Figure of the Earth and the Reference Ellipsoid - Exam Answer Templates
Proper answer writing is the single most controllable factor in your PRC board exam score. Many examinees lose marks not because they lack knowledge, but because they write incomplete, unstructured, or imprecise answers. These templates show you exactly how a perfect answer looks for each mark level — the precise vocabulary, the correct formula citations, the step-by-step numerical layout, and the key phrases that evaluators reward. Study these templates, internalize the structure, and replicate the format under exam conditions. For Geodesy problems on the Figure of the Earth, a systematic presentation (Given → Required → Formula → Solution → Answer) is non-negotiable for full marks.
Answer templates
Define geodesy. [1 mark]
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.
Answer structure
- Single sentence: define the science with its two core objectives (size/shape + point location) [1 mark]
Scoring breakdown
Criteria
Correct, complete one-sentence definition identifying geodesy as a science dealing with Earth's size, shape, and point location.
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
What geometric shape is used to model the Earth in geodetic computations? [1 mark]
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.
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
Criteria
Correctly identifies the shape as an oblate ellipsoid of revolution or oblate spheroid.
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
State the two defining parameters of a reference ellipsoid. [1 mark]
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).
Answer structure
- Name both parameters with their symbols and brief descriptions [1 mark]
Scoring breakdown
Criteria
Correctly names both defining parameters: semi-major axis a AND flattening f (or semi-minor axis b as alternative).
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
Write the formula for the flattening f of a reference ellipsoid in terms of the semi-major axis a and semi-minor axis b. [1 mark]
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.
Answer structure
- Write the formula with correct symbols [1 mark]
- Define symbols (improves clarity but not required for the mark)
Scoring breakdown
Criteria
Correct formula f = (a − b) / a with appropriate symbols.
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
Differentiate between the prime-vertical radius of curvature N and the meridian radius of curvature M of a reference ellipsoid. [2 marks]
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.
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
Criteria
Correct definition and formula for N (prime vertical, E–W direction, N = a/√(1−e²sin²φ)).
Criteria
Correct definition and formula for M (meridian, N–S direction, M = a(1−e²)/(1−e²sin²φ)^(3/2)) and the statement that N ≥ M.
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
State the WGS84 ellipsoid parameters: semi-major axis a, inverse flattening 1/f, and first eccentricity squared e². [2 marks]
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.
Answer structure
- State a and 1/f correctly [1 mark]
- State e² correctly (or derive it from f) [1 mark]
Scoring breakdown
Criteria
Correct values of a = 6,378,137 m and 1/f = 298.257223563.
Criteria
Correct value of e² = 0.00669438 (accepted range: 0.006694 to 0.006695).
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
The Clarke 1866 ellipsoid has a semi-major axis a = 6,378,206.4 m and semi-minor axis b = 6,356,583.8 m. Compute the flattening f and the first eccentricity squared e². [2 marks]
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.
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
Criteria
Correct computation of f = 0.003390 (or 1/295.0) using f = (a−b)/a.
Criteria
Correct computation of e² = 0.006769 using e² = 1−(b/a)² or equivalent.
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
Compute the prime-vertical radius of curvature N at geodetic latitude φ = 45°N using the WGS84 ellipsoid. [3 marks]
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).
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
Criteria
Correctly writes the formula N = a / √(1 − e²sin²φ) with e² identified.
Criteria
Correctly evaluates the denominator: 1 − e²sin²45° = 0.996653.
Criteria
Correct final answer N = 6,388,838 m (±5 m acceptable rounding).
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
Compute the meridian radius of curvature M at geodetic latitude φ = 30°N using the WGS84 ellipsoid (a = 6,378,137 m; e² = 0.00669438). [3 marks]
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).
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
Criteria
Correct formula M = a(1−e²)/(1−e²sin²φ)^(3/2) stated explicitly.
Criteria
Correct numerator a(1−e²) = 6,335,439 m and correct denominator base 0.998326.
Criteria
Correct final answer M = 6,351,378 m (±10 m acceptable).
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
Derive the relationship e² = 2f − f² starting from the definitions of flattening f and first eccentricity squared e². [3 marks]
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 ∎
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
Criteria
Correctly starts from e² = 1 − (b/a)² and states b = a(1−f).
Criteria
Correctly substitutes to get e² = 1 − (1−f)².
Criteria
Correctly expands and simplifies to e² = 2f − f².
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
Compute the second eccentricity squared e'² for the WGS84 ellipsoid, given a = 6,378,137 m, b = 6,356,752.31 m, and e² = 0.00669438. [3 marks]
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).
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
Criteria
Correct formula e'² = (a²−b²)/b² or e'² = e²/(1−e²).
Criteria
Correct substitution of WGS84 values.
Criteria
Correct final answer e'² = 0.006740 (±0.000005 acceptable).
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
At a point P on the WGS84 ellipsoid at geodetic latitude φ = 14°N (approximately the latitude of Manila), compute the prime-vertical radius N, meridian radius M, and the Gaussian mean radius of curvature R = √(MN). [5 marks]
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).
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
Criteria
All three formulas correctly written: N, M, and R = √(MN).
Criteria
Correct computation of sin²14° = 0.058526 and 1−e²sin²φ = 0.999608.
Criteria
Correct N = 6,378,388 m (±5 m).
Criteria
Correct M = 6,339,161 m (±10 m), requiring correct 3/2 exponent.
Criteria
Correct Gaussian mean radius R = √(MN) = 6,358,764 m (±10 m).
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
Explain the concept of the geoid and distinguish it from the reference ellipsoid. [3 marks]
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.
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
Criteria
Correct definition of the geoid as an equipotential (gravity) surface coinciding with mean sea level.
Criteria
Correct definition of the reference ellipsoid as a smooth mathematical (geometric) surface defined by a and f.
Criteria
Correct identification of geoid undulation as the vertical separation, with any relevant Philippine context (PRS92/GRS80).
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
From the WGS84 ellipsoid parameters (a = 6,378,137 m; 1/f = 298.257223563), derive the semi-minor axis b and verify using the formula e² = 2f − f². Compare your e² with the adopted WGS84 value of e² = 0.00669438. [5 marks]
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.
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
Criteria
Correctly computes f = 1/298.257223563 = 0.0033528 (at least 6 significant figures).
Criteria
Correctly computes b = a(1−f) = 6,356,752.31 m.
Criteria
Correctly states the formula e² = 2f − f² and applies it.
Criteria
Correctly derives e² = 0.006694380.
Criteria
Explicit comparison showing agreement with adopted WGS84 e² = 0.00669438 and a concluding statement.
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
In the context of Philippine geodetic surveys under RA 8560 (RA of Geodetic Engineers), briefly explain why the reference ellipsoid is essential for cadastral surveys governed by PD 1529 (Property Registration Decree). [2 marks]
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.
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
Criteria
Correctly identifies the ellipsoid as the mathematical reference surface for computing geographic coordinates.
Criteria
Correctly connects to Philippine geodetic law/standards: PD 1529 (land registration), PRS92, PPCS/UTM, or cadastral survey requirements.
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
Mark wise strategy
Recall and state. One-mark questions test pure recall. Write the definition, formula, or name precisely using correct technical terminology. Do not over-explain.
Do
- Write the exact technical term (e.g., 'oblate ellipsoid of revolution' not just 'ellipsoid')
- Include the symbol and unit where the question asks for a parameter (e.g., 'a = 6,378,137 m')
- Use one clean, complete sentence
- Memorize WGS84 constants: a, 1/f, e²
Don't
- Do not write more than 2–3 lines for a 1-mark answer — wastes time without gaining marks
- Do not use vague terms like 'round shape' or 'Earth measurement'
- Do not confuse similar terms (e, e², e'²; N vs. M; f vs. 1/f)
Define + Distinguish or Formula + Substitution. Two-mark questions typically test two related concepts or a simple one-step computation. Structure your answer in two visibly distinct parts.
Do
- Use bullet points or line breaks to separate the two marks visually
- For comparison questions, use a table or parallel sentences ('N is... M is...')
- For numerical questions, write Given–Required–Formula even if computation is simple
- State WGS84 parameters explicitly even if they are given in the question
Don't
- Do not merge two concepts into one run-on sentence — the examiner needs to identify two separate credited points
- Do not skip formula citation for numerical questions
- Do not round intermediate values in a 2-step computation
Three-mark answers require three distinct credited elements. For derivations: state → substitute → simplify. For numerical: formula → computation → answer. For explanations: define → explain → apply.
Do
- Number your steps (Step 1, Step 2, Step 3) for full numerical solutions
- For derivations, write each algebraic transformation on a new line
- Include a concluding statement or final boxed answer
- For curvature questions, always state which direction (N–S or E–W) the radius applies to
Don't
- Do not skip the (3/2) exponent in M formula — this is the most costly single error in 3-mark curvature questions
- Do not use approximate WGS84 constants — use full precision values
- Do not present a wall of text — use line breaks and labels
Five-mark questions are mini-essays or multi-step problems. Every sub-task earns one mark. Maximize partial-credit capture by showing every intermediate result even if you are unsure of the final answer. Include a labeled sketch for curvature or ellipsoid problems.
Do
- Write GRFSA heading: Given / Required / Formula / Solution / Answer
- Show every intermediate computation on its own labeled line
- Draw a simple ellipse with a, b, N, M labeled if the question involves curvature
- End with a summary box listing all final answers
- Include a sanity check (e.g., 'N > M as expected; R lies between M and N')
- Reference Philippine context (latitude, PRS92, PPCS) where applicable
Don't
- Do not skip directly to the final answer — showing work is mandatory for partial credit
- Do not omit units on any intermediate or final answer
- Do not use the mean Earth radius (6,371,000 m) instead of WGS84 a (6,378,137 m)
- Do not compute (base)^(3/2) as (base)^3/2 without careful bracket — use (base)^1 × √(base)
General answer writing tips
- Always begin concept questions with a one-sentence definition using precise geodetic terminology (e.g., 'oblate ellipsoid of revolution', 'semi-major axis', 'first eccentricity squared').
- For numerical problems, always use the Given–Required–Formula–Solution–Answer (GRFSA) format; examiners award partial credit for each correctly identified step.
- State the reference ellipsoid explicitly when computing radii of curvature — WGS84 values (a = 6,378,137 m, 1/f = 298.257223563, e² = 0.00669438) must be written out for full marks.
- Never write the eccentricity formula using e alone when the problem requires e²; confusing e and e² is one of the top reasons for mark deduction in Geodesy boards.
- When distinguishing between N (prime vertical) and M (meridian radius), always state which direction each applies to (E–W for N, N–S for M) — this earns the identification mark.
- For 5-mark and long-answer questions, include a labeled sketch of the ellipse showing a, b, N, M, and the geodetic latitude φ — a diagram earns the 'illustration' mark even if computations have minor errors.
- Express final answers to at least 3 significant figures for radii (in metres) and at least 6 significant figures for dimensionless parameters like f and e².
- Check unit consistency throughout; all axes and radii are in metres for WGS84 and Clarke 1866 problems in Philippine geodetic practice.
Diagrams
Figure of the Earth — Key Concepts Mind Map
Hierarchical mind map showing all major topics in the chapter: defining parameters (a, b, f), derived quantities (e², e'²), radii of curvature (N, M, R), reference surfaces (geoid vs. ellipsoid), and Philippine geodetic context (PRS92, PPCS/UTM, legal framework). Use this as a pre-exam overview checklist.
Step-by-Step Algorithm for Computing N and M at a Given Latitude
Decision flowchart showing the systematic algorithm for computing prime-vertical radius N, meridian radius M, and Gaussian mean radius R. The shared intermediate step W² = 1 − e²sin²φ is computed once and reused. The final sanity check (N > M always) is built into the algorithm. Use this flowchart to structure 3-mark and 5-mark numerical answers.
Ellipsoid Parameter Derivation Chain: From a and f to All Derived Quantities
Logical flow diagram showing how starting from the two defining parameters a and f, all derived quantities (b, e², e'², N, M, R) are computed in sequence. Arrows show dependencies — e² is the central derived quantity that feeds into N, M, and e'². This diagram makes it clear why knowing e² precisely (WGS84: 0.00669438) is critical to all curvature computations.
Historical Development of Reference Ellipsoids Used in the Philippines
Timeline showing the progression of reference ellipsoids used in the Philippines from Clarke 1866 (old cadastral surveys) to GRS80/WGS84 (modern GNSS-based surveys) and the adoption of PRS92 in 1992 as the current national datum. Understanding this history is essential for PRC board questions that reference 'old' versus 'new' survey datums and the legal implications under PD 1529 and RA 8560.