Q1 — Accumulation: Switch to Flow accumulation. Hover over a ridge crest, then a valley floor. What values do you get? Now look at Flow direction in the inspector: which neighbour does D8 pick, and why that one?
- Ridge crests: 0 (or single digits just below the crest). Nothing drains into a ridge cell. About 5% of land cells in this DEM have accumulation 0.
- Valley floors: hundreds to tens of thousands. The main stream reaching Charteris Bay collects 82,469 cells ≈ 5.3 km², the highest value in the DEM.
D8 rule: the inspector highlights the neighbour with the
steepest downhill slope, i.e. the largest drop divided by distance. Diagonal neighbours are 8√2 ≈ 11.3 m away, not 8 m, so a diagonal cell needs a bigger drop to win. All the water goes to that one neighbour.
Note: zoomed in, students see long parallel arrows on smooth slopes. That is a known D8 artefact: with only 8 directions, flow on planar slopes runs in straight parallel lines instead of spreading. Multiple-flow-direction methods (D∞, MFD) address this.
Must give ridge ≈ 0 and valley in the thousands, and state that D8 sends all flow to the steepest downhill of the 8 neighbours. Bonus for the √2 diagonal distance or the parallel-flow artefact.
Q2 — Threshold: Turn on the OpenStreetMap streams. Find the threshold where the extracted network best matches them. Write down the threshold and the network length. Where does the DEM put a stream that the map doesn't have, and which one is wrong?
There is no single right answer; a
good match is roughly 500–2,000 cells (3–13 ha upstream). Reference values from the demo:
| Threshold (cells) | Network length | Drainage density |
| 100 | 95.5 km | 6.6 km/km² |
| 500 | 45.1 km | 3.1 km/km² |
| 840 (default) | 35.1 km | 2.4 km/km² |
| 2,000 | 24.0 km | 1.7 km/km² |
| 10,000 | 7.6 km | 0.5 km/km² |
Below ~200 cells the DEM draws a "stream" in almost every gully; above ~5,000 only the main stems survive and mapped tributaries disappear. The threshold is an
analyst choice, which is why it must be reported and validated.
Who is wrong? Usually neither is simply right. Many extra DEM streams are
ephemeral gullies that run only in storms and that OSM volunteers never mapped; OSM is incomplete and hand-digitised. The DEM knows nothing about infiltration, springs, culverts or drains, so it can also put a channel where there is none. Validation needs field data or aerial imagery; neither layer is ground truth.
Must give a threshold and length consistent with the table, and explain that low thresholds over-extract and high thresholds under-extract. Full marks for recognising that the OSM layer is itself incomplete and that the threshold is an analyst decision.
Q3 — Sinks: Untick Fill sinks and note the number of sinks. Now switch the precision to 0.1 m, then 1 m. It is the same terrain, so where did the extra sinks come from? Place a pour point near the coast on the largest stream and compare its area with filling on and off. Are most sinks real depressions or DEM artefacts?
Sinks: 5 at full (float) precision,
260 at 0.1 m,
11,647 at 1 m.
The extra sinks come purely from
rounding the elevations. Neighbouring cells rounded to the same value form a flat patch with no downhill neighbour, so D8 has nowhere to send the water. Integer-metre storage (as in SRTM and many older DEM products) is a classic source of artefact sinks and flats.
Catchment at the main stream's coastal outlet (with snapping):
| Precision | Filled | Not filled |
| Float | 5.28 km² | 4.33 km² |
| 0.1 m | 5.36 km² | 0.09 km² |
| 1 m | 5.28 km² | ≈ 0 |
Even the 5 full-precision sinks cut the catchment by 18%, because water reaching a sink stops there and never arrives at the outlet. Filling makes the result stable at about 5.3 km² whatever the precision. The deepest fill at full precision is only
0.18 m, so these are tiny
artefacts, not real ponds. Real sinks do exist (volcanic craters, quarries, karst) and should be kept, not filled.
Must give the three sink counts, explain rounding → flats → no downhill neighbour, and show the unfilled catchment is much smaller. Bonus for noting that real depressions exist and filling is a judgement.
Q4 — Pour points: Drop a pour point on a stream. Then untick Snap and click one or two cells beside the channel. Why does the catchment collapse? How does this explain a delineated catchment boundary that runs straight through a gully?
Example on the main stream 1 km above the coast: on the channel the catchment is 4.82 km². One to five cells (8–40 m) to the side, without snapping, it is only 0.06–0.2 ha, a patch of hillslope.
A cell beside the channel sits on the valley side: its D8 arrow points into the channel, so only the strip of slope above it drains through it. All the upstream water flows past in the neighbouring channel cells without touching the pour point.
That is the "boundary through a gully" error: an outlet a few metres off the true channel captures only part of the drainage, and the boundary cuts across valleys it should contain. The fix is pour-point snapping (move the outlet to the highest-accumulation cell within a small radius, 40 m here) and always checking the result visually.
Must explain that off-channel cells have tiny accumulation and that the outlet must sit on the high-accumulation channel cell. Full marks for linking this to snapping and to the gully-boundary symptom.
Q5 — Culvert design: Imagine a coastal road crossing the main stream. Delineate its catchment and read the rational-method peak flow. Move the culvert 500 m upstream: how do the area and Q change? Which input (C, i or A) would you trust least, and why?
With the defaults (C = 0.35, i = 40 mm/h) at the coast: A ≈ 5.28 km², Q = 0.35 × 40 × 5.28 / 3.6 ≈ 20.5 m³/s. Exact values depend on where students click.
500 m upstream the area barely changes (≈ 5.28 km²) because no big tributary joins on that reach; by 1 km upstream it is 4.82 km² and Q ≈ 18.7 m³/s. Q scales linearly with A, so moving a structure past a tributary junction gives a step change in design flow.
Least trustworthy: generally C and i. C is a lumped guess for land cover, soil and wetness (about 0.2 for forest vs 0.9 for pavement). i depends on the return period and on a time of concentration that the rational method needs but this demo does not compute. A is the most defensible input, provided sinks were filled and the pour point snapped (Q3, Q4). The rational method is normally limited to small catchments (a few km²), so 5 km² is at its edge.
Must calculate Q with the formula and correct units, describe how A and Q change upstream, and justify which input is least certain. Bonus for time of concentration or the small-catchment limit of the rational method.