Aerial Photogrammetry: Photographs, Scale, Relief Displacement, Stereoscopy, Parallax and the DEM
1. Types of photographs
| Type | Camera axis | Character |
|---|---|---|
| Vertical | within about 3° of the plumb line (unintentional tilt only) | nearly uniform scale over flat ground; the basis of mapping |
| Tilted | unintentionally tilted beyond about 3° | scale varies across the photo; needs rectification |
| Low oblique | intentionally inclined; horizon not visible | natural view, reconnaissance |
| High oblique | strongly inclined; horizon visible | large area, strongly varying scale |
Photographs are also terrestrial (camera on the ground) or aerial; taken with metric cameras (calibrated focal length and lens distortion, fiducial marks that locate the principal point where the optical axis meets the image) or non-metric ones; and film or digital. Three points matter on a photograph: the principal point; the nadir (plumb point), vertically below the lens; and the isocentre, halfway between them on the principal line. On a truly vertical photograph all three coincide. Relief displacement is radial from the nadir; tilt displacement is radial from the isocentre. A standard aerial frame is 23 cm × 23 cm, and a common wide-angle focal length is about 152 mm.
2. Flying height, scale, coverage and overlap
By similar triangles through the lens, the scale of a vertical photograph at a point of elevation h is S = f/(H − h), where f is the focal length and H the flying height above the datum. Scale is therefore larger over hills than over valleys; the average scale uses the average terrain height. With f = 150 mm, H = 2000 m and terrain at 500 m, S = 0.15/1500 = 1:10 000. Rearranged, the flying height for a desired scale 1:n over terrain at h is H = f·n + h: 1:12 000 with f = 152 mm over terrain at 300 m needs H = 1824 + 300 = 2124 m. Scale can also be found by comparing a photo distance with the same distance on a map: 8 cm on the photo against 2 cm on a 1:50 000 map (1 km) gives 1:12 500.
Coverage and overlap. A 23 cm frame at 1:10 000 covers 2.3 km × 2.3 km = 5.29 km². Consecutive photographs overlap along the strip by about 60 % (forward overlap or end lap), so every point appears on at least two photos for stereo viewing; adjacent strips overlap by about 30 % (side lap). The air base (distance between exposures) is B = (1 − 0.60) × 2.3 km = 0.92 km, and the spacing of flight lines is W = (1 − 0.30) × 2.3 km = 1.61 km. At a ground speed of 200 km/h (55.6 m/s) the exposure interval is 920/55.6 = 16.6 s. The ground sampling distance of a digital frame camera is pixel size × scale number: 6 µm at 1:10 000 is 6 cm.
3. Relief (height) displacement
On a vertical photograph, the image of the top of an object of height h stands farther from the nadir than the image of its base. The shift, the relief displacement, is d = r·h/H, where r is the radial distance of the displaced (top) image from the nadir and H the flying height above the base of the object. It is radial from the nadir, outward for points above the datum and inward for points below; zero at the nadir; larger for taller objects, lower flying heights and points farther from the centre. Worked: a tower 60 m high, its top imaged 80 mm from the nadir, from 1200 m above its base: d = 80 × 60/1200 = 4 mm. Turned round, it measures height: h = d·H/r; a displacement of 2.5 mm at r = 75 mm from 1500 m gives h = 50 m.
4. Stereoscopy and heights from parallax
Stereoscopic vision fuses two views taken from different positions into one three-dimensional impression, because the brain reads depth from the difference between them. Two overlapping photographs viewed so that each eye sees one — with a lens (pocket) stereoscope for small separations or a mirror stereoscope for full frames — give a stereo model. The model appears vertically exaggerated, because the air base-to-height ratio B/H of the photography (about 0.6 for 60 % overlap with a wide-angle lens: 920/1500) is much larger than the equivalent ratio of the eyes at viewing distance.
The absolute stereoscopic parallax (x-parallax) of a point is p = x − x′, the difference of its x-coordinates on the two photos measured parallel to the flight line. Higher points have larger parallax. With air base B and focal length f, the height of a point is h = H − B·f/p. The parallax bar (stereometer) measures parallax differences Δp precisely, and the height difference between a point and a reference (base) point of parallax p_b is Δh = (H − h_b)·Δp/(p_b + Δp). Worked: H = 1500 m above a base point at the datum, B = 900 m, f = 150 mm; the base point has p_b = Bf/(H − 0) = 90 mm. A point with Δp = 2.4 mm is Δh = 1500 × 2.4/92.4 = 38.96 m higher — and h = 1500 − 900 × 0.15/0.0924 = 38.96 m gives the same answer. For small Δp, Δh ≈ (H − h_b)Δp/p_b is a useful approximation.
5. The digital elevation model
A DEM represents terrain elevations digitally, usually as a regular grid (raster), sometimes as a TIN or contours. A DSM (digital surface model) records the top of everything — canopy, buildings; a DTM (digital terrain model) records the bare ground, and often includes breaklines and spot heights. Sources: stereo photogrammetry (automatic image matching of overlapping photos or satellite stereo pairs such as those from India's Cartosat-1), LiDAR (laser ranging; multiple returns let the last return reach the ground under vegetation), InSAR (radar interferometry; the Shuttle Radar Topography Mission of 2000 produced near-global elevation at about 30 m spacing), ground survey and contour digitising. DEMs feed orthophoto production, slope and drainage analysis, volumes and visibility (Part A's terrain modelling). Accuracy is checked against independent check points by the vertical RMSE √(Σe²/n): errors of 0.3, −0.4, 0.1, −0.2 and 0.2 m give √(0.34/5) = 0.26 m.
Key takeaways
- Vertical (tilt under about 3°), tilted, low oblique (no horizon), high oblique (horizon); relief displacement is radial from the nadir, tilt displacement from the isocentre.
- Scale S = f/(H − h); flying height H = fn + h; coverage = format × scale number; air base B = (1 − 0.6) × coverage; strip spacing (1 − 0.3) × coverage.
- Relief displacement d = rh/H, outward for points above datum; h = dH/r.
- Parallax p = x − x′; h = H − Bf/p; parallax-bar height difference Δh = (H − h_b)Δp/(p_b + Δp).
- DSM includes canopy and buildings, DTM is bare earth; DEMs come from stereo matching, LiDAR, InSAR and survey; accuracy is the RMSE at check points.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A vertical photograph is taken with a 150 mm lens from 2000 m above datum over terrain at 500 m. The photo scale is 1 : ____.
Numerical answer — type the value.
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Answer: 10000
S = f/(H − h) = 0.150/(2000 − 500) = 0.150/1500 = 1/10,000. Using H alone gives 1:13,333, the scale at the datum, not at the ground actually photographed.Photography at 1:12 000 is required with a 152 mm camera over terrain of average elevation 300 m. The flying height above datum, in metres, is ____.
Numerical answer — type the value.
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Answer: 2124
H − h = f × n = 0.152 × 12,000 = 1824 m above the ground, so H = 1824 + 300 = 2124 m above datum. Stopping at 1824 m flies too low by the terrain height and gives a larger scale than asked.A road measures 8.0 cm on an aerial photograph and 2.0 cm on a 1:50 000 map. The scale of the photograph is 1 : ____.
Numerical answer — type the value.
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Answer: 12500
Ground length = 2.0 cm × 50,000 = 100,000 cm = 1 km. Photo scale = 8.0 cm/100,000 cm = 1/12,500. Dividing the map scale by 4 is the same thing; multiplying by 4 (1:200,000) inverts the ratio.Vertical photographs of 23 cm × 23 cm format are taken at 1:10 000 with 60 % forward overlap. The air base, in metres, is ____.
Numerical answer — type the value.
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Answer: 920
Ground side of one photo = 0.23 m × 10,000 = 2300 m. Each new photo advances by the non-overlapping part: B = (1 − 0.60) × 2300 = 920 m. Using 0.60 instead of 0.40 gives 1380 m, the overlapped length rather than the advance.For the same photography (23 cm format at 1:10 000), the spacing between adjacent flight lines with 30 % side lap, in metres, is ____.
Numerical answer — type the value.
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Answer: 1610
W = (1 − 0.30) × 2300 = 0.70 × 2300 = 1610 m. Side lap is smaller than forward overlap because it only ensures no gaps between strips, whereas forward overlap must give every point a stereo pair.The top of a 60 m tower is imaged 80 mm from the nadir on a vertical photograph taken from 1200 m above the tower's base. The relief displacement, in millimetres, is ____.
Numerical answer — type the value.
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Answer: 4
d = r h/H = 80 × 60/1200 = 4 mm, directed radially outward from the nadir. Using the flying height above sea level instead of above the base would understate it for a tower on high ground.On a vertical photograph taken from 1500 m above the base of a building, the image of its top is 75 mm from the nadir and is displaced 2.5 mm from the image of its base. The height of the building, in metres, is ____.
Numerical answer — type the value.
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Answer: 50
h = d H/r = 2.5 × 1500/75 = 50 m. The radial distance r belongs to the displaced (top) image; using the base image's distance, 72.5 mm, gives 51.7 m.On a truly vertical aerial photograph, relief displacement is
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Answer: A — radial from the nadir, and zero at the nadir
d = rh/H is zero when r = 0, so an object directly below the camera shows no lean, and every other displacement points along a radius from the nadir. It grows with h and with r, so it is largest at the edges, not at the principal point.A stereo pair is flown at 1500 m above a base point at datum, whose absolute parallax is 90.0 mm. A parallax bar gives a parallax difference of +2.4 mm between the base point and a hilltop. The height of the hilltop above the base point, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 38.96
Δh = (H − h_b)Δp/(p_b + Δp) = 1500 × 2.4/(90.0 + 2.4) = 3600/92.4 = 38.96 m. The approximation (H − h_b)Δp/p_b = 40.00 m overstates it because it ignores the Δp in the denominator.Photographs with f = 150 mm are taken from 1500 m above datum with an air base of 900 m. A point has an absolute stereoscopic parallax of 95.0 mm. Its elevation above datum, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 78.95
h = H − Bf/p = 1500 − (900 × 0.150)/0.095 = 1500 − 135/0.095 = 1500 − 1421.05 = 78.95 m. A point at the datum would have p = Bf/H = 90 mm, so a larger parallax means a higher point, as found.Objects in a stereo model of aerial photographs usually appear taller than they are because
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Answer: A — the air base-to-height ratio of the photography exceeds the eye base-to-viewing-distance ratio
Vertical exaggeration is roughly (B/H)/(b_e/h_e): the camera stations are hundreds of metres apart, a much wider baseline relative to height than the eyes' 6–7 cm at viewing distance, so depth is amplified. Lens distortion and tilt displace images but do not make every object look taller.A DEM is checked at five independent points, with elevation errors of +0.3, −0.4, +0.1, −0.2 and +0.2 m. Its vertical RMSE, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 0.26
Σe² = 0.09 + 0.16 + 0.01 + 0.04 + 0.04 = 0.34 m²; RMSE = √(0.34/5) = √0.068 = 0.26 m. The mean error, 0.0 m, would wrongly suggest a perfect DEM, because positive and negative errors cancel in a mean but not in a mean square.A model of elevations that includes the tops of trees and buildings is a
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Answer: A — digital surface model (DSM)
A DSM records the first surface a sensor meets — canopy, roofs — while a DTM is the bare earth beneath. Photogrammetric matching over forest therefore produces a DSM unless the vegetation is filtered out.Which of the following statements about vertical aerial photographs are correct?
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Answer: A — The scale is larger over high ground than over low ground; B — Forward overlap of about 60 % is used so that every point appears in stereo; C — The principal point is located from the fiducial marks
S = f/(H − h) increases as h increases; with 60 % overlap the middle 20 % of each photo is also on the photos either side, so the whole strip is in stereo; and joining opposite fiducial marks locates the principal point. A high oblique is strongly inclined and shows the horizon — the opposite of a vertical photograph.Relief displacement of a given object on a vertical photograph increases when
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Answer: A — the object is imaged farther from the nadir; B — the flying height is reduced; C — the object is taller
d = rh/H grows with the radial distance r and the object height h, and shrinks as the flying height H grows. Flying higher therefore reduces relief displacement, which is one reason small-scale photography from high altitude shows buildings leaning less.Which of the following can be used to generate a DEM?
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Answer: A — Automatic matching of a stereo pair of images; B — Airborne LiDAR; C — Radar interferometry (InSAR)
Stereo matching measures parallax, LiDAR measures laser ranges, and InSAR measures the phase difference between two radar acquisitions — each yields elevation. A single image has no parallax, so on its own it cannot give heights (except by assumptions such as shadow length for individual objects).