GELE Geodesy — Satellite Geodesy and GNSSExam Answer Templates

Marks

1

Topic

Observables — Pseudorange

Difficulty

easy

Template Id

T1

Examiner Tip

The single mark is awarded for the keyword 'receiver clock bias' combined with the formula. If you include both in one sentence, you secure the mark even if your wording is not perfect.

Model Answer

A pseudorange is the apparent distance from a GNSS receiver to a satellite, computed as ρ = c Δt, where c is the speed of light and Δt is the measured signal travel time. It is called 'pseudo' because it contains the receiver clock bias error.

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition and formula ρ = c Δt [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'true range' instead of emphasizing the clock bias makes the answer incomplete
• Omitting the formula or not defining the variables loses the technical precision mark
• Writing 'pseudo because of satellite clock' is incorrect — the bias is in the receiver clock

Key Phrases To Include

• pseudorange
• ρ = c Δt
• receiver clock bias
• speed of light
• signal travel time

Marks

1

Topic

Positioning Principle

Difficulty

easy

Template Id

T2

Examiner Tip

Board exams often ask this as a one-mark item. The single mark is always tied to naming the clock bias — never leave out that specific phrase.

Model Answer

A minimum of four (4) satellites is required. Three satellites would solve three position unknowns (X, Y, Z), but the receiver clock bias Δt is a fourth unknown, necessitating a fourth pseudorange equation.

Question Type

very_short_answer

Answer Structure

• Line 1: State the number (4) [0.5 mark]
• Line 2: Name the fourth unknown as receiver clock bias Δt [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing '3 satellites' as the answer — this is the most common and fatal error
• Saying '4 satellites because of accuracy' does not earn the mark — the reason must be the clock bias
• Writing 'satellite clock bias' instead of 'receiver clock bias' is technically incorrect

Key Phrases To Include

• four satellites
• receiver clock bias
• fourth unknown
• three coordinates X Y Z

Marks

1

Topic

Errors and DOP

Difficulty

easy

Template Id

T3

Examiner Tip

Remember: DOP and accuracy move in the same direction — both high means poor result, both low means good result. The examiner wants to see you connect DOP to satellite geometry explicitly.

Model Answer

DOP stands for Dilution of Precision. A low DOP value indicates good satellite geometry (well-spread satellites), which results in better (more accurate) positioning.

Question Type

very_short_answer

Answer Structure

• Line 1: Full expansion of acronym DOP [0.5 mark]
• Line 2: Interpretation — low DOP = good geometry = better accuracy [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that high DOP is better — this is reversed and loses the mark
• Expanding DOP incorrectly (e.g., 'Degree of Precision')
• Not relating DOP to geometry — saying only 'low DOP = more accurate' without mentioning geometry is incomplete

Key Phrases To Include

• Dilution of Precision
• satellite geometry
• low DOP
• well-spread satellites
• better accuracy

Marks

2

Topic

Observables — Pseudorange

Difficulty

easy

Template Id

T4

Examiner Tip

Always write c = 299,792,458 m/s explicitly. The examiner awards one mark for the method (formula + substitution) and one mark for the correct answer with units. Show all steps even for simple multiplication.

Model Answer

Given: Δt = 0.072 s, c = 299,792,458 m/s Using the pseudorange formula: ρ = c × Δt ρ = 299,792,458 × 0.072 ρ = 21,585,057 m ρ ≈ 21,585 km ∴ The pseudorange is approximately 21,585 km.

Question Type

numerical

Answer Structure

• Line 1: State given data with correct units [0.5 mark]
• Line 2: Write the formula ρ = c Δt [0.5 mark]
• Line 3: Substitute and compute [0.5 mark]
• Line 4: State the final answer with correct SI unit (m or km) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using c = 3 × 10^8 m/s (approximate) instead of the exact value — may cause minor rounding but generally acceptable if stated
• Reporting the answer without units (metres or km) loses the unit mark
• Not showing the formula step loses the method mark even if the final number is correct

Key Phrases To Include

• ρ = c Δt
• c = 299,792,458 m/s
• 21,585 km
• pseudorange

Marks

2

Topic

Errors and DOP

Difficulty

easy

Template Id

T5

Examiner Tip

The question specifies horizontal accuracy, so use HDOP specifically. Always match the DOP type to the accuracy dimension being computed. The formula must appear explicitly on your paper.

Model Answer

Given: σ = 2.5 m, HDOP = 1.8 Using the accuracy formula: Horizontal Accuracy ≈ HDOP × σ Horizontal Accuracy ≈ 1.8 × 2.5 Horizontal Accuracy ≈ 4.5 m ∴ The estimated horizontal positioning accuracy is 4.5 m.

Question Type

numerical

Answer Structure

• Line 1: Identify given values with labels [0.5 mark]
• Line 2: Write Accuracy ≈ DOP × σ [0.5 mark]
• Line 3: Substitute HDOP and σ correctly [0.5 mark]
• Line 4: State final answer in metres [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing HDOP with PDOP — using the wrong DOP type for horizontal accuracy loses a mark
• Dividing σ by HDOP instead of multiplying — a conceptual error that loses both marks
• Not writing the formula before substituting

Key Phrases To Include

• HDOP
• Accuracy ≈ DOP × σ
• 4.5 m
• horizontal accuracy

Marks

2

Topic

Observables — Pseudorange and Carrier Phase

Difficulty

medium

Template Id

T6

Examiner Tip

For any 'differentiate' question, structure your answer as a clear two-part comparison. The examiner is specifically checking for 'integer ambiguity' as the distinguishing concept of carrier phase — do not omit it.

Model Answer

Pseudorange is measured by correlating the code signal timing, giving a range estimate ρ = c Δt with metre-level accuracy (~1–3 m), but it requires no ambiguity resolution. Carrier-phase measures the phase of the carrier wave (L1 = 19 cm wavelength for GPS), providing millimetre-level accuracy (~1–10 mm). However, it includes an unknown integer number of whole wavelengths called the integer ambiguity, which must be resolved before precise positioning is possible.

Question Type

short_answer

Answer Structure

• Point 1: Define pseudorange with accuracy level (metre) [1 mark]
• Point 2: Define carrier phase with accuracy level (mm) and mention integer ambiguity [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not mentioning integer ambiguity for carrier phase — this is the key distinguishing feature
• Stating carrier phase has metre accuracy — this is incorrect and shows conceptual confusion
• Omitting accuracy levels entirely

Key Phrases To Include

• code correlation
• pseudorange ρ = c Δt
• metre-level
• carrier phase
• millimetre-level
• integer ambiguity

Marks

2

Topic

Errors and DOP

Difficulty

easy

Template Id

T7

Examiner Tip

Board exams often accept 'any four of five common errors.' Always list the classic four (ionosphere, troposphere, satellite clock, multipath) to be safe. Add a brief parenthetical description to each for maximum credit.

Model Answer

The four major GNSS error sources are: 1. Satellite clock and orbit errors — inaccuracies in the broadcast satellite clock correction and ephemeris. 2. Ionospheric delay — signal retardation by free electrons in the ionosphere (~5–15 m). 3. Tropospheric delay — signal slowing due to atmospheric water vapor and dry gas (~2.5 m zenith). 4. Multipath — reflection of signals off buildings or terrain before reaching the antenna. (Receiver noise is also acceptable as a fifth source.)

Question Type

short_answer

Answer Structure

• Item 1: Satellite clock/orbit errors [0.5 mark]
• Item 2: Ionospheric delay [0.5 mark]
• Item 3: Tropospheric delay [0.5 mark]
• Item 4: Multipath [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Listing only names without any brief description — some marking schemes require at least a phrase of explanation
• Repeating similar errors (e.g., 'ionosphere' and 'atmosphere') as separate items
• Writing 'weather' instead of 'tropospheric delay' — use precise geodetic terminology

Key Phrases To Include

• ionospheric delay
• tropospheric delay
• satellite clock
• orbit errors
• multipath
• receiver noise

Marks

3

Topic

Differential GNSS and RTK

Difficulty

medium

Template Id

T8

Examiner Tip

The 3-mark question tests Definition (1) + Mechanism (1) + Accuracy (1). Each sentence should target exactly one mark. Write 'base on known coordinates' and 'common error cancellation' — these are the two examiner triggers.

Model Answer

Differential GNSS (DGPS) is a positioning technique that uses two receivers simultaneously: a base (reference) station placed on a point of known coordinates, and a rover receiver at the unknown point. Principle: The base station computes the difference between its known position and its GNSS-derived position, generating range corrections. These corrections are transmitted to the rover, which applies them to its own measurements. Because both receivers are in proximity, they experience nearly the same satellite clock, orbit, ionospheric, and tropospheric errors — applying the base corrections cancels these common errors. Achievable accuracy: DGPS (code-based differential) achieves centimetre-to-decimetre accuracy (~0.1–1 m), significantly better than standalone GNSS (~3–5 m).

Question Type

short_answer

Answer Structure

• Sentence 1: Define DGPS and its two components (base + rover) [1 mark]
• Sentence 2: Explain the correction mechanism — base computes corrections, rover applies them [1 mark]
• Sentence 3: State achievable accuracy with value (~0.1–1 m) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing DGPS (code-based, decimetre) with RTK (carrier-phase, centimetre) — they have different accuracy levels
• Not explaining the mechanism of error cancellation — just saying 'it uses two receivers' is insufficient
• Not stating a numerical accuracy value

Key Phrases To Include

• reference (base) station
• known coordinates
• range corrections
• common errors cancelled
• rover
• 0.1–1 m accuracy

Marks

3

Topic

Differential GNSS and RTK

Difficulty

medium

Template Id

T9

Examiner Tip

The keyword 'integer ambiguity resolution' is the single most important phrase in any RTK answer. Never write an RTK answer without it. Examiners will check for it specifically.

Model Answer

RTK (Real-Time Kinematic) GNSS is a carrier-phase differential positioning method in which a base station on a known point transmits carrier-phase observations and corrections to a rover in real time via a radio or cellular data link. Comparison with Static Differential GNSS: - RTK achieves centimetre-level accuracy (~1–3 cm) because it resolves the integer ambiguity of the carrier phase (wavelength ~19 cm for GPS L1). - Static differential GNSS (DGPS) uses code pseudoranges and achieves decimetre-to-metre accuracy (~0.1–1 m). Applications: RTK is used for construction stake-out, topographic surveys, and control densification where real-time cm accuracy is required. Static DGPS suits navigation and GIS data collection where decimetre accuracy suffices.

Question Type

short_answer

Answer Structure

• Sentence 1: Define RTK — carrier-phase differential, real-time, base + rover [1 mark]
• Sentence 2: State RTK accuracy (~1–3 cm) with reason (carrier phase + integer ambiguity resolution) [1 mark]
• Sentence 3: Compare with DGPS accuracy (~0.1–1 m) and give one application each [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating RTK accuracy as 'metre-level' — this is incorrect; RTK is centimetre-level
• Omitting 'integer ambiguity resolution' as the reason for RTK's high accuracy
• Not distinguishing between code-based (DGPS) and phase-based (RTK)

Key Phrases To Include

• carrier-phase differential
• integer ambiguity resolution
• real-time
• 1–3 cm
• base station
• rover
• stake-out

Marks

3

Topic

Errors and DOP

Difficulty

medium

Template Id

T10

Examiner Tip

For 'compute and assess' questions, always complete the numerical part first, then use that result to justify your engineering judgment. State a threshold (PDOP ≤ 3) and compare — this earns the assessment mark.

Model Answer

(a) Position Accuracy: Given: PDOP = 4.0, σ = 3 m Accuracy ≈ PDOP × σ Accuracy ≈ 4.0 × 3.0 = 12 m ∴ Estimated 3-D position accuracy ≈ 12 m. (b) Acceptability: A PDOP of 4.0 is generally considered marginal to poor for precise engineering surveys. For geodetic and engineering control surveys, PDOP should ideally be ≤ 3 (or PDOP ≤ 2 for highest precision). A PDOP of 4.0 produces ~12 m accuracy with this range error, which is unacceptable for stake-out or control densification requiring centimetre accuracy. The surveyor should wait for better satellite geometry (PDOP ≤ 3) or use RTK with carrier-phase processing.

Question Type

numerical

Answer Structure

• Part (a) Step 1: Write formula Accuracy = PDOP × σ [0.5 mark]
• Part (a) Step 2: Substitute and compute = 12 m [0.5 mark]
• Part (b) Step 1: State the acceptable PDOP threshold (≤ 3 or ≤ 2 for high precision) [1 mark]
• Part (b) Step 2: Conclude that PDOP = 4.0 is unacceptable and give practical recommendation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not stating a threshold value for acceptable PDOP — the conclusion must be supported by a number
• Computing accuracy without showing the formula step
• Concluding PDOP = 4.0 is 'acceptable' — this is incorrect for engineering precision work

Key Phrases To Include

• PDOP × σ
• 12 m
• PDOP ≤ 3
• marginal geometry
• wait for better satellite geometry

Marks

3

Topic

Reference Frames

Difficulty

hard

Template Id

T11

Examiner Tip

Any question about reference frames in the Philippine context should mention NAMRIA, PRS92, and at least one relevant law (RA 8560 or PD 1529). This shows legal and professional awareness, which is rewarded in the board exam.

Model Answer

RTK GNSS positions are computed in the WGS84 geocentric reference frame (or more precisely in the ITRF realization used by the satellite system). The WGS84 ellipsoid and its origin, orientation, and scale are defined globally and are not identical to the Philippine Reference System of 1992 (PRS92). PRS92 is the national geodetic datum of the Philippines, defined by a best-fit ellipsoid orientation and scale for the Philippine archipelago. WGS84 and PRS92 differ by datum transformation parameters (translation, rotation, and scale factor) derived by NAMRIA. If RTK coordinates in WGS84 are directly used with PRS92 control monuments without transformation, position discrepancies of several metres can result, rendering surveys non-compliant with RA 8560 (Philippine Geodetic Engineering Act) and inconsistent with existing cadastral records under PD 1529.

Question Type

short_answer

Answer Structure

• Sentence 1: State that RTK outputs coordinates in WGS84/ITRF, not PRS92 [1 mark]
• Sentence 2: Explain that WGS84 and PRS92 differ due to different datum definition (origin, orientation, scale) [1 mark]
• Sentence 3: State the consequence of not transforming — metre-level discrepancy and legal non-compliance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'WGS84 and PRS92 are the same' — they are not; this is a fundamental conceptual error
• Not mentioning a consequence (the reason why the transformation matters)
• Omitting the Philippine legal framework — references to RA 8560 and PD 1529 demonstrate professional awareness expected in the board exam

Key Phrases To Include

• WGS84
• ITRF
• PRS92
• datum transformation
• NAMRIA
• RA 8560
• geocentric frame

Marks

5

Topic

Positioning Principle and Observables

Difficulty

hard

Template Id

T12

Examiner Tip

Five-mark long answers are graded on completeness and structure. Follow the four-part structure: Principle → Justification → Computation → Application. Each part targets one mark. The fifth mark is often an 'integration' mark for showing how the parts connect — here it is the reference frame application.

Model Answer

I. Trilateration Principle in GNSS GNSS determines the position of a receiver by measuring its distance (range) to multiple satellites whose positions are precisely known from broadcast ephemeris data. This is called trilateration — the simultaneous intersection of range spheres. Each satellite i broadcasts its position (Xi, Yi, Zi) and the receiver measures the travel time Δti of the signal. The measured range is: ρi = c × Δti where c = 299,792,458 m/s. The geometric relationship is: ρi = √[(Xi − X)² + (Yi − Y)² + (Zi − Z)²] + c·Δt_clock where (X, Y, Z) are the receiver coordinates and Δt_clock is the receiver clock bias. II. Why Four Satellites are Required A 3-D fix requires solving for four unknowns: • X — receiver X-coordinate • Y — receiver Y-coordinate • Z — receiver Z-coordinate • Δt_clock — receiver clock bias Three unknowns (X, Y, Z) could theoretically be solved with three satellites. However, receiver clocks are imperfect (inexpensive quartz oscillators), introducing a clock bias Δt_clock as a fourth unknown. Therefore, a minimum of four simultaneous pseudorange equations — one per satellite — is required to uniquely solve the system of four equations in four unknowns. III. Pseudorange Computation Given: Δt = 0.064 s, c = 299,792,458 m/s ρ = c × Δt ρ = 299,792,458 × 0.064 ρ = 19,186,717 m ρ ≈ 19,187 km ∴ The pseudorange to the satellite is approximately 19,187 km. This value is consistent with the GPS orbital altitude of ~20,200 km, confirming the order of magnitude. IV. Reference Frame Note In the Philippines, GNSS positions are first computed in the WGS84 geocentric frame. For use with existing control monuments and cadastral records, these must be transformed to PRS92 using NAMRIA-published transformation parameters.

Question Type

long_answer

Answer Structure

• Part I: Define trilateration and write the range equation ρ = √[ΣΔX²] + c·Δt_clock [1 mark]
• Part II: List all four unknowns — X, Y, Z, and receiver clock bias Δt — and explain why each requires one equation [1 mark]
• Part III Step 1: Write ρ = c Δt with given values [1 mark]
• Part III Step 2: Compute ρ = 19,187 km with unit, and sanity-check against orbital altitude [1 mark]
• Part IV: Mention WGS84 output and PRS92 transformation requirement for Philippine surveys [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not showing all four unknowns — listing only X, Y, Z and omitting clock bias loses the key mark
• Computing ρ = 3×10^8 × 0.064 = 19,200 km (minor rounding) — acceptable, but using the exact c value is preferred
• Not writing the general range equation before the pseudorange formula
• Omitting Part IV (WGS84 to PRS92) — this mark is easily earned and often skipped
• Failing to perform a sanity check on the computed range against the known GPS orbital altitude

Key Phrases To Include

• trilateration
• ρ = c Δt
• c = 299,792,458 m/s
• receiver clock bias
• four unknowns
• four equations
• 19,187 km
• WGS84
• PRS92
• NAMRIA

Marks

5

Topic

Errors, DOP, and Differential GNSS

Difficulty

hard

Template Id

T13

Examiner Tip

In a 5-mark discuss question, breadth and structure both matter. Use numbered lists for error sources (saves time and is clear), then switch to prose for the DGPS mechanism explanation. Always end with the Philippine professional context — this is the integrative fifth mark that many reviewees miss.

Model Answer

I. Major GNSS Error Sources GNSS positioning is affected by the following systematic and random errors: 1. Satellite clock and orbit errors — broadcast clock corrections and ephemeris have residual errors (~1–2 m equivalent). 2. Ionospheric delay — the ionosphere retards the code signal and advances the carrier phase, introducing position errors of 5–15 m (single-frequency). 3. Tropospheric delay — the neutral atmosphere causes a zenith delay of ~2.3 m (dry component) and ~0.1–0.5 m (wet/humidity component). 4. Multipath — reflected signals from buildings, terrain, or water mix with the direct signal, biasing the pseudorange (~0.3–3 m). 5. Receiver noise — thermal noise in the tracking loop (~0.1–0.3 m). The combined effect is quantified as the User Equivalent Range Error (UERE), denoted σ. II. How Differential GNSS (DGPS) Reduces Errors DGPS places a base (reference) station on a point of known geodetic coordinates. The base measures the difference between its GNSS-derived position and its surveyed position to generate pseudorange corrections. These corrections are transmitted to the rover in real time. Because both base and rover are within the same atmospheric environment (typically < 50 km baseline), they share nearly identical ionospheric, tropospheric, and satellite clock errors. Applying the base-station corrections at the rover cancels these common-mode errors, reducing the effective UERE from ~3–5 m (standalone) to ~0.1–1 m (DGPS code) or ~0.01–0.03 m (RTK carrier-phase). III. Role of DOP in Accuracy Estimation Dilution of Precision (DOP) quantifies the amplification of ranging errors due to satellite geometry. When satellites are well-spread across the sky, the geometry is strong and DOP is low (good). When satellites are clustered, DOP is high (poor). The achievable accuracy is estimated as: Accuracy ≈ DOP × σ Example: If σ = 2 m (after DGPS corrections) and PDOP = 2.0: Accuracy ≈ 2.0 × 2 = 4.0 m Types of DOP: PDOP (3-D position), HDOP (horizontal), VDOP (vertical), TDOP (time). For engineering surveys, PDOP ≤ 3 and HDOP ≤ 2 are recommended. In Philippine practice, GNSS surveys under RA 8560 require proper observation planning — using software to predict DOP and UERE to ensure compliance with NAMRIA accuracy standards before fieldwork.

Question Type

long_answer

Answer Structure

• Part I: List 4–5 GNSS error sources with brief description of each [1.5 marks]
• Part II: Explain DGPS mechanism — base on known point, corrections transmitted, common errors cancelled [1.5 marks]
• Part III: State Accuracy ≈ DOP × σ, define DOP types, give numerical example, recommend threshold [1 mark]
• Integration: Mention Philippine standards (RA 8560, NAMRIA) [0.5–1 bonus/integration mark depending on rubric]

Scoring Breakdown

Common Mark Deductions

• Listing error sources without explanation — examiner expects a brief impact magnitude for each
• Not explaining the mechanism of error cancellation in DGPS — just saying 'corrections are applied' is insufficient
• Omitting the formula Accuracy ≈ DOP × σ when the question explicitly asks for it
• Not providing a numerical example for Part III
• Omitting Philippine standards and professional context — misses the integration mark

Key Phrases To Include

• ionospheric delay
• tropospheric delay
• multipath
• satellite clock
• UERE
• common-mode error cancellation
• Accuracy ≈ DOP × σ
• PDOP ≤ 3
• RA 8560
• NAMRIA

Marks

1

Topic

Reference Frames

Difficulty

easy

Template Id

T14

Examiner Tip

WGS84 is the GNSS broadcast frame; PRS92 is the Philippine national datum. The examiner will award the mark for WGS84 + geocentric/ECEF. These two concepts are frequently tested as a contrast pair.

Model Answer

GPS satellites broadcast positions in the World Geodetic System 1984 (WGS84), which is a geocentric, Earth-centered, Earth-fixed (ECEF) reference frame maintained by the US National Geospatial-Intelligence Agency (NGA).

Question Type

very_short_answer

Answer Structure

• Line 1: Name the frame — WGS84 [0.5 mark]
• Line 2: Brief characterization — geocentric/ECEF [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'PRS92' as the GNSS reference frame — PRS92 is the Philippine local datum, not the GNSS output frame
• Writing 'ITRF' only — while closely aligned to WGS84, the specific GPS broadcast reference is WGS84
• Not characterizing the frame as geocentric

Key Phrases To Include

• WGS84
• geocentric
• Earth-centered Earth-fixed
• ECEF

Marks

2

Topic

Reference Frames — Philippine Context

Difficulty

medium

Template Id

T15

Examiner Tip

Questions about PRS92 are almost always paired with WGS84. The key relationship is: GNSS → WGS84 → transform using NAMRIA parameters → PRS92. Memorize this workflow and cite PD 1529 for legal context.

Model Answer

PRS92 (Philippine Reference System of 1992) is the official national geodetic datum of the Philippines, adopted by NAMRIA to provide a consistent horizontal reference for all land surveys, cadastral records, and engineering works in the country. It is a local, best-fit reference system defined using ground control established in 1992. GNSS (GPS) provides coordinates in WGS84, a global geocentric datum. Because WGS84 and PRS92 are not identical (they differ in origin, orientation, and scale), GNSS-derived WGS84 coordinates must be transformed to PRS92 using NAMRIA-derived transformation parameters before they can be tied to existing PRS92 control monuments or used in cadastral surveys under PD 1529.

Question Type

short_answer

Answer Structure

• Sentence 1: Define PRS92 as the Philippine national geodetic datum established by NAMRIA [1 mark]
• Sentence 2: Explain that GNSS outputs WGS84 which must be transformed to PRS92 — not the same datum [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that PRS92 and WGS84 are the same or equivalent — they are not
• Not mentioning NAMRIA as the authority for PRS92
• Not explaining that transformation is needed before tying to existing control

Key Phrases To Include

• PRS92
• Philippine Reference System of 1992
• NAMRIA
• national geodetic datum
• WGS84
• transformation parameters
• PD 1529