GELE Geodesy — Geodetic and Cartesian CoordinatesExam Answer Templates

Marks

1

Topic

Prime-Vertical Radius of Curvature

Difficulty

easy

Template Id

T1

Examiner Tip

At 1-mark level, one correct formula or one precise definition is enough. Do not over-explain — keep it to 2–3 lines maximum.

Model Answer

The prime-vertical radius of curvature N is the radius of curvature of the ellipsoid in the plane perpendicular to the meridian at a given point. Its formula is: N = a / √(1 − e² sin²φ) where a is the semi-major axis and e² is the first eccentricity squared.

Question Type

very_short_answer

Answer Structure

• Line 1: State what N represents physically [0.5 mark]
• Line 2: Write the correct formula with all symbols defined [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing N = a(1 − e²) — that is the semi-latus rectum, not the prime-vertical radius.
• Placing sin²φ outside the square root.
• Confusing N with the radius of curvature in the meridian (M).

Key Phrases To Include

• prime-vertical radius of curvature
• N = a / √(1 − e² sin²φ)
• semi-major axis
• first eccentricity squared

Marks

1

Topic

Forward Conversion Formulas

Difficulty

easy

Template Id

T2

Examiner Tip

Memorise the Z equation with its (1 − e²) factor — it is the most tested equation at the 1-mark level because it is the most commonly written incorrectly.

Model Answer

The forward conversion equations are: X = (N + h) cos φ cos λ Y = (N + h) cos φ sin λ Z = [N(1 − e²) + h] sin φ where N is the prime-vertical radius of curvature, φ is geodetic latitude, λ is geodetic longitude, and h is ellipsoidal height.

Question Type

very_short_answer

Answer Structure

• Line 1–3: Write all three equations correctly [1 mark — all three must be correct for the full mark]

Scoring Breakdown

Common Mark Deductions

• Writing Z = (N + h) sin φ — omitting (1 − e²) converts the ellipsoid to a sphere.
• Swapping sin and cos assignments for X and Y.
• Using orthometric height H instead of ellipsoidal height h.

Key Phrases To Include

• (N + h) cos φ cos λ
• (N + h) cos φ sin λ
• [N(1 − e²) + h] sin φ
• ellipsoidal height h

Marks

2

Topic

Z-Component and Ellipsoidal Flattening

Difficulty

medium

Template Id

T3

Examiner Tip

This is a conceptual 2-mark question. Earn the first mark with a factual statement and the second with a geometric or quantitative elaboration. Two clear sentences are enough — do not write a paragraph.

Model Answer

The factor (1 − e²) in the Z-component accounts for the ellipsoidal flattening of the Earth. On a sphere, the distance from the axis to the surface at any latitude would be simply N sin φ; however, the Earth is an oblate spheroid — it is compressed at the poles. The quantity N(1 − e²) equals the radius of curvature in the meridian plane, reduced to reflect the shorter polar semi-axis b relative to the equatorial semi-axis a. Without this factor, the Z formula would treat the reference surface as a perfect sphere, introducing a systematic error of several kilometres in the Z-coordinate near the mid-latitudes.

Question Type

short_answer

Answer Structure

• Sentence 1: State what (1 − e²) corrects for — ellipsoidal flattening [1 mark]
• Sentence 2: Explain the geometric consequence — error without the factor or link to the polar semi-axis b [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying (1 − e²) is just a 'correction factor' without explaining what physical property it corrects.
• Not mentioning that omission leads to a spherical approximation.
• Confusing e² (first eccentricity squared) with e'² (second eccentricity squared).

Key Phrases To Include

• ellipsoidal flattening
• oblate spheroid
• (1 − e²)
• polar semi-axis b
• systematic error
• sphere vs ellipsoid

Marks

2

Topic

Inverse Conversion — Longitude from Cartesian

Difficulty

easy

Template Id

T4

Examiner Tip

Always explicitly state the signs of X and Y and name the resulting quadrant. This one-line check is exactly what earns the second mark and is exactly what most students skip.

Model Answer

Step 1 — Compute the raw arctan: λ_raw = arctan(Y / X) = arctan(5,305,814.4 / −3,188,054.9) = arctan(−1.6641) = −59.00° Step 2 — Resolve the quadrant: X < 0 and Y > 0 → the point is in the second quadrant (90° < λ < 180°). λ = 180° + (−59°) = 121°E ∴ λ = 121°E

Question Type

numerical

Answer Structure

• Step 1: Compute arctan(Y/X) and state the raw result [1 mark]
• Step 2: Identify the quadrant from signs of X and Y; apply the quadrant correction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reporting λ = −59° without quadrant correction — loses the second mark.
• Using arctan(X/Y) instead of arctan(Y/X).
• Not stating the directional suffix (°E or °W) in the final answer.

Key Phrases To Include

• arctan(Y/X)
• quadrant resolution
• X < 0, Y > 0
• second quadrant
• 180° + (−59°) = 121°E

Marks

2

Topic

Prime-Vertical Radius of Curvature N

Difficulty

easy

Template Id

T5

Examiner Tip

Write out the intermediate value of sin²φ explicitly — it shows the examiner your method is correct even if you make a subsequent arithmetic slip.

Model Answer

Given: a = 6,378,137 m; e² = 0.00669438; φ = 14° Step 1 — Compute sin²φ: sin 14° = 0.24192; sin²14° = 0.058526 Step 2 — Compute the denominator: 1 − e²sin²φ = 1 − (0.00669438)(0.058526) = 1 − 0.000391935 = 0.999608065 √(0.999608065) = 0.999804028 Step 3 — Compute N: N = 6,378,137 / 0.999804028 = 6,379,386.8 m ∴ N = 6,379,386.8 m

Question Type

numerical

Answer Structure

• Step 1: Compute sin²φ [0.5 mark]
• Step 2: Evaluate the denominator √(1 − e²sin²φ) [0.5 mark]
• Step 3: Divide a by the denominator to obtain N [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using sin 14° = 0.2419 but forgetting to square it before multiplication by e².
• Dividing by (1 − e²sin²φ) without taking the square root.
• Using incorrect WGS84 constants (e.g., using a = 6,371,000 m, which is the mean spherical radius).

Key Phrases To Include

• N = a / √(1 − e² sin²φ)
• sin²14° = 0.058526
• e²sin²φ = 0.000392
• N = 6,379,387 m

Marks

3

Topic

Ellipsoidal Height vs. Orthometric Height

Difficulty

medium

Template Id

T6

Examiner Tip

For 3-mark conceptual questions, examiners look for definition → distinction → application. One mark per component. A Philippine-specific example (PRS92, flood mapping, PD 1529) almost always earns the third mark.

Model Answer

Ellipsoidal height h is the vertical distance from a point on or above the Earth's surface to the surface of the reference ellipsoid (WGS84 or PRS92/GRS80), measured along the normal to the ellipsoid. It is purely geometric — a product of GNSS positioning. Orthometric height H is the vertical distance from a point to the geoid (the equipotential surface approximating mean sea level), measured along the curved plumb line. It is the physically meaningful height used in engineering and levelling. The relationship is: h = H + N_geoid, where N_geoid is the geoid undulation (geoid height). In Philippine geodetic surveys under PRS92, GNSS receivers output ellipsoidal heights. To obtain usable orthometric heights for engineering projects — which are referenced to mean sea level — surveyors must apply the geoid undulation from the Philippine geoid model (PHGEOID). Failure to apply this correction introduces height errors of 10–50 m in the Philippine archipelago, which can critically affect flood-risk mapping, infrastructure design, and property boundary determination under PD 1529 and CA 141.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define ellipsoidal height h — geometric, GNSS-derived, measured from ellipsoid surface along the normal [1 mark]
• Sentence 3–4: Define orthometric height H — physical, geoid-referenced, used in engineering [1 mark]
• Sentence 5–6: State the relationship h = H + N_geoid and explain the practical implication for PRS92 surveys [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining h as 'height above sea level' — that is H, not h.
• Not mentioning the geoid undulation N_geoid as the link between h and H.
• Failing to connect to any Philippine context (PRS92, PHGEOID, or Philippine laws).

Key Phrases To Include

• ellipsoidal height
• orthometric height
• geoid undulation N_geoid
• h = H + N_geoid
• PRS92
• GNSS
• geoid
• mean sea level

Marks

3

Topic

Inverse Conversion — Iterative Method

Difficulty

medium

Template Id

T7

Examiner Tip

Present the iteration as a numbered list — it makes the sequence crystal clear and examiners can award marks step by step. Always mention convergence tolerance to show you understand the algorithm, not just the formulas.

Model Answer

Given geocentric Cartesian coordinates (X, Y, Z), the iterative inverse conversion proceeds as follows: Step 1 — Compute the horizontal distance p: p = √(X² + Y²) Step 2 — Compute longitude λ directly (no iteration needed): λ = arctan(Y / X), resolved for the correct quadrant using signs of X and Y. Step 3 — Compute initial latitude estimate φ₀: φ₀ = arctan[Z / p(1 − e²)] Step 4 — Iterate until convergence: (a) Compute N(φ) = a / √(1 − e² sin²φ) (b) Compute h = p / cos φ − N (c) Update φ = arctan{(Z/p) · [1 − e²N / (N + h)]⁻¹} (d) Repeat steps (a)–(c) until |φ_new − φ_old| < 0.0001" (or equivalent tolerance). Note: Bowring's closed-form formula avoids iteration and is preferred when computational efficiency is needed.

Question Type

short_answer

Answer Structure

• Step 1: Formula for p = √(X² + Y²) [0.5 mark]
• Step 2: Formula for λ with quadrant note [0.5 mark]
• Step 3: Initial latitude estimate φ₀ = arctan[Z / p(1 − e²)] [1 mark]
• Step 4: Iteration loop — N(φ), h, updated φ, convergence criterion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Starting the iteration without first computing p.
• Using φ₀ = arctan(Z / p) instead of arctan[Z / p(1 − e²)] — the (1 − e²) in the initial estimate is critical.
• Not mentioning a convergence criterion or stopping condition.
• Skipping the h update formula within the iteration loop.

Key Phrases To Include

• p = √(X² + Y²)
• λ = arctan(Y/X)
• φ₀ = arctan[Z / p(1 − e²)]
• N(φ) = a / √(1 − e² sin²φ)
• h = p / cos φ − N
• convergence
• Bowring

Marks

3

Topic

Inverse Conversion — Ellipsoidal Height Check

Difficulty

easy

Template Id

T8

Examiner Tip

In verification questions, the expected answer is already implied. Show every arithmetic step clearly — the marks are in the method, not just the final number.

Model Answer

Given: p = 6,189,940.3 m φ = 14°N → cos 14° = 0.970296 N = 6,379,386.8 m Formula for ellipsoidal height from horizontal component: h = p / cos φ − N Substitution: h = 6,189,940.3 / 0.970296 − 6,379,386.8 h = 6,379,436.8 − 6,379,386.8 ∴ h = 50.0 m This confirms the ellipsoidal height of the point is 50 m above the WGS84 ellipsoid, consistent with the forward-conversion input.

Question Type

numerical

Answer Structure

• Line 1–3: List the given values (p, φ, cos φ, N) [0.5 mark]
• Line 4: State the formula h = p / cos φ − N [1 mark]
• Line 5–6: Substitute and compute the intermediate p / cos φ = 6,379,436.8 m [1 mark]
• Line 7: Subtract N and state final answer h = 50 m [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using cos φ in the denominator but computing cos in degrees with calculator in radian mode.
• Subtracting h from N instead of subtracting N from p/cosφ.
• Not stating the units (m) in the final answer.

Key Phrases To Include

• h = p / cos φ − N
• p / cos 14° = 6,379,436.8 m
• h = 50 m
• WGS84 ellipsoid

Marks

5

Topic

Forward Conversion — Full Worked Problem

Difficulty

hard

Template Id

T9

Examiner Tip

For 5-mark numerical questions, each step earns one mark. Show every step in full — do not combine steps. A computation error in Step 2 (N) that propagates to Steps 3–5 will only cost you one mark if subsequent steps are correctly structured (follow-through marking).

Model Answer

Given: φ = 10°, λ = 123°, h = 100 m, a = 6,378,137 m, e² = 0.00669438 Step 1 — Compute sin φ and cos φ: sin 10° = 0.173648; cos 10° = 0.984808 sin²10° = 0.030154 Step 2 — Compute N: N = a / √(1 − e²sin²φ) = 6,378,137 / √(1 − 0.00669438 × 0.030154) = 6,378,137 / √(1 − 0.000201919) = 6,378,137 / √(0.999798081) = 6,378,137 / 0.999899027 = 6,378,781.8 m Step 3 — Compute X: cos λ = cos 123° = −0.544639 X = (N + h) cos φ cos λ = (6,378,781.8 + 100)(0.984808)(−0.544639) = 6,378,881.8 × 0.984808 × (−0.544639) = −3,419,319.5 m Step 4 — Compute Y: sin λ = sin 123° = 0.838671 Y = (N + h) cos φ sin λ = 6,378,881.8 × 0.984808 × 0.838671 = 5,260,979.3 m Step 5 — Compute Z: N(1 − e²) = 6,378,781.8 × (1 − 0.00669438) = 6,378,781.8 × 0.993306 = 6,336,073.0 m Z = [N(1 − e²) + h] sin φ = (6,336,073.0 + 100)(0.173648) = 6,336,173.0 × 0.173648 = 1,099,910.4 m Final Answers: X = −3,419,319.5 m Y = 5,260,979.3 m Z = 1,099,910.4 m

Question Type

numerical

Answer Structure

• Step 1: Compute trigonometric values (sin φ, cos φ, sin²φ, cos λ, sin λ) [1 mark]
• Step 2: Compute N using the prime-vertical radius formula — show full working [1 mark]
• Step 3: Compute X = (N + h) cos φ cos λ [1 mark]
• Step 4: Compute Y = (N + h) cos φ sin λ [1 mark]
• Step 5: Compute N(1 − e²), then Z = [N(1 − e²) + h] sin φ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing N without showing the denominator step — loses the method mark.
• Using cos 123° = +0.544 (positive) — sign error loses the X mark.
• Writing Z = (N + h) sin φ instead of [N(1 − e²) + h] sin φ — loses the Z mark.
• Mixing radians and degrees in the calculator.
• Not clearly labelling which computed value is X, Y, and Z.

Key Phrases To Include

• N = a / √(1 − e² sin²φ)
• (N + h) cos φ cos λ
• (N + h) cos φ sin λ
• [N(1 − e²) + h] sin φ
• cos 123° is negative
• N(1 − e²)

Marks

2

Topic

Quadrant Resolution for Longitude

Difficulty

medium

Template Id

T10

Examiner Tip

In board exams, a Philippine context (Cebu, Manila) is sometimes given with coordinates that test whether you can correctly identify a non-Philippine quadrant — this tests your understanding of the global Cartesian system, not just local coordinates. Flagging a geographic inconsistency earns extra credit if the rubric allows it.

Model Answer

Step 1 — Compute the raw arctan: λ_raw = arctan(Y / X) = arctan(−4,500,000 / −2,100,000) = arctan(2.14286) = 65.0° Step 2 — Resolve the quadrant: X < 0 and Y < 0 → the point is in the third quadrant (180° < |λ| < 270°, or equivalently, the point is in the Western Hemisphere south of the equator). λ = 180° + 65.0° = 245° In conventional notation: λ = 245° − 360° = −115°, or λ = 115°W Note: Since Cebu City is at approximately 123°E, this result (115°W) indicates either an error in the given Cartesian coordinates or a data transposition issue. A proper Philippine site would have X < 0 and Y > 0 (second quadrant), yielding an easterly longitude. Candidates should flag this inconsistency. ∴ λ = 115°W (245° from the positive X-axis) based on the given data.

Question Type

numerical

Answer Structure

• Step 1: Compute arctan(Y/X) and state the raw result [1 mark]
• Step 2: Identify quadrant from signs of X and Y; apply correction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reporting λ = 65° without any quadrant correction.
• Adding 90° instead of 180° for the third-quadrant correction.
• Not identifying the directional suffix (°E or °W).

Key Phrases To Include

• arctan(Y/X)
• X < 0, Y < 0
• third quadrant
• 180° + 65° = 245°
• 115°W
• quadrant resolution

Marks

3

Topic

Geocentric Cartesian vs. Geodetic Coordinate Systems

Difficulty

medium

Template Id

T11

Examiner Tip

When comparing two systems, use a parallel structure: describe System 1 fully, then System 2, then connect them. This gives the examiner a clear three-part answer matching the three marks.

Model Answer

The geocentric Cartesian coordinate system (ECEF — Earth-Centred Earth-Fixed) uses three mutually perpendicular axes: X (pointing toward the intersection of the equatorial plane and the prime meridian), Y (pointing 90° east in the equatorial plane), and Z (pointing toward the North Pole). Coordinates are expressed in linear units (metres). The reference surface is the centre of mass of the Earth (geocentre). GNSS receivers compute positions directly in this system. The geodetic coordinate system expresses position as geodetic latitude φ (angle between the ellipsoid normal and the equatorial plane), geodetic longitude λ (angle in the equatorial plane from the prime meridian), and ellipsoidal height h (distance from the ellipsoid surface along the normal). The reference surface is the reference ellipsoid — for the Philippines, this is GRS80 as realised by PRS92 (practically identical to WGS84). The two systems are mathematically equivalent and related by the forward/inverse conversion equations involving N (the prime-vertical radius of curvature) and the ellipsoidal parameters a and e². The geodetic system is preferred for mapping, survey computations, and legal land descriptions under PD 1529.

Question Type

short_answer

Answer Structure

• Paragraph 1: Describe the ECEF Cartesian system — axes, units, reference surface (geocentre) [1 mark]
• Paragraph 2: Describe the geodetic system — φ, λ, h definitions, reference surface (ellipsoid), Philippine reference (PRS92/GRS80) [1 mark]
• Paragraph 3: State the mathematical relationship and practical context [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining φ as the angle from the centre of the Earth (that is geocentric latitude, not geodetic).
• Not mentioning that the Z-axis points to the North Pole.
• Not distinguishing between the geocentre (Cartesian origin) and the ellipsoid surface (geodetic reference).

Key Phrases To Include

• ECEF
• geocentre
• ellipsoid normal
• geodetic latitude φ
• ellipsoidal height h
• PRS92
• GRS80
• WGS84
• N (prime-vertical radius)
• forward/inverse conversion

Marks

3

Topic

WGS84, GRS80, and PRS92 Reference Systems

Difficulty

medium

Template Id

T12

Examiner Tip

Questions combining technical parameters with Philippine laws (RA 8560, PD 1529) are very common in PRC board exams. Always memorise the key law numbers and their subject matter alongside the technical content.

Model Answer

WGS84 (World Geodetic System 1984) is the global geocentric reference system established by the United States Department of Defense and used as the basis for GPS/GNSS positioning worldwide. It defines both a reference ellipsoid and an ECEF coordinate frame. Principal ellipsoidal parameters of WGS84 (and the underlying GRS80 ellipsoid): - Semi-major axis: a = 6,378,137.0 m - Flattening: f = 1/298.257223563 - First eccentricity squared: e² = 0.00669437999014 PRS92 (Philippine Reference System of 1992), established under RA 8560 (amending RA 4374), uses the GRS80 ellipsoid, which is geometrically virtually identical to WGS84 (the semi-major axis is the same; the flattening differs only in the last few decimal places). In practice, WGS84 and PRS92/GRS80 parameters are interchangeable for licensure-exam computations. PRS92 is realised through a network of geodetic control monuments (Luzon Datum control points transformed to GRS80) and is the legally mandated reference datum for all cadastral surveys under PD 1529 and CA 141 in the Philippines.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define WGS84 — global geocentric reference system, GPS basis [1 mark]
• Sentence 3: List the key ellipsoidal parameters a, f (or e²) [1 mark]
• Sentence 4–5: Relate to PRS92 — GRS80 ellipsoid, RA 8560, PD 1529 legal mandate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating a = 6,371,000 m (mean spherical radius, not the WGS84 semi-major axis).
• Confusing PRS92 with the old Luzon Datum of 1911 (Clarke 1866 ellipsoid).
• Not citing any Philippine law related to the adoption of PRS92.

Key Phrases To Include

• WGS84
• GRS80
• a = 6,378,137 m
• f = 1/298.257
• PRS92
• RA 8560
• RA 4374
• PD 1529
• geocentric
• GPS

Marks

2

Topic

Longitude Quadrant Resolution

Difficulty

medium

Template Id

T13

Examiner Tip

Memorise the four-quadrant rule: Q1 (X+,Y+): λ = arctan; Q2 (X−,Y+): λ = 180° + arctan (negative angle); Q3 (X−,Y−): λ = 180° + arctan (positive angle); Q4 (X+,Y−): λ = 360° + arctan (negative angle). This rule works without memorising signed angle corrections.

Model Answer

Step 1 — Compute arctan(Y/X): λ_raw = arctan(Y/X) = arctan(−4,000,000 / −2,500,000) = arctan(1.6000) = 58.0° Step 2 — Resolve the quadrant: Sign check: X = −2,500,000 (negative); Y = −4,000,000 (negative) X < 0 and Y < 0 → Point lies in the third quadrant (southwest of the prime meridian, below the equator in the XY plane). Quadrant correction: λ = 180° + 58.0° = 238.0° Equivalent conventional longitude: 238° − 360° = −122°, i.e., λ = 122°W ∴ λ = 238° (or equivalently 122°W)

Question Type

numerical

Answer Structure

• Step 1: Compute arctan(Y/X) = arctan(1.6) = 58.0° [1 mark]
• Step 2: Sign analysis (X < 0, Y < 0) → third quadrant; add 180° → 238° or 122°W [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reporting λ = 58° without any quadrant correction — a very common error.
• Applying a 90° correction instead of 180° for the third quadrant.
• Using arctan(X/Y) instead of arctan(Y/X).

Key Phrases To Include

• arctan(Y/X)
• X < 0 and Y < 0
• third quadrant
• 180° + 58° = 238°
• 122°W
• quadrant correction

Marks

5

Topic

Forward Conversion with PPCS/UTM Zone Identification

Difficulty

hard

Template Id

T14

Examiner Tip

Five-mark questions in PRC board exams frequently combine a numerical computation with a Philippine-law or mapping-system question. Prepare for this hybrid format — the last mark is often the easiest (a definition or zone identification) but is lost by candidates who run out of time.

Model Answer

Given: φ = 8°, λ = 125°, h = 200 m, a = 6,378,137 m, e² = 0.00669438 Part A — Compute N: sin 8° = 0.139173; sin²8° = 0.019369 1 − e²sin²φ = 1 − (0.00669438)(0.019369) = 1 − 0.000129634 = 0.999870366 √(0.999870366) = 0.999935177 N = 6,378,137 / 0.999935177 = 6,378,551.4 m Part B — Compute X: cos 8° = 0.990268; cos 125° = −0.573576 X = (N + h) cos φ cos λ = (6,378,551.4 + 200)(0.990268)(−0.573576) = 6,378,751.4 × 0.990268 × (−0.573576) = −3,620,564.5 m Part C — Compute Y: sin 125° = 0.819152 Y = (N + h) cos φ sin λ = 6,378,751.4 × 0.990268 × 0.819152 = 5,172,067.8 m Part D — Compute Z: N(1 − e²) = 6,378,551.4 × 0.993306 = 6,335,847.6 m Z = [N(1 − e²) + h] sin φ = (6,335,847.6 + 200)(0.139173) = 6,336,047.6 × 0.139173 = 881,756.2 m Final Answers: X = −3,620,564.5 m Y = 5,172,067.8 m Z = 881,756.2 m Part E — UTM Zone Identification: UTM zone number = ⌊(λ + 180°) / 6°⌋ + 1 = ⌊(125 + 180) / 6⌋ + 1 = ⌊305/6⌋ + 1 = 50 + 1 = 51 λ = 125°E falls in UTM Zone 51N (covering 120°E–126°E, Northern Hemisphere). Under the Philippine Plane Coordinate System (PPCS), this corresponds to Zone IV.

Question Type

numerical

Answer Structure

• Part A: Compute N ≈ 6,378,551 m — show full denominator working [1 mark]
• Part B: Compute X = −3,620,565 m with correct negative sign [1 mark]
• Part C: Compute Y = 5,172,068 m [1 mark]
• Part D: Compute Z = 881,756 m with explicit (1 − e²) factor [1 mark]
• Part E: Identify UTM Zone 51N / PPCS Zone IV [1 mark]

Scoring Breakdown

Common Mark Deductions

• Positive sign for X — cos 125° is negative (second quadrant in standard trigonometry).
• Omitting (1 − e²) in Z formula.
• Using the wrong UTM zone formula or not converting to PPCS zone nomenclature.
• Not showing the N computation — loses the first mark even if X, Y, Z are correct.

Key Phrases To Include

• N = a / √(1 − e² sin²φ)
• cos 125° is negative
• [N(1 − e²) + h] sin φ
• UTM Zone 51N
• PPCS Zone IV
• RA 8560

Marks

2

Topic

Common Errors in Coordinate Conversion

Difficulty

medium

Template Id

T15

Examiner Tip

For 'describe errors' questions, always pair each error with its consequence. One mark is for naming/identifying the error; the partial credit within that mark is for the consequence. Two-sentence answers — one per error — work perfectly here.

Model Answer

Error 1 — Omitting (1 − e²) in the Z formula: If Z is computed as (N + h) sin φ instead of [N(1 − e²) + h] sin φ, the Z-coordinate is overestimated by approximately N × e² × sin φ. For WGS84 at mid-latitudes (φ ≈ 14°), this error is approximately 6,379,387 × 0.006694 × 0.2419 ≈ 10,330 m — more than 10 km. This effectively treats the Earth as a sphere rather than an oblate ellipsoid. Error 2 — Quadrant ambiguity in longitude: When computing λ = arctan(Y/X) for the inverse conversion, the standard arctan function returns values only in the range (−90°, +90°). If X < 0 (second or third quadrant), the longitude is wrong by ±180° unless the quadrant is resolved using the signs of both X and Y. In the forward conversion, this error manifests as using the wrong sign for cos λ or sin λ, causing X or Y to have an incorrect sign and the computed distance from the origin to differ from the true value.

Question Type

short_answer

Answer Structure

• Error 1: Name the error (omit (1 − e²) in Z), explain the numerical consequence [1 mark]
• Error 2: Name the error (longitude quadrant ambiguity), explain the consequence [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing the same error twice in different words.
• Describing an error without explaining its numerical or practical consequence.
• Vague answers like 'wrong formula' without specifying which formula or which part is wrong.

Key Phrases To Include

• (1 − e²) in Z
• oblate ellipsoid vs sphere
• 10 km error
• arctan quadrant ambiguity
• signs of X and Y
• 180° error