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7.2 Area Computations (DMD, DPD, Coordinates) and Subdivision/Consolidation Survey Standards

Key Takeaways

  • Double Meridian Distance (DMD) of a succeeding course equals preceding DMD + preceding departure + current departure, yielding area via ½ |Σ(DMD × Lat)|.

  • Gauss's Shoelace formula computes polygon lot areas directly from ordered grid coordinates (E_i, N_i).

  • Circular segment corrections (A = ½ R² (θ - sin θ)) are added for convex outward lot boundaries and subtracted for concave inward boundaries.

  • DENR DAO 2007-29 requires Private Survey (P.S.) cylindrical concrete monuments measuring 15 cm diameter by 50 cm length with 45 cm buried below ground.

Last updated: July 2026

7.2 Area Computations (DMD, DPD, Coordinates) and Subdivision/Consolidation Survey Standards

Accurate area calculation of boundary lots is a primary legal responsibility of Geodetic Engineers. Under Philippine property law and land registration statutes (PD 1529), land titles specify lot areas to the nearest square meter. This section covers classical numerical methods (DMD and DPD), modern coordinate geometry methods, treatment of circular curved boundaries, and DENR administrative regulations for lot subdivision and consolidation surveys.


Double Meridian Distance (DMD) Method

The Double Meridian Distance (DMD) of a course is defined as the sum of the meridian distances of its two endpoints. The meridian distance of a point is the perpendicular distance from that point to the reference north-south meridian passing through the most westerly station of the traverse.

Rules for Computing DMDs:

  1. Rule 1 (First Course): The DMD of the first course equals the departure of the first course itself. DMD1=Dep1\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. DMDi=DMDi−1+Depi−1+Depi\text{DMD}_i = \text{DMD}_{i-1} + Dep_{i-1} + Dep_i
  3. Rule 3 (Check / Last Course): The DMD of the final course must be equal in magnitude to the departure of the final course, but opposite in sign. DMDn=−Depn\text{DMD}_n = - Dep_n

Area Formula using DMD:

Double Area (2A)=∑i=1n(DMDi⋅Ladj,i)\text{Double Area } (2A) = \sum_{i=1}^{n} (\text{DMD}_i \cdot L_{adj,i}) Area A=12∣∑i=1n(DMDi⋅Ladj,i)∣\text{Area } A = \frac{1}{2} \left| \sum_{i=1}^{n} (\text{DMD}_i \cdot L_{adj,i}) \right|

Double Parallel Distance (DPD) Method

The Double Parallel Distance (DPD) method is the orthogonal counterpart to DMD, using an East-West reference parallel (passing through the most southerly station) and working with latitudes instead of departures.

Rules for Computing DPDs:

  1. First Course: DPD1=L1\text{DPD}_1 = L_1
  2. Succeeding Courses: DPDi=DPDi−1+Li−1+Li\text{DPD}_i = \text{DPD}_{i-1} + L_{i-1} + L_i
  3. Check (Last Course): DPDn=−Ln\text{DPD}_n = - L_n

Area Formula using DPD:

Double Area (2A)=∑i=1n(DPDi⋅Depadj,i)\text{Double Area } (2A) = \sum_{i=1}^{n} (\text{DPD}_i \cdot Dep_{adj,i}) Area A=12∣∑i=1n(DPDi⋅Depadj,i)∣\text{Area } A = \frac{1}{2} \left| \sum_{i=1}^{n} (\text{DPD}_i \cdot Dep_{adj,i}) \right|

Coordinate Method (Gauss's Shoelace Formula)

With computer automation and total station/GNSS field data, the Coordinate Method is the standard approach for area determination. Given a closed polygon with nn vertices having ordered grid coordinates (E1,N1),(E2,N2),…,(En,Nn)(E_1, N_1), (E_2, N_2), \dots, (E_n, N_n) listed sequentially in either clockwise or counter-clockwise order around the perimeter:

2A=∣(E1N2+E2N3+⋯+EnN1)−(N1E2+N2E3+⋯+NnE1)∣2A = |(E_1 N_2 + E_2 N_3 + \dots + E_n N_1) - (N_1 E_2 + N_2 E_3 + \dots + N_n E_1)|

In matrix / shoelace expansion form:

2A=∣∑i=1n(EiNi+1−Ei+1Ni)∣where (En+1,Nn+1)=(E1,N1)2A = \left| \sum_{i=1}^{n} (E_i N_{i+1} - E_{i+1} N_i) \right| \quad \text{where } (E_{n+1}, N_{n+1}) = (E_1, N_1)

Alternatively, using coordinate differences:

2A=∑i=1nEi(Ni+1−Ni−1)or2A=∑i=1nNi(Ei+1−Ei−1)2A = \sum_{i=1}^{n} E_i (N_{i+1} - N_{i-1}) \quad \text{or} \quad 2A = \sum_{i=1}^{n} N_i (E_{i+1} - E_{i-1})

Area of Irregular and Curved Boundaries

Lots bounded by circular curves (such as road right-of-ways, cul-de-sacs, or river arcs) require combining polygon traverse area with circular segment corrections.

1. Circular Segment Correction

For a lot boundary following a circular curve with radius RR and central angle II (or Δ\Delta):

  • Central Angle in Radians: Irad=I∘×π180∘I_{rad} = I^\circ \times \frac{\pi}{180^\circ}
  • Area of Circular Sector: Asector=12R2Irad=πR2I∘360∘A_{\text{sector}} = \frac{1}{2} R^2 I_{rad} = \frac{\pi R^2 I^\circ}{360^\circ}
  • Area of Sector Triangle: Atriangle=12R2sin⁡(I)A_{\text{triangle}} = \frac{1}{2} R^2 \sin(I)
  • Area of Circular Segment: Asegment=Asector−Atriangle=12R2(Irad−sin⁡(I))A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = \frac{1}{2} R^2 \left( I_{rad} - \sin(I) \right)

Rule of Application:

  • If the circular arc lies outside the straight chord joining the two lot corners (convex lot curve), ADD the segment area AsegmentA_{\text{segment}} to the polygon area calculated via DMD or Coordinates.
  • If the circular arc curves inward toward the lot interior (concave lot curve, like a setback curve), SUBTRACT the segment area AsegmentA_{\text{segment}} from the polygon area.

2. Irregular Water Boundaries (Trapezoidal & Simpson's 1/3 Rules)

When a boundary follows a natural shoreline or river bank, perpendicular offsets h0,h1,h2,…,hnh_0, h_1, h_2, \dots, h_n are measured at uniform intervals dd along a base line:

  • Trapezoidal Rule: A=d(h0+hn2+h1+h2+⋯+hn−1)A = d \left( \frac{h_0 + h_n}{2} + h_1 + h_2 + \dots + h_{n-1} \right)
  • Simpson's 1/3 Rule (requires an even number of intervals nn, i.e., odd number of offsets): A=d3[(h0+hn)+4(h1+h3+⋯+hn−1)+2(h2+h4+⋯+hn−2)]A = \frac{d}{3} \left[ (h_0 + h_n) + 4(h_1 + h_3 + \dots + h_{n-1}) + 2(h_2 + h_4 + \dots + h_{n-2}) \right]

DENR Standards for Subdivision and Consolidation Surveys (DAO 2007-29)

Under the Philippine Department of Environment and Natural Resources (DENR) Memorandum Circulars and DAO 2007-29 (Revised Regulations on Land Surveys):

1. Classification of Land Surveys

Survey TypeCode PrefixDescription
Private Subdivision SurveyPsdSubdivision of titled land surveyed by a private Geodetic Engineer
Private Consolidation SurveyPcnAmalgamation of two or more titled parcels into one lot
Consolidation-SubdivisionPcsCombining multiple titled parcels and re-subdividing into smaller lots
Cadastral SubdivisionCsdSubdivision of untitled public agricultural land within a cadastre

2. Corner Markers & Monumentation Rules

All lot boundary corners established or re-established during cadastral or property surveys must be marked on the ground with standard monuments:

  • Standard Corner Monument: Private Survey (P.S.) cylindrical concrete monument.
  • Dimensions: 15 cm in diameter by 50 cm in length.
  • Installation: Set vertically in the ground with 45 cm buried below ground surface and 5 cm projecting above ground level.
  • Material Composition: Cast concrete with a volumetric mix ratio of 1 part Portland cement, 2 parts clean sand, and 4 parts gravel (1:2:4 mix).
  • Center Mark: The exact corner point is marked on top of the monument by a galvanized iron (G.I.) nail/spike centered in a shallow cross mark.
  • Bureau of Lands Monuments (B.L.): Primary control corners use 15×15×60 cm15 \times 15 \times 60\text{ cm} square or 20 cm20\text{ cm} diameter cylindrical monuments.

3. Tie-Line Requirements

Every isolated lot or subdivision survey plan submitted to the Land Management Bureau (LMB) or Land Registration Authority (LRA) must be mathematically connected by a Tie-Line (bearing and distance) from Corner 1 of the lot to an official reference monument, such as a Bureau of Lands Location Monument (BLLM), City Boundary Monument (CBM), or established triangulation station.


Worked Example: DMD Area Computation

Using the adjusted traverse data from Section 7.1:

CourseAdjusted Latitude (LL, m)Adjusted Departure (DepDep, m)
1-2+86.585+86.585+50.024+50.024
2-3−75.027-75.027+129.940+129.940
3-4−84.875-84.875−84.824-84.824
4-1+73.317+73.317−95.140-95.140

Step 1: Compute DMD for each Course

  • DMD1−2=Dep1−2=+50.024 m\text{DMD}_{1-2} = Dep_{1-2} = \mathbf{+50.024\text{ m}}
  • DMD2−3=DMD1−2+Dep1−2+Dep2−3=50.024+50.024+129.940=+229.988 m\text{DMD}_{2-3} = \text{DMD}_{1-2} + Dep_{1-2} + Dep_{2-3} = 50.024 + 50.024 + 129.940 = \mathbf{+229.988\text{ m}}
  • DMD3−4=DMD2−3+Dep2−3+Dep3−4=229.988+129.940+(−84.824)=+275.104 m\text{DMD}_{3-4} = \text{DMD}_{2-3} + Dep_{2-3} + Dep_{3-4} = 229.988 + 129.940 + (-84.824) = \mathbf{+275.104\text{ m}}
  • DMD4−1=DMD3−4+Dep3−4+Dep4−1=275.104+(−84.824)+(−95.140)=+95.140 m\text{DMD}_{4-1} = \text{DMD}_{3-4} + Dep_{3-4} + Dep_{4-1} = 275.104 + (-84.824) + (-95.140) = \mathbf{+95.140\text{ m}} (Check: DMD4−1=+95.140\text{DMD}_{4-1} = +95.140, which equals −Dep4−1=−(−95.140)=+95.140 m-Dep_{4-1} = -(-95.140) = +95.140\text{ m}. Check passes!)

Step 2: Compute Double Areas (2A=DMD×Ladj2A = \text{DMD} \times L_{adj})

CourseDMD (m)Adjusted Latitude (LL, m)Double Area (2A=DMD×L2A = \text{DMD} \times L, m2\text{m}^2)
1-2+50.024+50.024+86.585+86.585+4,331.328+4,331.328
2-3+229.988+229.988−75.027-75.027−17,255.310-17,255.310
3-4+275.104+275.104−84.875-84.875−23,349.452-23,349.452
4-1+95.140+95.140+73.317+73.317+6,975.379+6,975.379
SUM∑2A=−29,298.055 m2\sum 2A = -29,298.055\text{ m}^2

Step 3: Compute Final Lot Area

Double Area 2A=∣−29,298.055 m2∣=29,298.055 m2\text{Double Area } 2A = |-29,298.055\text{ m}^2| = 29,298.055\text{ m}^2 Area A=29,298.0552=14,649.03 m2or1.4649 hectares\text{Area } A = \frac{29,298.055}{2} = \mathbf{14,649.03\text{ m}^2} \quad \text{or} \quad \mathbf{1.4649\text{ hectares}}
Test Your Knowledge

In calculating lot area using the Double Meridian Distance (DMD) method, what is the correct rule for finding the DMD of any succeeding course?

A

DMD of preceding course minus departure of preceding course plus departure of current course

B

DMD of preceding course plus departure of preceding course plus departure of current course

C

Latitude of preceding course plus departure of current course multiplied by two

D

Departure of current course plus latitude of current course divided by total perimeter

Test Your Knowledge

Under DENR Administrative Order No. 2007-29, what are the standard physical dimensions and buried depth for a Private Survey (P.S.) cylindrical concrete corner monument?

A

15 cm diameter by 50 cm length, with 45 cm buried below ground surface

B

20 cm diameter by 60 cm length, with 30 cm buried below ground surface

C

10 cm diameter by 40 cm length, with 35 cm buried below ground surface

D

15 cm x 15 cm square by 60 cm length, flush with ground surface

Test Your Knowledge

A triangular parcel has grid coordinates A (100.00 m, 100.00 m), B (160.00 m, 100.00 m), and C (160.00 m, 140.00 m). What is the area of the lot calculated via the Coordinate Method?

A

2,400 sq m

B

1,800 sq m

C

1,200 sq m

D

600 sq m

Test Your Knowledge

When computing the total area of a lot where one boundary follows a circular arc that curves outwards (convex relative to the lot interior), how is the circular segment area treated relative to the polygon traverse area?

A

Subtracted from the polygon traverse area

B

Multiplied by the total perimeter prior to area summation

C

Ignored if the arc length is less than 50 meters

D

Added to the polygon traverse area

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