle (Double): $180^\circ 08'$

  • Mean Observed Interior Angle: $90^\circ 04'$

  • Observed Bearing: $\text{N } 11^\circ 30' \text{ W}$

  • Station $E$:

  • Backsight to $D$, Foresight to $A$

  • Distance $EA = 74.46\,m$

  • Observed Interior Angle (Single): $90^\circ 16'$

  • Observed Interior Angle (Double): $180^\circ 32'$

  • Mean Observed Interior Angle: $90^\circ 16'$

  • Observed Bearing: $\text{N } 78^\circ 25' \text{ E}$

• Closure Computations and Adjustment Summary:

  • Number of sides: $n = 5$

  • Theoretical sum: $(5 - 2) \times 180^\circ = 540^\circ 00'$

  • Sum of mean observed angles:     $\sum \text{Int } \angle = 166^\circ 52' + 103^\circ 15' + 89^\circ 28' + 90^\circ 04' + 90^\circ 16' = 539^\circ 55'$

  • Angular misclosure error:     $\text{Error} = 540^\circ 00' - 539^\circ 55' = +05'$

  • Correction per angle:     $\text{Correction} = \frac{+05'}{5} = +01'$

  • Adjusted Angles:

  • Station $A$: $166^\circ 52' + 01' = 166^\circ 53'$

  • Station $B$: $103^\circ 15' + 01' = 103^\circ 16'$

  • Station $C$: $89^\circ 28' + 01' = 89^\circ 29'$

  • Station $D$: $90^\circ 04' + 01' = 90^\circ 05'$

  • Station $E$: $90^\circ 16' + 01' = 90^\circ 17'$

  • Sum of Adjusted Angles: $540^\circ 00'$ (checks)

Deflection Angle Traverse Principles and Geometry

• Definition and Characteristics:

  • A deflection angle is a horizontal angle measured from the prolongation (extension) of the preceding traverse line to the succeeding traverse line.

  • Direction of turn: measured either clockwise (to the right) or counterclockwise (to the left).

  • Angular range: varies continuously between $0^\circ$ and $180^\circ$.

  • Designation requirement: every deflection angle must explicitly include its numerical magnitude and direction of turn designated as Right (R) or Left (L).

  • In route surveying:

  • At Station $B$ connecting course $AB$ ($165.30\,m$) to course $BC$ ($185.15\,m$), the deflection angle is $20^\circ 10' \text{ (R)}$.

  • At Station $C$ connecting course $BC$ to course $CD$ ($120.25\,m$), the deflection angle is $85^\circ 30' \text{ (L)}$.

• Mathematical Rules for Closed Deflection Traverses:

  • Non-Crossing Closed Traverse (Standard Polygon):

  • In any closed traverse whose sides do not cross one another, the algebraic sum of the deflection angles must equal $360^\circ$, assigning opposite signs to angles turned to the left versus those turned to the right:       $\left| \sum \text{Deflection Angles}_R - \sum \text{Deflection Angles}_L \right| = 360^\circ$

  • Self-Crossing Closed Traverse:

  • If the traverse lines cross each other once, or any odd number of times, the sum of the right deflection angles must equal the sum of the left deflection angles:       $\sum \text{Deflection Angles}_R = \sum \text{Deflection Angles}_L$

Deflection Angle Field Procedures and Best Practices

• Standard Step-by-Step Field Operation:

  • 1. Set up and level the transit/theodolite over the traverse station (e.g., station $A$).

  • 2. Set the horizontal circle reading to $0^\circ 00'$.

  • 3. Take a backsight to the rear station (e.g., station $E$) with the telescope in the direct position.

  • 4. Plunge (transit) the telescope about its horizontal axis so it points along the forward prolongation of line $EA$.

  • 5. Loosen the upper clamp and rotate the upper motion to sight the forward station (station $B$).

  • 6. Read and record the single deflection angle on the horizontal circle, noting whether the turn was Right (clockwise) or Left (counterclockwise).

  • 7. Loosen the lower clamp (retaining the angle reading on the circle) and rotate the transit about its vertical axis to backsight on station $E$ a second time.

  • 8. Plunge the telescope a second time (restoring it to the direct position).

  • 9. Loosen the upper clamp and sight on station $B$.

  • 
    

  10. The horizontal circle now reads double the original angle value.

  • 
    

  11. Divide the total accumulated reading by $2$ to yield the mean (average) deflection angle.

• Elimination of Instrumental Errors:

  • Measuring deflection angles by double deflection (or preferably quadrupling the angle) eliminates:

  • Imperfect adjustment of the line of collimation.

  • Horizontal axis inclination errors.

  • Reading mistakes and observational slips.

  • The second observed value provides an immediate mathematical verification of the first single reading.

Deflection Angle Field Notes and Bearings

• Field Record Metadata (Sample Route Exercise):

  • Title: Deflection Angle Traverse

  • Date: November 10, 1986

  • Time / Temperature: 1–4 PM / $30^\circ\text{C}$

  • Weather: Heavy Overcast

  • Location: Rimando Road, Baguio

  • Lab Exercise: No. 16

  • Party: Group No. 5 / CE 221

  • Known Base Bearing: Course $AB = \text{N } 66^\circ 30' \text{ E}$

• Field Observations and Closure Tabulation:

  • Station $A$:

  • Occupied: $A$, Backsight: $E$, Foresight: $B$

  • Distance $AB = 139.11\,m$

  • Single Deflection: $40^\circ 15' \text{ (L)}$; Double: $80^\circ 30'$; Mean: $40^\circ 15' \text{ (L)}$

  • Observed Bearing: $\text{N } 66^\circ 30' \text{ E}$

  • Correction: $+15'$

  • Adjusted Deflection Angle: $40^\circ 30' \text{ (L)}$

  • Calculated Bearing: $\text{N } 66^\circ 30' \text{ E}$

  • Station $B$:

  • Occupied: $B$, Backsight: $A$, Foresight: $C$

  • Distance $BC = 64.25\,m$

  • Single Deflection: $101^\circ 21' \text{ (R)}$; Double: $202^\circ 40'$; Mean: $101^\circ 20' \text{ (R)}$

  • Observed Bearing: $\text{S } 12^\circ 00' \text{ E}$

  • Correction: $-15'$

  • Adjusted Deflection Angle: $101^\circ 05' \text{ (R)}$

  • Calculated Bearing: $\text{S } 12^\circ 25' \text{ E}$

  • Station $C$:

  • Occupied: $C$, Backsight: $B$, Foresight: $D$

  • Distance $CD = 210.46\,m$

  • Single Deflection: $90^\circ 20' \text{ (R)}$; Double: $180^\circ 40'$; Mean: $90^\circ 20' \text{ (R)}$

  • Observed Bearing: $\text{S } 78^\circ 45' \text{ W}$

  • Correction: $-15'$

  • Adjusted Deflection Angle: $90^\circ 05' \text{ (R)}$

  • Calculated Bearing: $\text{S } 77^\circ 40' \text{ W}$

  • Station $D$:

  • Occupied: $D$, Backsight: $C$, Foresight: $E$

  • Distance $DE = 80.23\,m$

  • Single Deflection: $89^\circ 30' \text{ (R)}$; Double: $179^\circ 00'$; Mean: $89^\circ 30' \text{ (R)}$

  • Observed Bearing: $\text{N } 11^\circ 30' \text{ W}$

  • Correction: $-15'$

  • Adjusted Deflection Angle: $89^\circ 15' \text{ (R)}$

  • Calculated Bearing: $\text{N } 13^\circ 05' \text{ W}$

  • Station $E$:

  • Occupied: $E$, Backsight: $D$, Foresight: $A$

  • Distance $EA = 87.30\,m$

  • Single Deflection: $120^\circ 21' \text{ (R)}$; Double: $240^\circ 40'$; Mean: $120^\circ 20' \text{ (R)}$

  • Observed Bearing: $\text{S } 72^\circ 30' \text{ E}$

  • Correction: $-15'$

  • Adjusted Deflection Angle: $120^\circ 05' \text{ (R)}$

  • Calculated Bearing: $\text{S } 73^\circ 00' \text{ E}$

• Closure and Summation Checks:

  • Sum of mean observed deflection angles: $361^\circ 15'$

  • Algebraic closure target: $360^\circ 00'$

  • Total correction applied across all stations: $-1^\circ 15'$

  • Sum of adjusted deflection angles: $360^\circ 00'$ (checks)

Advantages, Disadvantages, and Precision in Deflection Traversing

• Principal Operational Advantages:

  • Forward azimuths are calculated directly from deflection angles by simple addition and subtraction:     $\text{Forward Azimuth}_{\text{succeeding}} = \text{Forward Azimuth}_{\text{preceding}} + \text{Deflection Angle}_R$     $\text{Forward Azimuth}_{\text{succeeding}} = \text{Forward Azimuth}_{\text{preceding}} - \text{Deflection Angle}_L$

  • Deflection angles are straightforward to plot during drafting operations.

  • Deflection angles translate directly into the calculations of circular horizontal curves utilized in highway and railway alignment design.

• Inherent Disadvantages and Risks:

  • Requires extreme care when measuring small deflection angles, where pointings can easily introduce large relative angular errors.

  • High susceptibility to notekeeper blunders regarding the recording of the direction (confusing Right and Left).

  • Severe vulnerability to instrument collimation error if the instrument is not in adjustment, causing readings to systematically deviate high or low depending on whether the telescope is turned right or left.

Illustrative Problem 1: Interior Angle Traverse and Course Bearings

• Problem Statement:

  • A five-sided closed interior angle traverse has the following observed interior angles:

  • $A = 118^\circ 30'$

  • $B = 95^\circ 33'$

  • $C = \text{Not Measured}$

  • $D = 134^\circ 10'$

  • $E = 140^\circ 50'$

  • Given that all observed angles are correct, compute the interior angle at station $C$.

  • Given that the bearing of line $AB$ is $\text{N } 15^\circ 30' \text{ E}$, determine the bearings of all remaining courses and tabulate the results.

• Step (a): Determining the Interior Angle at Station $C$:

  • Number of sides: $n = 5$

  • Theoretical sum of interior angles:     $\text{Sum} = (n - 2) \times 180^\circ = (5 - 2) \times 180^\circ = 540^\circ 00'$

  • Summation of all angles:     $\text{Sum} = \angle A + \angle B + \angle C + \angle D + \angle E$     $540^\circ 00' = 118^\circ 30' + 95^\circ 33' + \angle C + 134^\circ 10' + 140^\circ 50'$     $540^\circ 00' = 489^\circ 03' + \angle C$     $\angle C = 540^\circ 00' - 489^\circ 03' = 50^\circ 57'$

• Step (b): Determining the Bearing of Each Course:

• Bearing of Course $BC$ (Station $B$):

  • Back bearing of line $AB$ from $B$ is $\text{S } 15^\circ 30' \text{ W}$.

  • Interior angle at $B = 95^\circ 33'$

  • Angle with South-East quadrant:       $\alpha_{bc} = 95^\circ 33' - 15^\circ 30' = 80^\circ 03'$

  • Bearing of line $BC$ = $\text{S } 80^\circ 03' \text{ E}$

• Bearing of Course $CD$ (Station $C$):

  • Back bearing of line $BC$ from $C$ is $\text{N } 80^\circ 03' \text{ W}$.

  • Interior angle at $C = 50^\circ 57'$

  • South-West quadrant calculation:       $\alpha_{cd} = 180^\circ 00' - (80^\circ 03' + 50^\circ 57') = 180^\circ 00' - 131^\circ 00' = 49^\circ 00'$

  • Bearing of line $CD$ = $\text{S } 49^\circ 00' \text{ W}$

• Bearing of Course $DE$ (Station $D$):

  • Back bearing of line $CD$ from $D$ is $\text{N } 49^\circ 00' \text{ E}$.

  • Interior angle at $D = 134^\circ 10'$

  • North-West quadrant calculation:       $\alpha_{de} = 134^\circ 10' - 49^\circ 00' = 85^\circ 10'$

  • Bearing of line $DE$ = $\text{N } 85^\circ 10' \text{ W}$

• Bearing of Course $EA$ (Station $E$):

  • Back bearing of line $DE$ from $E$ is $\text{S } 85^\circ 10' \text{ E}$.

  • Interior angle at $E = 140^\circ 50'$

  • North-West quadrant calculation:       $\alpha_{ea} = 85^\circ 10' + 140^\circ 50' - 180^\circ 00' = 226^\circ 00' - 180^\circ 00' = 46^\circ 00'$

  • Bearing of line $EA$ = $\text{N } 46^\circ 00' \text{ W}$

  • Solution Check (Station $A$):

• Back bearing of line $EA$ from $A$ is $\text{S } 46^\circ 00' \text{ E}$.

• Interior angle at $A = 118^\circ 30'$

• North-East quadrant calculation:     $\alpha_{ab} = 180^\circ 00' - (118^\circ 30' + 46^\circ 00') = 180^\circ 00' - 164^\circ 30' = 15^\circ 30'$

• Bearing of line $AB$ = $\text{N } 15^\circ 30' \text{ E}$

• The calculated bearing matches the initial known bearing exactly, confirming all angular and bearing calculations are correct.

  • Tabulated Summary of Solution:

• Station $A$ | Interior Angle: $118^\circ 30'$ | Line $AB$ | Bearing: $\text{N } 15^\circ 30' \text{ E}$

• Station $B$ | Interior Angle: $95^\circ 33'$ | Line $BC$ | Bearing: $\text{S } 80^\circ 03' \text{ E}$

• Station $C$ | Interior Angle: $50^\circ 57'$ | Line $CD$ | Bearing: $\text{S } 49^\circ 00' \text{ W}$

• Station $D$ | Interior Angle: $134^\circ 10'$ | Line $DE$ | Bearing: $\text{N } 85^\circ 10' \text{ W}$

• Station $E$ | Interior Angle: $140^\circ 50'$ | Line $EA$ | Bearing: $\text{N } 46^\circ 00' \text{ W}$

Illustrative Problem 2: Interior Angle Traverse Angular Closure and Adjustment

• Problem Statement:

  • A closed five-sided traverse has the following observed interior angles:

  • $A = 153^\circ 30' 00''$

  • $B = 58^\circ 20' 20''$

  • $C = 139^\circ 19' 00''$

  • $D = 78^\circ 21' 20''$

  • $E = 110^\circ 28' 00''$

  • Determine the angular closure error and adjust the measured angles assuming that the error is distributed equally among all stations. Tabulate the values.

• Step (a): Determining the Angular Error of Closure:

  • Number of interior angles: $n = 5$

  • Sum of observed interior angles ($\text{Sum}_1$):     $\text{Sum}_1 = \angle A + \angle B + \angle C + \angle D + \angle E$     $\text{Sum}_1 = 153^\circ 30' 00'' + 58^\circ 20' 20'' + 139^\circ 19' 00'' + 78^\circ 21' 20'' + 110^\circ 28' 00'' = 539^\circ 58' 40''$

  • Theoretical sum ($\text{Sum}_2$):     $\text{Sum}_2 = (n - 2) \times 180^\circ = (5 - 2) \times 180^\circ = 540^\circ 00' 00''$

  • Angular misclosure error:     $\text{Closure} = \text{Sum}_2 - \text{Sum}_1 = 540^\circ 00' 00'' - 539^\circ 58' 40'' = +1' 20''$

  • Correction per angle ($\text{Corr}$):     $\text{Corr} = \frac{\text{Closure}}{n} = \frac{+1' 20''}{5} = \frac{+80''}{5} = +16''$

• Step (b): Adjusting the Interior Angles:

  • $\angle A' = \angle A + \text{Corr} = 153^\circ 30' 00'' + 16'' = 153^\circ 30' 16''$

  • $\angle B' = \angle B + \text{Corr} = 58^\circ 20' 20'' + 16'' = 58^\circ 20' 36''$

  • $\angle C' = \angle C + \text{Corr} = 139^\circ 19' 00'' + 16'' = 139^\circ 19' 16''$

  • $\angle D' = \angle D + \text{Corr} = 78^\circ 21' 20'' + 16'' = 78^\circ 21' 36''$

  • $\angle E' = \angle E + \text{Corr} = 110^\circ 28' 00'' + 16'' = 110^\circ 28' 16''$

• Solution Check:

  • Sum of adjusted angles:     $\text{Sum} = 153^\circ 30' 16'' + 58^\circ 20' 36'' + 139^\circ 19' 16'' + 78^\circ 21' 36'' + 110^\circ 28' 16'' = 540^\circ 00' 00''$

  • Matches the theoretical sum of $540^\circ 00' 00''$ exactly.

• Tabulated Solution:

  • Station $A$ | Observed Interior Angle: $153^\circ 30' 00''$ | Correction: $+16''$ | Adjusted Interior Angle: $153^\circ 30' 16''$

  • Station $B$ | Observed Interior Angle: $58^\circ 20' 20''$ | Correction: $+16''$ | Adjusted Interior Angle: $58^\circ 20' 36''$

  • Station $C$ | Observed Interior Angle: $139^\circ 19' 00''$ | Correction: $+16''$ | Adjusted Interior Angle: $139^\circ 19' 16''$

  • Station $D$ | Observed Interior Angle: $78^\circ 21' 20''$ | Correction: $+16''$ | Adjusted Interior Angle: $78^\circ 21' 36''$

  • Station $E$ | Observed Interior Angle: $110^\circ 28' 00''$ | Correction: $+16''$ | Adjusted Interior Angle: $110^\circ 28' 16''$

  • Sums | Observed Sum: $539^\circ 58' 40''$ | Total Correction: $+1' 20''$ | Adjusted Sum: $540^\circ 00' 00''$

Illustrative Problem 3: Deflection Angle Traverse Closure and Angular Adjustment

• Problem Statement:

  • The observed deflection angles of a closed traverse are:

  • $A = 28^\circ 25' 00'' \text{ (L)}$

  • $B = 68^\circ 03' 30'' \text{ (L)}$

  • $C = 120^\circ 34' 00'' \text{ (L)}$

  • $D = 58^\circ 30' 00'' \text{ (R)}$

  • $E = 108^\circ 13' 30'' \text{ (L)}$

  • $F = 16^\circ 50' 00'' \text{ (R)}$

  • $G = 110^\circ 00' 30'' \text{ (L)}$

  • Compute the error of closure and adjust the angular values assuming the error is distributed equally among all angles. Tabulate the results.

• Step (a): Determining the Error of Closure:

  • Total number of deflection angles: $n = 7$

  • Sum of Left Deflection Angles ($\sum \text{DEFL}_L$):     $\sum \text{DEFL}_L = A + B + C + E + G$     $\sum \text{DEFL}_L = 28^\circ 25' 00'' + 68^\circ 03' 30'' + 120^\circ 34' 00'' + 108^\circ 13' 30'' + 110^\circ 00' 30'' = 435^\circ 16' 30''$

  • Sum of Right Deflection Angles ($\sum \text{DEFL}_R$):     $\sum \text{DEFL}_R = D + F$     $\sum \text{DEFL}_R = 58^\circ 30' 00'' + 16^\circ 50' 00'' = 75^\circ 20' 00''$

  • Difference between Left and Right sums ($\text{DIFF}_1$):     $\text{DIFF}_1 = \sum \text{DEFL}_L - \sum \text{DEFL}_R = 435^\circ 16' 30'' - 75^\circ 20' 00'' = 359^\circ 56' 30''$

  • Angular Error of Closure:     $\text{ERROR} = 360^\circ 00' 00'' - \text{DIFF}_1 = 360^\circ 00' 00'' - 359^\circ 56' 30'' = 3' 30''$

  • Correction per observed deflection angle:     $\text{CORR} = \frac{\text{ERROR}}{n} = \frac{3' 30''}{7} = \frac{210''}{7} = 30'' = 0' 30''$

• Step (b): Adjusting the Deflection Angles:

  • Because $\text{DIFF}_1 = 359^\circ 56' 30''$ is less than $360^\circ$, the net difference must be increased to reach $360^\circ$.

  • Consequently, all Left deflection angles must be increased by $+30''$, and all Right deflection angles must be decreased by $-30''$:

  • $A' = 28^\circ 25' 00'' \text{ (L)} + 30'' = 28^\circ 25' 30'' \text{ (L)}$

  • $B' = 68^\circ 03' 30'' \text{ (L)} + 30'' = 68^\circ 04' 00'' \text{ (L)}$

  • $C' = 120^\circ 34' 00'' \text{ (L)} + 30'' = 120^\circ 34' 30'' \text{ (L)}$

  • $D' = 58^\circ 30' 00'' \text{ (R)} - 30'' = 58^\circ 29' 30'' \text{ (R)}$

  • $E' = 108^\circ 13' 30'' \text{ (L)} + 30'' = 108^\circ 14' 00'' \text{ (L)}$

  • $F' = 16^\circ 50' 00'' \text{ (R)} - 30'' = 16^\circ 49' 30'' \text{ (R)}$

  • $G' = 110^\circ 00' 30'' \text{ (L)} + 30'' = 110^\circ 01' 00'' \text{ (L)}$

• Solution Check:

  • Sum of adjusted Left deflection angles ($\sum \text{Adj DEFL}_L$):     $\sum \text{Adj DEFL}_L = 28^\circ 25' 30'' + 68^\circ 04' 00'' + 120^\circ 34' 30'' + 108^\circ 14' 00'' + 110^\circ 01' 00'' = 435^\circ 19' 00''$

  • Sum of adjusted Right deflection angles ($\sum \text{Adj DEFL}_R$):     $\sum \text{Adj DEFL}_R = 58^\circ 29' 30'' + 16^\circ 49' 30'' = 75^\circ 19' 00''$

  • Difference between adjusted sums ($\text{DIFF}_2$):     $\text{DIFF}_2 = \sum \text{Adj DEFL}_L - \sum \text{Adj DEFL}_R = 435^\circ 19' 00'' - 75^\circ 19' 00'' = 360^\circ 00' 00''$

  • Matches the theoretical closed traverse criterion ($360^\circ 00' 00''$) exactly.

• Tabulated Solution:

  • Station $A$ | Observed Left: $28^\circ 25' 00''$ | Observed Right: — | Correction: $+30''$ | Adjusted Left: $28^\circ 25' 30''$ | Adjusted Right: —

  • Station $B$ | Observed Left: $68^\circ 03' 30''$ | Observed Right: — | Correction: $+30''$ | Adjusted Left: $68^\circ 04' 00''$ | Adjusted Right: —

  • Station $C$ | Observed Left: $120^\circ 34' 00''$ | Observed Right: — | Correction: $+30''$ | Adjusted Left: $120^\circ 34' 30''$ | Adjusted Right: —

  • Station $D$ | Observed Left: — | Observed Right: $58^\circ 30' 00''$ | Correction: $-30''$ | Adjusted Left: — | Adjusted Right: $58^\circ 29' 30''$

  • Station $E$ | Observed Left: $108^\circ 13' 30''$ | Observed Right: — | Correction: $+30''$ | Adjusted Left: $108^\circ 14' 00''$ | Adjusted Right: —

  • Station $F$ | Observed Left: — | Observed Right: $16^\circ 50' 00''$ | Correction: $-30''$ | Adjusted Left: — | Adjusted Right: $16^\circ 49' 30''$

  • Station $G$ | Observed Left: $110^\circ 00' 30''$ | Observed Right: — | Correction: $+30''$ | Adjusted Left: $110^\circ 01' 00''$ | Adjusted Right: —

  • Sums:

  • Observed Left Sum ($\sum \text{DEFL}_L$): $435^\circ 16' 30''$

  • Observed Right Sum ($\sum \text{DEFL}_R$): $75^\circ 20' 00''$

  • Adjusted Left Sum ($\sum \text{Adj DEFL}_L$): $435^\circ 19' 00''$

  • Adjusted Right Sum ($\sum \text{Adj DEFL}_R$): $75^\circ 19' 00''$