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GELE Geodesy — Figure of the Earth and the Reference EllipsoidDetailed Explanation

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If the summary was not enough, this is the deep dive. Detailed explanations for Figure of the Earth and the Reference Ellipsoid in the GELE Geodesy context, written to turn surface familiarity into genuine understanding. Professional Regulation Commission (PRC) — Board of Geodetic Engineering's toughest GELE questions on this chapter are answered by the reasoning built here.

Exam context

For the Geodetic Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests Geodesy under a "Core" label, with Figure of the Earth and the Reference Ellipsoid in the 1st slot across 6 chapters. GELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Geodesy questions. Date to watch: September 2026.

Figure of the Earth and the Reference Ellipsoid - Detailed Explanation

Geodesy is the science that deals with the measurement and representation of the Earth, including its gravitational field, in a three-dimensional, time-varying space. Before any precise computation of positions, distances, or areas on the Earth's surface can be performed, the geodesist must define a mathematical model of the Earth. This model is the reference ellipsoid — an oblate spheroid of revolution that closely approximates the true shape of the Earth. For Filipino geodetic engineers, mastery of ellipsoid parameters is foundational: all geodetic control surveys in the Philippines under the Philippine Reference System of 1992 (PRS92) are referenced to the WGS84 ellipsoid, and all cadastral, topographic, and hydrographic surveys are performed on this mathematical surface. The PRC Geodetic Engineer Licensure Examination consistently tests candidates on ellipsoid parameters, derived quantities, and radii of curvature. This chapter provides a rigorous, board-exam-focused treatment of these essential concepts.

Concepts

The Figure of the Earth: From Sphere to Ellipsoid

The Earth is not a perfect sphere. Rotational forces cause it to bulge at the equator and flatten at the poles — a shape described as an oblate ellipsoid of revolution (also called an oblate spheroid). Understanding the progression of Earth models is important both historically and conceptually. 1. SPHERE MODEL: The simplest approximation. Uses a mean radius R ≈ 6,371 km. Adequate only for small-scale mapping (scales smaller than 1:5,000,000). Error introduced is about 1 part in 297 (approximately 0.34%), which is unacceptable for geodetic precision. 2. ELLIPSOID MODEL: The standard model for geodetic work. An ellipse is rotated about its minor (polar) axis to generate the oblate spheroid. This is the model used for all national geodetic reference systems, including PRS92 in the Philippines. 3. GEOID: The equipotential surface of the Earth's gravity field that coincides with mean sea level. The geoid is the physical reference surface for heights (orthometric heights H). It is irregular and cannot be described by a simple equation. The separation between the geoid and the ellipsoid is the geoid undulation N_geoidal (not to be confused with the prime-vertical radius N). 4. TOPOGRAPHIC SURFACE: The actual physical surface where measurements are made. The relationship chain is: Topographic Surface → Geoid (physical height reference) → Ellipsoid (geometric computation surface). In the Philippines, the Bureau of Lands surveys and the NAMRIA control network are all referenced to the PRS92/WGS84 ellipsoid, making ellipsoid parameter knowledge directly applicable to professional practice under RA 8560 (Geodetic Engineering Law).

Examples

The flattening f ≈ 1/297 represents the fractional difference between the equatorial and polar radii. Using a sphere instead of an ellipsoid introduces a systematic error of about 1 part in 297 in distances — far exceeding the allowable tolerance for geodetic control surveys (typically better than 1 part in 100,000 for third-order, 1 part in 1,000,000 for first-order). This illustrates why the ellipsoid model is essential for geodetic engineering practice.

Scenario

A surveyor uses a spherical Earth model with R = 6,371,000 m to compute the length of a geodetic line that is actually 100,000 m long on the ellipsoid. Estimate the maximum error introduced.

Solution

Error ≈ (1/297) × 100,000 m ≈ 336.7 m

Applications

Defining the reference surface for all geodetic control surveys in the Philippines (PRS92/WGS84).
GNSS positioning: All GPS/GNSS receivers output positions on the WGS84 ellipsoid.
Converting between ellipsoidal heights (from GNSS) and orthometric heights (for engineering work) requires geoid models.
Datum transformation between old (Luzon Datum, Clarke 1866) and new (PRS92, WGS84) systems.
Base for all PPCS/UTM coordinate computations in the Philippines.

Misconceptions

MISCONCEPTION: The geoid and ellipsoid are the same surface. REALITY: They differ by the geoid undulation, which ranges from about -105 m to +85 m globally, and about +15 m to +25 m over the Philippines.
MISCONCEPTION: GPS gives orthometric heights directly. REALITY: GPS/GNSS gives ellipsoidal heights h; orthometric heights H require subtraction of the geoid undulation.
MISCONCEPTION: The ellipsoid is the physical Earth. REALITY: It is a purely mathematical model; the physical surface is the topographic surface.

Related Concepts

Geoid undulation (N_geoidal)
Orthometric vs. ellipsoidal heights
Datum transformation
PRS92 and WGS84
PPCS/UTM coordinate system

Common Exam Questions

Example

Which surface is the reference for orthometric heights? Answer: The Geoid.

Approach

Identify which surface (geoid, ellipsoid, topographic) corresponds to a given description. Remember: geoid = gravity/MSL, ellipsoid = mathematical/GNSS, topographic = physical land surface.

Question Type

Conceptual identification

Example

A GNSS survey gives h = 87.23 m. The geoid undulation is N_geoidal = 21.5 m. Find the orthometric height H. Solution: H = h - N_geoidal = 87.23 - 21.5 = 65.73 m.

Approach

Use the formula h = H + N_geoidal, where h = ellipsoidal height, H = orthometric height, N_geoidal = geoid undulation. Be careful not to confuse geoid undulation N_geoidal with prime-vertical radius N.

Question Type

Relationship between height types

Key Points To Remember

The Earth is an oblate ellipsoid — flattened at poles, bulging at equator.
The ellipsoid is a mathematical surface; the geoid is a physical (gravity-defined) surface.
PRS92 uses the WGS84 ellipsoid as its reference surface.
Ellipsoidal height h = orthometric height H + geoid undulation N_geoidal.
Sphere model is adequate only for scales smaller than 1:5,000,000.
The oblate ellipsoid is generated by rotating an ellipse about its MINOR (polar) axis.
Under RA 8560, only licensed Geodetic Engineers may perform first-order geodetic surveys in the Philippines.

Ellipsoid Parameters: Semi-Major Axis, Semi-Minor Axis, and Flattening

The reference ellipsoid is fully defined by exactly TWO independent parameters. The most common choice for geodetic work is the semi-major axis a and the flattening f. SEMI-MAJOR AXIS (a): The equatorial radius of the ellipsoid. This is the longest radius, measured from the center to the equatorial surface. SEMI-MINOR AXIS (b): The polar radius. This is the shortest radius, measured from the center to either pole. It is always less than a for an oblate ellipsoid. FLATTENING (f): A dimensionless ratio that quantifies how much the ellipsoid departs from a sphere. f = (a - b) / a Derived relationships: b = a(1 - f) ... (Eq. 1) f = (a - b) / a ... (Eq. 2) FIRST ECCENTRICITY SQUARED (e²): e² = (a² - b²) / a² = 2f - f² ... (Eq. 3) SECOND ECCENTRICITY SQUARED (e'²): e'² = (a² - b²) / b² ... (Eq. 4) Relationship between e² and e'²: e'² = e² / (1 - e²) ... (Eq. 5) WGS84 PARAMETERS (memorize these for the board exam): a = 6,378,137.000 m (exact by definition) f = 1/298.257223563 b = 6,356,752.3142 m e² = 0.00669437999014 CLARKE 1866 PARAMETERS (used for old Philippine surveys, Luzon Datum): a = 6,378,206.4 m b = 6,356,583.8 m f = 1/294.978698 IMPORTANT MAGNITUDE CHECK: f is a very small number (approximately 1/298). A common board-exam error is writing f ≈ 0.298 instead of f ≈ 0.00335. Always check that 1/f ≈ 298 for WGS84.

Examples

Step (a) uses the fundamental relationship b = a(1-f). Step (b) uses e² = 2f - f², which avoids computing b first. Note that f² is very small (about 1.12 × 10⁻⁵) but must be included for accuracy. Step (c) uses the relationship between first and second eccentricities. These three computed values — b, e², e'² — are the most frequently tested derived quantities on the PRC board examination.

Scenario

BOARD-TYPE PROBLEM: Given WGS84 parameters a = 6,378,137 m and f = 1/298.257223563, compute (a) b, (b) e², and (c) e'².

Solution

(a) f = 1/298.257223563 = 0.0033528107 b = a(1 - f) = 6,378,137 × (1 - 0.0033528107) b = 6,378,137 × 0.9966471893 b = 6,356,752.314 m (b) e² = 2f - f² e² = 2(0.0033528107) - (0.0033528107)² e² = 0.0067056214 - 0.0000112413 e² = 0.0066943801 ≈ 0.00669438 (c) e'² = e²/(1 - e²) e'² = 0.00669438 / (1 - 0.00669438) e'² = 0.00669438 / 0.99330562 e'² = 0.00673966

When given both a and b, use f = (a-b)/a and e² = 1 - (b/a)². This form of e² avoids the need to compute f first and minimizes rounding errors. The Clarke 1866 ellipsoid is significant in Philippine geodesy because it was the basis for the Luzon Datum (old cadastral surveys). Knowing both WGS84 and Clarke 1866 parameters is essential for datum transformation problems — a frequent board exam topic.

Scenario

BOARD-TYPE PROBLEM: Given the Clarke 1866 ellipsoid with a = 6,378,206.4 m and b = 6,356,583.8 m, compute f and e².

Solution

f = (a - b)/a = (6,378,206.4 - 6,356,583.8) / 6,378,206.4 f = 21,622.6 / 6,378,206.4 f = 0.003390075 = 1/294.978 e² = 1 - (b/a)² e² = 1 - (6,356,583.8/6,378,206.4)² e² = 1 - (0.996610779)² e² = 1 - 0.993234853 e² = 0.006768347 ≈ 0.0068

Applications

Computing geodetic positions in PRS92/WGS84 for NAMRIA control surveys.
Datum transformation from Luzon Datum (Clarke 1866) to PRS92 (WGS84) for updating old cadastral maps.
GNSS data processing: all GNSS computations use WGS84 ellipsoid parameters.
Computing the scale factor and convergence for PPCS/UTM projections.
Reduction of observed distances to the ellipsoid surface for geodetic network adjustment.

Misconceptions

MISCONCEPTION: f ≈ 0.298 (confusing f with 1/f). REALITY: f ≈ 0.00335, while 1/f ≈ 298.
MISCONCEPTION: e and e² can be used interchangeably in formulas. REALITY: The radius of curvature formulas use e² explicitly; using e (not squared) gives wrong results.
MISCONCEPTION: WGS84 and Clarke 1866 give the same positions. REALITY: They differ by up to 200 m in the Philippine region — critical for cadastral boundary surveys.
MISCONCEPTION: The ellipsoid has a unique semi-major axis value. REALITY: Different ellipsoids have different values of a; WGS84 and Clarke 1866 differ by about 69 m in their a values.

Related Concepts

Radii of curvature (M and N)
Geocentric and geodetic latitude
Geodetic datum
Datum transformation parameters
PPCS/UTM scale factor computation

Common Exam Questions

Example

Given a = 6,378,137 m and b = 6,356,752.314 m, find f. Answer: f = (6,378,137 - 6,356,752.314)/6,378,137 = 21,384.686/6,378,137 = 0.003352811 = 1/298.257.

Approach

Given a and b, compute f using f = (a-b)/a; compute e² using e² = 1-(b/a)². Given a and f, compute b using b = a(1-f); compute e² using e² = 2f-f².

Question Type

Compute derived parameter from given axes

Example

Which ellipsoid is used as the basis for PRS92? Answer: WGS84 (a = 6,378,137 m, f = 1/298.257223563).

Approach

PRS92 and GNSS surveys use WGS84. Old cadastral surveys (pre-1990s) use Clarke 1866 (Luzon Datum). Memorize both sets of parameters.

Question Type

Identify the correct ellipsoid for Philippine surveys

Key Points To Remember

WGS84: a = 6,378,137 m; f = 1/298.257223563; b = 6,356,752.314 m; e² = 0.00669438.
Clarke 1866 (Luzon Datum): a = 6,378,206.4 m; b = 6,356,583.8 m.
b = a(1 - f) — derive b from a and f.
e² = 2f - f² = (a² - b²)/a² — two equivalent forms, both appear in board exams.
e'² = e²/(1 - e²) = (a² - b²)/b².
Flattening f is dimensionless and approximately 1/298 for WGS84 (NOT 1/3 or 0.298).
The ellipsoid is oblate: a > b, so f > 0.
Two independent parameters are sufficient to define the ellipsoid completely.

Radii of Curvature: Meridian (M) and Prime Vertical (N)

Unlike a sphere, an ellipsoid does not have a single radius of curvature. The curvature varies with both latitude and direction. The two principal radii of curvature are the most important for geodetic computations. PRIME VERTICAL RADIUS OF CURVATURE (N): Also called the radius of curvature in the prime vertical, N is the radius of curvature in the east-west direction (perpendicular to the meridian plane). N = a / √(1 - e²sin²φ) ... (Eq. 6) Where φ is the geodetic latitude. The denominator is so important it gets its own symbol: W = √(1 - e²sin²φ) So N = a/W. MERIDIAN RADIUS OF CURVATURE (M): Also called the radius of curvature in the meridian, M is the radius of curvature in the north-south direction. M = a(1 - e²) / (1 - e²sin²φ)^(3/2) ... (Eq. 7) M = a(1 - e²) / W³ IMPORTANT RELATIONSHIPS: 1. At the equator (φ = 0°): N_eq = a (the semi-major axis) M_eq = a(1 - e²) = b²/a 2. At the poles (φ = 90°): N_pole = a/√(1 - e²) = a²/b M_pole = a(1 - e²)/(1 - e²)^(3/2) = a/√(1 - e²) = a²/b Therefore: M_pole = N_pole = a²/b (the two are equal at the poles!) 3. General relationship: N ≥ M everywhere on the ellipsoid. N > M at all latitudes except the poles. MEAN RADIUS OF CURVATURE (Gaussian Mean Radius, R): R = √(M × N) ... (Eq. 8) This is used for approximating the ellipsoid locally as a sphere. RADIUS OF CURVATURE IN AN ARBITRARY AZIMUTH (Euler's Formula): R_α = MN / (M sin²α + N cos²α) ... (Eq. 9) Where α is the azimuth of the direction. This is tested less frequently but is important for advanced geodetic computations. NUMERICAL MAGNITUDES (WGS84, approximate): At φ = 0° (equator): N = 6,378,137 m; M = 6,335,439 m At φ = 14°N (approximate latitude of Central Philippines): N ≈ 6,379,853 m; M ≈ 6,337,455 m At φ = 45°: N ≈ 6,388,838 m; M ≈ 6,367,489 m At φ = 90° (pole): N = M ≈ 6,399,594 m

Examples

The key computational strategy is: (1) compute sin²φ, (2) compute W² and then W, (3) compute N = a/W, (4) compute M = a(1-e²)/W³ using the already-computed W. Note that a(1-e²) is a constant for a given ellipsoid and can be precomputed. For WGS84, a(1-e²) = 6,335,439.3 m — memorize this value! The latitude 14°18'N is representative of the Metro Manila/Central Luzon area, making this computation directly relevant to Philippine geodetic practice.

Scenario

BOARD-TYPE PROBLEM: Using WGS84 (a = 6,378,137 m, e² = 0.00669438), compute N and M at latitude φ = 14°18'N (approximate latitude of Manila).

Solution

Step 1: Compute sin²φ φ = 14°18' = 14.3° sin(14.3°) = 0.24714 sin²(14.3°) = 0.06108 Step 2: Compute W W² = 1 - e²sin²φ = 1 - 0.00669438 × 0.06108 W² = 1 - 0.000408906 = 0.999591094 W = √0.999591094 = 0.999795528 Step 3: Compute N N = a/W = 6,378,137 / 0.999795528 N = 6,379,441.6 m ≈ 6,379,442 m Step 4: Compute M M = a(1-e²)/W³ a(1-e²) = 6,378,137 × (1 - 0.00669438) = 6,378,137 × 0.99330562 = 6,335,439.3 m W³ = W × W² = 0.999795528 × 0.999591094 = 0.999386784 M = 6,335,439.3 / 0.999386784 M = 6,339,326.5 m ≈ 6,339,327 m Check: N > M ✓ (6,379,442 > 6,339,327)

This verification problem tests whether candidates understand the special values at the equator and can algebraically show that a(1-e²) = b²/a. This is derived from b = a(1-f) and e² = 2f-f², giving 1-e² = (1-f)² = b²/a². The result b²/a = 6,335,439 m is an important benchmark value to remember.

Scenario

BOARD-TYPE PROBLEM: Verify that at the equator (φ = 0°) on WGS84, N = a and M = b²/a.

Solution

At φ = 0°: sin φ = 0, so W = √(1 - e²×0) = √1 = 1 N = a/W = a/1 = a = 6,378,137 m ✓ M = a(1-e²)/W³ = a(1-e²)/1 = a(1-e²) Now check: b²/a = [a(1-f)]²/a = a(1-f)² ≈ a(1-2f) Also: a(1-e²) and b²/a — are they equal? b² = a²(1-f)² and b²/a = a(1-f)² a(1-e²) = a(1-2f+f²) = a(1-f)² ✓ Therefore M_equator = b²/a = 6,335,439.3 m ✓

The Gaussian mean radius R = √(MN) gives the radius of the sphere that most closely approximates the ellipsoid at a given latitude. Note that at φ = 45°, R ≈ 6,377,638 m, very close to the commonly used mean Earth radius of 6,371 km (which is a global average). The Gaussian radius is used when reducing observed distances to the reference surface and in small-area computations where a local sphere approximation is acceptable.

Scenario

BOARD-TYPE PROBLEM: Compute the Gaussian mean radius R at φ = 45° on WGS84.

Solution

First, compute M and N at φ = 45°: sin²(45°) = 0.5 W² = 1 - 0.00669438 × 0.5 = 1 - 0.00334719 = 0.99665281 W = √0.99665281 = 0.998325010 N = 6,378,137 / 0.998325010 = 6,388,838.5 m M = 6,335,439.3 / W³ W³ = 0.998325010 × 0.99665281 = 0.994994 (approx) Actual W³ = (0.998325010)³ = 0.994985... Let's compute carefully: W³ = W × W² = 0.998325010 × 0.99665281 = 0.994985... M = 6,335,439.3 / 0.994985 = 6,367,489 m Mean radius R = √(M × N) R = √(6,367,489 × 6,388,838.5) R = √(40,672,882,000,000) ≈ 6,377,638 m ≈ 6,377,638 m

Applications

Reduction of slope distances to the ellipsoid surface: ΔS_ellipsoid = ΔS_topographic × (R/(R+h)).
Computing arc-to-chord corrections for long geodetic lines.
Computing the scale factor in the UTM/PPCS projection (involves N at the survey latitude).
Geodetic line computations (direct and inverse problems on the ellipsoid).
Computing latitude departures and longitude departures for geodetic traverses.
Estimating the radius of the Earth for local survey computations in specific provinces of the Philippines.

Misconceptions

MISCONCEPTION: N is the north-south radius and M is the east-west radius. REALITY: N (prime vertical) is EAST-WEST; M (meridian) is NORTH-SOUTH. The letters are counterintuitive — remember 'N for normaL to the meridian, M for Meridian'.
MISCONCEPTION: M and N are constant across latitudes. REALITY: Both increase from equator to pole, but at different rates.
MISCONCEPTION: At the equator, N = M. REALITY: N = a and M = b²/a at the equator; they are most DIFFERENT at the equator and equal only at the poles.
MISCONCEPTION: The mean radius R = (M+N)/2. REALITY: R = √(MN) — it is the GEOMETRIC mean, not the arithmetic mean.

Related Concepts

Ellipsoid parameters (a, b, f, e²)
Geodetic latitude φ
Reduction of distances to the ellipsoid
UTM/PPCS scale factor
Geodetic direct and inverse problems

Common Exam Questions

Example

Find N at φ = 30°N on WGS84. sin²30° = 0.25; W² = 1 - 0.00669438(0.25) = 0.998326; W = 0.999163; N = 6,378,137/0.999163 = 6,383,480 m.

Approach

Step 1: Compute sin²φ. Step 2: Compute W = √(1-e²sin²φ). Step 3: N = a/W or M = a(1-e²)/W³. Always check N > M for latitudes between 0° and 90°.

Question Type

Direct computation of N or M at a given latitude

Example

What is the prime vertical radius N at the North Pole on WGS84? Answer: N_pole = a²/b = (6,378,137)²/6,356,752.314 = 6,399,593.6 m.

Approach

Memorize: At equator (φ=0°): N=a, M=b²/a. At poles (φ=90°): N=M=a²/b. These make excellent fill-in or multiple-choice questions.

Question Type

Special latitude values

Example

True or False: The meridian radius M is always greater than the prime vertical radius N. Answer: FALSE. N ≥ M always.

Approach

N > M everywhere except at the poles. If a question asks which is larger, always answer N (prime vertical) unless specifically asked about the poles.

Question Type

Comparing N and M

Key Points To Remember

N = a/W where W = √(1 - e²sin²φ) — prime vertical, E-W direction.
M = a(1-e²)/W³ — meridian, N-S direction.
N ≥ M everywhere; N = M only at the poles.
At the equator: N = a, M = a(1-e²) = b²/a.
At the poles: M = N = a²/b.
Mean radius R = √(MN) used for local sphere approximation.
Both M and N increase from equator to pole.
N is always the LARGER radius (east-west); M is the SMALLER radius (north-south).
The formula denominator W = √(1-e²sin²φ) appears in both M and N — compute it once and reuse.

Relationships Between Ellipsoid Parameters: Key Formulas

The board examination frequently tests the ability to derive one parameter from others. The following is a systematic summary of all critical inter-relationships. FUNDAMENTAL PARAMETER SET (choose any two to define the ellipsoid): (a, b), (a, f), (a, e²), (b, f), (b, e²) COMPLETE FORMULA TABLE: From (a, b): f = (a-b)/a e² = 1-(b/a)² = (a²-b²)/a² e'² = (a²-b²)/b² = (a/b)²-1 n = (a-b)/(a+b) [third flattening] From (a, f): b = a(1-f) e² = 2f-f² = f(2-f) e'² = f(2-f)/(1-f)² From (e²): e'² = e²/(1-e²) f = 1 - √(1-e²) [exact] f ≈ e²/2 [approximate, since f is small] ADDITIONAL USEFUL IDENTITIES: b² = a²(1-e²) a² = b²(1+e'²) (1-f)² = 1-e² b/a = √(1-e²) = 1-f a/b = √(1+e'²) = 1/(1-f) THE THIRD FLATTENING n (appears in some advanced series): n = (a-b)/(a+b) ≈ f/2 for small f For WGS84: n = (6,378,137 - 6,356,752.314)/(6,378,137 + 6,356,752.314) n = 21,384.686/12,734,889.314 = 0.001679222 MEMORY AID — FORMULA HIERARCHY: Given a and f → find b (b=a(1-f)) → find e² (e²=2f-f²) → find e'² (e'²=e²/(1-e²)) Given a and b → find f (f=(a-b)/a) → find e² (e²=1-(b/a)²) → find e'²

Examples

This problem tests the ability to move between different parameter representations. The formula f = 1-√(1-e²) is exact but requires a square root. The approximation f ≈ e²/2 gives f ≈ 0.003347, which is close but not exact enough for precise geodetic work. Always use the exact formula for board exam computations unless told otherwise.

Scenario

BOARD-TYPE PROBLEM: Given e² = 0.00669438 for WGS84, compute (a) f, (b) e'², and (c) verify using b = a√(1-e²).

Solution

(a) f = 1 - √(1-e²) f = 1 - √(1-0.00669438) f = 1 - √(0.99330562) f = 1 - 0.99664725 f = 0.00335275 ≈ 1/298.26 ✓ (b) e'² = e²/(1-e²) e'² = 0.00669438/(1-0.00669438) e'² = 0.00669438/0.99330562 e'² = 0.00673966 (c) b = a√(1-e²) b = 6,378,137 × √(0.99330562) b = 6,378,137 × 0.99664725 b = 6,356,752.3 m ✓

Applications

Converting between different geodetic software packages that use different parameter sets (some use a and f, others use a and e²).
Checking the consistency of given ellipsoid parameters.
Deriving parameters for the Clarke 1866 ellipsoid when converting old Philippine cadastral data.
Computing the series expansion coefficients for the meridian arc length formula.

Misconceptions

MISCONCEPTION: e'² = 1/e². REALITY: e'² = e²/(1-e²), which is only slightly larger than e².
MISCONCEPTION: The formula e² = 2f-f² is approximate. REALITY: This is exact — it follows algebraically from b = a(1-f) and e² = (a²-b²)/a².
MISCONCEPTION: Different ellipsoids can have the same f but different a values with the same e². REALITY: e² depends only on f (through e²=2f-f²), so same f always means same e² regardless of a.

Related Concepts

Ellipsoid definition parameters
Radii of curvature (M and N)
Meridian arc computation
Series expansions in geodesy

Common Exam Questions

Example

Given f = 1/298.257, find e². Solution: e² = 2f-f² = 2/298.257 - (1/298.257)² = 0.006705522 - 0.0000112 = 0.006694322 ≈ 0.006694.

Approach

Start from the given parameters and systematically apply the formulas. Always verify with an independent check (e.g., compute e² two ways).

Question Type

Parameter derivation chain

Key Points To Remember

e² = 2f - f² = f(2-f) — use when given f.
e² = 1 - (b/a)² — use when given a and b.
e'² = e²/(1-e²) — always derives from e².
(1-f)² = 1-e² — a fundamental identity linking f and e².
b/a = 1-f = √(1-e²) — three equivalent forms.
Third flattening n = (a-b)/(a+b) ≈ f/2.
WGS84: e² = 0.00669438; e'² = 0.00673966; n = 0.001679.
All derived parameters can be expressed in terms of any two primary parameters.

Practice Problems

This is the most fundamental computation in the chapter. Part (a) uses the basic relationship b=a(1-f). Part (b) avoids computing b by using e²=2f-f² directly — faster and more accurate. Part (c) converts from first to second eccentricity. Part (d) computes the third flattening, which appears in series expansions for the meridian arc. Note: verify part (b) independently using e² = (a²-b²)/a² = 1-(b/a)² ≈ 1-(0.9966472)² = 1-0.9933056 = 0.0066944 ✓

Problem

PROBLEM 1 (WGS84 Parameters): Using the WGS84 defining parameters a = 6,378,137 m and f = 1/298.257223563, compute: (a) the semi-minor axis b, (b) the first eccentricity squared e², (c) the second eccentricity squared e'², and (d) the third flattening n.

Solution

Given: a = 6,378,137 m; f = 1/298.257223563 = 0.0033528107 (a) b = a(1-f) b = 6,378,137 × (1 - 0.0033528107) b = 6,378,137 × 0.9966471893 b = 6,356,752.314 m (b) e² = 2f - f² e² = 2(0.0033528107) - (0.0033528107)² e² = 0.0067056214 - 0.0000112413 e² = 0.0066943801 ≈ 0.00669438 (c) e'² = e²/(1-e²) e'² = 0.00669438/(1-0.00669438) e'² = 0.00669438/0.99330562 e'² = 0.00673967 (d) n = (a-b)/(a+b) n = (6,378,137 - 6,356,752.314)/(6,378,137 + 6,356,752.314) n = 21,384.686/12,734,889.314 n = 0.001679222

The computation follows the standard sequence: sin²φ → W → N → M → R. The latitude 10°N corresponds to the Zamboanga/Davao region — relevant to Philippine geodetic practice. Note that R ≈ 6,358,042 m at this latitude, about 13 km less than the equatorial radius. This mean radius is used when reducing distances to the ellipsoid or when setting up a local approximating sphere for cadastral computations in southern Philippine provinces.

Problem

PROBLEM 2 (Radii of Curvature): Compute the prime vertical radius N, the meridian radius M, and the Gaussian mean radius R at latitude φ = 10°N on WGS84 (a = 6,378,137 m, e² = 0.00669438). This latitude is representative of southern Philippines (Mindanao region).

Solution

Step 1: sin φ = sin(10°) = 0.173648 sin²φ = 0.030154 Step 2: Compute W W² = 1 - e²sin²φ = 1 - 0.00669438 × 0.030154 W² = 1 - 0.000201907 = 0.999798093 W = √0.999798093 = 0.999899042 Step 3: Prime vertical radius N N = a/W = 6,378,137/0.999899042 N = 6,378,779.5 m ≈ 6,378,780 m Step 4: Meridian radius M a(1-e²) = 6,378,137 × 0.99330562 = 6,335,439.3 m W³ = W × W² = 0.999899042 × 0.999798093 = 0.999697168 M = 6,335,439.3/0.999697168 M = 6,337,356.0 m ≈ 6,337,356 m Step 5: Gaussian mean radius R R = √(M × N) = √(6,337,356 × 6,378,780) R = √(40,434,007,000,000) R = 6,358,042 m ≈ 6,358,042 m Verification: N > M ✓ (6,378,780 > 6,337,356)

This problem is directly relevant to Philippine geodetic practice — the Clarke 1866 ellipsoid underpins the Luzon Datum used in all pre-1990s cadastral surveys. When a Geodetic Engineer encounters old titles or plans using the Luzon Datum (as authorized under CA 141 and PD 1529), knowledge of Clarke 1866 parameters is essential for datum transformation to PRS92/WGS84. The difference in N values between Clarke 1866 and WGS84 at the same latitude is typically a few hundred meters — significant for precision boundary surveys under RA 8560.

Problem

PROBLEM 3 (Clarke 1866 Ellipsoid): The old cadastral maps of a municipality in Cagayan Valley were surveyed using the Luzon Datum based on the Clarke 1866 ellipsoid (a = 6,378,206.4 m, b = 6,356,583.8 m). Compute: (a) flattening f, (b) e², and (c) N at the approximate latitude of the area, φ = 18°N.

Solution

(a) Flattening f f = (a-b)/a = (6,378,206.4 - 6,356,583.8)/6,378,206.4 f = 21,622.6/6,378,206.4 f = 0.0033900753 = 1/294.978 (b) First eccentricity squared e² e² = 1-(b/a)² b/a = 6,356,583.8/6,378,206.4 = 0.99660932 (b/a)² = 0.993230557 e² = 1 - 0.993230557 = 0.006769443 ≈ 0.0067694 (c) Prime vertical radius N at φ = 18°N sin(18°) = 0.309017 sin²(18°) = 0.095491 W² = 1 - 0.0067694 × 0.095491 = 1 - 0.000646478 = 0.999353522 W = √0.999353522 = 0.999676734 N = 6,378,206.4/0.999676734 = 6,380,269.3 m ≈ 6,380,269 m

These special values are high-frequency board exam items. Key facts: (1) At equator: N=a (maximum is a), M=b²/a (minimum of M). (2) At pole: N=M=a²/b. (3) The value a²/b = 6,399,593.6 m > a = 6,378,137 m — the polar radius of curvature EXCEEDS the equatorial radius. This seems counterintuitive but is correct: the Earth curves more sharply at the equator (smaller M) and less sharply at the poles (larger M=N). This is consistent with the oblate shape — flatter at poles means less curvature at poles.

Problem

PROBLEM 4 (Equator and Pole Special Cases): For the WGS84 ellipsoid (a = 6,378,137 m, b = 6,356,752.314 m, e² = 0.00669438), verify the special values: (a) N and M at the equator, and (b) N and M at the pole.

Solution

(a) At the equator (φ = 0°): sin²(0°) = 0; W = √(1-0) = 1 N_eq = a/W = 6,378,137/1 = 6,378,137 m = a ✓ M_eq = a(1-e²)/W³ = 6,335,439.3/1 = 6,335,439 m Check using b²/a: b²/a = (6,356,752.314)²/6,378,137 b² = 40,408,299,984,484.5 b²/a = 40,408,299,984,484.5/6,378,137 = 6,335,439.3 m ✓ (b) At the pole (φ = 90°): sin²(90°) = 1; W = √(1-e²) = √(0.99330562) = 0.99664723 N_pole = a/W = 6,378,137/0.99664723 = 6,399,593.6 m a²/b = (6,378,137)²/6,356,752.314 = 40,680,631,590,769/6,356,752.314 = 6,399,593.6 m ✓ M_pole = a(1-e²)/W³ W³ = (0.99664723)³ = 0.98997... Actually W³ = W × W² = 0.99664723 × 0.99330562 = 0.98997... M_pole = 6,335,439.3/0.99000562... Alternatively use: M_pole = a²/b = 6,399,593.6 m Confirm: M_pole = N_pole = a²/b ✓

This problem connects ellipsoid theory to practical geodetic survey work — a common board exam application. The elevation factor k_elev = R/(R+h) reduces measured distances to the reference ellipsoid surface. Since the survey is above the ellipsoid (h > 0), the ellipsoidal distance is shorter than the surface distance. For a 15 km line at 150 m elevation, the reduction is 0.353 m. This is significant for first-order and second-order geodetic control surveys under NAMRIA specifications. Note: in the complete distance reduction process, a projection scale factor (from the UTM/PPCS projection) is also applied: S_grid = S_ellipsoid × k_projection.

Problem

PROBLEM 5 (Applied Problem — Distance Reduction): A geodetic survey line measured on the topographic surface at an average ellipsoidal height h = 150 m has a slope distance of 15,000.000 m (after slope correction). Using the Gaussian mean radius R = 6,370,000 m at the survey location, reduce this distance to the ellipsoid. Also compute the scale factor k = R/(R+h).

Solution

Given: S_surface = 15,000.000 m (distance at elevation h above ellipsoid) h = 150 m (average ellipsoidal height) R = 6,370,000 m (Gaussian mean radius) Scale factor (elevation factor) for distance reduction: k_elev = R/(R+h) = 6,370,000/(6,370,000+150) k_elev = 6,370,000/6,370,150 k_elev = 0.999976439 Distance reduced to ellipsoid: S_ellipsoid = S_surface × k_elev S_ellipsoid = 15,000.000 × 0.999976439 S_ellipsoid = 14,999.647 m Correction to distance: ΔS = S_ellipsoid - S_surface = 14,999.647 - 15,000.000 = -0.353 m This is a shortening of 0.353 m or about 1 part in 42,500.

Exam Preparation Tips

MEMORIZE WGS84 VALUES: a = 6,378,137 m; b = 6,356,752.314 m; f = 1/298.257223563; e² = 0.00669438. These appear in almost every quantitative geodesy problem on the board exam.
MEMORIZE CLARKE 1866 VALUES: a = 6,378,206.4 m; b = 6,356,583.8 m. These are needed for old Philippine cadastral datum (Luzon Datum) problems.
KNOW THE FORMULA PAIRS: For (a,f) use b=a(1-f) and e²=2f-f². For (a,b) use f=(a-b)/a and e²=1-(b/a)². Practice both directions.
NEVER CONFUSE N AND M: N (prime vertical) = E-W direction; M (meridian) = N-S direction; N ≥ M always. Associate N with the Normal to the meridian.
SPECIAL LATITUDE BENCHMARKS: Equator: N=a, M=b²/a. Pole: N=M=a²/b. These are high-probability multiple-choice items.
COMPUTATION STRATEGY FOR RADII: Always compute W = √(1-e²sin²φ) first. Then N = a/W and M = a(1-e²)/W³. Compute W only once per problem.
WATCH THE FLATTENING MAGNITUDE: f ≈ 0.00335, NOT 0.335 or 0.298. The reciprocal 1/f ≈ 298. A quick sanity check: is your f between 0.003 and 0.004?
CALCULATOR SETTINGS: Always verify your calculator is in DEGREE mode when computing sin φ. A common error is computing sin(φ in radians) when φ is given in degrees.
CONNECT TO PHILIPPINE LAWS: Under RA 8560, only licensed Geodetic Engineers can perform first-order surveys. PD 1529 (Property Registration Decree) requires surveys to be tied to the national geodetic network. CA 141 (Public Land Act) requires approval of cadastral surveys by NAMRIA/DENR.
UNDERSTAND THE DATUM CONTEXT: PRS92 is the current legal datum in the Philippines (NAMRIA Order 1992). Older surveys used the Luzon Datum (Clarke 1866). Datum transformation problems require knowing both ellipsoid parameter sets.
PRACTICE DIMENSIONAL ANALYSIS: All radii of curvature (M, N, R) are in meters. Flattening f and eccentricities e, e' are dimensionless. Check units before submitting answers.
USE THE ANSWER CHOICES AS CHECKS: In multiple-choice problems, if your answer for N at φ=14°N is 6,379,000 m, eliminate choices below 6,378,137 m (less than a, which is impossible for N at any latitude) and choose the closest reasonable value.
LINK THEORY TO PRACTICE: Reduction of distances to the ellipsoid (for geodetic traverse), UTM scale factors, and GNSS baseline components all use M and N. Understanding where these formulas come from makes application easier.
REVIEW PAST BOARD EXAMS: PRC board exam problems on this topic typically include: (1) compute b given a and f, (2) compute e² given f, (3) compute N or M at a given latitude, (4) identify the correct formula for M vs N, (5) identify WGS84 parameters.

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In summary

The figure of the Earth and the reference ellipsoid form the geometric foundation upon which all geodetic computations in the Philippines — and globally — are built. The key takeaways for the PRC Geodetic Engineer Licensure Examination are: (1) The Earth is modeled as an oblate ellipsoid of revolution, defined by two parameters (conventionally a and f); (2) WGS84 (a = 6,378,137 m, f = 1/298.257223563, e² = 0.00669438) is the current Philippine reference ellipsoid under PRS92, adopted by NAMRIA and mandated under RA 8560; (3) The old Luzon Datum used the Clarke 1866 ellipsoid (a = 6,378,206.4 m, b = 6,356,583.8 m), relevant for interpreting old cadastral surveys under PD 1529 and CA 141; (4) The prime vertical radius N (east-west) and meridian radius M (north-south) vary with latitude, with N ≥ M always, and N = M = a²/b only at the poles; (5) The Gaussian mean radius R = √(MN) provides a local sphere approximation for practical distance reduction computations. Mastery of these parameters, their inter-relationships, and their board-exam computation procedures will give candidates a strong foundation not only for the geodesy portion of the PRC examination but also for professional practice in cadastral, topographic, and geodetic control surveys throughout the Philippine archipelago.

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Geodetic Datums and Coordinate Systems

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