Remote Sensing, GIS and Urban Analytics: Imagery, Photogrammetry, Spatial Data, SQL, Python and Machine Learning

Practice questions (20)

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.

1. Why are most land-observation satellites placed in sun-synchronous orbits?

   1. They stay fixed above one point on the equator and so can watch a single city continuously through the day
   2. They orbit at about 36,000 km, which is the altitude that gives the finest pixel size obtainable from space
   3. They cross a latitude at about the same local solar time on each pass, so illumination is comparable
   4. They give a swath wider than any other orbit could, so that every place on Earth is imaged once every day
   Show answer

   Answer: C — They cross a latitude at about the same local solar time on each pass, so illumination is comparable

   A sun-synchronous orbit makes the satellite cross each latitude at about the same local solar time, so shadows and brightness are comparable when images of different dates are compared. Fixed position above the equator and an altitude of about 36,000 km describe a geostationary satellite, which gives continuous but coarse views, not fine pixels.

2. Healthy green vegetation has low reflectance in the red band but high reflectance in the near-infrared because

   1. chlorophyll absorbs red strongly, while the internal structure of the leaf scatters near-infrared strongly
   2. water in the leaf absorbs near-infrared most strongly, while red light is scattered back by the cell walls
   3. the atmosphere is opaque in the red region of the spectrum and transparent in the near-infrared region
   4. chlorophyll reflects red light strongly, while the outer layer of the leaf surface absorbs near-infrared energy
   Show answer

   Answer: A — chlorophyll absorbs red strongly, while the internal structure of the leaf scatters near-infrared strongly

   Chlorophyll absorbs blue and red light for photosynthesis and the spongy cell structure of the leaf scatters near-infrared strongly, giving the sharp rise at about 0.7 µm. Leaf water absorbs mainly in the shortwave infrared, not in the near-infrared, so the water-based explanation is wrong; the atmosphere is not opaque in red.

3. A city wants to reduce losses of treated water in its distribution network. Which smart-city IoT deployment fits this aim best?

   1. Air-quality and noise sensors at the main road junctions, feeding a public dashboard of pollution levels
   2. Adaptive signal controllers at busy junctions that respond to the traffic queues measured by detectors
   3. Fill-level sensors in street bins that reroute the waste-collection vehicles to the full bins first
   4. Flow and pressure sensors on pipes in metered districts, raising leak alerts at a control centre
   Show answer

   Answer: D — Flow and pressure sensors on pipes in metered districts, raising leak alerts at a control centre

   Comparing inflow and outflow, and watching pressure, in each metered district shows where water is leaking, so the alert can be sent to a control centre. The other deployments are sound smart-city uses (air and noise monitoring, traffic management, waste logistics), but none measures the water network.

4. A land-cover map of a region that is 90 per cent vegetation has an overall accuracy of 92 per cent but a kappa of only 0.45. The best explanation is that

   1. kappa measures only the accuracy of the dominant class, which must therefore have been mapped poorly since it covers 90 per cent
   2. kappa removes chance agreement, which is high when one class dominates, so the map is weaker than 92 per cent suggests
   3. kappa and overall accuracy must always be equal for the same error matrix, so one of the two figures has been computed wrongly
   4. overall accuracy was computed on the training pixels while kappa was computed on independent test pixels of the same map
   Show answer

   Answer: B — kappa removes chance agreement, which is high when one class dominates, so the map is weaker than 92 per cent suggests

   If 90 % of the area is vegetation, a map that labelled nearly everything vegetation would already agree with the reference most of the time by chance (pe of about 0.85 here, since 0.45 = (0.92 − pe)/(1 − pe) gives pe = 0.85). Kappa = (po − pe)/(1 − pe) discounts that, so a 92 % overall accuracy can correspond to a modest kappa. Kappa uses the whole error matrix, not just the dominant class.

5. A study area of 12 km × 9 km is stored as a single-band raster with 30 m cells and 16 bits per cell, uncompressed. Taking 1 KB = 1,000 bytes, the file size is about

   1. 120 KB
   2. 480 KB
   3. 240 KB
   4. 1,920 KB
   Show answer

   Answer: C — 240 KB

   Rows × columns = (9,000/30) × (12,000/30) = 300 × 400 = 120,000 cells. At 16 bits = 2 bytes each, the file holds 240,000 bytes = 240 KB. The figure 120 KB forgets the second byte, 480 KB is the size at 15 m cells with 8 bits or the 30 m case at 32 bits, and 1,920 KB treats 16 bits as 16 bytes.

6. A planner needs the parcels within 500 m of a proposed metro alignment, with each parcel cut at the 500 m limit and the attributes of both layers kept. The appropriate sequence is to

   1. intersect the alignment line directly with the parcel layer, since the line already marks the zone
   2. union the parcel layer with the alignment line and then dissolve the result on parcel identifier
   3. build Thiessen polygons around the parcels and then reclassify them by their distance to the line
   4. buffer the alignment by 500 m and then intersect the buffer polygon with the parcel layer
   Show answer

   Answer: D — buffer the alignment by 500 m and then intersect the buffer polygon with the parcel layer

   A buffer turns the line into a 500 m wide zone, and intersecting that zone with the parcels keeps the part of each parcel inside it with the attributes of both layers. Intersecting the bare line with the parcels returns only the parcels it crosses, because a line has no width: this is the tempting shortcut.

7. Which statement about terrain models is correct?

   1. A TIN joins irregularly spaced elevation points into triangles, so points can be dense on rugged ground and sparse on flat
   2. A raster DEM joins irregularly spaced points into triangles and therefore cannot represent flat areas efficiently at all
   3. A DTM includes the tops of buildings and trees in its heights, whereas a DSM shows only the bare ground surface
   4. A DEM has a uniform cell size and for that reason cannot be used to derive slope or aspect in terrain analysis
   Show answer

   Answer: A — A TIN joins irregularly spaced elevation points into triangles, so points can be dense on rugged ground and sparse on flat

   A TIN is a vector surface of non-overlapping triangles whose vertices are the sampled (x, y, z) points, which can be placed densely where the terrain is complex. A DEM is the raster of uniform cells, and slope and aspect are derived from it routinely. The DSM/DTM statement is reversed: the DSM includes roofs and trees, the DTM is bare ground.

8. Table wards(ward_id, zone, pop) holds (1, N, 12000), (2, N, 8000), (3, S, 15000), (4, S, 5000), (5, E, 9000). The query
   SELECT zone, COUNT(*) FROM wards WHERE pop >= 8000 GROUP BY zone HAVING COUNT(*) >= 2;
   returns

   1. two rows: N with 2 and S with 2
   2. one row: N with 2
   3. three rows: N with 2, S with 1 and E with 1
   4. no rows at all
   Show answer

   Answer: B — one row: N with 2

   WHERE runs first and drops ward 4 (5000 is below 8000), leaving wards 1, 2, 3 and 5. Grouping by zone gives N 2, S 1, E 1, and HAVING COUNT(*) >= 2 then keeps only N. Counting S as 2 forgets that WHERE removed ward 4 before the grouping; showing all three groups forgets the HAVING.

9. A model flags 50 buildings as unauthorised. On inspection 40 of them are unauthorised, and a further 20 unauthorised buildings were not flagged. Because every field inspection is costly, the planner wants fewer false alarms. Which statement is correct?

   1. Precision is 0.67 and recall is 0.80, and the planner should aim to raise recall
   2. Precision and recall are both 0.80, so neither needs to be improved
   3. Precision is 0.80 and recall is 0.67, and recall is the figure that measures false alarms
   4. Precision is 0.80 and recall is 0.67, and the planner should aim to raise precision
   Show answer

   Answer: D — Precision is 0.80 and recall is 0.67, and the planner should aim to raise precision

   TP = 40 and FP = 10 (50 flagged), FN = 20. Precision = 40/50 = 0.80 and recall = 40/60 = 0.67. False alarms lower precision, not recall, so when inspections are costly the target is higher precision. Swapping the two ratios, or tying false alarms to recall, are the usual errors.

10. Which description best fits a Spatial Data Infrastructure (SDI)?

    1. A framework of policies, standards, institutions and metadata that lets geospatial data be found and shared
    2. A single desktop GIS software package that holds all the maps, layers and tables of one city planning office
    3. A constellation of Earth-observation satellites that collects imagery for every department of a government
    4. A national register of land ownership and tenure records, kept and updated by the revenue departments alone
    Show answer

    Answer: A — A framework of policies, standards, institutions and metadata that lets geospatial data be found and shared

    An SDI is the enabling framework (policy, standards, institutional arrangements, metadata, access technology and people), not any single product. A GIS package or a satellite programme may feed an SDI, and a land register may be one of its datasets, but none of them is the infrastructure itself.

11. Which of the following statements about energy sources, sensors and the atmosphere are correct?

    1. Synthetic aperture radar can image the ground through cloud and at night
    2. The thermal infrared window at about 8 to 14 µm is used to estimate land surface temperature
    3. Rayleigh scattering affects red light more strongly than blue light
    4. A passive sensor records reflected sunlight or emitted radiation instead of supplying its own energy
    Show answer

    Answer: A — Synthetic aperture radar can image the ground through cloud and at night; B — The thermal infrared window at about 8 to 14 µm is used to estimate land surface temperature; D — A passive sensor records reflected sunlight or emitted radiation instead of supplying its own energy

    Radar uses microwaves that pass through cloud and carries its own illumination; the 8 to 14 µm window records emitted heat, hence surface temperature; and passive sensors rely on the Sun or on the surface’s own emission. Rayleigh scattering is stronger at shorter wavelengths, so blue is scattered more than red, which is why the sky is blue.

12. Which of the following statements about vertical aerial photographs are correct?

    1. The image of a tall building is displaced radially outward from the nadir
    2. For the same object and radial distance, a greater flying height gives a smaller relief displacement
    3. The scale is the same at every point of the photograph whatever the terrain relief
    4. Stereoscopic viewing needs two overlapping photographs taken from different exposure stations
    Show answer

    Answer: A — The image of a tall building is displaced radially outward from the nadir; B — For the same object and radial distance, a greater flying height gives a smaller relief displacement; D — Stereoscopic viewing needs two overlapping photographs taken from different exposure stations

    Relief displacement d = r·h/H is outward from the nadir for features above the datum and falls as H increases; stereo models need two overlapping exposures. The scale S = f/(H − h) changes with ground elevation, so it is not uniform over hilly terrain; only over flat ground is it constant.

13. Which of the following statements about spatial data storage and georeferencing are correct?

    1. Vector topology records connectivity and adjacency, which makes network routing and neighbour queries possible
    2. Halving the cell size of a raster multiplies the number of cells by four
    3. Nearest-neighbour resampling keeps the original class codes, so it suits classified rasters
    4. A raster is the natural model for crisp parcel boundaries whose exact length and area must be reported
    Show answer

    Answer: A — Vector topology records connectivity and adjacency, which makes network routing and neighbour queries possible; B — Halving the cell size of a raster multiplies the number of cells by four; C — Nearest-neighbour resampling keeps the original class codes, so it suits classified rasters

    Topology stores connectivity and adjacency; halving the cell size doubles rows and columns, so cells quadruple; and nearest-neighbour resampling does not invent new values, so class codes survive. Parcels with exact boundaries, lengths and areas are the natural job of the vector model, as a raster only approximates edges by its cell size.

14. Which of the following statements about machine learning in urban planning are correct?

    1. Training accuracy far above test accuracy points to overfitting
    2. K-means clustering is an unsupervised method because it needs no labelled examples
    3. The test set should be used repeatedly to tune the model, because that gives the most honest accuracy
    4. Supervised land-cover classification needs labelled training samples such as ground-truth polygons
    Show answer

    Answer: A — Training accuracy far above test accuracy points to overfitting; B — K-means clustering is an unsupervised method because it needs no labelled examples; D — Supervised land-cover classification needs labelled training samples such as ground-truth polygons

    A large gap between training and test accuracy is the signature of overfitting; k-means groups unlabelled data; supervised classification needs labelled samples. Using the test set repeatedly to tune the model leaks information from it into the model, so its score is no longer an honest estimate for unseen data. A separate validation set or cross-validation should be used for tuning.

15. A vertical aerial photograph is taken with a camera of focal length 150 mm from a flying height of 3,000 m above flat ground. Two points are 62 mm apart on the photograph. Their ground distance, in metres, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1240

    Scale S = f/H = 0.150/3,000 = 1/20,000. Ground distance = photo distance × 20,000 = 62 mm × 20,000 = 1,240,000 mm = 1,240 m. Check: 62 mm = 0.062 m, and 0.062 × 20,000 = 1,240 m.

16. On a vertical photograph the image of the top of a 50 m tall tower is displaced radially by 4.0 mm. The top of the tower lies 80 mm from the nadir on the photograph. The flying height above the base of the tower, in metres, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1000

    Relief displacement d = r·h/H, so H = r·h/d = 80 mm × 50 m / 4.0 mm = 1,000 m. The units of r and d cancel, leaving metres from h. Check: with H = 1,000 m, d = 80 × 50/1,000 = 4.0 mm.

17. An accuracy assessment of a two-class map (urban, non-urban) uses 100 independent samples. Of the 60 samples the map calls urban, 50 are urban in the reference and 10 are not. Of the 40 samples the map calls non-urban, 20 are urban in the reference and 20 are not. Cohen’s kappa, correct to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.35

    Observed agreement po = (50 + 20)/100 = 0.70. Row totals (map): urban 60, non-urban 40. Column totals (reference): urban 50 + 20 = 70, non-urban 10 + 20 = 30. Chance agreement pe = (60 × 70 + 40 × 30)/100² = (4,200 + 1,200)/10,000 = 0.54. κ = (0.70 − 0.54)/(1 − 0.54) = 0.16/0.46 = 0.348, i.e. 0.35.

18. A pixel has surface reflectance 0.52 in the near-infrared band and 0.08 in the red band. Its NDVI, correct to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.73

    NDVI = (NIR − Red)/(NIR + Red) = (0.52 − 0.08)/(0.52 + 0.08) = 0.44/0.60 = 0.733, i.e. 0.73. A value this high indicates dense, healthy vegetation. Using the simple ratio NIR/Red instead gives 6.5, far outside the −1 to +1 range that NDVI always lies in.

19. Table wards(ward_id) holds the five ward ids 1, 2, 3, 4, 5. Table schools(school_id, ward_id) holds (1, 1), (2, 1), (3, 3), (4, 3), (5, 3), (6, 5). The query
    SELECT COUNT(*) FROM wards w LEFT JOIN schools s ON w.ward_id = s.ward_id;
    returns the value ____.

    Numerical answer — type the value.

    Show answer

    Answer: 8

    Rows per ward in the LEFT JOIN: ward 1 has 2 schools (2 rows), ward 2 has none (1 row with NULLs), ward 3 has 3 (3 rows), ward 4 has none (1 row), ward 5 has 1 (1 row). Total = 2 + 1 + 3 + 1 + 1 = 8. An INNER JOIN would give only the 6 matching rows; COUNT(*) counts the NULL-padded rows too.

20. What does this Python code print?
    def f(n):
s = 0
for i in range(1, n + 1):
if i % 3 != 0:
s += i
return s
print(f(7))

    Numerical answer — type the value.

    Show answer

    Answer: 19

    range(1, 8) gives i = 1 to 7. The condition i % 3 != 0 skips the multiples of 3 (3 and 6). The loop adds 1 + 2 + 4 + 5 + 7 = 19, and the function returns 19. A total of 28 would be the sum of all seven values, forgetting the filter.