GELE Photogrammetry & Cartography — Aerial Photography and Camera GeometryDetailed Explanation
Detailed explanation of Aerial Photography and Camera Geometry for the GELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests this chapter with applied or scenario-based questions in the GELE Photogrammetry & Cartography subtest.
Exam context
For the Geodetic Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests Photogrammetry & Cartography under a "Core" label, with Aerial Photography and Camera Geometry in the 1st 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.
Aerial Photography and Camera Geometry - Detailed Explanation
Aerial photography and camera geometry form the geometric foundation of photogrammetry — the science of extracting reliable measurements from photographs. For the PRC Geodetic Engineer Licensure Examination, this topic is consistently tested through numerical problems involving photo scale, ground coverage computation, flight planning parameters, and camera geometry relationships. A solid grasp of the similar-triangle principle that governs the vertical aerial photograph will allow you to solve virtually every board-exam problem in this chapter quickly and confidently. This chapter also connects directly to practical geodetic workflows in the Philippines, where aerial photogrammetry supports cadastral surveys under PD 1529 (Property Registration Decree), land classification under CA 141 (Public Land Act), and topographic mapping projects of NAMRIA and LMB.
Concepts
The Metric Aerial Camera and Its Principal Elements
A metric aerial camera is a precision optical-mechanical instrument designed to produce photographs of known, stable geometry. Unlike ordinary cameras, its interior orientation parameters — focal length (f), principal point location, and fiducial marks — are calibrated and certified by the manufacturer. These parameters are essential for photogrammetric computation. **Focal Length (f):** The perpendicular distance from the rear nodal point of the lens to the image plane (film or sensor). Common values used in Philippine aerial survey practice are 152 mm (6-inch, wide-angle) and 210 mm (normal-angle). A shorter focal length gives a wider field of view and smaller scale at the same flying height. **Format Size:** The standard film format is 230 mm × 230 mm (9 in × 9 in). Digital aerial cameras may have varying sensor sizes, but 230 mm is the classic board-exam standard. **Principal Point (PP):** The point where the optical axis (perpendicular from the lens) intersects the image plane. It is the geometric center of the photo, located at the intersection of lines connecting opposite fiducial marks. **Fiducial Marks:** Four (or eight) small reference marks fixed in the camera body at the edges or corners of the photo. Their precise coordinates are given in the camera calibration certificate. They define the image coordinate system and locate the principal point. **Nadir Point:** The point on the photo that is directly below the camera's perspective center (the point directly below the aircraft on the ground, projected onto the photo). For a truly vertical photo, the nadir and principal point coincide. **Angular Field of View:** Determined by the format size and focal length. For a 230 mm format and 152 mm focal length: half-angle = arctan(115/152) ≈ 37.1°, giving a total field of about 74°. This is classified as wide-angle.
Examples
Camera classification by angular FOV: Normal-angle < 60°; Wide-angle 60°–90°; Super-wide-angle > 90° (along the diagonal). The 152 mm lens on a 230 mm format is the most common wide-angle configuration in Philippine photogrammetric surveys.
Scenario
A metric aerial camera has a focal length of 152 mm and a format of 230 mm × 230 mm. Compute the total angular field of view (across the diagonal) and classify the camera type.
Solution
Half-diagonal of format = √(115² + 115²) = 115√2 = 162.6 mm. Half-angle (diagonal) = arctan(162.6 / 152) = arctan(1.0697) ≈ 46.9°. Total diagonal FOV ≈ 93.8°. Along one side: half-angle = arctan(115/152) = arctan(0.7566) ≈ 37.1°; total = 74.2°. Camera classification: Wide-angle (total angular FOV along side > 60°).
Applications
- Selecting appropriate focal length for a given mapping scale and flying height in NAMRIA aerial survey projects.
- Checking camera calibration certificates before photogrammetric control work.
- Establishing the photo coordinate system using fiducial marks in analytical stereoplotters.
- Determining whether a photo is truly vertical by comparing nadir and principal point positions.
Misconceptions
- Confusing focal length with the distance from the lens to the ground — focal length is measured to the image plane, not to the terrain.
- Assuming the principal point is always exactly at the geometric center — in practice it may be slightly off-center (calibrated eccentricity), which is why the calibration certificate is used.
- Thinking fiducial marks are optional — they are mandatory reference points for all photogrammetric measurements.
- Confusing nadir (ground point directly below camera) with the isocenter (used in tilted photo analysis) — for vertical photos, all three special points coincide.
Related Concepts
- Photo Scale and Flying Height
- Interior vs. Exterior Orientation
- Tilted Photo Geometry
- Ground Control Points
Common Exam Questions
Example
On a truly vertical aerial photograph, the _____ and _____ coincide. Answer: principal point and nadir point.
Approach
Know the definitions of each camera element and their functional roles. Board exams may ask 'which point coincides with the nadir on a vertical photo?' or 'what marks define the image coordinate system?'
Question Type
Conceptual identification
Example
f = 210 mm, format = 230 mm. Half-angle = arctan(115/210) = 28.7°; total FOV = 57.4°. Classification: Normal-angle.
Approach
Apply trigonometry: half-angle = arctan((format side / 2) / f). Classify result as normal, wide, or super-wide angle.
Question Type
Angular FOV computation
Key Points To Remember
- Metric camera parameters (f, principal point, fiducial marks) are fixed and calibrated — this is called interior orientation.
- Standard format: 230 mm × 230 mm; common focal lengths: 152 mm (wide-angle) and 210 mm (normal-angle).
- Principal point = intersection of optical axis with image plane; nadir = directly below the camera center projected on the photo.
- On a perfectly vertical photo, nadir ≡ principal point.
- Fiducial marks define the photo coordinate system — always used as reference in analytical photogrammetry.
- Shorter focal length → wider angle → smaller scale at same H (but covers more ground).
Photo Scale of a Vertical Photograph
The photo scale is the ratio of a distance on the photograph to the corresponding distance on the ground. For a vertical photo, the geometry is a simple similar-triangle (central projection) relationship between the image plane and the ground plane. **Fundamental Scale Formula:** $$\text{Scale} = \frac{f}{H}$$ where: - $f$ = calibrated focal length of the camera (metres) - $H$ = flying height above the ground (metres) This can also be written as: $$\frac{1}{S} = \frac{f}{H} \quad \Rightarrow \quad S = \frac{H}{f}$$ where $S$ is the scale denominator (e.g., for 1:10 000, $S$ = 10 000). **Key relationship between photo distance and ground distance:** $$\text{Ground distance} = \text{Photo distance} \times S = \frac{\text{Photo distance} \times H}{f}$$ $$\text{Photo distance} = \frac{\text{Ground distance}}{S} = \frac{\text{Ground distance} \times f}{H}$$ **Flying Height H — Critical Clarification:** - $H$ is always measured above the terrain (ground), NOT above mean sea level (MSL). - If the aircraft altimeter reads altitude above MSL = $A$, and the average ground elevation = $h$, then $H = A - h$. - Over varying terrain, the scale varies across the photo. The average (nominal) scale uses the mean ground elevation. **Example of scale variation:** If one hill is 200 m higher than the mean ground, the effective H over that hill is reduced by 200 m, making the scale larger (more detail) over the hill and smaller in the valley. **Expressing scale:** Always express as a unit fraction: 1:10 000 or 1/10 000, never just '10 000'.
Examples
Step 1: Convert f to same units as H (0.152 m). Step 2: Apply Scale = f/H. Step 3: For ground distance, multiply photo measurement by S. Always verify unit conversion: 450 000 mm ÷ 1 000 = 450 m.
Scenario
BOARD-TYPE PROBLEM: A vertical aerial photo is taken with a camera having f = 152 mm from a flying height of 1 520 m above mean ground level. Determine: (a) the photo scale, and (b) the ground distance between two road intersections that are 45 mm apart on the photo.
Solution
(a) Scale = f/H = 0.152 m / 1 520 m = 1/10 000 (b) Ground distance = photo distance × S = 45 mm × 10 000 = 450 000 mm = 450 m
The critical step is computing H above ground, not using the altimeter reading directly. H = 3 150 m (not 3 500 m). Scale = 1:15 000.
Scenario
BOARD-TYPE PROBLEM: An aircraft flies at an altitude of 3 500 m above MSL. The terrain below has a mean elevation of 350 m above MSL. The camera focal length is 210 mm. What is the photo scale?
Solution
H (above ground) = Altitude (MSL) − Terrain elevation = 3 500 m − 350 m = 3 150 m Scale = f/H = 0.210 m / 3 150 m = 1/15 000
Straightforward application of the ground-distance formula. Scale denominator S = 12 000. Convert mm to metres by dividing by 1 000.
Scenario
BOARD-TYPE PROBLEM: A road appears 62 mm long on a 1:12 000 scale vertical photo. What is the actual ground length of the road?
Solution
Ground distance = photo distance × S = 62 mm × 12 000 = 744 000 mm = 744 m
Applications
- Determining the appropriate flying height to achieve a required mapping scale.
- Measuring ground distances from aerial photos for preliminary engineering surveys.
- Computing photo scale for NAMRIA topographic mapping and LMB cadastral photogrammetry.
- Estimating map accuracy from photo scale (e.g., 1:50 000 topo maps from 1:40 000 photos).
Misconceptions
- Using the altitude above MSL as H instead of the height above the ground — this is the most common board-exam trap.
- Forgetting to convert units (e.g., f in mm, H in m) — always use consistent units before computing the ratio.
- Inverting the formula: writing Scale = H/f instead of f/H.
- Treating scale as a pure number without the fraction — scale must always be expressed as 1:S or 1/S.
- Assuming scale is constant over the entire photo when terrain is hilly — scale varies with local ground elevation.
Related Concepts
- Camera Focal Length
- Flying Height Determination
- Ground Coverage Computation
- Relief Displacement
Common Exam Questions
Example
f = 152 mm, H = 2 280 m. Scale = 0.152/2280 = 1/15 000.
Approach
Given f and H, directly apply Scale = f/H. Watch for unit inconsistency (mm vs. m).
Question Type
Direct scale computation
Example
Required scale 1:20 000, f = 152 mm. H = 0.152 × 20 000 = 3 040 m above ground.
Approach
Given desired scale 1/S and focal length f, solve for H: H = f × S.
Question Type
Required flying height
Example
A 1.2 km road at 1:8 000 scale. Photo distance = 1 200 m / 8 000 = 0.15 m = 150 mm.
Approach
Photo distance = Ground distance / S. Always match units.
Question Type
Photo distance from ground distance
Key Points To Remember
- Scale = f/H where H is height above the GROUND (not MSL).
- Larger H → smaller scale denominator S is LARGER → smaller features on photo.
- Ground distance = photo distance × S (scale denominator).
- H above ground = altitude (MSL) − mean ground elevation.
- Over hilly terrain, scale is not uniform — use mean elevation for average scale.
- Units must be consistent: both f and H in metres (or both in mm, etc.).
- A scale of 1:10 000 means 1 mm on photo = 10 000 mm = 10 m on the ground.
Ground Coverage and Photo Format
Ground coverage refers to the area of the Earth's surface captured in a single aerial photograph. It is directly computed from the photo format size and the photo scale. **Ground Coverage Formulas:** For a square format of side $d$ (in metres) at scale $1/S$: $$\text{Ground side length} = d \times S$$ $$\text{Ground area covered} = (d \times S)^2$$ For a rectangular format $d_1 \times d_2$: $$\text{Ground area} = (d_1 \times S) \times (d_2 \times S)$$ **Standard Format Reminder:** 230 mm = 0.230 m **Worked Example Logic:** At scale 1:10 000 with 230 mm format: - Ground side = 0.230 m × 10 000 = 2 300 m = 2.3 km - Ground area = 2 300² = 5 290 000 m² = 529 ha **Unit Conversions for Area:** - 1 ha = 10 000 m² - 1 km² = 100 ha = 1 000 000 m² **Effect of Scale on Coverage:** A smaller scale (larger S) covers MORE ground per photo but with less resolution. A larger scale (smaller S) covers LESS ground but with more detail. **Effective Coverage with Overlap:** Because adjacent photos overlap, the net (non-redundant) area per photo in a stereo mission is less than the full photo coverage. With 60% forward overlap (p) and 30% sidelap (q): $$\text{Net forward advance per photo} = (1 - p) \times \text{ground side (along flight)}$$ $$\text{Net sidelap spacing} = (1 - q) \times \text{ground side (across flight)}$$
Examples
Always convert format size from mm to metres first (230 mm = 0.230 m). Multiply by scale denominator to get ground side. Square for area. Convert m² to ha by dividing by 10 000.
Scenario
BOARD-TYPE PROBLEM: A 230 mm × 230 mm format camera is used at a scale of 1:8 000. Calculate: (a) the ground side length covered and (b) the total ground area in hectares.
Solution
(a) Ground side = 0.230 m × 8 000 = 1 840 m (b) Ground area = 1 840² = 3 385 600 m² = 3 385 600 / 10 000 = 338.56 ha ≈ 338.6 ha
Net advance per photo = (1 − overlap fraction) × ground side. Always round UP the number of photos and strips. The board exam typically asks for the approximate number — use the formula approach and round up.
Scenario
BOARD-TYPE PROBLEM: A photogrammetric project requires coverage of a 50 km × 30 km area. If the photo scale is 1:20 000 and the camera format is 230 mm × 230 mm, how many photos (approximately) are needed assuming 60% forward overlap and 30% sidelap?
Solution
Ground side per photo = 0.230 × 20 000 = 4 600 m = 4.6 km Net forward advance = (1 − 0.60) × 4.6 = 0.40 × 4.6 = 1.84 km Net sidelap spacing = (1 − 0.30) × 4.6 = 0.70 × 4.6 = 3.22 km Number of photos along flight direction (50 km): 50 / 1.84 = 27.17 → 28 photos per strip (round up) Number of strips (across 30 km): 30 / 3.22 = 9.32 → 10 strips (round up) Total photos ≈ 28 × 10 = 280 photos (Note: Add 2 photos per strip for end-lap clearance in precise planning: adjusted = 30 × 10 = 300, but the basic board answer is 280.)
Applications
- Flight planning for aerial photogrammetric projects covering Philippine municipalities or provinces.
- Estimating the number of photos and cost of an aerial survey for a DPWH road corridor.
- Computing film/storage requirements for a NAMRIA mapping mission.
- Verifying that specified coverage meets cadastral requirements under PD 1529.
Misconceptions
- Forgetting to convert the format from mm to metres before multiplying by the scale denominator.
- Computing area in m² and forgetting to convert to hectares for the final answer.
- Using overlap percentage directly without applying (1 − p) for net advance — i.e., using 60% instead of 40%.
- Rounding DOWN the number of photos — always round UP to ensure complete coverage.
- Using the total area and dividing by the full (non-overlapped) photo area — this ignores the required overlap.
Related Concepts
- Photo Scale
- Forward Overlap and Sidelap
- Flight Line Planning
- Stereo Coverage
Common Exam Questions
Example
Scale 1:15 000. Ground side = 0.230 × 15 000 = 3 450 m. Area = 3 450² = 11 902 500 m² = 1 190.25 ha.
Approach
Ground side = 0.230 × S (if format is standard 230 mm). Area = (ground side)². Convert to ha.
Question Type
Ground area computation
Example
See worked example above.
Approach
Compute net forward advance = (1−p) × ground side. Net sidelap spacing = (1−q) × ground side. Divide project dimensions by net spacings. Round UP. Multiply strips × photos per strip.
Question Type
Number of photos required
Key Points To Remember
- Ground side = format side (in metres) × scale denominator S.
- Ground area = (ground side)² for square format.
- Standard format is 230 mm = 0.230 m — memorize this conversion.
- 1 ha = 10 000 m²; convert m² to ha by dividing by 10 000.
- Increasing S (smaller scale number like 1:20 000 vs 1:5 000) increases ground coverage.
- Net effective coverage per photo is reduced by the overlap percentages in actual stereo missions.
Overlap, Sidelap, and Stereo Coverage
In a standard aerial photogrammetric mission, consecutive photos along a flight strip overlap each other and adjacent strips overlap laterally. This overlap is essential for: 1. **Stereo viewing** — two overlapping photos form a stereopair, enabling 3D perception and height measurement. 2. **Continuous coverage** — no ground gaps exist even at photo edges. 3. **Redundancy** — ties the strips together for block adjustment. **Forward Overlap (p):** The percentage of a photo's ground coverage that is shared with the next photo in the same strip. Standard value: **60%** (minimum 55%, up to 80% for close-range or precision mapping). **Sidelap (q):** The percentage of a photo's ground coverage shared with the adjacent parallel strip. Standard value: **30%** (minimum 20%). **Air Base (B):** The ground distance between successive exposure stations (camera positions) along the flight line. $$B = (1 - p) \times \text{ground side (along flight direction)}$$ **Strip Spacing (W):** The distance between adjacent flight lines. $$W = (1 - q) \times \text{ground side (across flight direction)}$$ **Photo Base (b):** The image distance corresponding to the air base: $$b = (1 - p) \times d$$ where $d$ is the format side. **Example (standard mission):** At 1:10 000 scale, 230 mm format, 60% overlap, 30% sidelap: - Ground side = 2 300 m - Air base B = (1 − 0.60) × 2 300 = 0.40 × 2 300 = **920 m** - Strip spacing W = (1 − 0.30) × 2 300 = 0.70 × 2 300 = **1 610 m** - Photo base b = (1 − 0.60) × 230 = 92 mm **Why 60% forward overlap?** The 60% overlap ensures that every point on the ground appears in at least two consecutive photos (stereopair), which is the minimum required for stereo measurement. The 30% sidelap ensures that the side margins of adjacent strips are also covered, preventing ground gaps.
Examples
The air base is the ground distance the aircraft travels between exposures. At 60% overlap, only 40% of the ground side is 'new' ground per photo advance. The photo base is the image equivalent. These values are critical for stereo depth perception: a longer base → better depth resolution (base-height ratio).
Scenario
BOARD-TYPE PROBLEM: A vertical aerial survey is conducted at scale 1:10 000 using a 230 mm × 230 mm format camera with 60% forward overlap. Compute: (a) the air base in metres, and (b) the photo base in mm.
Solution
(a) Ground side = 0.230 × 10 000 = 2 300 m Air base B = (1 − 0.60) × 2 300 = 0.40 × 2 300 = 920 m (b) Photo base b = (1 − 0.60) × 230 mm = 0.40 × 230 = 92 mm
This problem combines scale, overlap, and timing. In real flight planning, the shutter interval is set so that the air base matches the required overlap. A slight overshoot of the base (933 m vs 920 m) means slightly less than 60% overlap — acceptable in practice.
Scenario
BOARD-TYPE PROBLEM: At what flying height above ground should an aircraft fly to achieve a scale of 1:10 000 with f = 152 mm, if the forward overlap must be 60% and the interval between exposures is fixed at 12 seconds at a ground speed of 280 km/h?
Solution
First, verify air base from ground speed: Ground speed = 280 km/h = 280 000 m / 3 600 s = 77.78 m/s Air base = speed × time interval = 77.78 × 12 = 933.3 m Check: Required air base at 1:10 000, 230 mm format, 60% overlap: B = (1 − 0.60) × (0.230 × 10 000) = 0.40 × 2 300 = 920 m 933.3 m ≈ 920 m (within 1.5% — acceptable for practical planning) Flying height: H = f × S = 0.152 × 10 000 = 1 520 m above ground.
Applications
- Designing flight plans for NAMRIA aerial mapping missions over Luzon, Visayas, and Mindanao.
- Computing the number of exposures per flight line and total film/storage requirements.
- Setting the automatic intervalometer on the aircraft for the correct exposure interval.
- Planning drone (UAV) photogrammetry missions for barangay-level cadastral surveys under PD 1529.
Misconceptions
- Using the overlap percentage directly as the air base fraction (e.g., B = 0.60 × ground side) — the correct factor is (1 − p), not p.
- Confusing forward overlap (along the flight line) with sidelap (perpendicular to the flight line).
- Ignoring the effect of wind drift on actual sidelap — the flight path may deviate from the planned line.
- Assuming the photo base (image distance) and air base (ground distance) are the same — they differ by the scale factor S.
Related Concepts
- Ground Coverage
- Flight Planning
- Stereo Photogrammetry
- Photo Scale
Common Exam Questions
Example
p = 60%, scale 1:10 000, format 230 mm. B = 0.40 × 2 300 = 920 m.
Approach
B = (1 − p) × (format side × S). Substitute numbers directly.
Question Type
Air base computation
Example
B = 920 m, speed = 200 km/h = 55.56 m/s. Interval = 920/55.56 = 16.56 s.
Approach
Time = Air base / ground speed. Convert speed to m/s first.
Question Type
Exposure interval given aircraft speed
Example
B = 800 m, ground side = 2 300 m. p = 1 − 800/2300 = 1 − 0.348 = 0.652 = 65.2%.
Approach
p = 1 − (B / ground side). Multiply by 100 for percentage.
Question Type
Actual overlap percentage from air base
Key Points To Remember
- Standard forward overlap p = 60%; standard sidelap q = 30%.
- Air base B = (1 − p) × ground side (along flight).
- Strip spacing W = (1 − q) × ground side (across flight).
- Photo base b = (1 − p) × format side d.
- 60% overlap ensures every ground point appears in at least two photos (stereo).
- Higher overlap (e.g., 80%) is used for precision aerial triangulation or dense urban mapping.
- Overlap is measured along the flight direction; sidelap is measured perpendicular to the flight.
Relief Displacement in Vertical Photos
Relief displacement is the radial outward shift of an elevated object's image from where it would appear if the terrain were flat. On a vertical photo, objects above the datum plane appear to be displaced radially away from the principal point (nadir point). **Relief Displacement Formula:** $$d_r = \frac{r \cdot h}{H}$$ where: - $d_r$ = relief displacement on the photo (mm or m) - $r$ = radial distance on the photo from the nadir point to the image of the top of the object (mm) - $h$ = height of the object above the datum (m) - $H$ = flying height above the datum (m) **Physical meaning:** The base of a tall building appears at its true map position, but the top of the building is displaced radially outward by $d_r$. This is why tall buildings appear to 'lean outward' from the center of a vertical photo. **Computing Object Height from Displacement:** $$h = \frac{d_r \cdot H}{r}$$ This is used to measure tree heights, building heights, or terrain relief from a single photo. **Key Observations:** - Relief displacement is zero at the principal/nadir point (r = 0). - Relief displacement increases with distance r from the photo center. - Relief displacement decreases with increasing flying height H. - Objects below the datum (depressions) are displaced toward the center. **Importance:** Relief displacement causes planimetric errors in map compilation from a single unrectified photo. Orthophoto production corrects for relief displacement using a DEM.
Examples
r is measured to the TOP of the object. d_r is the difference in radial distances between top and base images. This technique allows height estimation from a single vertical photo without stereoscopic equipment.
Scenario
BOARD-TYPE PROBLEM: On a vertical photo taken from H = 1 500 m above the base of a telecommunications tower, the image of the tower top is 78 mm from the principal point, and the tower base image is 72 mm from the principal point. The relief displacement (difference) is 78 − 72 = 6 mm. Find the height of the tower.
Solution
d_r = 6 mm (displacement of the top relative to the base) r = 78 mm (radial distance to the top) H = 1 500 m h = d_r × H / r = 6 × 1 500 / 78 = 9 000 / 78 = 115.4 m
Applications
- Estimating tree heights in forest inventory photogrammetry.
- Measuring building heights in urban planning photogrammetry.
- Understanding planimetric error in unrectified photo maps.
- Explaining why orthophoto production requires a DEM for differential rectification.
Misconceptions
- Using r as the radial distance to the BASE of the object instead of the TOP — r must be measured to the top.
- Thinking that the object appears to lean toward the center — elevated objects lean OUTWARD (away from the center).
- Confusing relief displacement with image tilt — these are different effects (tilt displacement is from photo inclination, not terrain).
- Using H above MSL instead of H above the datum/terrain base of the object.
Related Concepts
- Orthophoto Production
- Single-Photo Measurements
- Tilt Displacement
- Digital Elevation Models
Common Exam Questions
Example
r = 65 mm, h = 120 m, H = 1 800 m. d_r = 65 × 120 / 1800 = 7 800/1800 = 4.33 mm.
Approach
Apply d_r = r·h/H. Ensure r is the radial distance to the top of the object. Keep units consistent.
Question Type
Relief displacement computation
Example
d_r = 5 mm, r = 80 mm, H = 2 000 m. h = 5 × 2000 / 80 = 125 m.
Approach
Rearrange to h = d_r × H / r. Identify d_r as the displacement and r as the radial distance to the TOP.
Question Type
Object height from relief displacement
Key Points To Remember
- Relief displacement formula: d_r = r·h/H (always radially outward for elevated objects).
- r = radial distance from the nadir point to the TOP of the object on the photo.
- Object height from displacement: h = d_r·H/r.
- Relief displacement is zero at the nadir; maximum at the photo edges.
- Tall objects appear to 'lean outward' from photo center — called the 'falling-over' effect.
- Orthophotos are free of relief displacement because they are rectified using a DEM.
- Use consistent units: if d_r and r are in mm, h and H must be in same unit ratio.
Practice Problems
Key steps: (1) Always subtract mean ground elevation from MSL altitude to get H above ground. (2) Apply Scale = f/H. (3) Multiply photo measurement by scale denominator for ground distance. The computed scale of 1:16 579 is non-standard but correct — the board exam sometimes uses non-round values to test the formula, not memorization.
Problem
PROBLEM 1: A camera with f = 152 mm is used to photograph an area where the mean ground elevation is 180 m above MSL. The aircraft flies at an altitude of 2 700 m MSL. Compute: (a) the flying height above mean ground, (b) the photo scale, and (c) the ground distance between two points measured as 38.5 mm apart on the photo.
Solution
(a) Flying height above mean ground: H = Altitude (MSL) − Ground elevation H = 2 700 − 180 = 2 520 m (b) Photo scale: Scale = f/H = 0.152/2 520 = 1/16 579 ≈ 1:16 579 (Often rounded to nearest standard scale: ≈ 1:16 600) (c) Ground distance: Ground distance = photo distance × S = 38.5 mm × 16 579 = 638 291 mm = 638.3 m
This is a classic multi-part board problem. Follow the sequence: scale → ground side → area → air base → strip spacing. Remember: air base uses (1 − p) and strip spacing uses (1 − q). Area conversion: divide m² by 10 000 to get hectares.
Problem
PROBLEM 2: A standard 230 mm × 230 mm format camera is used at a scale of 1:12 000. Calculate: (a) the ground dimensions covered by one photo, (b) the ground area in hectares, (c) the air base assuming 60% forward overlap, and (d) the strip spacing assuming 30% sidelap.
Solution
(a) Ground dimensions: Ground side = 0.230 m × 12 000 = 2 760 m Dimensions: 2 760 m × 2 760 m (b) Ground area: Area = 2 760² = 7 617 600 m² = 7 617 600 / 10 000 = 761.76 ha (c) Air base (B) with 60% forward overlap: B = (1 − 0.60) × 2 760 = 0.40 × 2 760 = 1 104 m (d) Strip spacing (W) with 30% sidelap: W = (1 − 0.30) × 2 760 = 0.70 × 2 760 = 1 932 m
This is a complete flight planning problem. Key decisions: (1) use H = f × S for flying height, not H = f/S; (2) net advance per photo = (1−p) × ground side; (3) always round UP the number of photos and strips; (4) some examiners add 2 photos/strip for clearance — both 396 and 420 may be accepted depending on the key.
Problem
PROBLEM 3: A NAMRIA photogrammetric crew must cover a rectangular project area of 45 km (E-W) × 28 km (N-S). The camera has f = 210 mm, format = 230 mm × 230 mm. The required scale is 1:15 000 with 60% forward overlap (E-W flight direction) and 30% sidelap. Determine: (a) the flying height above ground, (b) the ground coverage per photo, (c) the number of photos per strip, (d) the number of strips, and (e) the approximate total number of photos.
Solution
(a) Flying height: H = f × S = 0.210 × 15 000 = 3 150 m above ground (b) Ground side per photo: Ground side = 0.230 × 15 000 = 3 450 m = 3.45 km (c) Air base: B = (1 − 0.60) × 3 450 = 0.40 × 3 450 = 1 380 m = 1.38 km Number of photos per strip (E-W direction, 45 km): No. = 45 / 1.38 = 32.6 → round up to 33 Add 2 for end-lap clearance: 33 + 2 = 35 photos per strip (Note: board exams may accept 33 or 35 — use 33 if end-lap is not mentioned) (d) Strip spacing: W = (1 − 0.30) × 3 450 = 0.70 × 3 450 = 2 415 m = 2.415 km Number of strips (N-S direction, 28 km): No. = 28 / 2.415 = 11.6 → round up to 12 strips (e) Total photos: Using 33 photos/strip: Total = 33 × 12 = 396 photos Using 35 photos/strip: Total = 35 × 12 = 420 photos
The relief displacement is the difference in radial distances of the top and base images (d_r = 6 mm). The radial distance used in the formula is that of the TOP of the object (r = 94 mm). Flying height is above the datum at the building's base. Result: building height ≈ 127.7 m.
Problem
PROBLEM 4: On a vertical aerial photo taken from H = 2 000 m above the ground, the image of the top of a building measures 94 mm from the principal point, while the building's base image measures 88 mm from the principal point. Compute the height of the building.
Solution
Relief displacement: d_r = r(top) − r(base) = 94 − 88 = 6 mm Radial distance to top: r = 94 mm Building height: h = (d_r × H) / r = (6 × 2 000) / 94 = 12 000 / 94 = 127.7 m ≈ 127.7 m
Total flight distance = (number of strips) × (length of each strip). This is often asked in flight planning problems. Add transit distance between strips if the problem specifies it. For board exams, use the basic formula: total = strips × strip length.
Problem
PROBLEM 5 (Integration): An aircraft photographs the Metro Manila area at a scale of 1:10 000 using a 152 mm camera. A township (roughly rectangular) is 8 km × 12 km. With 60% forward overlap and 30% sidelap, and the aircraft flying N-S along the 12 km dimension: (a) compute the air base, (b) the strip spacing, (c) the number of photos per strip, (d) the number of strips, and (e) the total flight distance in km.
Solution
(a) Ground side = 0.230 × 10 000 = 2 300 m = 2.3 km Air base B = (1 − 0.60) × 2 300 = 920 m = 0.92 km (b) Strip spacing W = (1 − 0.30) × 2 300 = 1 610 m = 1.61 km (c) Photos per strip (N-S, 12 km): No. = 12 / 0.92 = 13.04 → 14 photos per strip (+2 for clearance → 16, but use 14 for basic answer) (d) Strips (E-W, 8 km): No. = 8 / 1.61 = 4.97 → 5 strips (e) Total flight distance: Each strip = 12 km (N-S) Total flight distance = number of strips × strip length = 5 × 12 = 60 km (Plus transit distance between strips — not typically required in board exams)
Exam Preparation Tips
- MEMORIZE the core formula: Scale = f/H, where H is ALWAYS above the GROUND, never above MSL. This single mistake causes the most errors in board exams.
- ALWAYS convert focal length to metres before computing scale (152 mm = 0.152 m). Keep units consistent throughout the solution.
- MEMORIZE the standard values: format = 230 mm × 230 mm, forward overlap p = 60%, sidelap q = 30%. These appear in almost every flight-planning problem.
- For air base: B = (1 − p) × ground side. Note the factor is (1 − p) = 0.40 for 60% overlap, NOT 0.60. This is the most common conceptual error.
- For relief displacement: d_r = r·h/H, where r is measured to the TOP of the object. The formula can be rearranged to find h when d_r and r are given.
- When the problem gives altitude above MSL and terrain elevation, ALWAYS subtract: H = altitude − terrain elevation. Write this as the first step.
- ROUND UP — never round down — the number of photos per strip and the number of strips. Rounding down leaves uncovered ground, which is unacceptable.
- Unit conversion drill: 1 m = 1 000 mm; 1 ha = 10 000 m²; 1 km = 1 000 m. Practice converting between these in under 5 seconds.
- Ground coverage area = (0.230 × S)² for standard format. At 1:10 000 this is (2 300)² = 5 290 000 m² = 529 ha. Memorize this benchmark value.
- On the board exam, if a problem combines scale, coverage, and overlap — solve in strict sequence: (1) flying height, (2) photo scale, (3) ground side, (4) air base/strip spacing, (5) number of photos/strips.
- Verify your answer with a reasonableness check: a 1:10 000 scale photo covers about 529 ha; a 1:50 000 photo covers about 132 000 ha. Does your answer make sense?
- For multiple-choice questions, if you compute a non-standard scale (e.g., 1:16 579), check if the choices are rounded — pick the closest standard value.
- The Philippine legal context: aerial photogrammetry for cadastral purposes is governed by PD 1529 (Property Registration Decree). NAMRIA is the primary government agency for national mapping and aerial survey. Knowing this context helps in applied/policy questions.
- Practice relief displacement problems by always identifying three variables: d_r (displacement), r (radial distance to TOP), and H (flying height). Given any two, solve for the third.
- In flight planning problems, the total number of photos is (photos/strip) × (number of strips) — always present this as the final answer even if intermediate steps are asked.
In summary
Aerial photography and camera geometry is a high-yield topic in the PRC Geodetic Engineer Licensure Examination. The entire chapter revolves around one elegant principle: the vertical aerial photo is a central projection, and all computations flow from the similar-triangle relationship Scale = f/H. From this single formula, you can derive photo scale, ground coverage, air base, strip spacing, number of photos, and even object heights through relief displacement. The three most common board-exam pitfalls are: (1) using altitude above MSL instead of H above ground; (2) using the overlap fraction p directly instead of (1 − p) for the air base; and (3) forgetting to convert focal length from mm to metres. Master these three points and you will correctly solve virtually every numerical problem in this chapter. In the context of Philippine geodetic practice, photogrammetry is governed by RA 8560 (Geodetic Engineering Act) and RA 4374, with NAMRIA as the primary government agency for aerial mapping. Cadastral photogrammetry operates within the framework of PD 1529 and CA 141. Understanding both the mathematical geometry and the legal-professional context will give you a complete picture for the licensure exam and for your career as a geodetic engineer in the Philippines. Practice the five worked problems in this chapter under exam conditions: no notes, time limit, show all unit conversions. Build speed and accuracy with the formulas, and you will approach this topic with full confidence on examination day.
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