Detailed ExplanationGELE · Photogrammetry & CartographyReal content

GELE Photogrammetry & Cartography — Remote Sensing and GISDetailed Explanation

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The Remote Sensing and GIS chapter rewards slow, careful thinking over quick pattern matching, especially on Professional Regulation Commission (PRC) — Board of Geodetic Engineering's scenario-based GELE items. This detailed explanation walks through the full derivation of every core idea, then links each one to a worked example pulled from recent GELE Photogrammetry & Cartography papers.

Exam context

For the Geodetic Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests Photogrammetry & Cartography under a "Core" label, with Remote Sensing and GIS in the 6th slot across 6 chapters. GELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Photogrammetry & Cartography questions. Date to watch: September 2026.

Remote Sensing and GIS - Detailed Explanation

Remote sensing and Geographic Information Systems (GIS) are indispensable tools in modern geodetic engineering practice. Remote sensing allows us to collect spatial data about the Earth's surface from a distance — using satellite or airborne platforms — without any physical contact with the terrain. GIS, on the other hand, is the digital framework that organizes, analyzes, and presents spatial and attribute data in a meaningful way. Together, they underpin land administration (PD 1529, CA 141), cadastral surveying (RA 8560), environmental monitoring, disaster risk reduction, and infrastructure planning across the Philippines. For PRC board examinees, mastery of resolution types, sensor classification, electromagnetic spectrum applications, and GIS data models is essential — these topics consistently appear as multiple-choice and computational items in the Photogrammetry and Cartography portion of the licensure examination.

Concepts

Electromagnetic Spectrum and Spectral Signatures

All remote sensing systems detect electromagnetic (EM) radiation — energy that travels as waves through space. The EM spectrum spans a wide range of wavelengths, from very short gamma rays to very long radio waves. In remote sensing, the most practically important regions are: (1) Visible light (0.4–0.7 µm) — the range detected by the human eye and standard optical cameras; (2) Near-Infrared (NIR, 0.7–1.3 µm) — strongly reflected by healthy vegetation, used in NDVI calculations; (3) Short-Wave Infrared (SWIR, 1.3–2.5 µm) — sensitive to soil moisture and mineral composition; (4) Thermal Infrared (TIR, 8–14 µm) — detects heat emitted by the Earth's surface, used for urban heat island studies and volcanic monitoring; (5) Microwave/Radar (1 mm–1 m) — penetrates clouds and vegetation, used by SAR (Synthetic Aperture Radar) systems. Every material on Earth's surface has a unique spectral signature — a distinctive pattern of reflectance or emittance across wavelengths. Healthy green vegetation, for instance, absorbs red and blue visible light for photosynthesis but strongly reflects NIR. Bare soil has a gradually increasing reflectance across the visible and infrared. Water absorbs most radiation beyond the visible, making it appear dark in NIR imagery. By analyzing spectral signatures, analysts can identify and classify land cover types without physical ground visits.

Examples

Healthy vegetation strongly reflects NIR and absorbs red light. In an FCC, healthy crops appear bright red (NIR assigned to the red display channel). NDVI mathematically isolates this contrast, producing a single-band index map. This is a standard board-exam scenario linking spectral bands to practical agricultural monitoring.

Scenario

A geodetic engineer is tasked with monitoring the health of rice fields in Nueva Ecija using satellite imagery. Which spectral band combination is most appropriate?

Solution

Use a False Color Composite (FCC) combining NIR (Band 4), Red (Band 3), and Green (Band 2) — displayed as Red, Green, Blue respectively. Alternatively, compute NDVI = (NIR − Red) / (NIR + Red). Pixels with NDVI > 0.4 indicate healthy, actively growing rice crops.

Passive optical sensors (visible, NIR) are blocked by ash and clouds. TIR detects emitted thermal energy from the hot lava. SAR, being an active microwave system, is unaffected by ash or clouds and can map surface deformation using InSAR techniques.

Scenario

A volcanic eruption at Mt. Mayon produces thick ash clouds. Which portion of the EM spectrum allows the satellite to image the lava flow beneath the ash cloud?

Solution

Thermal Infrared (TIR, ~8–14 µm) or Microwave/SAR. TIR detects the heat emitted by the lava flow even through thin ash. SAR (microwave) penetrates ash clouds entirely, providing surface mapping.

Applications

NDVI mapping for crop health monitoring and agricultural land classification under CA 141 surveys
Thermal band analysis for urban heat island studies in Metro Manila
SAR-based flood mapping during typhoon events (Ondoy, Yolanda disaster response)
Mineral exploration using SWIR bands to detect clay minerals and iron oxides
Coastal water quality assessment using visible and NIR bands

Misconceptions

MISCONCEPTION: Thermal infrared detects reflected sunlight like visible bands. FACT: TIR detects emitted thermal energy from the Earth's surface — it is effective even at night because hot objects emit radiation regardless of solar illumination.
MISCONCEPTION: A higher NDVI always means more vegetation. FACT: NDVI above ~0.8 can be saturated in very dense canopies and may not linearly represent leaf area index beyond a certain threshold.
MISCONCEPTION: All infrared bands are the same. FACT: NIR (0.7–1.3 µm), SWIR (1.3–2.5 µm), and TIR (8–14 µm) have completely different physical bases and applications.
MISCONCEPTION: Radar (microwave) can see through the ground. FACT: Radar penetrates clouds and vegetation but does NOT penetrate the solid ground to significant depths under normal conditions — it maps the surface or top-of-canopy.

Related Concepts

Four types of resolution (spatial, spectral, radiometric, temporal)
Passive vs. active sensors
Image classification and land-cover mapping
NDVI and vegetation indices

Common Exam Questions

Example

Which spectral band is most effective for detecting healthy vegetation? Answer: Near-Infrared (NIR, ~0.7–1.3 µm), because healthy vegetation strongly reflects NIR radiation.

Approach

Memorize which EM band is best suited for each application. The board exam commonly gives a scenario and asks you to identify the appropriate band or sensor type.

Question Type

Identification / Multiple Choice

Example

A pixel has NIR reflectance = 0.65 and Red reflectance = 0.10. Compute NDVI. Solution: NDVI = (0.65 − 0.10)/(0.65 + 0.10) = 0.55/0.75 = 0.733. This indicates dense, healthy vegetation.

Approach

NDVI problems are straightforward: NDVI = (NIR − Red)/(NIR + Red). Ensure you know the formula and can interpret the resulting value.

Question Type

Computation

Key Points To Remember

EM spectrum order (short to long wavelength): gamma → X-ray → UV → Visible → NIR → SWIR → TIR → Microwave → Radio
Visible range: 0.4–0.7 µm (blue 0.4–0.5, green 0.5–0.6, red 0.6–0.7 µm)
NIR (0.7–1.3 µm) is the key band for vegetation health — high NIR reflectance = healthy plants
Thermal IR detects emitted heat, NOT reflected sunlight — useful at night
Microwave (radar) penetrates clouds, smoke, and vegetation canopy
Spectral signature = unique reflectance/emittance curve of a material across wavelengths
NDVI = (NIR − Red) / (NIR + Red); values range from −1 to +1; healthy vegetation > 0.3

Four Types of Resolution in Remote Sensing

Resolution in remote sensing is NOT a single concept — it has four distinct dimensions, each measuring a different capability of the sensor system. Understanding all four is critical for exam success. (1) SPATIAL RESOLUTION: The smallest ground area represented by a single pixel. A sensor with 10 m spatial resolution means each pixel covers a 10 m × 10 m area on the ground. Finer (smaller number) = more detail. Philippine examples: Landsat = 30 m, SPOT = 10–20 m, Worldview = 0.5 m, drone imagery = cm-level. (2) SPECTRAL RESOLUTION: The number and width of spectral bands a sensor records. A panchromatic sensor records one broad band (grayscale). A multispectral sensor records 3–10 discrete bands (e.g., Landsat 8 has 11 bands). A hyperspectral sensor records hundreds of narrow, contiguous bands (e.g., AVIRIS). More bands = richer spectral discrimination. (3) RADIOMETRIC RESOLUTION: The sensor's ability to distinguish differences in energy intensity — expressed as bit depth. An 8-bit sensor records 2^8 = 256 grey levels per band. A 12-bit sensor records 2^12 = 4,096 grey levels. Higher bit depth = finer discrimination of subtle tonal differences. Modern satellites (e.g., Sentinel-2) use 12-bit or 16-bit encoding. (4) TEMPORAL RESOLUTION: The frequency with which a sensor revisits the same location — also called the revisit period. Landsat has a 16-day revisit cycle; Sentinel-2 = 5 days (with two satellites); commercial Planet Labs satellites = daily. High temporal resolution is critical for monitoring rapidly changing phenomena (floods, crop growth, disaster response). Note the fundamental trade-off: improving one resolution often degrades another (e.g., very fine spatial resolution usually means a narrower swath and less frequent revisit).

Examples

This is a standard board-exam computation. Always apply Grey Levels = 2^n. The 10-bit sensor records more than four times the tonal discrimination of an 8-bit (256-level) sensor, allowing detection of more subtle radiance differences — useful in atmospheric correction and land-cover classification.

Scenario

A satellite sensor uses 10-bit radiometric encoding. How many distinct grey levels can it record per band?

Solution

Grey Levels = 2^10 = 1,024 distinct grey levels per band.

IFOV (Instantaneous Field of View) is the angular measure of what a single detector element sees. Multiplying by altitude converts this to a linear ground dimension. This is Landsat-like geometry. The formula is a common exam item.

Scenario

A sensor with an IFOV of 0.0001 radians is mounted on a satellite at an altitude of 705 km. What is its spatial resolution (ground pixel size)?

Solution

Ground pixel size = IFOV × Altitude = 0.0001 rad × 705,000 m = 70.5 m ≈ 70 m.

This tests understanding of temporal resolution as a selection criterion. The 16-day Landsat cycle means flood waters may recede or conditions change before a cloud-free acquisition is obtained.

Scenario

For disaster response mapping after a major typhoon in Bicol, a disaster risk reduction agency needs imagery every 3–5 days. Which Landsat property limits its usefulness here?

Solution

Landsat's temporal resolution (revisit period) is 16 days, which is too infrequent for rapid post-typhoon damage assessment. Sentinel-2 (5-day revisit) or commercial daily satellites (Planet Labs) are more appropriate choices.

Applications

Spatial resolution selection for cadastral vs. regional mapping projects
Radiometric resolution choice for subtle vegetation stress detection
Temporal resolution planning for crop calendar monitoring in the Philippines (wet/dry season)
Spectral resolution for mineral mapping and lithological discrimination in Mindanao mining surveys
Trade-off analysis in sensor selection for national mapping by NAMRIA

Misconceptions

MISCONCEPTION: 'Higher resolution' always means better. FACT: It depends on the application. Fine spatial resolution has smaller swath and longer revisit. For regional mapping, moderate spatial resolution with wide swath is more efficient.
MISCONCEPTION: Spatial resolution is the only important resolution. FACT: All four resolutions are equally important depending on the task — the board exam tests all four equally.
MISCONCEPTION: More spectral bands always means higher spatial resolution. FACT: These are independent properties. A hyperspectral sensor may have hundreds of bands but only 30 m pixels (e.g., EO-1 Hyperion).
MISCONCEPTION: Radiometric resolution doesn't matter if spatial resolution is fine. FACT: An 8-bit sensor with 0.5 m pixels will still saturate (clip) over bright surfaces like concrete rooftops, losing tonal detail that a 12-bit sensor preserves.

Related Concepts

IFOV and sensor geometry
Swath width and pixels-per-line computation
Sensor selection criteria for specific mapping tasks
Image storage and data volume calculations

Common Exam Questions

Example

A sensor records 2,048 distinct grey levels. What is its bit depth? Solution: 2^n = 2,048 → n = log₂(2,048) = 11 bits.

Approach

Apply Grey Levels = 2^n. The board exam may give you bit depth and ask for grey levels, or vice versa (ask for bits given levels — use log₂).

Question Type

Computation — Radiometric

Example

IFOV = 0.00005 rad, altitude = 800 km → Ground size = 0.00005 × 800,000 = 40 m.

Approach

Ground pixel size = IFOV (radians) × Altitude (m). Ensure consistent units.

Question Type

Computation — Spatial (IFOV)

Example

Which resolution type determines how frequently a satellite images the same area? Answer: Temporal resolution.

Approach

Match a given scenario to the appropriate resolution type. Key cue words: 'how detailed' = spatial; 'how many bands' = spectral; 'how often' = temporal; 'how many grey levels/bit depth' = radiometric.

Question Type

Conceptual — Identification

Key Points To Remember

Spatial: pixel ground size (smaller number = finer detail)
Spectral: number and width of bands (more narrow bands = richer discrimination)
Radiometric: bit depth — 8-bit = 256 levels; 10-bit = 1,024; 12-bit = 4,096; 16-bit = 65,536
Temporal: revisit period in days (shorter = more frequent monitoring)
Formula for number of grey levels: Grey Levels = 2^n, where n = number of bits
Landsat 8/9 bands: Band 1 (coastal), 2 (blue), 3 (green), 4 (red), 5 (NIR), 6 (SWIR1), 7 (SWIR2), 8 (pan), 9 (cirrus), 10 (TIR1), 11 (TIR2) — 11 bands total
IFOV (Instantaneous Field of View) × flying altitude = pixel ground size (spatial resolution)

Passive vs. Active Remote Sensing

This is one of the most-tested conceptual distinctions in board examinations. The classification is based on whether the sensor supplies its own electromagnetic energy source or relies on an external source. PASSIVE SENSORS record electromagnetic energy that originates from an external source — primarily the Sun (reflected solar radiation in the visible and IR bands) or from the Earth itself (emitted thermal radiation in TIR). Because they depend on sunlight, passive optical sensors cannot operate at night (exception: TIR sensors, which detect Earth-emitted heat and can work at night, but are still 'passive'). More critically, passive sensors are blocked by cloud cover, smoke, and thick aerosols. Examples: Landsat (multispectral + thermal), SPOT, Sentinel-2, MODIS, digital cameras in UAVs, aerial film cameras. ACTIVE SENSORS generate and transmit their own electromagnetic pulse, then record the energy reflected back to the sensor. They are completely independent of solar illumination — they work day and night and in all weather conditions. The two dominant active sensor types are: (1) RADAR (Radio Detection And Ranging) / SAR (Synthetic Aperture Radar) — uses microwave pulses (1 mm to 1 m wavelength). Microwaves penetrate clouds, smoke, and even vegetation canopy. Used for surface deformation monitoring (InSAR), flood mapping, soil moisture estimation, and ship detection. Philippine application: Sentinel-1 SAR is routinely used for flood mapping during typhoon season. (2) LiDAR (Light Detection And Ranging) — uses laser pulses (near-IR wavelength). Measures precise time of pulse return to calculate distance. Produces extremely accurate 3D point clouds. Used for DTM/DSM generation, forest canopy height measurement, coastal mapping, and urban 3D modeling. NAMRIA has deployed airborne LiDAR extensively across the Philippines for the Phil-LiDAR program. Key distinction in practice: The Philippines has roughly 20 typhoons per year, making cloud cover a persistent challenge. Active SAR sensors are the only reliable all-weather option for real-time disaster monitoring.

Examples

This question combines active sensor knowledge with practical application. LiDAR's ability to separate canopy returns from ground returns makes it uniquely suited for forested terrain DTM extraction — a capability passive optical sensors lack. The board exam may ask you to justify sensor selection based on terrain type and environmental conditions.

Scenario

A geodetic engineer must produce a detailed Digital Terrain Model (DTM) of a heavily forested area in Palawan for a NIPAS management plan. The area is frequently cloud-covered. Recommend the most appropriate sensor and explain why.

Solution

Airborne LiDAR is the recommended sensor. Although it requires clear-air conditions (no thick cloud at flight altitude), it can penetrate forest canopy using multiple-return technology — the first return records the canopy top (DSM) while the last return reaches the bare ground (DTM). For cloud-penetrating mapping, L-band SAR (ALOS-2 PALSAR) can also provide structural information, though not bare-earth DTM directly.

This is the classic passive-vs-active scenario that appears in PRC board exams. The key phrase is 'cloud cover' combined with 'urgent mapping need' — this uniquely identifies an active microwave (SAR) solution. Always associate cloud-penetrating, all-weather, day-night imaging with active sensors.

Scenario

Typhoon Rolly makes landfall in Catanduanes. Relief operations need an updated flood extent map within 12 hours. Cloud cover is 100%. Which sensor do you use?

Solution

Sentinel-1 SAR (C-band, active microwave). Being an active sensor, it operates regardless of cloud cover and at night. ESA's Copernicus Emergency Management Service routinely activates Sentinel-1 for Philippine typhoon flood mapping within hours of a request.

Applications

SAR InSAR for ground subsidence monitoring in Metro Manila (land subsidence due to groundwater extraction)
LiDAR for cadastral mapping and boundary demarcation in forested areas (RA 8560 surveys)
SAR for agricultural area estimation under cloud cover during wet season planting
Thermal IR (passive) for volcanic monitoring at Philippine volcanoes (Taal, Mayon, Kanlaon)
UAV (passive optical) for rapid post-earthquake damage assessment in urban areas

Misconceptions

MISCONCEPTION: Thermal infrared (TIR) sensors are active because they work at night. FACT: TIR sensors are still passive — they detect energy emitted by the Earth itself (not generated by the sensor). Night-time operation is possible because Earth continuously emits thermal radiation.
MISCONCEPTION: LiDAR can penetrate cloud cover like radar. FACT: LiDAR uses near-IR laser light, which is scattered/blocked by clouds. LiDAR requires cloud-free conditions between sensor and ground (though it can penetrate thin vegetation). Radar (microwave) penetrates clouds.
MISCONCEPTION: Active sensors always give better data than passive. FACT: SAR imagery is geometrically complex (foreshortening, layover, shadow) and harder to interpret visually. Passive optical imagery is more intuitive. The best sensor depends entirely on the application.
MISCONCEPTION: All UAVs use active sensors. FACT: The overwhelming majority of UAV/drone systems carry passive optical cameras (RGB, multispectral) or passive thermal cameras. LiDAR-equipped UAVs exist but are specialized and expensive.

Related Concepts

Electromagnetic spectrum and spectral bands
SAR geometry — foreshortening, layover, shadow
LiDAR point cloud processing — first return, last return, DTM, DSM
Phil-LiDAR Program (DOST-DREAM)
InSAR for surface deformation monitoring

Common Exam Questions

Example

Which sensor type is most appropriate for flood mapping during a typhoon? Answer: Active microwave (SAR) — it penetrates cloud cover and operates at night.

Approach

Identify the key constraint in the scenario (cloud cover? night? need for 3D elevation? all-weather?). Match to the appropriate sensor type. Cloud + weather → SAR. 3D elevation + high accuracy → LiDAR. Optical detail in clear weather → passive.

Question Type

Conceptual — Scenario-based

Example

Classify Sentinel-1 and Landsat-8 as active or passive. Answer: Sentinel-1 = active (SAR/microwave); Landsat-8 = passive (optical + thermal).

Approach

Given a sensor name, classify as active or passive. Memorize lists: Landsat, SPOT, Sentinel-2 = passive optical; Sentinel-1, ALOS PALSAR, RADARSAT = active SAR; any LiDAR = active.

Question Type

Identification

Key Points To Remember

Passive: uses external energy (Sun or Earth's heat) — blocked by clouds, requires daylight (except TIR)
Active: generates own energy — works at night, all weather, penetrates clouds
Passive examples: Landsat, SPOT, Sentinel-2, MODIS, aerial cameras, UAV cameras, TIR sensors
Active examples: SAR/Radar (Sentinel-1, ALOS-2 PALSAR), LiDAR (airborne, terrestrial, mobile)
SAR wavelength bands: X-band (3 cm), C-band (5.6 cm, Sentinel-1), L-band (23 cm, ALOS PALSAR) — longer wavelength = deeper penetration
LiDAR gives direct 3D point clouds — highly accurate elevation data (cm-level vertical accuracy achievable)
Phil-LiDAR Program: DOST-funded nationwide LiDAR survey for flood hazard mapping — key Philippine context

Image Processing: Rectification, Enhancement, and Classification

Raw satellite imagery requires systematic processing before it can be used for mapping or analysis. The standard processing chain involves three major steps: (1) RECTIFICATION (Geometric Correction): Raw imagery contains geometric distortions caused by satellite orbit geometry, Earth's curvature, terrain relief, and sensor characteristics. Rectification removes these distortions by registering the image to a known coordinate system using Ground Control Points (GCPs) — surveyed points of known coordinates on the ground (ideally established in PRS92/WGS84). The process involves polynomial transformation or orthorectification (using a DEM to correct for terrain-induced relief displacement). The output is an orthoimage — geometrically corrected so that distances and areas can be accurately measured. (2) IMAGE ENHANCEMENT: Techniques applied to improve visual interpretability or prepare data for analysis. Common methods include: Contrast stretching (adjusting grey level range for better visual contrast), Histogram equalization, Spatial filtering (edge enhancement, smoothing), Band compositing (creating RGB composites from different bands — e.g., False Color Composite placing NIR in the red channel), and Pansharpening (fusing high-resolution panchromatic with lower-resolution multispectral to produce high-resolution color imagery). (3) CLASSIFICATION: The process of assigning each image pixel (or group of pixels) to a thematic category (e.g., forest, water, urban area, agricultural land). Two main approaches: SUPERVISED classification (analyst provides training samples for each class → algorithm learns and classifies; methods: Maximum Likelihood, Support Vector Machine, Random Forest) and UNSUPERVISED classification (algorithm groups pixels by spectral similarity without prior knowledge; method: K-means, ISODATA). After classification, accuracy assessment is mandatory — comparing classified pixels against ground truth data using a confusion matrix. Key metrics: Overall Accuracy, Producer's Accuracy, User's Accuracy, and Kappa coefficient.

Examples

The confusion matrix diagonal represents correctly classified pixels for each class. Overall Accuracy is the fraction of all sample pixels correctly classified. An 82.5% overall accuracy is generally acceptable for regional land-cover mapping, though individual class accuracies should also be checked.

Scenario

A classified land-cover map of Laguna de Bay has the following confusion matrix diagonal values: Water = 90, Fishpond = 80, Vegetation = 75, Urban = 85. Total pixels sampled = 400 (100 per class). Compute Overall Accuracy.

Solution

Overall Accuracy = (90 + 80 + 75 + 85) / 400 × 100% = 330/400 × 100% = 82.5%

RMSE is commonly expressed in pixels during rectification. Convert to meters by multiplying by the image spatial resolution. An RMSE of 1.8 pixels for a 10 m image gives 18 m positional accuracy — acceptable for small-scale mapping but insufficient for cadastral surveys requiring sub-meter accuracy.

Scenario

During rectification of a Sentinel-2 image over Cebu City, you collected 12 GCPs with a resulting RMSE of 1.8 pixels. The spatial resolution is 10 m. What is the positional error in meters?

Solution

Positional error = RMSE (in pixels) × pixel size = 1.8 × 10 m = 18 m

Applications

Orthorectification for NAMRIA topographic map production
Supervised classification for forest cover monitoring under DENR forest inventory programs
Land-use/land-cover mapping for municipal comprehensive land use plans (CLUP) under RA 7160
Change detection analysis: classified images from two dates compared to identify deforestation or urban expansion
Flood inundation mapping using classified SAR imagery for NDRRMC disaster response

Misconceptions

MISCONCEPTION: Rectification and registration are the same thing. FACT: Registration aligns one image to another (image-to-image). Rectification aligns an image to a geographic coordinate system (image-to-map) using GCPs.
MISCONCEPTION: More GCPs always give better rectification accuracy. FACT: Poorly distributed or inaccurately located GCPs can worsen accuracy. Well-distributed GCPs across the image extent, with accurate field-surveyed coordinates, are more important than quantity alone.
MISCONCEPTION: Supervised classification is always more accurate than unsupervised. FACT: Supervised classification is only as good as the quality and representativeness of training samples. Poor training samples → poor classification, even with sophisticated algorithms.
MISCONCEPTION: Kappa > 0.8 is always required for publication. FACT: Kappa thresholds depend on the application. Emergency mapping may accept lower accuracy; detailed cadastral or legal boundary mapping requires higher standards.

Related Concepts

Ground Control Points (GCPs) and accuracy standards
Spectral signatures in classification
Change detection methods
Accuracy assessment and confusion matrix
Digital Elevation Models (DEM, DTM, DSM)

Common Exam Questions

Example

Confusion matrix: 4 classes, 25 samples each (100 total). Diagonal: 22, 20, 19, 23. OA = (22+20+19+23)/100 = 84/100 = 84%.

Approach

Extract diagonal values from a given confusion matrix, sum them, divide by total sample size, multiply by 100%. Know the formula perfectly.

Question Type

Computation — Overall Accuracy

Example

An analyst delineates areas of known forest, water, and urban land on the image and uses these to train a Maximum Likelihood classifier. This is supervised classification.

Approach

If the problem states the analyst provides training areas/samples → supervised. If no prior knowledge and the algorithm clusters automatically → unsupervised. Know examples of each algorithm.

Question Type

Conceptual — Classification Type

Key Points To Remember

Rectification: geometric correction using GCPs → orthoimage in known coordinate system (PRS92, WGS84)
GCP requirement: minimum 3 GCPs for affine transformation; more GCPs = better accuracy
RMSE (Root Mean Square Error): standard metric for geometric accuracy of rectification
Image enhancement: contrast stretch, histogram equalization, spatial filter, pansharpening
Classification types: supervised (needs training samples) vs. unsupervised (no prior knowledge needed)
Confusion matrix: rows = reference classes; columns = classified classes; diagonal = correct classifications
Kappa coefficient: accounts for chance agreement; values 0.8–1.0 = excellent accuracy
Overall Accuracy = (sum of diagonal) / (total pixels in confusion matrix) × 100%

GIS Data Models: Vector and Raster

GIS stores spatial information using two fundamental data models, each suited to different types of geographic phenomena. VECTOR DATA MODEL: Represents geographic features as discrete geometric objects — points, lines, and polygons — each associated with attribute records in a database table. Points represent zero-dimensional features (e.g., survey monuments, BM locations, well locations). Lines (polylines) represent one-dimensional linear features (e.g., road centerlines, rivers, property boundaries as lines). Polygons represent two-dimensional area features (e.g., cadastral parcels, municipal boundaries, building footprints, forest cover areas). Vector data is well-suited for discrete, precisely-bounded features. It is the dominant model in cadastral GIS and legal boundary databases (PD 1529, RA 8560). Storage format examples: Shapefile (.shp), GeoJSON, File Geodatabase, KML. RASTER DATA MODEL: Represents geographic information as a regular grid of cells (pixels), where each cell holds a single value. Raster is ideal for continuous spatial phenomena — surfaces that vary smoothly across space. Examples: Digital Elevation Models (DEM/DTM), slope maps, rainfall surfaces, temperature grids, remotely sensed imagery, interpolated water table surfaces. Each raster cell's location is implicitly defined by the grid's origin, cell size (resolution), and number of rows/columns — no coordinate stored per cell. Storage formats: GeoTIFF, IMG, ASCII Grid. KEY DIFFERENCES: Vector preserves precise boundaries; raster approximates boundaries using cell edges. Vector stores complex attribute tables; raster stores a single value per cell (though multi-band rasters exist). Vector is inefficient for continuous surfaces; raster is inefficient for complex linear networks. CRITICAL RULE FOR EXAM: A GIS project can — and usually does — contain both vector and raster layers simultaneously. However, ALL layers must share a common geographic coordinate system (datum and projection) to overlay correctly. In the Philippines, PRS92 (the national geodetic datum) or WGS84, projected in the appropriate PPCS (Philippine Plane Coordinate System) zone or UTM zone, must be consistently applied across all layers.

Examples

This is a classic board-exam classification question. The key rule: discrete, precisely-bounded features = vector. Continuous surfaces varying smoothly across space = raster. Cadastral lots have legally defined boundaries → vector polygons. Rainfall and elevation vary continuously → raster.

Scenario

A GIS analyst is building a spatial database for a municipality in Iloilo. Classify the following as vector (and geometry type) or raster: (a) road centerlines, (b) elevation model, (c) building footprints, (d) monthly rainfall surface, (e) cadastral lot polygons.

Solution

(a) Road centerlines → Vector, Line; (b) Elevation model (DEM) → Raster; (c) Building footprints → Vector, Polygon; (d) Monthly rainfall surface → Raster; (e) Cadastral lot polygons → Vector, Polygon.

This is a critical exam pitfall. Even small datum differences (WGS84 vs. PRS92 differ by approximately 1–2 m, but projection differences between UTM and PPCS can cause hundreds of meters of offset if parameters are incorrectly applied). Always confirm the CRS of every layer before analysis.

Scenario

A GIS project has a road network layer in WGS84 UTM Zone 51N and a land-use classification raster in PRS92 PPCS Zone III. An analyst overlays them and notices misalignment of ~500 m. What is the cause and solution?

Solution

The misalignment is caused by using two different coordinate systems (WGS84 UTM Zone 51N vs. PRS92 PPCS Zone III) in the same project without proper reprojection. The solution is to reproject one or both layers to a single, consistent coordinate system — ideally PRS92 projected in PPCS Zone III (or another appropriate zone) for Philippine cadastral work, using the correct datum transformation parameters from WGS84 to PRS92.

Applications

Cadastral parcel management systems under PD 1529 (Property Registration Decree) — vector polygons
NAMRIA topographic database — integrated vector (features) and raster (imagery, DEM) layers
Watershed delineation using raster DEM analysis for DENR water resource planning
Municipal GIS for CLUP preparation — vector zoning polygons and raster hazard maps
Road network analysis (routing, buffer zones) using vector line topology

Misconceptions

MISCONCEPTION: Raster data is less accurate than vector. FACT: Accuracy depends on cell size (resolution). A 0.1 m raster DEM derived from LiDAR is far more accurate than a 1:50,000 vector contour map for elevation.
MISCONCEPTION: You can freely overlay any GIS layers regardless of their coordinate system. FACT: Layers MUST share an identical datum, projection type, and projection zone. Overlaying layers with mismatched CRS produces incorrect spatial relationships.
MISCONCEPTION: WGS84 and PRS92 are interchangeable in GIS. FACT: PRS92 and WGS84 differ by a small but significant offset (~1–2 m). For cadastral-level work, proper datum transformation is essential. In PPCS vs. UTM projections, offsets can be much larger if projection parameters differ.
MISCONCEPTION: A GIS can only use one data model per project. FACT: Professional GIS projects routinely integrate both vector and raster layers — for example, overlaying vector cadastral parcels on a raster satellite image or DEM.

Related Concepts

Philippine Plane Coordinate System (PPCS) zones and parameters
PRS92 vs. WGS84 datum differences
Topology in vector GIS
Spatial data infrastructure — National Spatial Data Infrastructure (NSDI) Philippines
Attribute tables and relational database concepts in GIS

Common Exam Questions

Example

Which GIS data model is most appropriate for a property boundary database? Answer: Vector (polygon), because cadastral boundaries are discrete, legally defined, and require precise coordinate representation.

Approach

Match geographic features to data model. Remember: discrete, bounded, legal, countable features = vector. Continuous, smoothly varying, surface data = raster.

Question Type

Classification

Example

Two GIS layers fail to align properly when overlaid. What is the most likely cause? Answer: The layers are in different coordinate reference systems (datums or projections) — they must be reprojected to a single common CRS.

Approach

Any question about GIS layer overlay problems almost always involves mismatched coordinate systems. The solution is always to reproject all layers to a single common CRS.

Question Type

Conceptual — Coordinate System

Key Points To Remember

Vector: points (0D), lines (1D), polygons (2D) — for discrete, precisely-bounded features
Raster: regular grid of cells — for continuous surfaces
Vector examples: survey monuments, road network, cadastral parcels, building footprints
Raster examples: DEM, slope, rainfall, temperature, satellite imagery, interpolated surfaces
All GIS layers in a project MUST share the same datum and projection for correct overlay
Philippine coordinate systems: PRS92 (datum), PPCS Zones (I–V, for local/cadastral work), UTM Zones 51N/52N (for national mapping)
Shapefile (.shp) = common vector format; GeoTIFF = common raster format
Topology rules (vector): polygons must not overlap, lines must not dangle — important for cadastral databases

GIS Spatial Analysis: Overlay, Buffer, Network, and Interpolation

GIS is far more than a digital map — its power lies in spatial analysis: extracting new information by combining, manipulating, and querying spatial data. Core analysis types tested in the board exam are: (1) OVERLAY ANALYSIS: Combines two or more layers to produce a new layer that contains information from all input layers. Used for land suitability analysis, zoning conflict identification, risk mapping. Types: Union (combines all features from both layers), Intersect (retains only overlapping areas), Difference/Erase, and Clip. Example: Overlaying a typhoon storm surge hazard map (polygon) with a land-use map (polygon) to identify settlements within the high-hazard zone. (2) BUFFER ANALYSIS: Creates a zone of specified distance around a feature (point, line, or polygon). Used for setback computations, easement delineation, and impact zone identification. Philippine application: Creating a 3 m riparian buffer along rivers as required by DAO 2000-20 for watershed protection, or delineating easements under PD 1067 (Water Code). (3) NETWORK ANALYSIS: Analyzes connectivity along linear networks (roads, rivers, utilities). Applications: shortest-path routing, service area delineation (e.g., 5 km hospital service radius), flow direction in drainage networks. (4) INTERPOLATION: Estimates surface values at unsampled locations based on measured values at known sample points. Common methods: IDW (Inverse Distance Weighting) — values weighted by inverse of distance to sample points; Kriging — geostatistical method that accounts for spatial autocorrelation; TIN (Triangulated Irregular Network) — connects sample points into triangles to model terrain. Used for: groundwater level surface, rainfall distribution, soil property mapping, DEM generation from survey points. (5) RASTER ANALYSIS: Raster Calculator performs cell-by-cell mathematical operations. Slope, aspect, curvature derived from DEM raster. Watershed delineation using flow direction and flow accumulation rasters.

Examples

This combines buffer analysis (50 m coastal buffer) with overlay analysis (intersect with parcels). This is a direct application of GIS to land administration and zoning enforcement — highly relevant to RA 8560 (geodetic survey practice) and local government planning under RA 7160.

Scenario

A municipality in Leyte requires a 50 m setback from the coastline per their CLUP zoning ordinance. Describe the GIS operation to identify all residential parcels that violate this setback.

Solution

Step 1: Create a 50 m buffer polygon along the coastline vector layer. Step 2: Perform an Intersect overlay between the buffer polygon and the residential parcel vector layer. Step 3: The resulting layer contains all residential parcels (or portions thereof) that fall within the 50 m coastal setback zone — these are the violating parcels.

Interpolation converts point measurements to a continuous raster surface. For hydrogeological data, Kriging is scientifically preferred. IDW is simpler and commonly tested in board exams because of its straightforward distance-weighting formula.

Scenario

You have 15 groundwater well measurements (water table depth in meters below ground) distributed across a municipal area. You need to produce a continuous groundwater depth surface for the entire municipality. Which interpolation method would you use and why?

Solution

Kriging is the preferred geostatistical method if spatial autocorrelation is present (nearby wells tend to have similar values). For simpler implementation, IDW is acceptable — values at unsampled locations are estimated as a weighted average of nearby well measurements, with weights inversely proportional to distance. IDW Formula: Z* = Σ(Zi / di^p) / Σ(1/di^p), where Zi = measured value, di = distance from sample point, p = power parameter (commonly p = 2).

Applications

Land suitability analysis for agricultural expansion using weighted overlay of slope, soil, and rainfall rasters
Flood risk mapping: overlay of flood hazard raster with residential area polygons
Shortest evacuation route computation using road network analysis for DRRM plans
Rainfall surface generation using IDW from PAGASA rain gauge stations for hydrologic modeling
Slope and aspect maps from DEM for landslide susceptibility assessment (MGB, PHIVOLCS)

Misconceptions

MISCONCEPTION: Buffer analysis by itself identifies affected features. FACT: Buffering creates a zone polygon. You must follow it with an Intersect or Clip overlay operation to identify which features fall within the buffer.
MISCONCEPTION: Kriging always gives the best interpolation. FACT: Kriging requires a sufficient sample size and spatial autocorrelation. With very few sample points or purely random spatial distributions, simpler methods like IDW or nearest-neighbor may be equally effective.
MISCONCEPTION: Union and Intersect overlays produce the same result. FACT: Union keeps ALL features from both layers (total extent). Intersect keeps ONLY features common to (overlapping in) both layers (smaller extent).
MISCONCEPTION: Raster overlay is the same as vector overlay. FACT: Raster overlay uses cell-by-cell arithmetic (Raster Calculator). Vector overlay involves computational geometry operations on polygon/line features. They can produce similar analytical results but are technically different processes.

Related Concepts

GIS data models (vector, raster)
Spatial autocorrelation and Tobler's First Law of Geography
Topology and network connectivity
DEM-derived terrain analysis (slope, aspect, watershed)
Weighted overlay for multi-criteria analysis

Common Exam Questions

Example

A GIS analyst creates a 100 m zone around all rivers to identify affected farmlands. What operation is this? Answer: Buffer analysis (100 m buffer around river lines, then overlay/intersect with farmland polygons).

Approach

Match scenario to GIS operation: 'zone around feature' = buffer; 'combine layers, keep common area' = intersect overlay; 'estimate surface from points' = interpolation; 'connectivity along road network' = network analysis.

Question Type

Conceptual — Operation Identification

Example

Two sample points: Point A (value = 10 m, distance = 2 km), Point B (value = 20 m, distance = 4 km). IDW estimate at query location (p=2): Weight A = 1/2² = 0.25; Weight B = 1/4² = 0.0625. Z* = (10×0.25 + 20×0.0625)/(0.25+0.0625) = (2.5+1.25)/0.3125 = 3.75/0.3125 = 12 m.

Approach

Apply IDW formula with given distances and measured values, typically with p = 2 (inverse-square weighting). Compute weights, compute weighted average.

Question Type

Computation — IDW

Key Points To Remember

Overlay types: Union (all areas), Intersect (common areas), Clip (one layer cuts another)
Buffer: fixed or variable distance zone around point/line/polygon features
Network analysis: requires topologically correct (connected) vector line layer
IDW interpolation: closer sample points have more weight — simple but ignores spatial trends
Kriging: geostatistical interpolation — best linear unbiased estimator; uses variogram
TIN: irregular triangulation from survey points — good for terrain modeling with variable point density
Raster Calculator: pixel-by-pixel algebra — e.g., NDVI = (Band5 − Band4)/(Band5 + Band4)
Slope = rise/run expressed in degrees or percent; derived from DEM in GIS

Swath Width and Pixels-per-Line Computation

One of the most directly computational topics in this chapter is calculating the number of pixels across a single image scan line (across-track pixels), given the sensor's spatial resolution and swath width. This is a straightforward division, but it appears regularly in PRC board examinations as a numerical problem. SWATH WIDTH: The width of the ground strip imaged by the satellite in a single orbital pass. For example, Landsat has a 185 km swath; Sentinel-2 has a 290 km swath; SPOT has a 60 km swath. SPATIAL RESOLUTION: The ground dimension of one pixel (e.g., 30 m for Landsat multispectral). FORMULA: Number of pixels per scan line = Swath Width (m) / Spatial Resolution (m/pixel). Note: Ensure units are consistent — convert km to m before dividing. This formula also links to IFOV (Instantaneous Field of View): Ground pixel size = IFOV (radians) × Orbital altitude (m). And the swath angle × altitude gives swath width. For total image data volume: Data volume (bits) = Rows × Columns × Bands × Bit depth. This scales to MB or GB for modern high-resolution, multi-band satellite images.

Examples

Direct application of the core formula. Convert 290 km → 290,000 m, then divide by pixel size. This is the exact type of computation the board exam presents — set up the units carefully and perform the division.

Scenario

A satellite sensor has a spatial resolution of 10 m and a swath width of 290 km (Sentinel-2-like). How many pixels span one across-track scan line?

Solution

Pixels per scan line = 290,000 m ÷ 10 m = 29,000 pixels

Two-part problem: first derive spatial resolution from IFOV and altitude, then compute pixels per scan line. Part (a) uses the IFOV formula; Part (b) uses the swath-to-pixel formula. Both parts are individually testable in the board exam.

Scenario

A sensor with an IFOV of 0.000143 radians orbits at 705 km altitude. (a) What is the spatial resolution? (b) If the swath width is 185 km, how many pixels per scan line?

Solution

(a) Spatial resolution = IFOV × Altitude = 0.000143 rad × 705,000 m = 100.815 m ≈ 100 m. (b) Pixels per scan line = 185,000 m ÷ 100 m = 1,850 pixels.

This multi-step problem integrates the pixel computation with data volume calculation. Note: 1 MB = 1,048,576 bytes (2^20). Some questions use 1 MB = 1,000,000 bytes (SI prefix) — read the problem carefully. Landsat 8 with 11 bands at 16-bit generates approximately 1 GB per scene.

Scenario

A 30 m resolution, 185 km swath, 7-band, 8-bit satellite image covers one complete pass. (a) How many pixels per scan line? (b) If the image has 6,000 scan lines, what is the total data volume in megabytes (MB)?

Solution

(a) Pixels per scan line = 185,000 ÷ 30 ≈ 6,167 pixels. (b) Total bits = 6,167 × 6,000 × 7 × 8 = 6,167 × 6,000 × 56 = 2,072,136,000 bits. Convert: ÷ 8 = 259,017,000 bytes ÷ 1,048,576 ≈ 247 MB.

Applications

Storage planning for satellite image archives at NAMRIA
Bandwidth estimation for satellite downlink systems
Image product specification for government procurement of satellite data
IFOV-based sensor design for future Philippine microsatellites (DIWATA-1, DIWATA-2)
Scene-coverage planning for large-area mapping projects

Misconceptions

MISCONCEPTION: Swath width divided by resolution gives the number of scan lines. FACT: Swath width / resolution gives pixels per ACROSS-TRACK scan line (columns). The number of scan lines (rows) depends on how long the sensor images — a separate parameter.
MISCONCEPTION: Higher resolution always means more data than needed. FACT: Data volume is also determined by swath width, number of bands, and bit depth. A narrow-swath, single-band, 8-bit sensor at 0.5 m may generate less data per scene than a wide-swath, 11-band, 16-bit sensor at 30 m.
MISCONCEPTION: IFOV is the same as Field of View (FOV). FACT: FOV is the total angular sweep of the entire sensor array. IFOV is the angle subtended by a single detector element. FOV = IFOV × number of detector elements across track.

Related Concepts

Spatial resolution and IFOV
Swath width and orbital parameters
Image data volume and storage
Philippine microsatellites DIWATA-1 and DIWATA-2
Four types of resolution

Common Exam Questions

Example

Swath = 185 km, resolution = 30 m → 185,000/30 = 6,166.7 ≈ 6,167 pixels.

Approach

Pixels per scan line = Swath Width ÷ Spatial Resolution. Always check units (km vs m). Round as appropriate — usually the answer is a whole number or the question asks to round to nearest integer.

Question Type

Direct Computation

Example

500 rows × 500 cols × 3 bands × 8 bits = 6,000,000 bits = 750,000 bytes = ~0.715 MB.

Approach

Compute total pixels (rows × columns), multiply by number of bands and bit depth to get total bits, convert to bytes then MB or GB. Watch the unit conversion: 1 byte = 8 bits; 1 KB = 1,024 bytes; 1 MB = 1,048,576 bytes.

Question Type

Multi-step — Data Volume

Key Points To Remember

Pixels per scan line = Swath Width (m) ÷ Spatial Resolution (m)
Always convert swath width from km to m before dividing
IFOV (rad) × Altitude (m) = Ground pixel size (m)
Landsat: 185 km swath, 30 m multispectral, 15 m panchromatic
Sentinel-2: 290 km swath, 10 m (RGB+NIR), 20 m (SWIR+Red-Edge), 60 m (atmospheric)
Total data volume = Rows × Columns × Bands × Bit-depth (in bits) — divide by 8 for bytes
Swath width = 2 × Altitude × tan(half-angle of sensor field of view)

Practice Problems

Straightforward application of the swath-to-pixel formula. Always convert swath width from km to m (290 km = 290,000 m) before dividing by the pixel size in meters. Answer: 29,000 pixels per scan line.

Problem

PROBLEM 1 (Pixels per Scan Line): A 10 m-resolution sensor has a 290 km swath width. How many pixels span one across-track scan line?

Solution

Number of pixels = Swath Width ÷ Spatial Resolution = 290,000 m ÷ 10 m = 29,000 pixels.

Formula: Grey Levels = 2^n where n = bit depth. The 12-bit sensor resolves 16 times more tonal levels than an 8-bit sensor (4,096/256 = 16×), making it significantly better for detecting subtle radiance differences — important for vegetation stress detection and atmospheric correction.

Problem

PROBLEM 2 (Radiometric Resolution): A satellite encodes each pixel with 12 bits per band. (a) How many grey levels can it distinguish? (b) A second satellite uses 8 bits per band. How many more grey levels does the 12-bit satellite resolve?

Solution

(a) Grey levels = 2^12 = 4,096 levels. (b) 8-bit grey levels = 2^8 = 256. Difference = 4,096 − 256 = 3,840 additional grey levels.

IFOV in radians multiplied by orbital altitude in meters yields the ground dimension of one pixel. The result (≈30 m) corresponds to Landsat multispectral spatial resolution. Ensure altitude is in meters (450 km = 450,000 m). This derivation is a common two-step board exam problem.

Problem

PROBLEM 3 (IFOV and Spatial Resolution): A pushbroom sensor has an IFOV of 0.0000667 radians and flies at an altitude of 450 km. Compute the ground pixel size (spatial resolution).

Solution

Ground pixel size = IFOV × Altitude = 0.0000667 rad × 450,000 m = 30.015 m ≈ 30 m.

The scenario presents two constraints: 100% cloud cover + nighttime. Both eliminate passive optical sensors (Landsat, Sentinel-2, UAV cameras). TIR passive sensors can work at night but not through thick clouds. The only solution is active microwave SAR. SAR flood mapping is based on the significantly lower backscatter of smooth open water surfaces (specular reflection away from sensor) compared to rough land surfaces.

Problem

PROBLEM 4 (Active vs. Passive): A disaster response team in Samar needs to map flood extent within 6 hours after Typhoon Rosita makes landfall. Cloud cover is estimated at 100% and it is nighttime. Identify the appropriate sensor type, give a specific sensor name, and justify your choice.

Solution

Sensor type: Active microwave (SAR). Specific sensor: Sentinel-1 (C-band SAR, ESA Copernicus) or ALOS-2 PALSAR-2 (L-band SAR, JAXA). Justification: SAR sensors generate their own microwave pulses, which penetrate dense cloud cover and are unaffected by the absence of sunlight. Flood water appears as dark low-backscatter areas in SAR imagery, clearly distinguishable from land. Sentinel-1 has a 12-day repeat cycle (6 days with two satellites) and can be tasked for emergency acquisitions. JAXA operates a disaster monitoring protocol (JAXA-DSMAC) that can provide ALOS-2 data within hours of a formal request.

The classification rule: Is the feature a discrete, countable, legally/physically bounded object? → Vector (choose point/line/polygon based on geometry). Does it represent a continuous quantity that varies smoothly across the landscape? → Raster. Note: municipal boundaries are vector polygons even though large — they have precise legal coordinates. Temperature is raster because temperature varies continuously at all spatial locations.

Problem

PROBLEM 5 (GIS Data Model Classification): Classify the following geographic data as Vector-Point, Vector-Line, Vector-Polygon, or Raster: (a) BM (benchmark) locations; (b) slope map of a watershed; (c) cadastral lot boundaries; (d) barangay road network; (e) monthly maximum temperature surface; (f) municipal boundary of Iloilo City.

Solution

(a) BM locations → Vector-Point (precise, discrete locations); (b) Slope map → Raster (continuous surface varying across space); (c) Cadastral lot boundaries → Vector-Polygon (legally defined area features); (d) Barangay road network → Vector-Line (linear connectivity network); (e) Monthly max temperature surface → Raster (continuously varying meteorological surface); (f) Municipal boundary → Vector-Polygon (administratively defined area).

This problem tests integration of buffer and overlay operations with Philippine legal context (CA 141, Commonwealth Act 141 — the Public Land Act governs alienable and disposable lands and forestland boundaries). The key procedural point: buffer first, then intersect. Note that both layers must share the same CRS — confirmed in the problem. The 30 m buffer represents the legal setback zone inside which agricultural activity may constitute encroachment.

Problem

PROBLEM 6 (Overlay and Buffer): A DENR forester needs to identify all agricultural parcels within a 30 m buffer of classified forestland boundaries in Bukidnon, to check for encroachment under CA 141. The GIS database has: Layer A (forestland polygon, vector) and Layer B (agricultural parcel polygon, vector), both in PRS92 PPCS Zone V. Describe the step-by-step GIS procedure.

Solution

Step 1: Verify both layers share the same CRS (PRS92 PPCS Zone V) — confirmed. Step 2: Apply a 30 m outward Buffer operation on Layer A (forestland polygons) to create a buffer zone polygon layer. Step 3: Perform an Intersect overlay between the buffer zone layer and Layer B (agricultural parcels). The output contains only those agricultural parcels (or portions) that overlap with the 30 m forestland buffer zone. Step 4: Export the intersect result and compute the area of each encroaching parcel (using the geometry attribute). Step 5: Generate a report and field verification list for DENR enforcement action.

Overall Accuracy = correct pixels / total samples. Assumes 200 samples / 4 classes = 50 samples per class. The Water class accuracy (38/50 = 76%) is notably lower, which would be flagged in a full accuracy assessment. Board exams typically test the OA computation formula only, but may also ask you to compute Producer's Accuracy for a specific class (correctly classified in class / total reference samples in that class).

Problem

PROBLEM 7 (Classification Accuracy): A land-cover classification of a Philippine island was validated using 200 ground truth samples. The confusion matrix diagonal (correctly classified samples) for 4 classes is: Forest = 48, Agricultural = 42, Water = 38, Urban = 41. (a) Compute Overall Accuracy. (b) Is this classification acceptable for a regional land-use planning map?

Solution

(a) Overall Accuracy = (48 + 42 + 38 + 41) / 200 × 100% = 169/200 × 100% = 84.5%. (b) Yes — an 84.5% overall accuracy is generally acceptable for regional land-use planning maps. ASPRS standards for medium-scale thematic mapping typically require ≥85%, so this result is marginally below that threshold. The analyst should review individual class accuracies (particularly for Water at 38/50 = 76%) and consider additional training samples for the Water class to improve its producer's and user's accuracy.

IDW formula: Z* = Σ(Zi × wi) / Σ(wi), where wi = 1/di^p. With p = 2 (inverse-square): closer points have much greater influence. Station C (1.0 km away, depth 9 m) has the largest weight (1.0) and pulls the estimate closest to its value of 9 m. Station B (farthest, 3.0 km) has the smallest weight and least influence. The estimated depth of 10.50 m reflects this — closer to Station C and Station A than to the more distant Station B.

Problem

PROBLEM 8 (IDW Interpolation): Three groundwater level observations near a proposed well site: Station A (depth = 12 m, distance to query = 1.5 km), Station B (depth = 18 m, distance = 3.0 km), Station C (depth = 9 m, distance = 1.0 km). Estimate the groundwater depth at the query point using IDW with power p = 2.

Solution

Weight computation: wA = 1/(1.5)² = 1/2.25 = 0.4444; wB = 1/(3.0)² = 1/9.0 = 0.1111; wC = 1/(1.0)² = 1/1.0 = 1.0000. Sum of weights = 0.4444 + 0.1111 + 1.0000 = 1.5556. IDW estimate Z* = (12×0.4444 + 18×0.1111 + 9×1.0000) / 1.5556 = (5.333 + 2.000 + 9.000) / 1.5556 = 16.333 / 1.5556 = 10.50 m.

Exam Preparation Tips

MEMORIZE THE FOUR RESOLUTIONS with distinct definitions and examples: Spatial (pixel size), Spectral (number/width of bands), Radiometric (bit depth = 2^n grey levels), Temporal (revisit period in days). The board exam routinely presents scenarios and asks you to identify which resolution type is being described or optimized.
MASTER THE PASSIVE vs. ACTIVE DISTINCTION: Create a mental checklist — 'clouds? night? all-weather?' → Active SAR or LiDAR. 'Clear sky, daylight, optical detail?' → Passive optical. 'Night, no clouds?' → Passive TIR or Active. 'Forest penetration, 3D point cloud?' → Active LiDAR. This decision logic covers 90% of sensor-selection exam questions.
KNOW YOUR FORMULAS COLD: (1) Grey Levels = 2^n; (2) Pixels per line = Swath Width (m) / Spatial Resolution (m); (3) Ground pixel size = IFOV (rad) × Altitude (m); (4) NDVI = (NIR − Red)/(NIR + Red); (5) IDW Z* = Σ(Zi/di^p) / Σ(1/di^p); (6) Overall Accuracy = Diagonal Sum / Total Samples × 100%. Practice applying each in 2 minutes.
VECTOR vs. RASTER — USE THE 'DISCRETE vs. CONTINUOUS' RULE: If a feature has legally or physically defined boundaries and can be individually counted or addressed → Vector. If a phenomenon varies continuously across space with no discrete boundaries → Raster. Cadastral lots, roads, buildings = Vector. Elevation, rainfall, temperature, satellite imagery = Raster.
COORDINATE SYSTEMS ARE A RECURRING PITFALL: In the Philippines, use PRS92 (national datum) with PPCS Zones I–V for cadastral/local work, or WGS84 with UTM Zones 51N/52N for national mapping. Any GIS overlay problem that mentions misalignment or positional error — the answer is almost always mismatched coordinate systems. Know the five PPCS zones and their coverage.
PHILIPPINE LEGAL CONTEXT: Memorize which laws relate to which GIS/RS applications — PD 1529 (Property Registration) = cadastral vector GIS; RA 8560 (Geodetic Engineering Act) = professional practice in RS/GIS surveys; CA 141 (Public Land Act) = forestland boundary mapping; PD 1067 (Water Code) = riparian buffer GIS analysis. These connections appear in scenario-based questions.
CONFUSION MATRIX PROBLEMS: Always check if the question gives you (a) the full matrix to compute from, or (b) only the diagonal values with sample sizes. For OA: sum the diagonal, divide by total samples. For Producer's Accuracy of class X: correctly classified in X / total reference samples in X. For User's Accuracy of class X: correctly classified in X / total classified as X.
SPECTRAL BANDS AND APPLICATIONS — BUILD A REFERENCE TABLE: Visible Blue → water depth, atmosphere; Visible Green → vegetation green peak; Visible Red → chlorophyll absorption, NDVI denominator; NIR → vegetation health, NDVI numerator, land/water boundary; SWIR → soil moisture, minerals; TIR → heat/volcanic/urban; Microwave → all-weather, SAR, soil moisture, flood. Reproduce this table from memory under exam conditions.
TIMING STRATEGY FOR NUMERICAL PROBLEMS: Remote sensing computation questions in the board exam are typically 2–4 step calculations. Allocate 2–3 minutes each. For pixel-count and radiometric problems: solve in under 90 seconds. For IDW or accuracy assessment: allow 3–4 minutes. Practice working without a calculator for simple 2^n and division problems.
PHIL-LIDAR AND NAMRIA CONTEXT: The Philippine government's Phil-LiDAR program (DOST-DREAM) generated nationwide LiDAR data for flood hazard mapping — know that this uses active LiDAR sensors for DTM generation. NAMRIA is the national mapping authority that produces official topographic maps, orthoimages, and maintains the national spatial data infrastructure. These institutional contexts regularly appear in exam scenarios.

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In summary

Remote sensing and GIS represent the technological backbone of 21st-century geodetic engineering in the Philippines and worldwide. For PRC board examination success, focus your review on six high-yield areas: (1) The four resolution types — spatial, spectral, radiometric, temporal — with associated formulas and satellite examples; (2) The passive/active sensor dichotomy — especially the all-weather, day-night capability of SAR and LiDAR; (3) The EM spectrum and spectral signatures — which band detects what, and why; (4) Image processing steps — rectification (GCPs, RMSE), enhancement, and classification (supervised vs. unsupervised, confusion matrix, overall accuracy); (5) GIS data models — vector (point/line/polygon for discrete features) vs. raster (grid for continuous surfaces); and (6) Spatial analysis operations — buffer, overlay (union/intersect), network analysis, and interpolation (IDW formula). Throughout your review, connect these technical concepts to Philippine institutional and legal contexts: NAMRIA's mapping mandate, Phil-LiDAR's nationwide LiDAR surveys, PRS92/PPCS coordinate systems, and the legal frameworks of PD 1529, RA 8560, CA 141, and PD 1067. Board exam questions increasingly present integrated scenarios requiring you to simultaneously apply technical knowledge (which sensor? which data model? which analysis?) and Philippine practice context (which law? which coordinate system?). Master the formulas through repeated problem-solving, and master the concepts through the decision frameworks presented in this chapter. Mabuting pagsubok — best of luck in your board examinations!

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