GELE Photogrammetry & Cartography — Scale, Relief Displacement and ParallaxDetailed Explanation
The Scale, Relief Displacement and Parallax 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 Scale, Relief Displacement and Parallax in the 2nd 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.
Scale, Relief Displacement and Parallax - Detailed Explanation
In photogrammetry, aerial photographs are not perfect maps — they contain systematic geometric distortions caused by the camera perspective, terrain elevation differences, and the overlap between adjacent photos. Three of the most fundamental and board-exam-critical concepts governing these distortions are: (1) photo scale, (2) relief displacement, and (3) stereoscopic parallax. Mastery of these topics allows a geodetic engineer to extract quantitative measurements — including object heights and terrain elevations — directly from aerial photography. These concepts appear consistently in the PRC Geodetic Engineer Licensure Examination under the Photogrammetry and Cartography subject area and require both conceptual understanding and computational proficiency.
Concepts
Photo Scale
The scale of a vertical aerial photograph expresses the ratio of a distance on the photo to the corresponding distance on the ground. For a truly vertical photo taken over flat terrain, the photo scale S is given by: S = f / H where f = focal length of the camera lens (in metres or millimetres) and H = flying height above the ground (in the same units as f). Because real terrain is not flat, photo scale varies from point to point across the image. A point higher in elevation (closer to the camera) appears at a larger scale than a point in a valley (farther from the camera). This variable scale is the root cause of relief displacement. For a specific ground point at elevation h above the datum: S_point = f / (H - h) The nominal or average scale of a project is computed using the average terrain elevation. In Philippine aerial survey projects referenced to PRS92 and flown for NAMRIA topographic mapping, the nominal scale is always computed against mean sea level (the vertical datum). Scale can also be expressed as a Representative Fraction (RF): e.g., 1:25 000 means 1 mm on the photo equals 25 000 mm (25 m) on the ground.
Examples
Convert f to metres first: 152.4 mm = 0.1524 m. The resulting scale of approximately 1:20 000 is a standard NAMRIA mapping scale. Note that at a hilltop 300 m above MSL, the local scale would be S = 0.1524 / (3000 - 300) = 0.1524 / 2700 = 1:17 717, which is a larger scale.
Scenario
A NAMRIA aerial survey is flown at H = 3000 m above mean sea level using a camera with f = 152.4 mm. Compute the nominal photo scale.
Solution
S = f / H = 0.1524 m / 3000 m = 1/19 685 ≈ 1:20 000 (nominal)
Ground distance must be in the same units as photo distance (convert 500 m to 500 000 mm). The focal length of 162.5 mm corresponds approximately to a wide-angle mapping camera.
Scenario
Two ground points A and B are 500 m apart. On the aerial photo, the image distance between A' and B' is 28.5 mm. Compute the photo scale and the focal length if the flying height is 2850 m.
Solution
Scale S = photo distance / ground distance = 28.5 mm / 500 000 mm = 1:17 544. Focal length f = S × H = (1/17544) × 2850 m = 0.1625 m = 162.5 mm
Applications
- Computing ground distances from photo measurements
- Designing aerial survey flight plans (determining flying height for a required map scale)
- Estimating planimetric accuracy of photogrammetric products
- Determining ground sampling distance (GSD) for UAV/drone surveys under CAAP regulations
- Scaling stereo models in analytical and digital photogrammetry workstations
Misconceptions
- Confusing flying height above ground (AGL) with flying height above datum (AMSL) — always clarify the vertical reference
- Forgetting to convert focal length from mm to m before computing scale as a simple ratio
- Assuming photo scale is constant across an image — it varies with terrain elevation
- Treating 1:50 000 as a larger scale than 1:25 000 — it is actually a smaller scale
Related Concepts
- Relief displacement
- Stereoscopic parallax
- Flight planning and overlap
- Ground sampling distance (GSD)
- Map projection and coordinate systems (PPCS/UTM, PRS92)
Common Exam Questions
Example
A camera with f = 88.9 mm is flown at 1500 m AGL. What is the photo scale? Answer: S = 0.0889/1500 = 1:16 872 ≈ 1:17 000
Approach
Apply S = f/H directly; ensure consistent units
Question Type
Direct computation of photo scale
Example
Required scale 1:20 000, f = 152.4 mm. H = 0.1524 × 20 000 = 3048 m
Approach
Rearrange S = f/H to get H = f/S
Question Type
Find flying height given scale and focal length
Example
H = 2000 m, h = 400 m, f = 100 mm. S = 0.100/(2000-400) = 0.100/1600 = 1:16 000
Approach
Use S = f/(H-h); treat H as flying height above datum, h as point elevation
Question Type
Scale at a specific elevated point
Key Points To Remember
- S = f / H for vertical photo over flat terrain
- Photo scale increases (larger scale) as H decreases or as terrain rises toward the camera
- S_point = f / (H - h) at a point with elevation h above datum
- Scale varies across an aerial photo due to terrain relief
- Always check units: f and H must be in the same unit before dividing
- Nominal scale uses average terrain elevation
- A larger RF denominator means a smaller (more reduced) scale
Relief Displacement
Relief displacement is the radial outward shift of an image point from its true planimetric position, caused by the elevation of the object above the datum. On a vertical photograph, a tall object (like a building, tower, or mountain peak) does not appear directly above its base — its top is displaced radially outward from the principal point. The fundamental relief displacement formula is: d = (r × h) / H where: d = relief displacement (in mm, the same unit as r) r = radial distance of the image TOP of the object from the principal point (mm) h = height of the object above the datum (m) H = flying height of the aircraft above the datum (m) Rearranged to solve for object height: h = (d × H) / r Key geometric facts: 1. Displacement is ZERO at the principal point (r = 0) — objects directly below the camera show no displacement. 2. Displacement INCREASES toward the photo edges (larger r). 3. The direction is strictly RADIAL and OUTWARD from the principal point. 4. A taller object at the same r has larger displacement (d ∝ h). 5. A lower flying height produces larger displacement (d ∝ 1/H). This is why vertical aerial photos make tall buildings appear to 'lean outward' like a falling tree — a phenomenon called the 'tip-of-tree effect' in forestry photogrammetry used in DENR mapping projects.
Examples
The top of the tower appears 2.667 mm farther from the principal point than its base on the photo. This is measurable with a precise ruler or comparator. Note that both r and d are in mm, while h and H are in m — the ratio h/H is dimensionless.
Scenario
A vertical aerial photo is taken from H = 1500 m above datum. A communications tower has its top imaged at r = 80 mm from the principal point. The tower is 50 m tall. Compute the relief displacement.
Solution
d = (r × h) / H = (80 mm × 50 m) / 1500 m = 4000/1500 = 2.667 mm
This is the inverse problem — given d and r measured on the photo, we solve for h. This is the standard board-exam form. Always identify which variable is unknown before selecting the formula form.
Scenario
On a photo taken from H = 1500 m above datum, a building image shows a relief displacement of d = 2.667 mm measured at radial distance r = 80 mm from the principal point. Determine the building height.
Solution
h = (d × H) / r = (2.667 mm × 1500 m) / 80 mm = 4000.5 / 80 = 50.0 m
Here H = 2400 m is measured above MSL (the datum), consistent with the building base elevation also referenced to MSL. The formula gives the height of the object (rooftop above base), not the elevation of the rooftop. This distinction is critical.
Scenario
An aerial photo over Metro Manila is taken at H = 2400 m above MSL. A high-rise building in BGC has its rooftop image at r = 112 mm from the principal point, and its base is at 30 m above MSL. The relief displacement between rooftop and base images is 4.5 mm. Find the building height.
Solution
h = (d × H) / r = (4.5 × 2400) / 112 = 10800 / 112 = 96.4 m ≈ 96 m
Applications
- Measuring heights of buildings, towers, trees, and terrain features from single aerial photos
- Forest inventory (tree height measurement) in DENR LiDAR and aerial survey programs
- Identifying and correcting geometric distortions in photogrammetric products
- Orthophoto production — relief displacement is eliminated to produce a true orthophoto
- Building footprint extraction for cadastral mapping under PD 1529 requirements
Misconceptions
- Using the radial distance to the BASE instead of the TOP of the object for r — the formula uses the top image position
- Confusing relief displacement (radial, due to elevation) with tilt displacement (due to camera tilt) — these are distinct phenomena
- Thinking displacement makes the image shorter — it shifts the top radially outward, making tall objects appear tilted
- Using H above ground level instead of H above datum when h is measured from the datum
- Assuming displacement is the same across the photo — it varies with r
Related Concepts
- Photo scale
- Principal point and principal plane
- Orthophoto and orthorectification
- Tilt displacement
- Digital Elevation Model (DEM)
Common Exam Questions
Example
r = 95 mm, h = 60 m, H = 1800 m → d = (95×60)/1800 = 5700/1800 = 3.167 mm
Approach
Direct substitution into d = rh/H; check that r and d will have same units
Question Type
Compute relief displacement d
Example
d = 3.0 mm, r = 100 mm, H = 2000 m → h = (3.0×2000)/100 = 60 m
Approach
Use h = dH/r; this is the practical measurement application
Question Type
Compute object height h from measured d
Example
A feature directly below the camera (at the principal point) shows zero relief displacement regardless of its height.
Approach
Recall geometric principle: zero at principal point, maximum at photo corners
Question Type
Conceptual: Where is displacement zero/maximum?
Key Points To Remember
- d = rh/H — relief displacement formula; r is measured to the TOP of the object
- h = dH/r — rearrangement to find object height
- Displacement is radially OUTWARD from the principal point
- Displacement is ZERO at the principal point (r = 0)
- Maximum displacement occurs at the photo corners (maximum r)
- H is flying height ABOVE THE DATUM (not above ground if datum ≠ ground)
- Units: r and d must be in the same unit (both mm); h and H in the same unit (both m)
- Relief displacement is exploited to measure heights of objects from single photos
Stereoscopic Parallax
Parallax is the apparent displacement of an object's image position when viewed from two different camera stations — the two successive exposure positions along the flight line. When adjacent photos in a stereopair are viewed together, a point on the ground appears to shift laterally between the left and right photos. This shift, measured parallel to the flight line (the x-axis of the photo), is called the absolute stereoscopic parallax. The absolute parallax P of a point is defined as the algebraic sum of the x-coordinates of that point's images on the left and right photos: P = x_L + |x_R| (or P = x_L - x_R when using the signed convention) For practical height determination, the parallax DIFFERENCE (Δp) between the top and base of an object is used. The height formula derived from the geometry of the stereo model is: h = (H × Δp) / (P + Δp) For small Δp relative to P (which is usually the case), the approximation is: h ≈ (H × Δp) / P where: h = height of the object (m) H = flying height above the datum (m) P = absolute parallax of the base of the object (mm) Δp = parallax difference between top and base (mm, always positive for elevated features) Parallax is also the basis of stereoscopic vision in photogrammetry — the human visual system (or a photogrammetric workstation) uses parallax differences to perceive depth and reconstruct 3D terrain models. In digital photogrammetry software (e.g., ERDAS Imagine, Agisoft Metashape), parallax matching is performed automatically by image correlation algorithms. The air base B is the distance between two successive camera stations. The relationship between B, H, f, and P is: P = B × f / H (for a simple stereo model) This shows that a longer air base gives more parallax and better height accuracy — which is why 60% forward overlap (not 100%) is used in standard photogrammetric flights.
Examples
The exact formula gives 24.6 m. Using the approximation: h ≈ (1500 × 1.5)/90 = 25.0 m — an overestimate of about 1.6%, acceptable for most practical purposes. For board exams, use the exact formula unless instructed otherwise.
Scenario
A stereopair taken from H = 1500 m has an absolute parallax of P = 90 mm at the base of a chimney. The parallax difference between the top and base of the chimney is Δp = 1.5 mm. Find the chimney height.
Solution
h = (H × Δp) / (P + Δp) = (1500 × 1.5) / (90 + 1.5) = 2250 / 91.5 = 24.59 m ≈ 24.6 m
This forest inventory application is highly practical in the Philippines under DENR aerial survey programs. The parallax method gives tree heights with ±1–2 m accuracy from aerial photography, comparable to field measurement for large-scale forest inventories.
Scenario
A stereopair over a forested area in Mindanao shows P = 88 mm at tree base level, and Δp = 2.2 mm between the tree crown and base. Flying height H = 1600 m. Find the tree height.
Solution
h = (1600 × 2.2) / (88 + 2.2) = 3520 / 90.2 = 39.02 m ≈ 39.0 m
Note: f must be in mm and B in m converted to the photo scale: alternatively, P = B × S × ... The cleaner approach is P = f × (B/H) where B and H are in metres, f in mm. P = 152.4 × (900/3000) = 152.4 × 0.30 = 45.72 mm.
Scenario
Two photos have a 60% forward overlap. The air base B = 900 m, focal length f = 152.4 mm, and flying height H = 3000 m. Compute the absolute parallax P.
Solution
P = (B × f) / H = (900 m × 0.1524 m) / 3000 m... wait, units: P = B×f/H in consistent units. P = (900 × 152.4 mm) / 3000 = 137160/3000 = 45.72 mm
Applications
- Height determination of buildings, towers, and natural features from stereopairs
- Digital Elevation Model (DEM) and Digital Terrain Model (DTM) generation
- Contour mapping from aerial photography for topographic maps
- Forest canopy height mapping for DENR carbon stock assessment
- Stereo compilation of cadastral and topographic maps under NAMRIA standards
- Automated point cloud generation in Structure-from-Motion (SfM) photogrammetry
Misconceptions
- Measuring parallax perpendicular to the flight line — parallax is always parallel to the flight line (x-direction)
- Confusing absolute parallax P with parallax difference Δp — P is for a single point, Δp is between two points (top and base)
- Using flying height above ground instead of above datum when h includes terrain elevation
- Thinking 100% overlap is needed for stereo viewing — 60% forward overlap is the standard, providing adequate parallax
- Assuming the approximate formula h ≈ HΔp/P always gives exact results — it overestimates h when Δp is significant
Related Concepts
- Stereoscopic viewing and depth perception
- Air base and photo overlap
- Digital Elevation Model (DEM) generation
- Relief displacement
- Photo scale
- Bundle adjustment in digital photogrammetry
Common Exam Questions
Example
P = 88 mm, Δp = 2.2 mm, H = 1600 m → h = (1600×2.2)/(88+2.2) = 3520/90.2 = 39.0 m
Approach
Apply h = HΔp/(P+Δp); identify P as base parallax and Δp as the difference
Question Type
Compute object height from parallax difference
Example
P = 90 mm, Δp = 1.5 mm, H = 1500 m → h ≈ (1500×1.5)/90 = 25.0 m (exact: 24.6 m)
Approach
Use approximation when Δp << P (less than 5% of P); note error vs. exact
Question Type
Approximate height using h ≈ HΔp/P
Example
If the flight line runs north-south, parallax differences are measured in the north-south direction on the photo pair.
Approach
Parallax is measured PARALLEL to the flight line, not perpendicular
Question Type
Conceptual: direction of parallax measurement
Key Points To Remember
- Parallax is measured PARALLEL to the flight line (x-direction)
- Δp = parallax of top minus parallax of base (positive for objects above datum)
- h = HΔp / (P + Δp) — exact formula; h ≈ HΔp/P — approximation for small Δp
- P is the absolute parallax at the BASE of the object
- Larger Δp means taller object; smaller P (higher flying) means greater height sensitivity
- Standard forward overlap is 60% to ensure adequate parallax for stereo viewing
- Parallax is measured using a parallax bar (stereoscopic comparator) or digitally via image correlation
- Parallax differences across a stereo model generate a DEM
Practice Problems
The flagpole tip appears 3.167 mm farther from the principal point than its base. This is measurable with a precision ruler. The answer confirms that at r = 95 mm (near the edge of a standard 230 mm × 230 mm format photo), displacement is significant enough to cause noticeable image lean.
Problem
Problem 1 — Relief Displacement (Board Level) An aerial photo is taken vertically from H = 1800 m above datum. A flagpole located at r = 95 mm from the principal point has a height of h = 60 m. Compute the relief displacement d of the flagpole's tip.
Solution
Given: H = 1800 m, r = 95 mm, h = 60 m Formula: d = rh/H d = (95 mm × 60 m) / 1800 m d = 5700 / 1800 d = 3.167 mm
This is the standard inverse problem: given measurements from the photo (d and r) and the flying height H, compute the real-world object height h. This approach is used in practice to measure heights of structures from aerial photography without field measurement.
Problem
Problem 2 — Object Height from Relief Displacement On a vertical photo from H = 2000 m above datum, a pole shows a relief displacement of d = 3.0 mm with its tip image at radial distance r = 100 mm from the principal point. Determine the height of the pole.
Solution
Given: H = 2000 m, d = 3.0 mm, r = 100 mm Formula: h = dH/r h = (3.0 mm × 2000 m) / 100 mm h = 6000 / 100 h = 60.0 m
The exact formula is preferred in board exams. The approximation (h ≈ HΔp/P) overestimates by 2.6% here because Δp/P = 2.2/88 = 2.5% — not negligible. As a rule of thumb, use the exact formula when Δp > 1% of P.
Problem
Problem 3 — Height from Parallax Difference A stereopair is taken from H = 1600 m above datum. The absolute parallax at the base of a communications tower is P = 88 mm. The parallax difference between the top and base of the tower is Δp = 2.2 mm. Compute the tower height h using (a) the exact formula and (b) the approximation.
Solution
(a) Exact formula: h = (H × Δp) / (P + Δp) h = (1600 × 2.2) / (88 + 2.2) h = 3520 / 90.2 h = 39.02 m ≈ 39.0 m (b) Approximate formula: h ≈ (H × Δp) / P h ≈ (1600 × 2.2) / 88 h ≈ 3520 / 88 h ≈ 40.0 m Percentage error of approximation: (40.0 - 39.0)/39.0 × 100 = 2.6%
This problem illustrates why a single map scale cannot describe an uncorrected aerial photo over mountainous terrain. The 600 m elevation difference between Points A and B causes a 27.3% scale variation — a significant source of planimetric error. Orthorectification eliminates this variation by projecting all image points to a common datum using a DEM.
Problem
Problem 4 — Photo Scale Variation with Terrain An aerial camera has f = 152.4 mm and is flown at H = 3000 m above MSL. Two ground points are photographed: Point A at elevation 200 m MSL and Point B at elevation 800 m MSL. Compute the photo scale at each point and determine the percentage difference.
Solution
Scale at Point A: S_A = f / (H - h_A) = 0.1524 / (3000 - 200) = 0.1524 / 2800 = 1/18 373 ≈ 1:18 400 Scale at Point B: S_B = f / (H - h_B) = 0.1524 / (3000 - 800) = 0.1524 / 2200 = 1/14 435 ≈ 1:14 400 Percentage difference in scale: ΔS/S_A = (1/14435 - 1/18373) / (1/18373) × 100 = (18373 - 14435) / 14435 × 100 = 3938 / 14435 × 100 = 27.3% Point B (higher elevation) has a 27.3% larger scale than Point A.
Part (a) shows that even at moderate radial distance (75 mm), a 30 m building produces about 1 mm displacement — detectable on a precision photo. Part (b) demonstrates that the parallax method independently estimates height from a stereopair. Small differences arise from measurement errors in Δp (at the sub-millimetre level). For board exams, apply formulas consistently using H above the same datum for both methods.
Problem
Problem 5 — Combined Relief Displacement and Parallax A stereopair over Baguio City is flown at H = 2200 m above MSL. A building at 1400 m MSL has its rooftop image at r = 75 mm from the principal point on the left photo. (a) Compute the relief displacement of the rooftop relative to its base if the building is 30 m tall. (b) If the absolute parallax at the building base is P = 62 mm and Δp = 1.0 mm, compute the building height using the parallax method.
Solution
(a) Relief Displacement: H above datum = 2200 m h = 30 m (building height), r = 75 mm d = rh/H = (75 × 30) / 2200 = 2250 / 2200 = 1.023 mm (b) Parallax Height: h = HΔp / (P + Δp) = (2200 × 1.0) / (62 + 1.0) = 2200 / 63 = 34.9 m Note: The parallax method gives 34.9 m, slightly different from the 30 m, illustrating measurement sensitivity. In practice, both methods should agree within instrument precision.
Exam Preparation Tips
- MEMORIZE THE THREE CORE FORMULAS: d = rh/H (relief displacement), h = dH/r (object height from displacement), h = HΔp/(P+Δp) (height from parallax). These appear in almost every Photogrammetry board exam.
- UNIT CONSISTENCY IS CRITICAL: In relief displacement problems, r and d must be in the same unit (typically mm); h and H must be in the same unit (typically m). The ratio h/H is dimensionless — this is what makes the formula work.
- IDENTIFY THE UNKNOWN FIRST: Board exam problems give you three of the four variables (d, r, h, H) and ask for the fourth. Identify what is given and what is asked before selecting the formula form.
- USE THE EXACT PARALLAX FORMULA: h = HΔp/(P+Δp) is the correct formula. The approximation h ≈ HΔp/P overestimates h and may lead to wrong multiple-choice answers. Use the exact form unless the problem specifies the approximation.
- KNOW THE GEOMETRY: Relief displacement is (1) radial from the principal point, (2) outward for elevated features, (3) zero at the principal point, (4) maximum at the photo corners. Parallax is (1) parallel to the flight line, (2) larger for higher objects, (3) the basis of stereoscopic height determination.
- PHOTO SCALE: S = f/H for flat terrain; S = f/(H-h) for elevated points. A larger denominator means a smaller scale. Flying height H is always above the same datum as the terrain elevation h.
- DISTINGUISH d FROM Δp: Relief displacement d is a single-photo measurement (radial). Parallax difference Δp is a stereopair measurement (parallel to flight line). They are different phenomena measured differently.
- PRACTICE UNIT CONVERSIONS: Convert f from mm to m when computing S as a ratio (e.g., 152.4 mm = 0.1524 m). Keep P and Δp in mm throughout parallax computations.
- CONCEPTUAL QUESTIONS: Know WHY relief displacement is zero at the principal point (r = 0 in d = rh/H). Know WHY heights are measured near photo edges for relief displacement (larger r → larger d → easier to measure). Know that 60% forward overlap provides the air base needed for parallax.
- REVIEW NAMRIA AND DENR CONTEXT: Philippine board exams often set problems in the context of NAMRIA aerial surveys, DENR forest mapping, or cadastral surveys under PD 1529. Recognizing these contexts helps interpret the problem setup correctly.
In summary
Relief displacement and stereoscopic parallax are the two geometric pillars of quantitative photogrammetry. Relief displacement (d = rh/H) operates on a single photo, displacing elevated image points radially outward from the principal point — a distortion that can be exploited to measure object heights or must be removed to produce accurate orthophotos. Stereoscopic parallax (h = HΔp/[P+Δp]) operates on overlapping stereopairs, allowing precise height determination of any feature visible in both images and forming the mathematical basis of all DEM and contour generation. For the PRC Geodetic Engineer Licensure Examination, mastery of these concepts requires three capabilities: (1) memorizing and correctly applying the core formulas, (2) maintaining strict unit consistency throughout all calculations, and (3) understanding the underlying geometry so that conceptual questions — about direction of displacement, location of zero displacement, measurement axis of parallax — can be answered with confidence. In the Philippine professional context, these principles underpin NAMRIA's topographic mapping programs, DENR's aerial forest surveys, and cadastral surveys conducted under PD 1529 and RA 8560. Whether working with traditional stereoplotters or modern digital photogrammetry software, the geodetic engineer who understands these fundamentals is equipped to produce, evaluate, and quality-control photogrammetric products to the highest professional standards.
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