GELE Geodesy — Geodetic and Cartesian CoordinatesRevision Notes
Revision notes for GELE Geodesy — Geodetic and Cartesian Coordinates. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests.
Exam context
On the GELE 2026, the Geodesy subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Geodetic and Cartesian Coordinates lands at position 3rd out of 6 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Geodesy on a typical GELE paper.
Geodetic and Cartesian Coordinates - Revision Notes
GNSS receivers compute positions in the Earth-Centred, Earth-Fixed (ECEF) geocentric Cartesian system (X, Y, Z). Field surveys, cadastral maps, and the Philippine Reference System of 1992 (PRS92) use geodetic coordinates (φ, λ, h). Every practising geodetic engineer must convert fluently between these two systems. This chapter covers the forward transformation (geodetic → Cartesian) and the inverse transformation (Cartesian → geodetic), both built on the reference ellipsoid parameters of WGS84 and PRS92. Mastery of these conversions is a recurring theme in the PRC Geodetic Engineer Licensure Examination.
Sections
Formulas
Example
WGS84: e² = 2(1/298.257223563)⁻¹ step — easier to use tabulated value 0.00669437999014.
Formula
e² = (a² − b²) / a² = 2f − f²
Variables
a = semi-major axis; b = semi-minor axis; f = flattening
Application
Compute first eccentricity squared from ellipsoid constants.
Example
At φ = 14°N (Metro Manila area): N = 6 378 137 / √(1 − 0.00669438 × sin²14°) = 6 378 137 / √(0.999607747) = 6 379 386.8 m
Formula
N(φ) = a / √(1 − e² sin²φ)
Variables
N = prime-vertical radius of curvature; a = 6 378 137 m (WGS84); e² = 0.00669438; φ = geodetic latitude
Application
Central intermediate value in both forward and inverse coordinate conversion. N varies from a at the equator to a(1−e²)^(−1/2) ≈ 6 399 594 m at the poles.
Exam Tips
- Memorise WGS84 constants: a = 6 378 137 m, e² = 0.00669438 (6 significant figures).
- PRS92 = GRS80 ellipsoid — a is identical to WGS84; only the 7th decimal of e² differs.
- Board exam questions typically provide a and e² — always double-check the given values before computing N.
Key Points
- The reference ellipsoid is defined by two primary constants: semi-major axis (a) and flattening (f), from which all other parameters are derived.
- WGS84 (used by GPS/GNSS worldwide): a = 6 378 137.000 m, 1/f = 298.257223563, first eccentricity squared e² = 0.00669437999014.
- PRS92 (official Philippine datum, defined under RA 8560 and administered by NAMRIA): uses the GRS80 ellipsoid — a = 6 378 137.000 m, 1/f = 298.257222101, e² = 0.00669438002290. The difference from WGS84 is negligible for most engineering work.
- Second eccentricity squared: e'² = e²/(1 − e²).
- The prime-vertical radius of curvature N(φ) is the key intermediate quantity linking the two coordinate systems.
- Ellipsoidal height h is measured along the ellipsoid normal — it is NOT the same as orthometric height H (measured along the plumb line). H = h − N_geoid, where N_geoid is the geoid undulation.
Definitions
Term
Geodetic Latitude (φ)
Definition
The angle between the ellipsoid normal through a point and the equatorial plane. Ranges from −90° (South Pole) to +90° (North Pole).
Importance
Primary angular coordinate. Differs from geocentric latitude except at the equator and poles.
Term
Geodetic Longitude (λ)
Definition
The angle in the equatorial plane between the Greenwich meridian plane and the meridian plane of the point. Ranges from 0° to 360° (or ±180°).
Importance
Philippine longitudes range approximately 116°E to 127°E (PPCS/UTM Zones 50N–52N).
Term
Ellipsoidal Height (h)
Definition
The distance from the ellipsoid surface to the point, measured along the ellipsoid normal (positive upward).
Importance
Directly obtained from GNSS. Must be converted to orthometric height for engineering elevations using geoid model (e.g., EGM2008 or NAMRIA's local geoid).
Term
Prime-Vertical Radius of Curvature (N)
Definition
Radius of curvature of the ellipsoid cross-section perpendicular to the meridian at geodetic latitude φ.
Importance
Appears in all three coordinate conversion equations. Must be re-computed at each iteration in the inverse problem.
Term
ECEF Cartesian Coordinates (X, Y, Z)
Definition
Right-handed geocentric system: X-axis through the Greenwich meridian and equatorial plane; Y-axis 90°E in the equatorial plane; Z-axis toward the North Pole.
Importance
Native output of GNSS receivers and the backbone of global geodetic computations.
Section Title
1. Reference Ellipsoid Parameters
Common Mistakes
- Confusing WGS84 and GRS80/PRS92 ellipsoid parameters — they are numerically very close but not identical.
- Using geocentric latitude instead of geodetic latitude in the conversion formulas.
- Treating ellipsoidal height as orthometric height — these can differ by up to ±90 m in the Philippines.
Formulas
Example
φ=14°N, λ=121°E, h=50 m, N=6 379 386.8 m: X = (6 379 386.8+50)cos14°cos121° = 6 379 436.8 × 0.970296 × (−0.515038) = −3 188 054.9 m
Formula
X = (N + h) cos φ cos λ
Variables
N = prime-vertical radius; h = ellipsoidal height; φ = geodetic latitude; λ = geodetic longitude
Application
Gives the Cartesian X-coordinate (positive toward Greenwich meridian/equator intersection).
Example
Same point: Y = 6 379 436.8 × 0.970296 × 0.857167 = 5 305 814.4 m
Formula
Y = (N + h) cos φ sin λ
Variables
N = prime-vertical radius; h = ellipsoidal height; φ = geodetic latitude; λ = geodetic longitude
Application
Gives the Cartesian Y-coordinate (positive toward 90°E meridian/equator intersection).
Example
N(1−e²) = 6 379 386.8 × (1 − 0.00669438) = 6 336 680.8 m; Z = (6 336 680.8 + 50)sin14° = 6 336 730.8 × 0.241922 = 1 532 993.9 m
Formula
Z = [N(1 − e²) + h] sin φ
Variables
N = prime-vertical radius; e² = first eccentricity squared; h = ellipsoidal height; φ = geodetic latitude
Application
Gives the Cartesian Z-coordinate (positive toward North Pole). The (1−e²) factor reduces the contribution of N to account for ellipsoidal flattening.
Exam Tips
- Always compute N first, then substitute into all three equations.
- Check signs: for the Philippines, X is negative (λ between 90° and 270°), Y is positive (λ between 0° and 180°), Z is positive (northern hemisphere).
- A quick sanity check: √(X²+Y²+Z²) should be close to the mean Earth radius (~6 371 km) ± a few km.
Key Points
- The forward conversion is direct (non-iterative) — a single computation yields exact X, Y, Z.
- Three equations use N(φ), the trigonometric functions of φ and λ, and the ellipsoidal height h.
- The Z-equation carries the factor (1 − e²) multiplied into N — this is the critical term that accounts for the ellipsoid's flattening. Omitting it gives a sphere.
- For a point on the ellipsoid surface (h = 0): X = N cos φ cos λ; Y = N cos φ sin λ; Z = N(1−e²) sin φ.
- The Philippine archipelago lies in the eastern hemisphere and northern hemisphere: λ ≈ 116°–127°E and φ ≈ 5°–21°N. Expect Y positive and X negative (2nd quadrant in XY-plane) for points with λ > 90°E.
Definitions
Term
Forward Conversion
Definition
The direct (non-iterative) computation of ECEF Cartesian coordinates (X, Y, Z) from geodetic coordinates (φ, λ, h) using the three standard equations.
Importance
Used whenever GNSS geodetic output must be transformed to a 3D Cartesian framework for datum transformation or network adjustment.
Section Title
2. Forward Conversion: Geodetic (φ, λ, h) → Cartesian (X, Y, Z)
Common Mistakes
- CRITICAL: Dropping (1−e²) from the Z equation — this is the most common board-exam pitfall. Z = (N+h)sinφ is the spherical formula, not the ellipsoidal one.
- Using degrees directly in the trigonometric functions without converting to radians when using a scientific calculator set to radians mode.
- Forgetting that cos 121° is negative — Philippine longitudes (>90°) give negative X values.
- Using the wrong quadrant for sin and cos of the latitude/longitude — always check signs before calculating.
- Rounding N prematurely, which introduces centimetre-level errors in X, Y, Z.
Formulas
Example
X = −3 188 054.9 m, Y = 5 305 814.4 m: arctan(5 305 814.4 / −3 188 054.9) = arctan(−1.6641) = −59°. Since X<0, Y>0 (2nd quadrant): λ = 180° − 59° = 121°E ✓
Formula
λ = atan2(Y, X)
Variables
Y = Cartesian Y-coordinate; X = Cartesian X-coordinate; atan2 = four-quadrant arctangent
Application
Direct computation of longitude. Use atan2 (four-quadrant inverse tangent) to resolve the correct quadrant automatically. If using standard arctan: if X<0,Y>0 → λ = 180° + arctan(Y/X); if X<0,Y<0 → λ = −180° + arctan(Y/X).
Example
X=−3 188 054.9, Y=5 305 814.4: p = √(3 188 054.9² + 5 305 814.4²) = 6 189 940.3 m
Formula
p = √(X² + Y²)
Variables
p = equatorial distance (distance from Z-axis); X, Y = Cartesian coordinates
Application
Intermediate quantity used in both the iterative and height recovery formulas.
Example
Z = 1 532 993.9, p = 6 189 940.3, e² = 0.00669438: φ₀ = arctan[1 532 993.9 / (6 189 940.3 × 0.993306)] = arctan(0.24924) = 13.988° ≈ 14°
Formula
φ₀ = arctan[ Z / (p(1 − e²)) ] (initial estimate)
Variables
Z = Cartesian Z-coordinate; p = equatorial distance; e² = first eccentricity squared
Application
First estimate of geodetic latitude to start the iteration. This is slightly less accurate than Bowring's seed but sufficient as a starting point.
Example
p = 6 189 940.3 m, φ = 14°, N = 6 379 386.8 m: h = 6 189 940.3/cos14° − 6 379 386.8 = 6 189 940.3/0.970296 − 6 379 386.8 = 6 379 436.8 − 6 379 386.8 = 50 m ✓
Formula
h = p / cos φ − N(φ) (valid when |φ| < ~75°)
Variables
p = equatorial distance; φ = converged geodetic latitude; N = prime-vertical radius at φ
Application
Recovery of ellipsoidal height once φ has converged. Use h = Z/sinφ − N(1−e²) in polar regions where cosφ ≈ 0.
Example
After first iteration with φ₀ ≈ 14°, N₀ = 6 379 386.8, h₀ = 50 m: correction term e²N/(N+h) = 0.00669438×6 379 386.8/6 379 436.8 = 0.006694325. φ₁ = arctan[1 532 993.9/6 189 940.3 × 1/(1−0.006694325)] = 14.000° → converged.
Formula
φ_{i+1} = arctan[ Z / p × (1 − e²N/(N+h))⁻¹ ]
Variables
Z, p = Cartesian intermediates; e², N, h = current iteration values
Application
Iterative update for geodetic latitude. Repeat until |φ_{i+1} − φᵢ| < 0.000001° (sub-millimetre convergence criterion).
Exam Tips
- For Philippine coordinates (λ ≈ 116°–127°E): X is always negative, Y is always positive. If your answer gives X>0 for a Philippine point, you have an error.
- The board exam typically asks for either λ (direct) or one iteration for φ — practise both starting from p and Z.
- If the problem gives X, Y, Z and asks for φ, λ, h: always compute λ first (direct), then iterate for φ and h.
- Quadrant rule summary: X>0,Y>0 → Q1 (0°–90°E); X<0,Y>0 → Q2 (90°–180°E); X<0,Y<0 → Q3 (180°–270°E or 90°–180°W); X>0,Y<0 → Q4 (270°–360°E or 0°–90°W).
Key Points
- Longitude λ is computed directly and unambiguously (after quadrant resolution) from X and Y.
- Latitude φ and ellipsoidal height h are coupled through N(φ) — they must be solved iteratively (or using a closed-form approximation such as Bowring's formula).
- The iterative process converges rapidly — typically 3–4 iterations give millimetre accuracy.
- The auxiliary quantity p = √(X²+Y²) is the distance from the Z-axis (equatorial distance) and is the starting point for the iteration.
- Bowring's closed-form formula is preferred in practice because it avoids iteration while achieving sub-millimetre accuracy.
- The board exam typically tests one iteration of the inverse process — know how to set up the first estimate of φ.
Definitions
Term
atan2(Y, X)
Definition
The four-quadrant inverse tangent function that returns the angle in the correct quadrant based on the individual signs of both Y and X. Returns values in (−180°, +180°].
Importance
Essential for unambiguous longitude recovery. Standard arctan only returns values in (−90°, +90°) and loses quadrant information.
Term
Bowring's Formula
Definition
A closed-form (non-iterative) approximation for the inverse conversion by Bowring (1985) that achieves sub-millimetre accuracy in a single computation step.
Importance
The preferred method in practice and in many geodetic software libraries. The board exam may reference it as an alternative to iteration.
Term
Equatorial Distance (p)
Definition
The perpendicular distance from the point to the Earth's rotation axis (Z-axis), computed as √(X²+Y²).
Importance
Key intermediate in the inverse conversion; relates directly to the cos φ term in the forward equations.
Section Title
3. Inverse Conversion: Cartesian (X, Y, Z) → Geodetic (φ, λ, h)
Common Mistakes
- CRITICAL: Resolving longitude quadrant incorrectly — bare arctan(Y/X) is ambiguous. Always use atan2 or manually check the signs of X and Y.
- Starting the iteration with φ₀ = arctan(Z/p) instead of arctan[Z/(p(1−e²))] — using the non-reduced denominator gives a spherical approximation, not an ellipsoidal one.
- Using the height formula h = p/cosφ near the poles (φ > 75°) where cosφ → 0. Use h = Z/sinφ − N(1−e²) instead.
- Stopping after only one iteration without checking convergence — for precise work, iterate until Δφ < 10⁻⁸ radians.
- Forgetting that N must be recomputed at each new φᵢ — using the initial N throughout introduces errors.
Formulas
Example
X = (6 379 433.7+100)cos10°cos123° = 6 379 533.7×0.984808×(−0.544639) = −3 419 490.2 m; Y = 6 379 533.7×0.984808×0.838671 = 5 263 244.9 m; Z = [6 379 433.7×0.993306+100]sin10° = [6 336 722.9+100]×0.173648 = 1 100 085.8 m
Formula
PROBLEM 1 SOLUTION: N = 6 378 137 / √(1 − 0.00669438 sin²10°) = 6 378 137 / √(0.999797635) = 6 379 433.7 m
Variables
sin10° = 0.173648; sin²10° = 0.030154; e²sin²10° = 0.000201828
Application
Step 1 of forward conversion for φ=10°, λ=123°, h=100 m.
Example
Final answer: λ = 238°E or equivalently 122°W. This point is in the Pacific Ocean west of South America — not a Philippine point.
Formula
PROBLEM 2 SOLUTION: λ = arctan(Y/X) + 180°×sign correction
Variables
X = −2 500 000 m; Y = −4 000 000 m; both negative → 3rd quadrant
Application
arctan(−4 000 000 / −2 500 000) = arctan(1.6) = 58.0°. Both X<0 and Y<0 → 3rd quadrant → λ = 180° + 58° = 238°E = 122°W.
Example
PRC exam context: A GNSS survey gives h = 75.320 m. The local geoid undulation is N_geoid = 22.5 m. Orthometric height H = 75.320 − 22.5 = 52.820 m. This is the elevation to use for engineering design.
Formula
PROBLEM 4 — Geoid relationship: H = h − N_geoid
Variables
H = orthometric height (from levelling); h = ellipsoidal height (from GNSS); N_geoid = geoid undulation
Application
In the Philippines, N_geoid ranges from approximately +10 m to +30 m (NAMRIA/EGM2008). A GNSS-derived h of 80 m may correspond to an orthometric elevation H ≈ 55 m.
Exam Tips
- When a board problem gives Cartesian coordinates with both X and Y negative, the longitude is in the 3rd quadrant (between 180° and 270°E, or 90°W and 180°W) — definitely NOT a Philippine point.
- If the problem asks to 'state why GNSS height differs from the elevation shown on topographic maps,' the answer is: GNSS gives ellipsoidal height h; topo maps show orthometric height H; they differ by the geoid undulation N_geoid.
- Philippine applications under RA 8560 (NAMRIA mandate) use PRS92/GRS80, but WGS84 values are acceptable since the difference is sub-centimetre.
Key Points
- Problem 1 (Forward): φ=10°N, λ=123°E, h=100 m on WGS84 — complete numerical solution.
- Problem 2 (Quadrant): X=−2 500 000, Y=−4 000 000 — longitude in 3rd quadrant.
- Problem 3 (Inverse, height): recovering h from p, φ, N.
- Problem 4 (Conceptual): why ellipsoidal height ≠ orthometric height — geoid undulation N_geoid.
Definitions
Term
Orthometric Height (H)
Definition
The height of a point above the geoid (mean sea level), measured along the plumb line. Used in engineering surveys, topographic maps, and flood elevation determinations in the Philippines.
Importance
Legally required for cadastral surveys (PD 1529) and public land surveys (CA 141). GNSS gives h; the engineer must convert to H.
Term
Geoid Undulation (N_geoid)
Definition
The vertical distance between the ellipsoid surface and the geoid surface at a given location. Can be positive or negative.
Importance
Links ellipsoidal heights (GNSS) to orthometric heights (levelling). NAMRIA provides the Philippine geoid model.
Section Title
4. Worked Board-Exam Problems
Common Mistakes
- In Problem 2-type questions: adding 180° to get the 3rd-quadrant longitude without checking the result — 238°E is the same as 122°W, verify it makes geographic sense.
- Confusing the geoid undulation symbol N (N_geoid) with the prime-vertical radius of curvature N(φ) — these are completely different quantities despite sharing the same letter. Context always clarifies which N is meant.
- Applying the geoid correction with the wrong sign: H = h − N_geoid (not h + N_geoid).
Formulas
Example
φ=14°N, λ=121°E, h=50m → X=−3188054.9m, Y=5305814.4m, Z=1532993.9m (see Section 2).
Formula
Complete Forward System: X=(N+h)cosφcosλ; Y=(N+h)cosφsinλ; Z=[N(1−e²)+h]sinφ
Variables
N=a/√(1−e²sin²φ); a=6378137m; e²=0.00669438 (WGS84)
Application
Master formula set — memorise this exactly for the board exam.
Example
From X=−3188054.9, Y=5305814.4, Z=1532993.9 → λ=121°E, φ=14°N, h=50m.
Formula
Complete Inverse System: λ=atan2(Y,X); p=√(X²+Y²); iterate φ and h
Variables
Initial: φ₀=arctan[Z/(p(1−e²))]; then h=p/cosφ−N; then update φ
Application
Starting set for inverse conversion on the board exam.
Exam Tips
- Allocate time wisely: the forward conversion is straightforward arithmetic — aim to complete it in under 5 minutes with a scientific calculator.
- For the inverse, if the board exam asks only for λ, compute it directly from atan2 and move on — do not iterate unnecessarily.
- Write down N(φ) first and box it — every subsequent calculation uses it.
- The PRC board exam frequently tests the conceptual distinction between ellipsoidal and orthometric height. Prepare a concise written answer: 'GNSS measures h (ellipsoidal); engineering surveys use H (orthometric); H = h − N_geoid where N_geoid is obtained from a geoid model such as EGM2008.'
Key Points
- Forward conversion: 3 equations, all use N(φ). Non-iterative. Remember (1−e²) in Z.
- Inverse conversion: λ direct via atan2; φ and h iterative. Start with p = √(X²+Y²).
- N(φ) = a/√(1−e²sin²φ) — compute this first in every problem.
- Quadrant of λ: determined by signs of X (cos λ component) and Y (sin λ component).
- Height hierarchy: h (ellipsoidal, GNSS) ≠ H (orthometric, levelling). H = h − N_geoid.
- Philippine context: GNSS and PPCS/UTM use WGS84/GRS80; PRS92 is the official national geodetic datum (RA 8560/NAMRIA).
- PD 1529 (Property Registration Decree) and CA 141 (Public Land Act) require accurate cadastral coordinates — coordinate conversion is the foundation of compliant surveys.
Definitions
Term
PRS92 (Philippine Reference System of 1992)
Definition
The official geodetic datum of the Philippines based on the GRS80 ellipsoid. Adopted under RA 8560 and administered by NAMRIA. Used in all official maps, cadastral surveys, and the PPCS/UTM coordinate system.
Importance
All licensed geodetic engineers in the Philippines must reference surveys to PRS92 as required by RA 8560 and implementing rules.
Term
PPCS/UTM (Philippine Plane Coordinate System / Universal Transverse Mercator)
Definition
The plane coordinate system used for large-scale mapping in the Philippines. Based on the Transverse Mercator projection of the PRS92 ellipsoid, divided into 5 zones (Zones I–V) covering the archipelago.
Importance
Used in all cadastral, topographic, and engineering surveys. Conversion from geodetic (φ, λ) to PPCS/UTM (Northing, Easting) requires knowing the reference meridian and false origin of each zone.
Section Title
5. Summary of Key Formulas and Exam Strategy
Common Mistakes
- Attempting the inverse problem by assuming φ = arctan(Z/p) — this is the GEOCENTRIC latitude, not the GEODETIC latitude.
- Solving for h before φ has fully converged — always complete the φ iteration first.
- Using a and b directly in the forward equations instead of the derived N(φ).
Connections
- Datum Transformation (Helmert/Bursa-Wolf): ECEF Cartesian coordinates (X,Y,Z) are the natural input/output of 3D similarity transformations between geodetic datums (e.g., from old LUZON Datum to PRS92). The 7-parameter transformation operates directly on X,Y,Z vectors.
- PPCS/UTM Projection: Converting geodetic (φ,λ) to plane coordinates (Northing N, Easting E) in the Philippine Plane Coordinate System. The geodetic-to-Cartesian conversion is a prerequisite step when transforming from GNSS (WGS84 X,Y,Z) all the way to PPCS grid coordinates.
- Geoid and Vertical Datum: The separation h (ellipsoidal) vs H (orthometric) connects coordinate conversion to the geoid chapter. The Philippine Vertical Datum is defined by mean sea level at Intramuros tide gauge; GNSS heights require geoid correction to match levelled elevations.
- Network Adjustment (Least Squares): 3D geodetic network adjustment is performed in ECEF Cartesian coordinates. Forward and inverse conversions are applied to bring observed angles/distances into the Cartesian framework and to recover geodetic results after adjustment.
- GNSS Processing: Every GNSS receiver internally computes X,Y,Z from pseudorange/phase observations, then applies the inverse conversion to output φ,λ,h. Understanding the conversion is fundamental to evaluating GNSS positioning accuracy.
- Cadastral Surveys under PD 1529 and CA 141: Cadastral lots require precise plane coordinates (PPCS/UTM). The chain: GNSS gives X,Y,Z → inverse conversion → φ,λ,h → UTM projection → N,E (cadastral plan coordinates). All steps must be consistent with PRS92.
- Meridian Convergence and Grid Bearing: When converting between geodetic azimuth and PPCS/UTM grid bearing, the meridian convergence γ depends on geodetic longitude λ — directly linked to the λ computed in the inverse conversion.
- Ellipsoid of Revolution Geometry: N(φ) and M(φ) (meridian radius of curvature) are derived from the same ellipsoid parameters. This chapter's N formula reappears in the arc-to-chord correction and in the normal section azimuth formulas.
Exam Strategy
For PRC Geodetic Engineer board exam problems on geodetic and Cartesian coordinates: (1) IDENTIFY the conversion direction — forward (geodetic→Cartesian) or inverse (Cartesian→geodetic). (2) WRITE DOWN the relevant formula set immediately. (3) COMPUTE N(φ) first — it is used in multiple equations, and a rounding error here propagates everywhere. (4) For forward problems: substitute directly into X, Y, Z — watch the (1−e²) in Z and the sign of cosλ for Philippine longitudes. (5) For inverse problems: compute λ first using atan2 with quadrant resolution, then compute p = √(X²+Y²), then iterate for φ and h. (6) VERIFY using the sanity check √(X²+Y²+Z²) ≈ 6 371–6 379 km. (7) For conceptual questions on h vs H: answer in three steps — define h (ellipsoidal, from GNSS), define H (orthometric, from levelling), state H = h − N_geoid. (8) Allocate 5–7 minutes for a complete numerical conversion — practise under timed conditions. (9) For Philippine-context questions, always mention PRS92/GRS80 and NAMRIA's role under RA 8560. (10) Never drop the (1−e²) factor — this single mistake converts an ellipsoidal result to a spherical one and will give a wrong answer.
Quick Review Questions
A GNSS receiver outputs X = −3 188 054.9 m, Y = 5 305 814.4 m, Z = 1 532 993.9 m. What is the geodetic longitude?
Compute arctan(Y/X) = arctan(5 305 814.4 / −3 188 054.9) = arctan(−1.6641) = −59°. Since X < 0 and Y > 0, the point is in the 2nd quadrant: λ = 180° − 59° = 121°E. Always resolve the quadrant from the signs of X and Y, never rely on the bare arctan result.
What is the prime-vertical radius of curvature N at φ = 14°N on WGS84 (a = 6 378 137 m, e² = 0.00669438)?
N = 6 378 137 / √(1 − 0.00669438 × sin²14°) = 6 378 137 / √(1 − 0.00669438 × 0.058526) = 6 378 137 / √(0.999607747) = 6 378 137 / 0.999803878 = 6 379 386.8 m. Note N > a because the denominator is less than 1.
A point has geodetic coordinates φ = 14°N, λ = 121°E, h = 50 m on WGS84. What is the Cartesian Z-coordinate?
First compute N(1−e²) = 6 379 386.8 × (1 − 0.00669438) = 6 379 386.8 × 0.993306 = 6 336 680.8 m. Then Z = [6 336 680.8 + 50] × sin14° = 6 336 730.8 × 0.241922 = 1 532 993.9 m. Critical: do NOT use Z = (N+h)sin φ — that omits the (1−e²) factor and treats the Earth as a sphere.
Why is the Z-equation written as Z = [N(1−e²) + h] sin φ rather than Z = (N + h) sin φ?
On the reference ellipsoid surface (h=0): Z = N(1−e²)sinφ. Note that b²/a = a(1−e²), so N(1−e²) is related to the b-axis geometry. The complete relationship is b² = a²(1−e²), and for a point on the ellipsoid, Z = (b²/a)sinφ / √(1−e²sin²φ) = N(1−e²)sinφ — exactly what the formula gives.
In the iterative inverse conversion, what is the initial estimate of geodetic latitude φ₀ from Cartesian coordinates?
This initial estimate accounts for the ellipsoid's flattening by dividing p by (1−e²) in the denominator, pushing the estimate closer to the true geodetic latitude. Using arctan(Z/p) instead gives the geocentric latitude, which differs from geodetic latitude by up to ~11.5 arcminutes at mid-latitudes. The correction factor (1−e²) ≈ 0.993306 is small but critical for accuracy.
A GNSS survey at a site in Quezon City gives ellipsoidal height h = 82.450 m. The EGM2008 geoid undulation at this location is N_geoid = 26.3 m. What is the orthometric height H?
H = h − N_geoid = 82.450 − 26.3 = 56.150 m. The geoid sits approximately 26.3 m above the WGS84 ellipsoid at this location, meaning a point on the geoid has an ellipsoidal height of +26.3 m. A point 56.150 m above the geoid therefore has h = 56.150 + 26.3 = 82.450 m. This distinction is critical for engineering design elevations required under PD 1529 and cadastral regulations.
Given p = 6 189 940.3 m, φ = 14°, N = 6 379 386.8 m on WGS84, compute the ellipsoidal height h.
h = p/cosφ − N = 6 189 940.3 / cos14° − 6 379 386.8 = 6 189 940.3 / 0.970296 − 6 379 386.8 = 6 379 436.8 − 6 379 386.8 = 50.0 m. This formula is valid for |φ| < approximately 75°. Near the poles, use h = Z/sinφ − N(1−e²) instead to avoid division by near-zero cosφ.
What is the official Philippine geodetic reference datum and what law governs it?
PRS92 is based on the GRS80 ellipsoid (a = 6 378 137 m, 1/f = 298.257222101), which is geometrically nearly identical to WGS84. RA 8560 defines the scope of geodetic engineering practice in the Philippines and mandates the use of PRS92 for official surveys. NAMRIA (National Mapping and Resource Information Authority) is the lead government agency for geodesy and mapping, maintaining control network benchmarks and geoid models.
For a point with X = −2 500 000 m, Y = −4 000 000 m, determine the longitude and identify the quadrant.
arctan(Y/X) = arctan(−4 000 000 / −2 500 000) = arctan(1.6) = +58.0°. Since both X and Y are negative (3rd quadrant), λ = 180° + 58° = 238°E. Note: 238°E = 360° − 238° = 122°W. This is in the Pacific Ocean east of the International Date Line, not a Philippine location. The Philippines has positive Y and negative X in ECEF (2nd quadrant), with λ ≈ 116°–127°E.
What are the WGS84 semi-major axis and first eccentricity squared values that must be memorised for the board exam?
These are the defining constants of WGS84, the global reference frame used by GPS. The semi-major axis a = 6 378 137 m is also the GRS80 value used by PRS92. The flattening f = 1/298.257223563 is the other defining constant; e² is derived as e² = 2f − f². For board exam calculations, use e² = 0.00669438 (6 significant figures) to maintain sufficient accuracy in N(φ).
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