A GPS receiver needs signals from at least four satellites to fix a position, although three spheres are enough to intersect at a point. The reason is that

the receiver clock bias is a fourth unknown, and the fourth satellite provides the fourth equation to solve for it

one satellite is always used as a spare in case another fails during the observation

three spheres intersect at two points, and the fourth resolves the ambiguity between them

the fourth satellite supplies the ionospheric correction, which the other three cannot

Which of the following statements about GIS data models and remote sensing are correct?

Vector suits discrete objects with sharp boundaries; raster suits continuous fields such as elevation

Radar can image through cloud, which passive optical sensors cannot

Vector suits discrete objects with sharp boundaries; raster suits continuous fields such as elevation

Radar can image through cloud, which passive optical sensors cannot

Improving spatial resolution costs nothing, since the same sensor simply records smaller pixels

Why can a height not be measured from a single vertical aerial photograph of flat-lying terrain, however sharp the image?

The film records only two dimensions, so the third is physically lost and unrecoverable by any method

Relief displacement is too small to measure at normal flying heights

The scale of a vertical photograph is constant, so no height information is encoded in it at all

An orthophoto differs from a raw vertical aerial photograph chiefly in that

it has been rectified with an elevation model so that relief displacement is removed and the scale is constant

it is taken with a longer focal length, which eliminates relief displacement optically

it records elevation as an extra band alongside the image

it is a composite of several photographs taken at different times of day

Photogrammetry, Remote Sensing, GIS and GPS

Everything in this chapter measures the ground without touching it, and the price of that convenience is a set of geometric distortions that must be understood before any measurement taken from an image can be trusted. A vertical aerial photograph is a central projection: rays converge on a single lens point, so a tall object does not appear in plan at its true position but leans outward from the centre of the photograph by an amount proportional to both its height and its distance from that centre. A map is an orthographic projection, where every ray is vertical and nothing leans. That single difference is the whole of photogrammetry: it is why a photograph is not a map, why the same building appears at two different positions on two overlapping photographs, and — turning the problem into the solution — why that difference of position, the parallax, is exactly what lets a height be computed from a pair of images. The second theme is that scale in a photograph is not a single number. S = f/H′ uses the flying height above the ground being photographed, so ground at 300 m and ground at 600 m under the same aircraft photograph at 1:13,026 and 1:11,053 — a difference of 18% within one frame. Remote sensing then generalises the camera to any part of the spectrum, and GIS and GPS supply what an image alone cannot: a coordinate to hang it on, and a structure that lets attributes be asked questions rather than merely displayed.

Photo scale, and the height that must be measured from the ground

For a vertical photograph the scale is the ratio of the focal length to the flying height above the ground being photographed: S = f/H′, where H′ = H − h, H is the flying height above datum and h the elevation of the terrain. Both halves of that sentence carry marks. Take f = 152 mm, an aircraft at 2280 m above datum, and ground at 300 m: H′ = 1980 m and S = 0.152/1980 = 1 in 13,026. Use 2280 by mistake and the answer is 1 in 15,000 — out by 15.15% — and the error is invisible because both numbers look like plausible scales.

⚠️ A photograph does not have "a" scale

Because H′ depends on the terrain, the scale changes within a single frame wherever the ground does. The same aircraft at 2280 m photographs ground at 0 m at 1:15,000, ground at 300 m at 1:13,026, and ground at 600 m at 1:11,053 — a spread of 35% across one image if the relief is that great. Three consequences follow, and each is examined. First, a "scale" quoted for a photograph is really the average scale, computed from the mean terrain elevation, and it is exact only where the ground happens to sit at that mean. Second, distances cannot be scaled off a photograph the way they can off a map: two lines of equal length on the photo represent different ground lengths if they lie over ground of different elevation. Third, this is precisely why orthophotos exist — an image that has been rectified using a digital elevation model so that every pixel is re-projected as if viewed vertically, giving a product with a single, constant scale that can be measured like a map. Note the direction of the effect, which candidates reverse: higher ground is closer to the camera, so it photographs at a larger scale — 1:11,053 is a larger scale than 1:15,000, even though the number is smaller. A hilltop looks bigger on the photograph than the same feature in a valley.

Relief displacement, and how a defect becomes a measurement

On a central projection a point at height h above datum is imaged not at its plan position but displaced radially outward from the principal point by d = rh/H, where r is the radial distance of the image from the principal point. Every term in that expression says something. The displacement is proportional to height, so a tall building leans more than a low one. It is proportional to r, so an object at the centre of the photograph does not lean at all and one at a corner leans most. And it is inversely proportional to flying height, so flying higher reduces the distortion — at the cost of scale and resolution, which is one of the real trade-offs in flight planning.

A 200 m tower photographed from 2000 m, at three positions on the frame
Radial distance r	Displacement d = rh/H	Comment
0 (the principal point)	exactly 0	The one point on the photo with no relief displacement — and therefore useless for measuring a height
80 mm	8.000 mm	At 1:13,026 that is 104 m of apparent shift on the ground
110 mm (a corner)	11.000 mm	Worst case — which is why control points are chosen away from corners

🎯 The two images disagree, and the disagreement is the answer

Relief displacement is a defect if you want a map and a measurement if you want a height, and photogrammetry is built entirely on the second reading. Photograph the same ground twice from two positions along the flight line and a tall point leans differently in the two images, because its radial distance from each principal point differs. The difference between its x-coordinates on the two photographs is its parallax p, and the height follows from h = H − Bf/p, where B is the air base — the ground distance between the two exposure stations. Put numbers to it: with B = 600 m, f = 152 mm and H = 2000 m, a parallax of 90.00 mm gives a point at 986.67 m and 92.50 mm gives 1014.05 m, so 2.5 mm of parallax difference is 27.4 m of height. Two things follow. Parallax is greater for higher points, which is the right way round to remember the sign — a hilltop is nearer the camera, so it shifts more between the two images. And the whole method needs stereo overlap: nothing can be measured in three dimensions from a single photograph, however good, because a single central projection cannot distinguish a tall object near the centre from a short one further out. That is the geometric reason aerial survey flies with 60% forward overlap rather than the 5% needed merely to avoid gaps.

That overlap is expensive, and the arithmetic is worth doing once. At 60% forward and 30% side lap, each photograph contributes only 0.40 × 0.70 = 28% of its gross coverage as new ground. A 230 mm frame at 1:13,026 covers 8.98 km² gross but only 2.51 km² effectively, so a 100 km² block needs 40 photographs, not the 12 the gross figure suggests. Overlap consumes nearly three-quarters of every image, and it buys two things that cannot be had otherwise: stereo coverage for height, and enough common points between adjacent photographs to tie the whole block together in adjustment.

Remote sensing: four resolutions that trade against each other

Remote sensing generalises the camera to the whole electromagnetic spectrum, and the useful mental model is that every surface has a spectral signature — a curve of reflectance against wavelength — so a sensor with several bands can distinguish materials that look identical in a black-and-white photograph. Healthy vegetation reflects strongly in the near infrared and weakly in the red, which is why the normalised difference of those two bands is the standard vegetation index and why stressed crops show up before any change is visible to the eye. Water absorbs in the infrared and so appears almost black there, making shorelines and flooding easy to map. Passive sensors use reflected sunlight or emitted thermal radiation and are stopped by cloud and night; active sensors — radar, lidar — supply their own energy, and radar’s long wavelengths pass through cloud, which is why flood mapping in a monsoon is a radar job.

The four resolutions, and what each one costs
Resolution	What it means	What improving it costs
Spatial	Ground size of one pixel	Less energy per pixel, so worse signal-to-noise; a narrower swath, so a longer revisit
Spectral	Number and narrowness of bands	Each narrow band collects less energy — traded against spatial resolution
Radiometric	Number of grey levels recorded (8-bit, 12-bit)	Data volume, and eventually the noise floor of the detector
Temporal	How often the same ground is revisited	A wider swath (coarser pixels) or more satellites

GIS, and the fourth satellite

A Geographical Information System is not a map on a screen. What distinguishes it is that geometry and attributes are held together and can be queried jointly — "every plot within 200 m of this alignment whose land use is agricultural and whose area exceeds one hectare" is a question a map cannot answer and a GIS answers by construction. Data is held in one of two models and the choice is not stylistic. Vector — points, lines, polygons — represents discrete objects with sharp boundaries (a plot, a pipeline, a building) compactly and at any scale, and supports network analysis. Raster — a grid of cells — represents continuous fields (elevation, rainfall, temperature, a satellite image) and makes overlay arithmetic trivial, at the cost of size and of a resolution fixed when the grid was made. Elevation is the standard example of the tension: a DEM is naturally raster, a contour map naturally vector, and the two are conversions of one another with loss in each direction.

🎯 Three satellites would be enough if receivers had atomic clocks

GPS positioning is trilateration: each satellite broadcasts its position and the time of transmission, the receiver measures the travel time, multiplies by the speed of light, and knows it lies on a sphere of that radius about the satellite. Three spheres intersect at a point, so three satellites should suffice — and would, if the receiver’s clock were perfect. It is not. Light travels 0.2998 m per nanosecond, so a receiver clock in error by one microsecond produces a range error of 299.79 m, and a quartz clock drifts by far more than that in a second. Putting an atomic clock in every receiver would solve it and would also make a handheld unit cost more than a car. The elegant alternative is to treat the clock error as a fourth unknown and solve for it: with four satellites there are four equations in four unknowns — x, y, z and the clock bias — so the receiver computes its own clock error along with its position, and a cheap oscillator becomes as good as an atomic one for this purpose. That is the entire reason the answer is four rather than three, and it is the most commonly asked conceptual question on GPS. Two corollaries. The geometry of the satellites matters as much as their number, which is what the dilution-of-precision figure reports — four satellites clustered in one part of the sky give a poor fix however strong the signals. And differential GPS improves accuracy by an entirely different route: a receiver at a known point measures the current error and broadcasts the correction, which removes the errors common to both receivers — satellite clock and ephemeris, and most of the atmospheric delay — leaving centimetre-level positioning from the same constellation.

Key takeaways

S = f/H′ uses the flying height above the ground photographed; using height above datum here was a 15.15% error.
A photograph has no single scale — higher ground is nearer the camera and images at a larger scale.
Relief displacement d = rh/H is radial and exactly zero at the principal point, greatest at the corners.
Parallax turns the defect into a measurement: h = H − Bf/p, and a higher point has the greater parallax.
At 60% forward and 30% side lap only 28% of each photo is new ground — 100 km² needs 40 photos, not 12.
GPS needs four satellites because the receiver clock bias is a fourth unknown: 1 μs of error is 300 m of range.

Practice questions (8)

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 152 mm focal length camera from a flying height of 2280 m above mean sea level. The average terrain elevation is 300 m. The scale of the photograph is 1 in ______ (round to the nearest whole number)
Numerical answer — type the value.
Show answer
Answer: 13026
1 in 13,026. S = f/H′ where H′ is the flying height above the terrain, not above datum: H′ = 2280 − 300 = 1980 m, so S = 0.152/1980 = 1/13,026.3. Using 2280 gives 1 in 15,000, an error of 15.15%, and it is the single commonest mistake in photogrammetry because both answers look like perfectly respectable scales — nothing about 1:15,000 announces itself as wrong. Push the idea one step and it produces the more important conclusion: since H′ depends on the terrain, the scale changes across the frame wherever the ground does. This same aircraft photographs ground at sea level at 1:15,000, at 300 m at 1:13,026, and at 600 m at 1:11,053 — a spread of 35% within one image over that much relief. So a scale quoted for a photograph is the average, computed from the mean terrain elevation, and it is exact only where the ground sits at that mean. Note the direction, which candidates routinely reverse: higher ground is closer to the camera and therefore images at a larger scale, even though the denominator is smaller. This variability is why distances cannot be scaled off a raw aerial photograph the way they can off a map, and why orthophotos are produced — an image rectified with a digital elevation model so that every pixel is re-projected as if seen vertically, yielding a single constant scale that can be measured directly.
A tower 200 m high appears on a vertical photograph taken from a flying height of 2000 m above its base. The image of the top of the tower is 80 mm from the principal point. The relief displacement in mm is ______
Numerical answer — type the value.
Show answer
Answer: 8
8 mm. d = rh/H = 80 × 200/2000 = 8 mm. Every factor in that expression carries a physical statement worth reading off. The displacement is proportional to h, so a tall object leans further than a short one — which is exactly what makes the effect a usable measurement rather than only a nuisance. It is proportional to r, the radial distance from the principal point, so the same tower at the principal point would show a displacement of exactly zero, and at a corner of the frame, say r = 110 mm, it would show 11 mm. And it is inversely proportional to H, so flying higher suppresses the distortion at the cost of scale and resolution — a real trade-off in flight planning rather than a free improvement. Two consequences are examined regularly. Because the displacement is radial and outward, its direction points away from the principal point, so a photograph of a city shows every building leaning outward from the centre like the spokes of a wheel; measuring that lean gives building heights from a single photograph, provided the base and top are both visible. And because d = 0 at the principal point, that one point is useless for measuring heights — a genuinely counter-intuitive result, since the centre of the frame is where the geometry is otherwise best behaved. This is also why relief displacement must be removed before a photograph can be treated as a map: at this scale, 1:13,026, an 8 mm displacement is 104 m of apparent horizontal error on the ground.
A GPS receiver needs signals from at least four satellites to fix a position, although three spheres are enough to intersect at a point. The reason is that
1. the receiver clock bias is a fourth unknown, and the fourth satellite provides the fourth equation to solve for it
2. one satellite is always used as a spare in case another fails during the observation
3. three spheres intersect at two points, and the fourth resolves the ambiguity between them
4. the fourth satellite supplies the ionospheric correction, which the other three cannot
Show answer
Answer: A — the receiver clock bias is a fourth unknown, and the fourth satellite provides the fourth equation to solve for it
The receiver clock bias is a fourth unknown. GPS ranging works by timing: the satellite broadcasts its position and the instant of transmission, the receiver notes the instant of arrival, multiplies the difference by the speed of light, and knows it lies somewhere on a sphere of that radius. Three such spheres would fix a point — if the receiver’s clock agreed exactly with the satellites’. It does not, and the sensitivity is brutal. Light travels 0.2998 m per nanosecond, so a receiver clock wrong by one microsecond yields a range error of 299.79 m, and a quartz oscillator of the kind found in a phone drifts far more than a microsecond in a second. The satellites carry atomic clocks; putting one in every receiver would fix the problem and would also cost more than the vehicle the receiver is fitted to. The elegant solution is to stop treating the clock error as an error and treat it as an unknown: with four satellites there are four equations in four unknowns — x, y, z and the clock bias Δt — so the receiver solves for its own clock error alongside its position, and a cheap oscillator becomes sufficient. That is why the answer is four, and it is also why a GPS receiver is an excellent clock as a by-product. Option (b) invents a redundancy rule; redundancy is valuable but is not why the minimum is four. Option (c) describes the genuine two-point ambiguity of trilateration, but it is resolved by discarding the solution that is not near the earth’s surface, not by a satellite. Option (d) misdescribes ionospheric correction, which comes from dual-frequency measurement or a model, not from an extra satellite.
Aerial photographs are taken with 60% forward overlap and 30% side lap. The effective ground area covered by each photograph, as a percentage of its gross coverage, is ______
Numerical answer — type the value.
Show answer
Answer: 28
28%. The new ground a photograph contributes along the flight line is 1 − 0.60 = 40% of its length, and across the strips it is 1 − 0.30 = 70% of its width, so the effective area is 0.40 × 0.70 = 0.28, or 28% of the gross. The number is startling the first time: overlap consumes nearly three-quarters of every image taken, and a flight plan that ignores it under-orders photographs by a factor of more than three. Work it through to see the cost. A 230 mm frame at 1:13,026 covers 2996 m a side, that is 8.98 km² gross but only 2.51 km² effectively, so a block of 100 km² needs 100/2.51 = 40 photographs rather than the 12 the gross figure suggests — and that is before the extra photographs at the ends of each strip. The overlap is not waste, though, and the two halves buy different things. Forward overlap of 60% exists so that every point on the ground appears on at least two successive photographs, which is what makes stereo viewing and parallax height measurement possible at all; 50% would in principle suffice but leaves no margin for crab, tilt and variation in flying height, so 60% is standard. Side lap of 30% exists mainly to guarantee no gaps between strips despite navigation error, and to provide common points that tie adjacent strips together when the block is adjusted. Reducing either saves photographs and costs either stereo coverage or block rigidity — which is why these figures are near-universal rather than a matter of preference.
Which of the following statements about GIS data models and remote sensing are correct?
1. Vector suits discrete objects with sharp boundaries; raster suits continuous fields such as elevation
2. Healthy vegetation reflects strongly in the near infrared and weakly in the red
3. Radar can image through cloud, which passive optical sensors cannot
4. Improving spatial resolution costs nothing, since the same sensor simply records smaller pixels
Show answer
Answer: A — Vector suits discrete objects with sharp boundaries; raster suits continuous fields such as elevation; B — Healthy vegetation reflects strongly in the near infrared and weakly in the red; C — Radar can image through cloud, which passive optical sensors cannot
(a), (b) and (c). (a) is the choice that decides how a GIS project is built, and it is not stylistic. Vector stores points, lines and polygons with explicit coordinates, which represents a plot boundary, a pipeline or a road network compactly, scales without becoming blocky, and supports topology and network analysis. Raster stores a grid of cells, which is the natural form for anything that varies continuously — elevation, rainfall, temperature, a satellite image — and makes overlay arithmetic trivial because cells align, at the cost of file size and of a resolution fixed when the grid was created. Elevation illustrates the tension exactly: a DEM is naturally raster, a contour map naturally vector, and converting either way loses something. (b) is the spectral signature that underlies most vegetation work: chlorophyll absorbs strongly in the red while leaf mesophyll reflects strongly in the near infrared, and the normalised difference of those two bands is the standard vegetation index — sensitive enough that a stressed crop shows up before any change is visible to the eye. (c) follows from wavelength: radar’s long wavelengths pass through cloud and, being active, need no sunlight, which is why flood mapping during a monsoon is a radar task and an optical satellite is useless for it. (d) is wrong, and the trade it denies is the central constraint of sensor design. A smaller pixel collects less energy, so the signal-to-noise ratio falls; recovering it means a wider spectral band (losing spectral resolution) or a longer dwell (narrowing the swath, and so lengthening the revisit interval). Spatial, spectral, radiometric and temporal resolution trade against one another, which is why no single satellite is best at everything and why missions are designed around one intended use.
Why can a height not be measured from a single vertical aerial photograph of flat-lying terrain, however sharp the image?
1. A central projection cannot distinguish a tall object near the centre from a short one further out — two views are needed
2. The film records only two dimensions, so the third is physically lost and unrecoverable by any method
3. Relief displacement is too small to measure at normal flying heights
4. The scale of a vertical photograph is constant, so no height information is encoded in it at all
Show answer
Answer: A — A central projection cannot distinguish a tall object near the centre from a short one further out — two views are needed
A single central projection is ambiguous. Relief displacement is d = rh/H, and the image records only the product rh — not r and h separately. A 200 m object at r = 80 mm and a 100 m object at r = 160 mm produce exactly the same 8 mm displacement, so from one photograph the two are indistinguishable. A second photograph taken from a different position along the flight line resolves the ambiguity, because the same point now has a different radial distance from the new principal point and therefore a different displacement. The difference between the point’s positions on the two images is its parallax p, and the height follows from h = H − Bf/p with B the air base. That is why aerial survey flies with 60% forward overlap rather than the small overlap needed merely to avoid gaps: the redundancy is not insurance, it is the measurement. Option (b) overstates the case — the third dimension is recoverable, and stereo photogrammetry is precisely the method of recovering it. Option (c) is wrong on the arithmetic: at these parameters the displacement is 8 mm on a 230 mm frame, easily measurable, and the question does not turn on precision at all. Option (d) is wrong twice over: the scale of a vertical photograph is not constant over varied terrain — that variation is exactly the height information — and even where the terrain is flat the displacement of objects standing on it still encodes their heights. What defeats a single photograph is ambiguity, not absence of information. A useful corollary: a single photograph can give a height when both the base and the top of an object are visible, since then r for the base is known and only h is unknown — which is how building heights are read from vertical imagery of a city.
Light travels at 3 × 10⁸ m/s. A GPS receiver clock is in error by 1 microsecond. The resulting error in the measured range to a satellite, in metres, is ______
Numerical answer — type the value.
Show answer
Answer: 300
300 m (299.79 m with the exact speed of light). error = c × Δt = 3 × 10⁸ × 10⁻⁶ = 300 m. The number is the whole reason GPS is engineered the way it is. One nanosecond of timing error is 0.3 m of range error, so metre-level positioning demands timing good to a few nanoseconds — a standard no ordinary quartz oscillator approaches, since a watch crystal drifts by microseconds within a second. The satellites solve this by carrying atomic clocks, and putting one in every receiver would work and would also cost more than the vehicle it is fitted to. Instead the receiver treats its own clock error as a fourth unknown and solves for it using a fourth satellite: four equations, four unknowns — x, y, z and Δt. The receiver therefore computes its own clock error as a by-product, which is why a GPS receiver is an excellent time source as well as a position source, and why timing for power grids and telecom networks is commonly taken from GPS. Two related points follow. Signals also slow in the ionosphere and troposphere, which is a range error of the same kind and is removed either by dual-frequency measurement (the ionospheric delay is frequency-dependent, so two frequencies reveal it) or by a model. And differential GPS removes almost everything else: a receiver at a known point measures the current total error and broadcasts the correction, cancelling the satellite clock, ephemeris and most atmospheric errors that are common to both receivers, and taking positioning from metres to centimetres without any change to the satellites.
An orthophoto differs from a raw vertical aerial photograph chiefly in that
1. it has been rectified with an elevation model so that relief displacement is removed and the scale is constant
2. it is taken with a longer focal length, which eliminates relief displacement optically
3. it records elevation as an extra band alongside the image
4. it is a composite of several photographs taken at different times of day
Show answer
Answer: A — it has been rectified with an elevation model so that relief displacement is removed and the scale is constant
It has been rectified with an elevation model. The two defects that stop a raw photograph being a map are exactly the two that this chapter has quantified. Relief displacement puts every raised point off its plan position by d = rh/H, radially outward, and variable scale means S = f/H′ changes wherever the terrain does — for the aircraft worked earlier, from 1:15,000 over sea level to 1:11,053 over ground at 600 m. Orthorectification removes both together by using a digital elevation model to compute, for every pixel, where that piece of ground would have appeared if it had been viewed vertically from directly above, and re-projecting it there. The result is an image with one constant scale and no relief displacement — a photograph that can be measured, overlaid on vector data and used as a map base, which is why orthophotos rather than raw imagery underlie most GIS work. Note what the process needs and what it therefore inherits: it is only as good as the DEM, so an orthophoto over terrain whose elevation model is coarse or dated carries residual displacement, and tall buildings not represented in a bare-earth DEM still lean (the reason "true orthophotos", which model structures as well, are produced for dense cities). Option (b) is wrong — a longer focal length raises the flying height for a given scale and so reduces d = rh/H, but no lens removes it, because it is a property of central projection itself rather than of the optics. Option (c) describes a DEM delivered beside an image, which is a different product. Option (d) describes a mosaic or a multi-temporal composite, neither of which corrects geometry at all.