4.3 Traverse Computations, Balancing, and Area Determinations

4.3 Traverse Computations, Balancing, and Area Determinations

A traverse consists of a series of connected survey lines whose lengths and directions are measured in the field. In civil engineering property surveys, highway alignments, and construction site layouts, closed traverses establish legal boundary coordinates and property land areas.

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1. Reference Meridians, Angles, and Bearings

Directions of survey courses are established relative to four reference meridians:

• True (Geographic) Meridian: The astronomic north-south line passing through the earth's geographic poles.
• Magnetic Meridian: The direction indicated by a freely suspended magnetic needle, deviating from true north by the local magnetic declination.
• Grid Meridian: A line parallel to the central meridian of a national map projection (such as the Philippine Transverse Mercator - PTM, PRS92).
• Assumed Meridian: An arbitrary baseline chosen for local project convenience.

Bearings vs. Azimuths

• Bearings: The acute horizontal angle between the reference meridian (North or South) and the line of interest, restricted to $0^\circ \le \theta \le 90^\circ$ within four quadrants: Northeast (NE), Southeast (SE), Southwest (SW), and Northwest (NW) (e.g., $\text{N } 42^\circ 30' \text{ E}$ or $\text{S } 18^\circ 45' \text{ W}$).
• Azimuths: Horizontal angles measured continuously clockwise from $0^\circ$ to $360^\circ$, referenced either from North (standard Philippine geomatics practice) or South (traditional astronomic practice).

Angular Geometric Closure for Closed Polygons

For an $n$-sided closed polygon traverse: $\sum \text{Interior Angles} = (n - 2) \times 180^\circ$ $\sum \text{Exterior Angles} = (n + 2) \times 180^\circ$ Before balancing latitudes and departures, any angular closure discrepancy within allowable limits must be distributed equally (or weighted inversely by sight distance) among all observed angles.

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2. Latitudes, Departures, and Error of Closure

Every traverse course of length $D$ and bearing $\theta$ represents a planar vector:

• Latitude ($L$): The north-south projection of the line: $L = D \cos \theta$ Sign Convention: Positive ($+$) for North; Negative ($-$) for South.
• Departure ($Dep$): The east-west projection of the line: $Dep = D \sin \theta$ Sign Convention: Positive ($+$) for East; Negative ($-$) for West.

Linear Error of Closure (LEC)

In a theoretically perfect closed loop traverse, returning to the origin requires: $\sum L = 0 \quad \text{and} \quad \sum Dep = 0$ Due to field measurement imperfections, non-zero residuals emerge: $E_L = \sum L \quad (\text{Closure error in latitude})$ $E_D = \sum Dep \quad (\text{Closure error in departure})$

The total Linear Error of Closure (LEC) is the magnitude of the closing vector: $\text{LEC} = \sqrt{E_L^2 + E_D^2}$

The bearing of the closing line is: $\tan \theta_{\text{closure}} = \left| \frac{E_D}{E_L} \right|$

Relative Precision (Precision Ratio)

The relative precision describes traverse quality as a fraction with a numerator of 1: $\text{Relative Precision} = \frac{\text{LEC}}{\text{Total Perimeter } P} = \frac{1}{P / \text{LEC}}$ Example: If $P = 1,200.00\text{ m}$ and $\text{LEC} = 0.24\text{ m}$, the relative precision is $\frac{1}{1,200 / 0.24} = \frac{1}{5,000}$ (expressed as $1:5,000$). Survey specifications set the minimum acceptable precision for each class of work; boundary and engineering control traverses are commonly required to close at $1:5,000$ or better, so check the governing project or land-survey specification.

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3. Balancing a Traverse: Compass Rule vs. Transit Rule

Balancing eliminates the closure error by adjusting latitudes and departures.

1. Compass Rule (Bowditch Method)

Formulated by Nathaniel Bowditch, the Compass Rule assumes that angle and distance measurements share equal relative precision. Corrections to a line's latitude and departure are directly proportional to the line's length relative to the total perimeter: $c_{L, i} = - E_L \left( \frac{D_i}{\sum D} \right)$ $c_{D, i} = - E_D \left( \frac{D_i}{\sum D} \right)$ Where:

• $c_{L, i}$, $c_{D, i}$ = corrections applied to course $i$
• $D_i$ = horizontal length of course $i$
• $\sum D = P$ = total traverse perimeter
• Adjusted latitude: $L_{\text{corr}} = L + c_{L, i}$
• Adjusted departure: $Dep_{\text{corr}} = Dep + c_{D, i}$

2. Transit Rule

The Transit Rule assumes angular measurements are made with significantly higher precision than taped or EDM distances. Corrections depend on the absolute magnitudes of the course's individual latitude and departure: $c_{L, i} = - E_L \left( \frac{|L_i|}{\sum |L|} \right)$ $c_{D, i} = - E_D \left( \frac{|Dep_i|}{\sum |Dep|} \right)$ Where $\sum |L|$ and $\sum |Dep|$ represent the arithmetic sum of latitudes and departures neglecting signs.

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4. Traverse Area Determinations

Once balanced, the enclosed boundary area is evaluated through three standard methods.

1. Coordinate Method (Gauss's Area / Shoelace Formula)

If points have known plane coordinates $(x_i, y_i)$ where $x$ = Departure/Easting and $y$ = Latitude/Northing, listed in clockwise or counterclockwise order: $2A = \left| \sum_{i=1}^n (x_i y_{i+1} - x_{i+1} y_i) \right| = \left| (x_1 y_2 + x_2 y_3 + \dots + x_n y_1) - (y_1 x_2 + y_2 x_3 + \dots + y_n x_1) \right|$ $A = \frac{1}{2} |\text{Cross-Products Difference}|$

2. Double Meridian Distance (DMD) Method

The Meridian Distance (MD) of a line is the shortest perpendicular distance from the midpoint of the line to the reference meridian. The Double Meridian Distance (DMD) is twice that value.

The Three DMD Rules

1. Rule 1 (First Course): The DMD of the first course departing from the reference meridian equals the departure of that course: $\text{DMD}_1 = Dep_1$
2. Rule 2 (Succeeding Courses): The DMD of any succeeding course equals the DMD of the preceding course, plus the departure of the preceding course, plus the departure of the course itself: $\text{DMD}_k = \text{DMD}_{k-1} + Dep_{k-1} + Dep_k$
3. Rule 3 (Verification Check on Last Course): The DMD of the final course must equal the departure of that course with opposite sign: $\text{DMD}_n = - Dep_n$

Area by DMD

The double area ($2A$) is the algebraic sum of the products of each course's DMD and its balanced latitude: $2A = \left| \sum_{i=1}^n (\text{DMD}_i \times L_i) \right| \implies A = \frac{1}{2} |\sum (\text{DMD}_i \times L_i)|$

3. Double Parallel Distance (DPD) Method

Analogous to DMD, but referenced to an east-west parallel axis using latitudes:

• $\text{DPD}_1 = L_1$
• $\text{DPD}_k = \text{DPD}_{k-1} + L_{k-1} + L_k$
• Check: $\text{DPD}_n = - L_n$
• Double area: $2A = \left| \sum (\text{DPD}_i \times Dep_i) \right|$

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5. Missing Data in Traverses

When field obstructions, rivers, or structures prevent direct observation of boundary elements, missing quantities are determined analytically by enforcing equilibrium: $\sum L = 0 \implies L_m = - \sum L_{\text{known}}$ $\sum Dep = 0 \implies Dep_m = - \sum Dep_{\text{known}}$

Case Classification

1. One Side Unknown (Both Length and Bearing Missing): Compute the closing line of the known courses: $L_m = -\sum L_{\text{known}}$, $Dep_m = -\sum Dep_{\text{known}}$. Then: $D_m = \sqrt{L_m^2 + Dep_m^2}, \quad \tan \theta_m = \left| \frac{Dep_m}{L_m} \right|$
2. Two Adjacent Sides with Missing Dimensions:
   • Sub-case A: Length of one side and bearing of another missing.
   • Sub-case B: Lengths of two sides missing.
   • Sub-case C: Bearings of two sides missing. Solution Protocol: Close the traverse across all fully known courses to establish an imaginary "closing baseline." This creates a closed triangle composed of the baseline and the two lines with missing data. Apply the Law of Sines or Law of Cosines to solve for the missing elements.

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6. Comprehensive Worked Example: Balancing & DMD Area

Problem

A 4-sided closed property traverse has the following raw latitudes and departures:

• Course AB: $L = +124.60\text{ m}, \quad Dep = +165.40\text{ m}, \quad D = 207.08\text{ m}$
• Course BC: $L = -82.40\text{ m}, \quad Dep = +210.80\text{ m}, \quad D = 226.33\text{ m}$
• Course CD: $L = -185.10\text{ m}, \quad Dep = -120.30\text{ m}, \quad D = 220.76\text{ m}$
• Course DA: $L = +142.50\text{ m}, \quad Dep = -256.30\text{ m}, \quad D = 293.25\text{ m}$

Perform:

1. Determine closure errors, Linear Error of Closure (LEC), and Relative Precision.
2. Balance the traverse using the Compass Rule.
3. Compute the area using the Double Meridian Distance (DMD) method.

Solution:

1. Closure and Precision Analysis: $\text{Perimeter } P = 207.08 + 226.33 + 220.76 + 293.25 = 947.42\text{ m}$ $E_L = \sum L = +124.60 - 82.40 - 185.10 + 142.50 = -0.40\text{ m}$ $E_D = \sum Dep = +165.40 + 210.80 - 120.30 - 256.30 = -0.40\text{ m}$ $\text{LEC} = \sqrt{(-0.40)^2 + (-0.40)^2} = \sqrt{0.32} = 0.5657\text{ m}$ $\text{Relative Precision} = \frac{0.5657}{947.42} = \frac{1}{1,675}$

2. Compass Rule Corrections: $c_L = -(-0.40) \left( \frac{D}{947.42} \right) = +0.40 \left( \frac{D}{947.42} \right)$ $c_D = -(-0.40) \left( \frac{D}{947.42} \right) = +0.40 \left( \frac{D}{947.42} \right)$

   • Course AB: $c_L = +0.09\text{ m}, \quad c_D = +0.09\text{ m} \implies L = +124.69\text{ m}, \quad Dep = +165.49\text{ m}$
   • Course BC: $c_L = +0.10\text{ m}, \quad c_D = +0.10\text{ m} \implies L = -82.30\text{ m}, \quad Dep = +210.90\text{ m}$
   • Course CD: $c_L = +0.09\text{ m}, \quad c_D = +0.09\text{ m} \implies L = -185.01\text{ m}, \quad Dep = -120.21\text{ m}$
   • Course DA: $c_L = +0.12\text{ m}, \quad c_D = +0.12\text{ m} \implies L = +142.62\text{ m}, \quad Dep = -256.18\text{ m}$ Check: $\sum L = 0.00\text{ m}, \quad \sum Dep = 0.00\text{ m}$.

3. DMD and Area Computation:

   • $\text{DMD}_{AB} = Dep_{AB} = +165.49\text{ m}$
   • $\text{DMD}_{BC} = 165.49 + 165.49 + 210.90 = +541.88\text{ m}$
   • $\text{DMD}_{CD} = 541.88 + 210.90 + (-120.21) = +632.57\text{ m}$
   • $\text{DMD}_{DA} = 632.57 + (-120.21) + (-256.18) = +256.18\text{ m}$ DMD Check: $\text{DMD}_{DA} = - Dep_{DA} = -(-256.18) = +256.18\text{ m}$ (Verified!).

   Course	Balanced L (m)	Balanced Dep (m)	DMD (m)	Double Area ($2A = \text{DMD} \times L$)
   AB	+124.69	+165.49	+165.49	$+20,634.95$
   BC	-82.30	+210.90	+541.88	$-44,596.72$
   CD	-185.01	-120.21	+632.57	$-117,031.78$
   DA	+142.62	-256.18	+256.18	$+36,536.39$
   SUM	0.00	0.00	—	$2A = -104,457.16\text{ m}^2$

   $\text{Area} = \frac{|-104,457.16|}{2} = 52,228.58\text{ m}^2 \quad (5.223\text{ hectares})$