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GELE GeodesyFigure of the Earth and the Reference EllipsoidStudy Notes

Detailed study notes for GELE Geodesy — Figure of the Earth and the Reference Ellipsoid. These are the kind of notes you would take if you were reviewing with someone who has already scored well on the GELE: organised by what Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests first, followed by the nice-to-knows, and ending with the traps to avoid.

Exam context

The Geodetic Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Geodetic Engineering and is scheduled for September 2026. The Geodesy subtest is marked as "Core" in the official pattern, and Figure of the Earth and the Reference Ellipsoid appears in position 1st of 6 in the GELE Geodesy review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent GELE 2026 papers have drawn roughly a meaningful share of questions from this subject.

Figure of the Earth and the Reference Ellipsoid - Study Notes

Geodesy is the science that determines the size, shape, and precise position of points on Earth. Unlike the common assumption that Earth is a perfect sphere, modern geodesy models Earth as an oblate ellipsoid of revolution—slightly flattened at the poles and bulging at the equator. This fundamental concept underpins all geodetic measurements, surveying calculations, and the Philippine Reference System (PRS92) used in national mapping and land registration under RA 4374 and RA 8560. Understanding the reference ellipsoid and its derived parameters is essential for the PRC Geodetic Engineer Licensure Examination, as these concepts form the mathematical foundation for coordinate transformations, map projections, and precise positioning work across the Philippine archipelago.

Summary

The reference ellipsoid is the mathematical model of Earth's size and shape, essential for all geodetic work and Philippine surveying under RA 4374, RA 8560, and PD 1529. The ellipsoid is uniquely defined by two parameters—the semi-major axis a and the flattening f—from which all other geometric properties are derived: the semi-minor axis b, the first eccentricity squared e², and the second eccentricity e'². The prime vertical radius of curvature N = a/√(1 − e² sin²φ) and the meridian radius M = a(1 − e²)/(1 − e² sin²φ)^(3/2) vary with latitude and are essential for distance reductions, map projections, and GNSS-based surveying. WGS84, the global standard and basis of the Philippine Reference System 1992 (PRS92), has parameters a = 6,378,137 m, 1/f = 298.257223563, and e² = 0.00669438. These values must be memorized for the PRC Geodetic Engineer Licensure Examination. Understanding the reference ellipsoid and correctly applying its parameters is fundamental to cadastral surveying, coordinate transformations, and the production of accurate maps in the Philippines. Common errors include confusing e with e², misreading the small flattening value, swapping N and M, and ignoring the latitude dependence of the radii of curvature. Mastery of this chapter ensures that surveyors and geodetic engineers can confidently perform the precise calculations required for legal compliance with Philippine surveying standards.

Sections

The Earth's true shape is an irregular geoid—a gravitational equipotential surface that would be occupied by seawater if the oceans covered the entire planet. However, for practical geodetic and surveying work, we model the Earth as a mathematical surface called the reference ellipsoid (also called the Earth ellipsoid or geodetic ellipsoid). This ellipsoid is a surface of revolution created by rotating an ellipse about its minor axis (the polar axis). The resulting shape is an oblate spheroid, slightly flattened at the North and South Poles and bulged outward at the equator due to Earth's rotation. The difference between the equatorial and polar radii is only about 21.4 km on a planet with a 6378 km equatorial radius—a remarkably small difference (about 0.335%), yet this small deviation is critically important in high-precision surveying and GNSS applications. The geoid deviates from the ellipsoid by as much as ±100 m, but the ellipsoid provides the best mathematical approximation for computational purposes. In the Philippines, the Philippine Reference System 1992 (PRS92) is based on the WGS84 ellipsoid, which is the standard for all GNSS positioning, land surveys, and official mapping under RA 4374 and PD 1529. Understanding why we use an ellipsoid rather than a sphere is crucial: a sphere cannot represent Earth's mass distribution and rotational characteristics as accurately, leading to significant errors in positioning, especially near the poles.

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1. The Shape of the Earth and Its Geometric Model

Examples

Problem

Explain why modern surveying and GNSS in the Philippines must use an ellipsoidal Earth model rather than a spherical model, even though the difference is only 21 km on a 6378 km radius.

Solution

A spherical Earth model would treat all latitudes identically, but the ellipsoidal geometry accounts for the centrifugal effect of Earth's rotation. At the equator, the centrifugal force causes a greater outward bulge. When converting between latitude-longitude coordinates and Cartesian coordinates (as done in GNSS), or when computing distances and areas in mapping projects, this ellipsoidal correction becomes essential. For example, a survey error of only 0.1% in linear distance would translate to roughly 6 m error per kilometer—unacceptable for cadastral surveys under RA 8560. The PRS92 standard mandates ellipsoidal calculations to ensure consistency across all Philippine mapping, land registration, and infrastructure projects.

Problem

A surveyor in Metro Manila observes that GNSS coordinates (WGS84) differ from older survey data referenced to a spherical Earth. What explains this discrepancy?

Solution

The difference arises from the ellipsoidal geometry of WGS84 versus a spherical approximation. WGS84's ellipsoid has a = 6,378,137 m (equatorial radius) and b = 6,356,752.31 m (polar radius). A sphere would use only one radius, typically the mean radius ≈ 6,371,000 m. At the latitude of Manila (≈14°N), the prime vertical radius of curvature N ≈ 6,389,000 m, which differs from a sphere. This causes coordinate shifts of 10–30 m depending on the old reference system used.

Key Points

  • Earth is modeled as an oblate ellipsoid of revolution (flattened at poles, bulged at equator)
  • The reference ellipsoid is a mathematical surface, distinct from the actual irregular geoid
  • The flattening is small but significant for precision surveying (≈0.335% of Earth's radius)
  • PRS92, used throughout the Philippines, is based on WGS84 ellipsoid parameters
  • High-precision positioning requires ellipsoidal geometry, not spherical approximations

A reference ellipsoid is uniquely defined by two independent geometric parameters. The most common choice is the semi-major axis (equatorial radius) denoted a, and the flattening denoted f. From these two parameters, all other ellipsoid properties can be derived mathematically. The semi-minor axis b is the polar radius, smaller than a by the amount a·f. Flattening f is defined as the relative difference between the equatorial and polar radii: f = (a − b)/a. For Earth, f is very small, approximately 1/298.257223563 for WGS84. The first eccentricity squared (e²) measures the deviation of the ellipse from a circle and is derived from a and b: e² = (a² − b²)/a² = 2f − f². There is also a second eccentricity e'² = (a² − b²)/b² used in some formulas. The relationship e² = 2f − f² shows how these parameters connect; since f is small, e² ≈ 2f for practical purposes. For WGS84, the standard ellipsoid of the Philippines' PRS92 system: a = 6,378,137 m exactly, 1/f = 298.257223563 (which means f ≈ 0.00335281), e² ≈ 0.00669438, and b ≈ 6,356,752.31 m. These precise values are fixed by international agreement and appear in all GNSS receivers and cadastral software used in the Philippines. Memorizing the WGS84 values is essential for exam success.

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2. Ellipsoid Parameters and Their Relationships

Examples

Problem

WGS84 Ellipsoid Parameters (PRC Exam Standard). Given a = 6,378,137 m and f = 1/298.257223563, compute the semi-minor axis b and the first eccentricity squared e².

Solution

Step 1: Calculate f as a decimal. f = 1 ÷ 298.257223563 = 0.00335281 Step 2: Compute the semi-minor axis b. b = a(1 − f) = 6,378,137 × (1 − 0.00335281) = 6,378,137 × 0.99664719 = 6,356,752.31 m Step 3: Calculate e² using the formula e² = 2f − f². e² = 2(0.00335281) − (0.00335281)² = 0.00670562 − 0.00001124 = 0.00669438 Verification using the alternative formula e² = (a² − b²)/a²: e² = (6,378,137² − 6,356,752.31²) / 6,378,137² = 0.00669438 ✓ Answer: b = 6,356,752.31 m; e² = 0.00669438

Problem

Inverse Calculation: From Axes to Parameters. A geodetic survey references the Clarke 1866 ellipsoid used in some older Philippine maps. Given a = 6,378,206.4 m and b = 6,356,583.8 m, find the flattening f and first eccentricity squared e².

Solution

Step 1: Calculate flattening f. f = (a − b) / a = (6,378,206.4 − 6,356,583.8) / 6,378,206.4 = 21,622.6 / 6,378,206.4 = 0.00338864 Step 2: Verify with the relationship e² = (a² − b²)/a². e² = 1 − (b/a)² = 1 − (6,356,583.8 / 6,378,206.4)² = 1 − (0.99664663)² = 1 − 0.99330396 = 0.00669604 Step 3: Check using e² = 2f − f². e² = 2(0.00338864) − (0.00338864)² = 0.00677728 − 0.00001149 = 0.00676579 (Note: The direct formula e² = 2f − f² gives a slightly different result due to rounding; the exact formula e² = (a² − b²)/a² is more reliable.) Answer: f ≈ 0.00338864 (or 1/f ≈ 295.24); e² ≈ 0.00669604

Problem

Working Backward from Eccentricity. If an old ellipsoid has e² = 0.00668 and a = 6,378,150 m, find b and f.

Solution

Step 1: Use e² = (a² − b²)/a² to solve for b. e² = (a² − b²)/a² ⟹ b² = a²(1 − e²) b² = 6,378,150² × (1 − 0.00668) = 40,681,897,422,500 × 0.99332 = 40,403,885,893,744 b = √40,403,885,893,744 = 6,356,416.6 m Step 2: Calculate flattening f. f = (a − b) / a = (6,378,150 − 6,356,416.6) / 6,378,150 = 21,733.4 / 6,378,150 = 0.003408 Verification: e² = 2f − f² = 2(0.003408) − (0.003408)² = 0.006816 − 0.0000116 = 0.006804 ≈ 0.00680 ✓ Answer: b ≈ 6,356,416.6 m; f ≈ 0.003408 (or 1/f ≈ 293.1)

Key Points

  • Ellipsoid is defined by two independent parameters: a (semi-major axis) and f (flattening)
  • Semi-minor axis b = a(1 − f); derived from the primary parameters
  • First eccentricity squared: e² = 2f − f² = (a² − b²)/a²
  • Second eccentricity squared: e'² = (a² − b²)/b²
  • WGS84 (PRS92 standard): a = 6,378,137 m, f = 1/298.257223563, e² = 0.00669438
  • All other ellipsoid properties are derived from these two primary parameters

Although the ellipsoid is a smooth, well-defined mathematical surface, its curvature is not uniform—it varies with latitude. This variation is crucial for geodetic calculations because many surveying operations depend on local radii of curvature. Two principal radii are used: the prime vertical radius of curvature (N, also called the radius of curvature in the prime vertical) and the meridian radius of curvature (M, also called the meridian radius). The prime vertical N represents the radius of curvature in the east–west direction (along a line of constant latitude). It is computed as N = a / √(1 − e²sin²φ), where φ is the geodetic latitude in radians or degrees. The meridian radius M represents the radius of curvature in the north–south direction (along a meridian, or line of constant longitude). It is computed as M = a(1 − e²) / (1 − e²sin²φ)^(3/2). A key observation is that N ≥ M everywhere on the ellipsoid. At the equator (φ = 0°), N equals a (the equatorial radius), while M = a(1 − e²) (the smallest meridian radius). At the poles (φ = ±90°), both N and M become infinite, but their ratio approaches 1. The mean radius of curvature R = √(M·N), often called the Gaussian radius, is used when approximating the ellipsoid locally as a sphere for simplified calculations. In Philippine surveying, these radii are essential for computing the convergence of meridians, scale factors in map projections (especially the Universal Transverse Mercator—UTM—system used in the Philippine Plane Coordinate System), and for reducing measured distances to the ellipsoid. Many surveying equations, particularly those for geodetic distance calculations and arc-length computations along parallels or meridians, require accurate knowledge of these radii at the specific latitude of the project.

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3. Radii of Curvature: Prime Vertical and Meridian

Examples

Problem

Prime Vertical Radius at Manila. Compute the prime vertical radius of curvature N at the latitude of Manila, which is approximately φ = 14°36'N = 14.6° on WGS84.

Solution

Step 1: Identify WGS84 parameters. a = 6,378,137 m e² = 0.00669438 Step 2: Compute sin²φ for φ = 14.6°. sin(14.6°) = 0.25228 sin²(14.6°) = 0.063645 Step 3: Calculate the denominator √(1 − e²sin²φ). 1 − e²sin²φ = 1 − 0.00669438 × 0.063645 = 1 − 0.000426 = 0.999574 √(1 − e²sin²φ) = √0.999574 = 0.999787 Step 4: Compute N. N = a / √(1 − e²sin²φ) = 6,378,137 / 0.999787 = 6,378,828 m Answer: N ≈ 6,378,828 m at Manila's latitude. (This is slightly larger than the equatorial radius a = 6,378,137 m, illustrating that N increases as we move from the equator toward the poles.)

Problem

Meridian Radius of Curvature at Davao. The city of Davao is located at approximately φ = 7°N. Compute the meridian radius M at this latitude on WGS84.

Solution

Step 1: Identify WGS84 parameters. a = 6,378,137 m e² = 0.00669438 1 − e² = 0.99330562 Step 2: Compute sin²φ for φ = 7°. sin(7°) = 0.121869 sin²(7°) = 0.014852 Step 3: Calculate 1 − e²sin²φ and its cube root. 1 − e²sin²φ = 1 − 0.00669438 × 0.014852 = 1 − 0.0000994 = 0.9999006 (1 − e²sin²φ)^(3/2) = (0.9999006)^1.5 = 0.9998509 Step 4: Compute M. M = a(1 − e²) / (1 − e²sin²φ)^(3/2) = 6,378,137 × 0.99330562 / 0.9998509 M = 6,336,990 / 0.9998509 = 6,337,626 m Answer: M ≈ 6,337,626 m at Davao's latitude.

Problem

Radii of Curvature at Special Latitudes. For WGS84, compute N and M at the equator (φ = 0°) and verify the relationships.

Solution

At the Equator (φ = 0°): sin(0°) = 0, so sin²(0°) = 0 1 − e²sin²(0°) = 1 − 0 = 1 N = a / √1 = a = 6,378,137 m M = a(1 − e²) / 1³/² = a(1 − e²) = 6,378,137 × 0.99330562 = 6,335,440 m At the Equator, N = 6,378,137 m and M = 6,335,440 m. Note N > M ✓ Difference: N − M = 42,697 m (this is roughly 21 km × 2, reflecting the polar flattening.) Mean radius: R = √(M·N) = √(6,335,440 × 6,378,137) = √(40,376,903,286,680) = 6,356,752 m This is very close to the semi-minor axis b = 6,356,752.31 m, which makes sense because the mean radius is a global average.

Problem

Scale Factor Calculation for PPCS/UTM. In the Philippine Plane Coordinate System (PPCS, which uses UTM), the scale factor k along a meridian at latitude φ = 12° depends on the radius of curvature. If the scale factor k = 0.99996 at the central meridian and increases away from it, estimate how the radius M affects the projection. (This is a conceptual question linking ellipsoid properties to map projections.)

Solution

The relationship between local curvature and scale factor is complex, but the concept is as follows: the UTM projection is a conformal (angle-preserving) transverse Mercator projection. Along a meridian at latitude φ = 12°, we have M ≈ 6,336,500 m (similar to the calculation above). The scale factor in UTM increases away from the central meridian because distances are stretched. The curvature M of the Earth at that latitude determines how much the projected surface deviates from the true ellipsoid. If we need to convert a measured ground distance to UTM coordinates, we must divide by the scale factor k. For example, if a measured distance is 1000 m at latitude 12°, the UTM grid distance would be approximately 1000 × 0.99996 ≈ 999.96 m at the central meridian. The radii N and M ensure that this projection is geometrically accurate for cadastral surveys under RA 8560 and RA 4374.

Key Points

  • Prime vertical radius N = a / √(1 − e²sin²φ) is the radius of curvature in the east–west direction
  • Meridian radius M = a(1 − e²) / (1 − e²sin²φ)^(3/2) is the radius of curvature in the north–south direction
  • N ≥ M everywhere; they are equal only at the poles
  • At equator: N = a, M = a(1 − e²); at poles: N and M become infinite
  • Mean radius of curvature R = √(M·N) is used for local spherical approximations
  • These radii are essential for map projection scale factors and distance reductions
  • In the Philippines, these calculations are mandatory for UTM/PPCS coordinate transformations

The World Geodetic System 1984 (WGS84) is the standard reference ellipsoid for all Global Positioning System (GPS) and modern GNSS receivers worldwide, including those used throughout the Philippines. The WGS84 ellipsoid is defined by two primary parameters: the semi-major axis a = 6,378,137 m (exact, by definition) and the inverse flattening 1/f = 298.257223563 (exact, by definition). From these, all other parameters are derived: f ≈ 0.00335281, b ≈ 6,356,752.31 m, and e² ≈ 0.00669438. In 1992, the Philippines officially adopted the Philippine Reference System 1992 (PRS92), which is explicitly based on the WGS84 ellipsoid. This was mandated under Republic Act 4374 (an act adopting the PRS as the standard datum for all mapping and surveying in the Philippines) and further reinforced by RA 8560 (the Property Boundaries Law) and Presidential Decree 1529 (the Property Registration Decree). All official cadastral maps, digital land records, infrastructure projects (roads, utilities, airports), and GNSS-based surveys in the Philippines must reference WGS84/PRS92. This ensures that a surveyor in Luzon and another in Mindanao are using the same geometric reference frame, critical for nationwide coordination, property disputes, and infrastructure planning. Any surveyor failing to use WGS84/PRS92 for official work is in violation of Philippine law. The PRC Geodetic Engineer Licensure Examination tests thorough knowledge of WGS84 parameters and their application to Philippine mapping and surveying standards. It is essential to memorize the WGS84 values and understand their role in the Philippine legal framework.

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4. The WGS84 Ellipsoid and Philippine Standards

Examples

Problem

Legal Compliance in Philippine Surveying. A surveyor is hired to conduct a cadastral survey of a 10-hectare property in Cavite province under RA 8560. What reference ellipsoid must they use, and why?

Solution

According to RA 8560 (the Property Boundaries Law) and its implementing rules, all boundary surveys in the Philippines must be referenced to the Philippine Reference System 1992 (PRS92), which is based on the WGS84 ellipsoid. Therefore, the surveyor must use WGS84 as their reference ellipsoid. All GNSS coordinates obtained from a GPS/GNSS receiver (which outputs WGS84 by default) must be used directly or verified against WGS84/PRS92 data. The surveyor must also ensure that all coordinate transformations, if any, are done using approved Philippine transformation parameters (e.g., from a local datum to WGS84/PRS92). Using any other ellipsoid (such as Clarke 1866 or Bessel 1841) would render the survey legally invalid and subject to rejection by the Land Registration Authority (LRA). This is not merely a technical choice but a legal requirement under Philippine law.

Problem

PRC Exam Question: WGS84 Ellipsoid Calculations. Verify that the first eccentricity squared for WGS84 is approximately 0.00669438. Use a = 6,378,137 m and 1/f = 298.257223563.

Solution

Step 1: Convert inverse flattening to flattening. f = 1 / 298.257223563 = 0.00335281 Step 2: Compute the semi-minor axis b. b = a(1 − f) = 6,378,137(1 − 0.00335281) = 6,378,137 × 0.99664719 = 6,356,752.31 m Step 3: Calculate e² = (a² − b²) / a². a² = 6,378,137² = 40,681,286,737,769 m² b² = 6,356,752.31² = 40,408,299,881,688 m² a² − b² = 272,986,856,081 m² e² = 272,986,856,081 / 40,681,286,737,769 = 0.00669438 ✓ Alternative formula: e² = 2f − f² e² = 2(0.00335281) − (0.00335281)² = 0.00670562 − 0.00001124 = 0.00669438 ✓ Both methods confirm e² ≈ 0.00669438 for WGS84.

Problem

Second Eccentricity for WGS84. Calculate the second eccentricity e' (or its square e'²) for WGS84. This parameter appears in some geodetic formulas.

Solution

Step 1: Use the definition e'² = (a² − b²) / b². From the previous problem: a² − b² = 272,986,856,081 m² b² = 40,408,299,881,688 m² e'² = 272,986,856,081 / 40,408,299,881,688 = 0.00673950 Step 2: Verify using the relationship e'² = e² / (1 − e²). e'² = 0.00669438 / (1 − 0.00669438) = 0.00669438 / 0.99330562 = 0.00673950 ✓ Therefore, e'² ≈ 0.00673950 and e' ≈ 0.08205 for WGS84. Note: The second eccentricity is used in some series expansions and alternative formulas for N and M, particularly in older geodetic texts. Modern computations prefer the first eccentricity e².

Key Points

  • WGS84 is the global standard for GNSS and the official ellipsoid of the Philippines
  • WGS84 parameters: a = 6,378,137 m (exact), 1/f = 298.257223563 (exact), e² = 0.00669438
  • The Philippine Reference System 1992 (PRS92) is based on WGS84
  • RA 4374, RA 8560, and PD 1529 mandate WGS84/PRS92 for all Philippine surveys and mapping
  • All official cadastral maps and land records in the Philippines reference WGS84/PRS92
  • Failure to use WGS84/PRS92 for official surveys is a legal violation in the Philippines
  • The PRC Geodetic Engineer Licensure Examination requires mastery of WGS84 values and applications

The reference ellipsoid and its derived parameters are not abstract theoretical concepts—they directly impact every surveying and mapping project in the Philippines. In cadastral surveying under RA 8560, surveyors use GNSS receivers that output WGS84 coordinates. These coordinates are then processed using N and M values at the local latitude to reduce measured distances to the ellipsoid and compute accurate grid coordinates in the Philippine Plane Coordinate System (PPCS). The PPCS is based on the Universal Transverse Mercator (UTM) projection, which divides the Philippines into zones. For each zone, a central meridian is selected, and the scale factor k at any point depends on the distance from the central meridian and the local radius of curvature N. Without accurate knowledge of N, the scale factor cannot be computed correctly, leading to errors in cadastral maps. For infrastructure projects (highways, railways, utilities), surveying networks must be established and adjusted using the WGS84 ellipsoid as the reference frame. Height measurements from GNSS are ellipsoidal heights (heights above the WGS84 ellipsoid), which differ from orthometric heights (heights above mean sea level) by the geoid undulation. Surveyors must understand this distinction to avoid mistakes in vertical data. In topographic mapping and the production of official maps by agencies like the National Mapping and Resource Information Authority (NAMRIA), the ellipsoid parameters determine the accuracy and consistency of the entire map. Moreover, under RA 4374, all public maps and surveys must be consistent with the official PRS92 datum—the ellipsoid is the foundation of this nationwide legal requirement. Finally, when converting between old survey data (which may reference Clarke 1866 or other historical ellipsoids) and modern WGS84/PRS92 data, knowledge of both ellipsoids and their geometric differences is essential. This conversion is a common task when modernizing old cadastral records or integrating historical survey data with new GNSS observations.

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5. Practical Applications in Philippine Surveying and Mapping

Examples

Problem

Cadastral Survey Workflow. A surveyor measures a distance of 500.000 m between two boundary monuments using a total station on the ground near latitude 15°N in Laguna province. The surveyor's GNSS receiver shows both points are at WGS84 latitude φ = 15°N. What steps are necessary to correctly record this distance in the cadastral map under RA 8560?

Solution

Step 1: Obtain the measured ground distance: D_ground = 500.000 m (at approximately 15°N latitude). Step 2: Compute the prime vertical radius N at φ = 15°N on WGS84. a = 6,378,137 m, e² = 0.00669438 sin(15°) ≈ 0.2588, sin²(15°) ≈ 0.0670 1 − e²sin²φ = 1 − 0.00669438 × 0.0670 = 1 − 0.000449 = 0.999551 N = a / √(0.999551) = 6,378,137 / 0.999776 = 6,378,863 m Step 3: Reduce the ground distance to the ellipsoid using the height h (obtained from GNSS). Assuming the surveyed monuments are at ellipsoidal height h ≈ 100 m (typical for lowlands): D_ellipsoid = D_ground × N / (N + h) = 500.000 × 6,378,863 / (6,378,863 + 100) D_ellipsoid = 500.000 × 6,378,863 / 6,378,963 = 500.000 × 0.9999843 = 499.992 m Step 4: Compute the scale factor k for the PPCS zone (e.g., Zone 3 for Laguna with central meridian 121°E). The scale factor depends on the distance from the central meridian; assume k = 0.99996 (typical for UTM central meridian). Step 5: Compute the grid distance in the PPCS. D_grid = D_ellipsoid × k = 499.992 × 0.99996 = 499.968 m Step 6: Record D_grid = 499.968 m in the cadastral map under RA 8560. This distance is now properly reduced to the WGS84 ellipsoid and the PPCS grid, ensuring legal compliance and consistency with the national coordinate system. Note: The full procedure includes converting WGS84 lat/lon to PPCS grid coordinates (Easting, Northing), which also requires N and the projection formulas.

Problem

Ellipsoidal vs. Orthometric Heights. A GNSS receiver at a point in Metro Manila shows an ellipsoidal height of h = 50.00 m above the WGS84 ellipsoid. A surveyor needs to know the orthometric height H (height above mean sea level) to design a water supply system. The geoid undulation at this location is N_geoid = −0.75 m. Compute H.

Solution

The relationship between ellipsoidal height h and orthometric height H is: h = H + N_geoid where N_geoid is the geoid undulation (negative if the geoid is below the ellipsoid). Rearranging: H = h − N_geoid = 50.00 − (−0.75) = 50.00 + 0.75 = 50.75 m Answer: The orthometric height is H = 50.75 m above mean sea level. This is important because engineering projects (like water systems, utilities, and buildings) are typically designed using orthometric heights (MSL reference), not ellipsoidal heights. The geoid undulation model for the Philippines is provided by NAMRIA or can be obtained from the EGM96 or EGM2008 global models.

Problem

Converting Between Ellipsoids: Clarke 1866 to WGS84. An old survey datum from the early 20th century used the Clarke 1866 ellipsoid (a = 6,378,206.4 m, b = 6,356,583.8 m). A coordinate from that survey is 10°N latitude and 120°E longitude. If this point is converted to WGS84 without accounting for the ellipsoid difference, what magnitude of error should be expected?

Solution

The prime vertical radii will differ because the ellipsoids have different parameters: Clarke 1866: a₁ = 6,378,206.4 m e₁² ≈ 0.00669604 At φ = 10°: N₁ = a₁ / √(1 − e₁²sin²10°) ≈ 6,378,700 m WGS84: a₂ = 6,378,137 m e₂² ≈ 0.00669438 At φ = 10°: N₂ = a₂ / √(1 − e₂²sin²10°) ≈ 6,378,340 m Difference in prime vertical radius: ΔN = N₂ − N₁ ≈ −360 m For a distance calculation or coordinate conversion error: In the north–south direction, an error in the meridian radius M would cause an arc-length error. As a rough estimate, if a 1° difference in latitude is computed using the wrong radius, the error scales as ΔM / R_earth ≈ 300 m / 6,371,000 m ≈ 5 cm per 1°. For more localized work, the error might be smaller but still significant (decimeters). Moreover, the Clarke 1866 ellipsoid's orientation and origin may differ from WGS84, introducing additional transformation errors of meters to tens of meters. A proper conversion requires a datum transformation (e.g., using Molodensky parameters), not just ellipsoid parameter changes. Conclusion: Naive conversion from Clarke 1866 to WGS84 without proper transformation can introduce errors of 1–10 meters, unacceptable for cadastral work under RA 8560. Modern surveys must use WGS84/PRS92 directly.

Key Points

  • GNSS receivers output WGS84 coordinates; these must be processed using local N and M values for cadastral surveys
  • The PPCS/UTM scale factor depends on the prime vertical radius N and the distance from the central meridian
  • Ellipsoidal heights (from GNSS) differ from orthometric heights by the geoid undulation
  • RA 8560 mandates that cadastral surveys reference WGS84/PRS92
  • Converting from historical ellipsoids (Clarke 1866) to WGS84 requires knowledge of both systems
  • Large-scale infrastructure projects depend on accurate ellipsoid parameters for surveying networks
  • Consistency with the WGS84 ellipsoid is a legal requirement under RA 4374, RA 8560, and PD 1529
  • The reference ellipsoid is the foundation of the Philippine national coordinate system

The topic of the reference ellipsoid and its parameters is a frequent source of errors among surveying students and even practicing surveyors. Understanding these common pitfalls is crucial for success on the PRC Geodetic Engineer Licensure Examination. First, students often confuse eccentricity e with eccentricity squared e². The formulas use e² (e-squared), not e. For WGS84, e² = 0.00669438, so e = √0.00669438 ≈ 0.0818. Many incorrect calculations result from using e instead of e². Second, the flattening f is very small—approximately 1/298 or 0.00335 for Earth. Students sometimes misread this as f = 0.298 or 0.335, leading to wildly incorrect results. Always double-check: for WGS84, f = 1 ÷ 298.257223563 ≈ 0.00335281. Third, students confuse the prime vertical radius N (east–west curvature) with the meridian radius M (north–south curvature). Remember: N is along a parallel (constant latitude), and M is along a meridian (constant longitude). At the equator, N = a (the equatorial radius) and M = a(1 − e²) (less than a). At the poles, both approach infinity, but M approaches infinity faster. Fourth, when calculating sin²φ, ensure your calculator is in degree mode, not radian mode, if φ is given in degrees. A common exam mistake is computing sin²(14°) with the calculator in radian mode, giving a completely wrong value. Fifth, the semi-minor axis b is very close to the semi-major axis a, differing by only about 0.335%. When computing b from f, use b = a(1 − f), not b = a − f. The latter would give a completely wrong answer. Sixth, students sometimes forget that the formulas for N and M depend on latitude φ. They compute N or M at the equator and mistakenly use that value for all latitudes, ignoring the sin²φ term in the denominator. This is incorrect; N and M vary significantly with latitude. Seventh, when solving problems, always include units (meters, not millimeters or kilometers, unless specified otherwise in SI units). A distance of 6,378,137 without units is ambiguous; 6,378,137 m makes it clear. Finally, for the PRC exam, memorize the WGS84 parameters exactly: a = 6,378,137 m, 1/f = 298.257223563, e² = 0.00669438. These values appear in nearly every surveying calculation and are the foundation of the Philippine Reference System 1992. Examiners expect you to know these values by heart.

Heading

6. Common Errors and Exam Tips

Examples

Problem

Spot the Error 1: A student calculates the flattening of WGS84 as f = 298.257223563 instead of f = 1/298.257223563. What is the resulting error, and why is this dangerous?

Solution

The student's value: f = 298.257223563 Correct value: f = 1/298.257223563 ≈ 0.00335281 The student's value is 298.257223563 ÷ 0.00335281 ≈ 88,900 times larger than the correct value! This is catastrophically wrong. If the student then uses this f to calculate b: b = a(1 − f) = 6,378,137(1 − 298.257223563) = 6,378,137 × (−297.257...) = −1.9 billion meters This negative, billion-meter value is nonsensical. The error arises from forgetting that the parameter is written as "1/f" in textbooks, meaning the denominator, not the numerator. Always compute f = 1 ÷ 298.257223563 ≈ 0.00335281, not f = 298.257223563. Danger: An exam grader would immediately recognize this as a fundamental misunderstanding of ellipsoid parameters and might award zero points for the entire question.

Problem

Spot the Error 2: A student forgets the latitude dependence and uses N = a for all latitudes. They compute a distance reduction assuming N = 6,378,137 m at latitude 45°N, where the actual N ≈ 6,388,838 m. What is the magnitude of the relative error?

Solution

Student's N: 6,378,137 m (for equator, φ = 0°) Actual N at 45°: 6,388,838 m Relative error: (6,388,838 − 6,378,137) / 6,388,838 ≈ 10,701 / 6,388,838 ≈ 0.00167 or 0.167% For a 1 km distance reduction at 45°: Error in distance: 1,000 m × 0.00167 ≈ 1.67 m This 1.67 m error is unacceptable in cadastral surveying. The student's mistake—ignoring the latitude dependence—is a common source of large systematic errors.

Problem

Spot the Error 3: A student computes e instead of e² and uses this in the formula N = a/√(1 − e sin²φ). For WGS84, they compute e = 0.0818 (correct value of √e²) and substitute into the formula as 1 − 0.0818 sin²φ instead of 1 − 0.00669438 sin²φ. What is the resulting error in N at φ = 30°?

Solution

Using e instead of e²: sin(30°) = 0.5 1 − 0.0818 × 0.5 = 1 − 0.0409 = 0.9591 N_wrong = 6,378,137 / √0.9591 = 6,378,137 / 0.9793 = 6,512,000 m (wildly wrong) Using correct e²: 1 − 0.00669438 × 0.5 = 1 − 0.00334719 = 0.99665281 N_correct = 6,378,137 / √0.99665281 = 6,378,137 / 0.998325 = 6,388,838 m Difference: 6,512,000 − 6,388,838 ≈ 123,000 m error! (about 2% of the radius) This is a catastrophic error stemming from using e (≈0.0818) instead of e² (≈0.00669438) in the denominator.

Problem

Exam Practice: Compute N and M for a point in the Cordillera region at latitude φ = 16°30'N = 16.5° on WGS84. Then check whether N > M.

Solution

Step 1: Convert parameters. a = 6,378,137 m, e² = 0.00669438, 1 − e² = 0.99330562 φ = 16.5° Step 2: Compute sin²φ. sin(16.5°) ≈ 0.2838 sin²(16.5°) ≈ 0.08054 Step 3: Compute N. 1 − e²sin²φ = 1 − 0.00669438 × 0.08054 = 1 − 0.000539 = 0.999461 √0.999461 ≈ 0.9997304 N = 6,378,137 / 0.9997304 ≈ 6,379,030 m Step 4: Compute M. (1 − e²sin²φ)^(3/2) = (0.999461)^1.5 ≈ 0.9991916 M = 6,378,137 × 0.99330562 / 0.9991916 = 6,335,440 / 0.9991916 ≈ 6,341,760 m Step 5: Verify N > M. N = 6,379,030 m > M = 6,341,760 m ✓ Difference: N − M ≈ 37,270 m Answer: At latitude 16.5°N in the Cordillera, N ≈ 6,379,030 m and M ≈ 6,341,760 m. The inequality N ≥ M is satisfied. Exam Tip: Always check that N ≥ M as a sanity check. If your calculations yield N < M, you have made an error.

Key Points

  • Do not confuse e with e². Always use e² in the formulas.
  • Flattening f is very small (≈0.00335), not 0.298 or 0.335. Always calculate f = 1/298.257... for WGS84.
  • Prime vertical radius N (east–west) and meridian radius M (north–south) are different. Do not swap them.
  • Ensure your calculator is in degree mode when computing sin²φ with φ in degrees.
  • When calculating b, use b = a(1 − f), not b = a − f.
  • N and M vary with latitude φ. Always include the sin²φ term; do not use equatorial values for all latitudes.
  • Include proper units (meters, not mm or km) in all answers.
  • Memorize WGS84 values: a = 6,378,137 m, 1/f = 298.257223563, e² = 0.00669438.
  • For the PRC exam, expect questions combining ellipsoid parameters, radii of curvature, and Philippine surveying standards.
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