GELE Geodesy — Figure of the Earth and the Reference EllipsoidExam Answer Templates

Marks

1

Topic

Introduction to Geodesy

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners reward the two-part structure: (1) size and shape + (2) point location. Include both elements in one clean sentence.

Model Answer

Geodesy is the science that deals with the determination of the size and shape of the Earth and the precise location of points on its surface.

Question Type

very_short_answer

Answer Structure

• Single sentence: define the science with its two core objectives (size/shape + point location) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining geodesy only as 'measurement of the Earth' without mentioning point location (-0 but incomplete for multi-part follow-ups)
• Writing 'geography' or 'surveying' as synonyms — these are incorrect and will score 0
• Omitting 'precise' — vague answers may not score if the examiner expects technical rigor

Key Phrases To Include

• science
• size and shape of the Earth
• precise location of points

Marks

1

Topic

The Reference Ellipsoid

Difficulty

easy

Template Id

T2

Examiner Tip

The exact term 'oblate ellipsoid of revolution' is the gold-standard phrase. 'Spheroid' is also accepted. Never write 'sphere' alone.

Model Answer

The Earth is modeled as an oblate ellipsoid of revolution (also called a spheroid), which is flattened at the poles and bulging at the equator.

Question Type

very_short_answer

Answer Structure

• Name the shape correctly: oblate ellipsoid of revolution / spheroid [1 mark]
• Optional qualifier (flattened at poles) — strengthens the answer but not required for the mark

Scoring Breakdown

Common Mark Deductions

• Writing 'sphere' — loses the mark because a perfect sphere ignores Earth's oblateness
• Writing 'ellipse' — an ellipse is a 2D curve, not a 3D solid
• Writing 'geoid' — the geoid is the equipotential surface, not the reference ellipsoid

Key Phrases To Include

• oblate ellipsoid of revolution
• spheroid
• flattened at the poles

Marks

1

Topic

The Reference Ellipsoid

Difficulty

easy

Template Id

T3

Examiner Tip

The examiner wants exactly two parameters. Either (a, f) or (a, b) is accepted. Specifying the symbol and its physical meaning in parentheses shows mastery.

Model Answer

A reference ellipsoid is defined by two parameters: the semi-major axis a (equatorial radius) and the flattening f (or equivalently, the semi-minor axis b).

Question Type

very_short_answer

Answer Structure

• Name both parameters with their symbols and brief descriptions [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only one parameter (e.g., only a) — earns 0 marks since both are required
• Listing eccentricity e² as a defining parameter — e² is derived, not defining
• Using incorrect symbols (e.g., r for radius instead of a)

Key Phrases To Include

• semi-major axis a
• flattening f
• equatorial radius
• semi-minor axis b

Marks

1

Topic

Ellipsoid Parameters

Difficulty

easy

Template Id

T4

Examiner Tip

Write the formula first, then define variables. This is the standard engineering exam format and signals methodical thinking to the examiner.

Model Answer

The flattening of a reference ellipsoid is given by: f = (a − b) / a, where a is the semi-major axis and b is the semi-minor axis.

Question Type

very_short_answer

Answer Structure

• Write the formula with correct symbols [1 mark]
• Define symbols (improves clarity but not required for the mark)

Scoring Breakdown

Common Mark Deductions

• Writing f = (b − a)/a — sign error, gives a negative flattening
• Writing f = (a − b)/b — this is the inverse flattening formula variant, not standard f
• Leaving out variable definitions when required by the question

Key Phrases To Include

• f = (a − b) / a
• semi-major axis a
• semi-minor axis b

Marks

2

Topic

Radii of Curvature

Difficulty

medium

Template Id

T5

Examiner Tip

The mnemonic 'N is for Normal (prime vertical)' helps. Write both formulas side by side and always state the direction of curvature to earn the identification mark.

Model Answer

The prime-vertical radius of curvature N is the radius of curvature of the ellipsoid in the east–west (prime vertical) direction at a given geodetic latitude φ, computed as: N = a / √(1 − e²sin²φ). The meridian radius of curvature M is the radius of curvature in the north–south (meridian) direction at latitude φ, computed as: M = a(1 − e²) / (1 − e²sin²φ)^(3/2). At any latitude, N ≥ M, with equality only at the poles.

Question Type

short_answer

Answer Structure

• Line 1–2: Define N with its direction (E–W) and formula [1 mark]
• Line 3–4: Define M with its direction (N–S) and formula, plus the inequality N ≥ M [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping N and M formulas — very common board exam error, loses both formula marks
• Omitting the directional association (which is E–W, which is N–S)
• Writing the exponent as 1/2 instead of 3/2 in the M formula

Key Phrases To Include

• prime-vertical radius N
• meridian radius M
• east–west direction
• north–south direction
• N = a / √(1 − e²sin²φ)
• M = a(1 − e²) / (1 − e²sin²φ)^(3/2)
• N ≥ M

Marks

2

Topic

WGS84 Ellipsoid

Difficulty

easy

Template Id

T6

Examiner Tip

Memorize WGS84 values by heart: a = 6,378,137 m; 1/f = 298.257223563; e² = 0.00669438. These appear on nearly every Geodesy board exam. Writing them without hesitation signals readiness.

Model Answer

The WGS84 reference ellipsoid parameters are: semi-major axis a = 6,378,137 m; inverse flattening 1/f = 298.257223563; first eccentricity squared e² = 0.00669438. The semi-minor axis derived from these values is b = 6,356,752.31 m.

Question Type

very_short_answer

Answer Structure

• State a and 1/f correctly [1 mark]
• State e² correctly (or derive it from f) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing f with 1/f — writing f = 298.257 instead of 1/f = 298.257 (a common and costly error)
• Rounding a to 6,378,000 m — insufficient precision for geodetic work
• Omitting the unit 'metres' for a

Key Phrases To Include

• a = 6,378,137 m
• 1/f = 298.257223563
• e² = 0.00669438
• WGS84

Marks

2

Topic

Ellipsoid Parameters — Clarke 1866

Difficulty

medium

Template Id

T7

Examiner Tip

Use two independent formulas (f from axes, e² from axes) as a self-check. If e² ≈ 2f, your answer is in the right ballpark. For Clarke 1866, e² ≈ 0.006769 while for WGS84 e² ≈ 0.006694 — know the difference.

Model Answer

Given: a = 6,378,206.4 m; b = 6,356,583.8 m. Required: 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 (≈ 1/294.98) e² = 1 − (b/a)² = 1 − (6,356,583.8/6,378,206.4)² (b/a) = 0.996609768 (b/a)² = 0.993231031 e² = 1 − 0.993231031 = 0.006768969 Answer: f = 0.003390 and e² = 0.006769.

Question Type

numerical

Answer Structure

• State Given and Required [0 marks — no penalty but shows structure]
• Correctly apply f = (a−b)/a to get f = 0.003390 [1 mark]
• Correctly apply e² = 1 − (b/a)² or e² = 2f − f² to get e² = 0.006769 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using e² = 2f − f² but computing f incorrectly first — cascading error loses both marks
• Expressing f as a percentage (0.339%) instead of a dimensionless fraction
• Arithmetic errors in subtraction (a − b); always double-check this subtraction

Key Phrases To Include

• f = (a − b)/a
• e² = 1 − (b/a)²
• Clarke 1866
• f = 0.003390
• e² = 0.006769

Marks

3

Topic

Prime-Vertical Radius of Curvature

Difficulty

medium

Template Id

T8

Examiner Tip

Show every intermediate step on a separate line. Examiners award partial marks for correct formula (even if computation is wrong) and for correct denominator (even if division is wrong). Showing work earns marks.

Model Answer

Given: WGS84 ellipsoid — a = 6,378,137 m; e² = 0.00669438; φ = 45°N. Required: Prime-vertical radius N. Formula: N = a / √(1 − e²sin²φ) Solution: Step 1: Compute sin²φ sin 45° = 0.707107; sin²45° = 0.500000 Step 2: Compute the denominator argument 1 − e²sin²φ = 1 − (0.00669438)(0.500000) = 1 − 0.003347190 = 0.996652810 Step 3: Take the square root √0.996652810 = 0.998325002 Step 4: Compute N N = 6,378,137 / 0.998325002 N = 6,388,838.5 m Answer: N = 6,388,838.5 m ≈ 6,388,838 m at φ = 45°N (WGS84).

Question Type

numerical

Answer Structure

• State Given data including WGS84 parameters [identification — sets up partial credit]
• Write the correct formula N = a / √(1 − e²sin²φ) [1 mark]
• Correctly compute the denominator 1 − e²sin²φ = 0.996653 [1 mark]
• Correctly compute final N = 6,388,838.5 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using e instead of e² in the formula (most common error — substituting 0.0818 instead of 0.006694)
• Computing sin 45° as 0.5 instead of sin²45° = 0.5 (forgetting to square)
• Rounding intermediate values too aggressively — carry at least 8 significant figures until the final step
• Using incorrect WGS84 value for a (e.g., 6,371,000 m which is the mean sphere, not WGS84)

Key Phrases To Include

• N = a / √(1 − e²sin²φ)
• e² = 0.00669438
• sin²45° = 0.5
• 1 − e²sin²φ = 0.996653
• N = 6,388,838 m

Marks

3

Topic

Meridian Radius of Curvature

Difficulty

hard

Template Id

T9

Examiner Tip

The 3/2 exponent is the key distinguishing feature of the M formula. Write it prominently and circle it if needed. Then compute (base)^(3/2) as (base)^1 × (base)^(1/2) to avoid calculator errors.

Model Answer

Given: a = 6,378,137 m; e² = 0.00669438; φ = 30°N. Required: Meridian radius M. Formula: M = a(1 − e²) / (1 − e²sin²φ)^(3/2) Solution: Step 1: Compute the numerator a(1 − e²) = 6,378,137 × (1 − 0.00669438) = 6,378,137 × 0.99330562 = 6,335,439.3 m Step 2: Compute sin²φ sin 30° = 0.500000; sin²30° = 0.250000 Step 3: Compute 1 − e²sin²φ 1 − (0.00669438)(0.250000) = 1 − 0.001673595 = 0.998326405 Step 4: Compute (1 − e²sin²φ)^(3/2) (0.998326405)^(3/2) = 0.998326405 × √0.998326405 √0.998326405 = 0.999162869 (0.998326405)^(3/2) = 0.998326405 × 0.999162869 = 0.997490895 Step 5: Compute M M = 6,335,439.3 / 0.997490895 M = 6,351,378 m Answer: M = 6,351,378 m at φ = 30°N (WGS84).

Question Type

numerical

Answer Structure

• Write the correct formula M = a(1−e²) / (1−e²sin²φ)^(3/2) [1 mark]
• Correctly compute numerator a(1−e²) and denominator argument [1 mark]
• Correctly compute final M (requires correct 3/2 exponent computation) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using exponent 1/2 instead of 3/2 — the M formula has (3/2) not (1/2); this is the most critical error
• Confusing M with N — writing N's formula for M loses all 3 marks
• Using a for the numerator without multiplying by (1−e²) — omitting the (1−e²) factor
• Calculator set to radians when computing sin 30° — always verify degree mode

Key Phrases To Include

• M = a(1 − e²) / (1 − e²sin²φ)^(3/2)
• numerator: a(1 − e²) = 6,335,439 m
• exponent 3/2
• M = 6,351,378 m

Marks

3

Topic

Relationship Between Ellipsoid Parameters

Difficulty

medium

Template Id

T10

Examiner Tip

Derivation questions reward every logical step. Write each substitution on a new line. Examiners check that the bridge from e² = 1−(b/a)² to e² = 2f−f² is logically continuous, not skipped.

Model Answer

Given the definitions: f = (a − b)/a → b = a(1 − f) e² = (a² − b²)/a² Derivation: Step 1: Express e² in terms of a and b: e² = (a² − b²)/a² = 1 − (b/a)² Step 2: Substitute b = a(1 − f): e² = 1 − [a(1 − f)/a]² = 1 − (1 − f)² Step 3: Expand (1 − f)²: (1 − f)² = 1 − 2f + f² Step 4: Subtract from 1: e² = 1 − (1 − 2f + f²) = 2f − f² Therefore: e² = 2f − f² ∎

Question Type

short_answer

Answer Structure

• State the starting definitions: f = (a−b)/a and e² = (a²−b²)/a² [1 mark]
• Correct substitution of b = a(1−f) into e² expression [1 mark]
• Correct algebraic expansion to arrive at e² = 2f − f² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Starting from e² = (a²−b²)/b² — this is the second eccentricity e'², not e²
• Skipping intermediate steps — derivation marks are awarded step-by-step
• Algebraic sign errors during expansion of (1−f)²
• Not writing the QED symbol or 'Therefore' to conclude the derivation

Key Phrases To Include

• b = a(1 − f)
• e² = 1 − (b/a)²
• e² = 1 − (1 − f)²
• (1 − f)² = 1 − 2f + f²
• e² = 2f − f²

Marks

3

Topic

Second Eccentricity

Difficulty

medium

Template Id

T11

Examiner Tip

The shortcut e'² = e²/(1−e²) is faster and less error-prone than computing a² and b² separately. Note that e'² is always slightly larger than e² for any oblate ellipsoid — use this as a sanity check.

Model Answer

Given: a = 6,378,137 m; b = 6,356,752.31 m; e² = 0.00669438. Required: Second eccentricity squared e'². Formula: e'² = (a² − b²)/b² Alternatively: e'² = e²/(1 − e²) Solution (Method 1 — using formula with e²): e'² = e²/(1 − e²) = 0.00669438/(1 − 0.00669438) = 0.00669438/0.99330562 = 0.00673966 Verification (Method 2 — using axes): a² = (6,378,137)² = 4.06805 × 10¹³ m² b² = (6,356,752.31)² = 4.04083 × 10¹³ m² a² − b² = 2.72228 × 10¹¹ m² e'² = 2.72228 × 10¹¹ / 4.04083 × 10¹³ = 0.006740 ✓ Answer: e'² = 0.006740 (WGS84).

Question Type

numerical

Answer Structure

• Write the correct formula for e'² [1 mark]
• Correctly substitute values [1 mark]
• Compute correct final value e'² = 0.006740 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using a² in the denominator instead of b² — confusing e² and e'² definitions
• Reporting e'² = 0.006694 — this is e², not e'²; note that e'² > e²
• Not simplifying to the shortcut formula e'² = e²/(1−e²), leading to rounding errors from large number arithmetic

Key Phrases To Include

• e'² = (a² − b²)/b²
• e'² = e²/(1 − e²)
• second eccentricity
• e'² = 0.006740

Marks

5

Topic

Radii of Curvature — Philippine Latitude Context

Difficulty

hard

Template Id

T12

Examiner Tip

This type of 5-mark problem is common on PRC Geodesy boards. Allocate 8–10 minutes. The Manila latitude of ~14°N is a recognized Philippine geodetic context — knowing this geographic reference demonstrates professional awareness. Always close with the inequality N > M as a self-check statement.

Model Answer

Given: WGS84: a = 6,378,137 m; e² = 0.00669438; φ = 14°N. Required: N, M, and R = √(MN). Formulas: N = a / √(1 − e²sin²φ) M = a(1 − e²) / (1 − e²sin²φ)^(3/2) R = √(MN) Solution: Step 1: Compute sin²φ sin 14° = 0.241922; sin²14° = 0.058526 Step 2: Compute 1 − e²sin²φ 1 − (0.00669438)(0.058526) = 1 − 0.000391958 = 0.999608042 Step 3: Compute N √0.999608042 = 0.999803997 N = 6,378,137 / 0.999803997 = 6,378,388.2 m Step 4: Compute M Numerator: a(1 − e²) = 6,378,137(0.99330562) = 6,335,439.3 m Denominator: (0.999608042)^(3/2) = 0.999608042 × 0.999803997 = 0.999412137 M = 6,335,439.3 / 0.999412137 = 6,339,161.4 m Step 5: Compute R = √(MN) MN = 6,378,388.2 × 6,339,161.4 = 4.04244 × 10¹³ m² R = √(4.04244 × 10¹³) = 6,358,764 m Answer: N = 6,378,388 m M = 6,339,161 m R = √(MN) = 6,358,764 m at φ = 14°N (WGS84). Note: N > M as expected (prime-vertical radius always exceeds meridian radius at any latitude).

Question Type

numerical

Answer Structure

• State Given data (WGS84 parameters + φ = 14°) and list all three Required quantities [sets up partial credit]
• Write all three formulas (N, M, R) [1 mark]
• Correctly compute sin²14° and the denominator factor 1−e²sin²φ [1 mark]
• Correctly compute N = 6,378,388 m [1 mark]
• Correctly compute M = 6,339,161 m (requires 3/2 exponent) [1 mark]
• Correctly compute R = √(MN) = 6,358,764 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using φ = 14° in radians in the calculator — always verify degree mode
• Forgetting to apply the (3/2) exponent in the M formula — this error cascades to the R calculation
• Computing R as (M+N)/2 instead of √(MN) — arithmetic mean is not the Gaussian mean
• Not checking that N > M at the end — this is a mandatory sanity check
• Using mean Earth radius 6,371,000 m for a instead of WGS84 a = 6,378,137 m

Key Phrases To Include

• N = a / √(1 − e²sin²φ)
• M = a(1 − e²) / (1 − e²sin²φ)^(3/2)
• R = √(MN)
• Gaussian mean radius
• sin²14° = 0.058526
• N = 6,378,388 m
• M = 6,339,161 m
• N > M

Marks

3

Topic

Geoid vs. Ellipsoid

Difficulty

medium

Template Id

T13

Examiner Tip

Mentioning PRS92 (Philippine Reference System 1992) and GRS80 in answers about Philippine geodetic reference frames earns bonus technical credibility. The PRC expects geodetic engineers to know national standards.

Model Answer

The geoid is the equipotential surface of the Earth's gravity field that coincides with mean sea level over the oceans and its hypothetical extension beneath the continents. It is an irregular, undulating surface that conforms to the actual distribution of Earth's mass. The reference ellipsoid, by contrast, is a smooth mathematical surface defined by two geometric parameters (semi-major axis a and flattening f) that best approximates the geoid globally. In the Philippines, the Geodetic Reference System 1980 (GRS80) is the basis for PRS92, adopted for PPCS mapping. The vertical separation between the geoid and the ellipsoid at any point is called the geoid undulation N (geoid height). In the Philippines, geoid undulations range approximately from +10 m to +40 m relative to WGS84.

Question Type

short_answer

Answer Structure

• Define the geoid — equipotential surface, mean sea level, irregular shape [1 mark]
• Define the reference ellipsoid — smooth mathematical surface, defined by a and f [1 mark]
• State the relationship: geoid undulation N separates them; Philippine context [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing geoid height N (geoid undulation) with prime-vertical radius N — use context to distinguish, or use the symbol N_geoid
• Describing the geoid as perfectly smooth — the geoid is irregular and undulating
• Stating that the Philippines uses WGS84 as the national datum without mentioning PRS92 — PRC exams test knowledge of the national reference system

Key Phrases To Include

• equipotential surface
• gravity field
• mean sea level
• irregular undulating surface
• smooth mathematical surface
• semi-major axis a and flattening f
• geoid undulation
• PRS92
• GRS80

Marks

5

Topic

WGS84 Parameter Derivation

Difficulty

hard

Template Id

T14

Examiner Tip

The instruction '1/f = 298.257223563' means the inverse flattening is given. Your first step must always be f = 1/(298.257...) not f = 298.257. This is PRC Board's most exploited trap question. Write 'f = 1/298.257223563 = 0.003353' boldly and clearly.

Model Answer

Given: a = 6,378,137 m; 1/f = 298.257223563. Required: b; e² (derived); comparison with e² = 0.00669438. Step 1: Compute f from 1/f f = 1/298.257223563 = 0.003352810664 Step 2: Compute b b = a(1 − f) = 6,378,137 × (1 − 0.003352810664) b = 6,378,137 × 0.996647189 b = 6,356,752.31 m Step 3: Compute e² using e² = 2f − f² e² = 2(0.003352810664) − (0.003352810664)² e² = 0.006705621328 − 0.000011241334 e² = 0.006694379994 e² ≈ 0.006694380 Step 4: Comparison Adopted WGS84 e² = 0.00669438 Derived e² = 0.00669438 ✓ (agreement to 8 significant figures) Answer: b = 6,356,752.31 m e² (derived) = 0.006694380 — in exact agreement with the adopted WGS84 value of e² = 0.00669438, confirming consistency of the ellipsoid parameters.

Question Type

numerical

Answer Structure

• Correctly compute f = 1/298.257223563 = 0.003352811 [1 mark]
• Correctly compute b = a(1−f) = 6,356,752.31 m [1 mark]
• Apply correct formula e² = 2f − f² [1 mark]
• Correctly compute e² = 0.006694380 [1 mark]
• Explicit comparison with adopted WGS84 e² = 0.00669438 and conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing f incorrectly (treating 1/f as f directly, i.e., writing f = 298.257) — this is the most catastrophic error, losing all subsequent marks
• Rounding f to fewer than 6 decimal places before computing b — causes significant error in b
• Not performing the comparison step — the question explicitly asks for comparison, so omitting it costs the 5th mark
• Expressing b in km rather than m

Key Phrases To Include

• f = 1/298.257223563
• b = a(1 − f)
• b = 6,356,752.31 m
• e² = 2f − f²
• e² = 0.006694380
• agreement with adopted WGS84 value

Marks

2

Topic

Reference Ellipsoid in Philippine Survey Practice

Difficulty

medium

Template Id

T15

Examiner Tip

PRC board questions frequently blend technical geodesy with Philippine law. RA 8560 governs the profession; PD 1529 governs land registration; CA 141 governs public lands; RA 4374 and its successor RA 8560 define GE practice. Weave legal references naturally into your answer for maximum marks.

Model Answer

The reference ellipsoid provides the precise mathematical surface on which the geographic coordinates (latitude and longitude) of cadastral boundaries are computed and expressed. Under PD 1529, land titles require unambiguous location using a national coordinate reference system. In the Philippines, PRS92 (based on GRS80/WGS84) is the adopted reference, ensuring that coordinates used in the PPCS/UTM projection for Torrens-system cadastral plans are consistent, reproducible, and legally defensible across different surveys and time periods.

Question Type

short_answer

Answer Structure

• Explain the mathematical role of the ellipsoid as the surface for coordinate computation [1 mark]
• Connect to Philippine legal/institutional context: PD 1529, PRS92, PPCS/UTM, Torrens system [1 mark]

Scoring Breakdown

Common Mark Deductions

• Answering in general geodetic terms without any Philippine legal reference (PD 1529, RA 8560, PRS92) — misses the context mark
• Stating that the ellipsoid is used 'for mapping' without explaining the coordinate computation role
• Confusing PD 1529 (property registration) with CA 141 (Public Land Act) — know which law governs which aspect

Key Phrases To Include

• mathematical reference surface
• geographic coordinates
• PRS92
• GRS80/WGS84
• PD 1529
• PPCS/UTM
• Torrens system
• legally defensible