Three rain gauges do not necessarily represent three equal pieces of a catchment. A simple average gives every instrument the same vote. A Thiessen estimate gives each gauge a weight based on the catchment area assigned to it.
We will work from a fictional catchment map with supplied polygon areas, calculate rainfall depth and volume, then remove one observation and redraw the boundary. The purpose is to understand what the weights mean before trusting the final average.
Continue from The Gold Standard of Hydrology, within the Gold Standard of Education: learning that transfers into life and the How X Works learning library. All numerical datasets below are invented teaching examples.
Find the explanation you need
Use the first route to connect the map and the calculation. Use the missing-gauge and changed-boundary tasks to see why an apparently routine average is a modelling decision.
Full chapter index · Practice and checking · Hydrology subject guide
Chapter contents
Define the spatial question · Chapters 1–2
Calculate a weighted depth · Chapters 3–4
Test the representation · Chapters 5–6
Handle incomplete evidence · Chapters 7–8
Change the setting · Chapters 9–10
Questions, sources and connected reading · Chapters 11–12
CHAPTER 1 OF 12 · Define the spatial question
1. Ask for an area average, not a typical gauge
A catchment rainfall estimate asks about precipitation over a defined area during a defined period. A gauge measures at a location. Moving from point observations to an area average requires an explicit representation rule.
Imagine a storm that passes over one side of a catchment while the other receives little rain. The arithmetic mean of three gauge totals could be useful under some conditions, but it does not automatically account for how much land each reading represents.
The US National Weather Service precipitation-measurement overview distinguishes arithmetic, isohyetal and Thiessen approaches. The useful choice depends on the available evidence and the question.
Start with the outlet, catchment boundary, observation period and unit. “Average rainfall” without those details is incomplete. A number describing yesterday’s whole basin cannot be substituted casually for a one-hour total over an upstream subcatchment.
Decide which average the reader needs
Imagine two requests arriving at the same desk. One asks what the typical instrument recorded. Another asks how much precipitation fell across a catchment. Giving each instrument one vote addresses the first request more naturally. Giving each represented area its proper share addresses the second under the spatial model. The word “average” does not choose between these questions for us.
An area mean has a physical interpretation: the estimated total precipitation volume divided by the area. It can also be described as the depth that would result if the estimated rainfall volume were spread evenly over the catchment. That imagined redistribution defines the summary; it does not say that the actual storm was uniform.
State the intended downstream use too. A water-budget exercise may need a volume-compatible mean. A local drainage question may need short-interval information over a smaller contributing area. A monthly basin average can be arithmetically correct while discarding the very location and timing needed for that drainage question. Choosing the spatial support is therefore part of formulating the problem, before the arithmetic begins.
Construct a region around each gauge containing locations closer to it than to another gauge under the chosen distance measure. In the usual planar Thiessen construction, neighbouring boundaries are perpendicular bisectors of lines joining gauge locations.
Clip those regions to the catchment boundary. The area inside each clipped polygon supplies that gauge’s weight: wᵢ = Aᵢ/A, where A is the whole catchment area. The weights should sum to one.
The rule assumes that each part of a polygon takes the depth measured at its assigned gauge. It is a transparent spatial approximation, not a claim that rain physically changes in steps at polygon borders.
A gauge outside the catchment can still have a region extending inside it. Do not exclude it solely because its point lies beyond the boundary. Construct the relevant network and inspect which cells intersect the area being estimated.
Work through an actual geometric construction
Consider a fictional rectangular catchment ten kilometres wide and ten kilometres high. Place three gauges on its horizontal centreline at x = 1, 5 and 9 kilometres, all with y = 5 kilometres. Straight-line distance is used in this flat teaching map. The boundaries between neighbouring gauges lie at x = 3 and x = 7 kilometres, halfway between their x positions.
The clipped cells are strips of width three, four and three kilometres. Multiplying by the catchment’s ten-kilometre height gives areas 30, 40 and 30 km², hence weights 0.30, 0.40 and 0.30. If their readings are 10, 40 and 100 mm, the mean is 3 + 16 + 30 = 49 mm. This is a second map with different weights from the main example, not a reconstruction of its supplied 50/30/20 partition.
This simple arrangement makes the nearest-gauge rule inspectable without specialist software. With more irregular locations and boundaries, the polygons become less rectangular, but the meaning of the area ratio stays the same. Keep the gauge coordinates, chosen distance convention and clipping boundary with the result so another person can understand which geometry produced the weights.
CHAPTER 3 OF 12 · Calculate a weighted depth
3. Calculate the fictional catchment depth
Our supplied map has already been partitioned into non-overlapping clipped polygons. Their areas are 50, 30 and 20 square kilometres. We assume the gauge readings have compatible timing and have passed the checks appropriate to this teaching exercise.
| Gauge | Assigned area | Weight | Rainfall | Weighted contribution |
|---|---|---|---|---|
| A | 50 km² | 0.50 | 10 mm | 5 mm |
| B | 30 km² | 0.30 | 40 mm | 12 mm |
| C | 20 km² | 0.20 | 100 mm | 20 mm |
| Total | 100 km² | 1.00 | — | 37 mm |
The weighted depth is P̄ = ΣwᵢPᵢ = 0.50 × 10 + 0.30 × 40 + 0.20 × 100 = 37 mm.
The arithmetic mean is (10 + 40 + 100)/3 = 50 mm. That calculation effectively assigns one third of the area to each gauge. Our supplied polygon map assigns different shares.
Neither calculation changes what the instruments recorded. The methods differ in how they represent space between the instruments. For this particular invented map and nearest-gauge rule, 37 mm is the consistent answer.
Check the meaning of each contribution
The 20 mm contributed by C is 0.20 × 100 mm, expressed as a share of the whole-catchment mean. It is not the rainfall measured at C, and it is not the mean over C’s own polygon. That polygon is assigned 100 mm by the method. The contribution becomes 20 mm only when its represented volume is spread over the whole 100 km² denominator.
This gives a useful denominator exercise. If someone reports that A and B together received 17 mm, ask “over which area?”. Their contributions to the whole-basin mean sum to 17 mm. Their own combined 80 km² has an assigned mean of 21.25 mm. Both numbers can describe the same represented volume, but their spatial denominators differ.
Do the arithmetic mean only after explaining its implicit equal weights. Three equal weights sum to one, just as the Thiessen weights do; that alone cannot decide which representation suits the question. The distinction is whether equal shares or the clipped polygon shares are justified for the intended estimate. A correct checksum is necessary, but it is not a substitute for an appropriate spatial model.
CHAPTER 4 OF 12 · Calculate a weighted depth
4. Convert depth to volume with the area still visible
One millimetre over one square kilometre corresponds to 1,000 cubic metres: 0.001 m × 1,000,000 m².
The 37 mm estimate over 100 km² therefore represents 37 × 100 × 1,000 = 3,700,000 m³ of precipitation. The arithmetic estimate would represent 5,000,000 m³.
The difference is 1,300,000 m³ of estimated rainfall volume. It is not an observed extra river flow. Infiltration, storage, evaporation and timing still separate rainfall from the outlet response.
Check the calculation by working with the polygon volumes directly: 10 × 50 × 1,000 + 40 × 30 × 1,000 + 100 × 20 × 1,000 = 3,700,000 m³. Dividing that sum by the total area returns 37 mm.
This second route is a useful verification because it exposes the physical meaning of the weights. The weighted average is total represented rainfall volume divided by the represented area.
Use units to locate a large error
A common conversion error is to treat one square kilometre as one thousand square metres. It is one thousand metres multiplied by one thousand metres, or one million square metres. Combining that area mistake with millimetres can produce an answer that looks neat but is wrong by orders of magnitude. Write the conversion before multiplying, especially when a result seems unexpectedly small.
For our main example, A represents 500,000 m³, B represents 1,200,000 m³ and C represents 2,000,000 m³. The wettest cell is the smallest by area yet contributes the largest represented rainfall volume. This is a reminder that area weight and total contribution are different quantities. Neither the largest polygon nor the wettest gauge alone determines the total.
There is also a useful reverse calculation. Divide the 3,700,000 m³ total by 100,000,000 m² to obtain 0.037 m, then convert to 37 mm. If the reverse calculation fails, inspect the area and depth conversions before interpreting the hydrology. The volume check tests dimensional consistency; it still cannot establish that the nearest-gauge approximation accurately represented the actual storm between instruments.
Proximity is not a rainfall mechanism. Elevation, storm movement and other spatial patterns can make the nearest gauge an imperfect representative of a location.
Adding a gauge changes the geometric partition even if the new instrument records the same storm. Moving a gauge changes it too. Keep a record of the network and weights used for each estimate.
Thiessen weighting does not automatically remove instrument bias, repair poor siting or detect a local storm missed by every gauge. It also does not produce a confidence interval merely because the polygons fit together exactly.
The National Weather Service training discussion of mean areal precipitation connects gauge observations, spatial weighting and gridded estimates. The method should remain visible alongside the result, especially when the rainfall pattern matters to the next model.
Add a gauge without pretending the storm changed
Return to the supplied 50/30/20 km² partition. Imagine a new network in which Gauge D takes ten square kilometres previously assigned to C, while all other assigned areas remain the same. Let D record 60 mm during the same period. The new areas are 50, 30, 10 and 10 km². The new mean is 5 + 12 + 10 + 6 = 33 mm.
The change from 37 to 33 mm is −4 mm. You can obtain it directly: the reassigned weight is 0.10 and the replacement depth differs by 60 − 100 = −40 mm, giving 0.10 × (−40) = −4 mm. This check isolates the effect of the representation change without suggesting that the historical rainfall itself changed when the network was redrawn.
A new estimate may be better informed, but that requires examining the added evidence and setting. More gauges do not guarantee improvement if an added reading is biased or poorly matched in time. Preserve the old estimate, network and method when a revised product is issued. Otherwise an analyst comparing two stored totals may mistake a change in observation support for a change in rainfall.
CHAPTER 6 OF 12 · Test the representation
6. Separate catchment volume from storm location
Two storms can have the same catchment mean while placing rain in different subareas. A whole-basin number removes that spatial arrangement.
In our example, Gauge C records the largest total but represents only one fifth of the area. If that wet region lies close to the outlet, the timing question differs from a similar region far upstream. A mean depth alone does not retain that information.
For a lumped volume calculation, an area mean may be a suitable input under the model’s assumptions. For questions about local peaks or spatially variable runoff, additional subdivision or spatial information may be needed.
Keep the inference in layers: first estimate rainfall over space; then estimate losses or excess; then transform and route the water. A more careful average helps the first layer. It does not establish the accuracy of the entire chain.
Compare two spatial patterns with the same mean
Keep the original weights 0.50, 0.30 and 0.20. Pattern One has gauge depths 10, 40 and 100 mm, giving 37 mm overall. Pattern Two has depths 50, 20 and 30 mm, giving 25 + 6 + 6 = 37 mm as well. Both represent 3,700,000 m³ over the catchment under the method, but they distribute it differently.
If subareas have different response times or runoff behaviour, a model that preserves these spatial patterns can produce different flow responses. The equal mean has not established equal hydrological effects. Conversely, an explicitly lumped model supplied only with the equal mean cannot recover the lost pattern without additional inputs or assumptions.
This is a precise limit of summarisation. The area average preserves the represented total volume for the selected period. It does not preserve which polygon was wettest, the movement of the storm, or short bursts within the period. Before passing the number onward, identify which of those features the next calculation requires. A more detailed output display cannot recreate information that the input summary already discarded.
CHAPTER 7 OF 12 · Handle incomplete evidence
7. Do not turn a missing gauge into zero rainfall
Now Gauge C has no valid reading. The other two contributions still total 5 + 12 = 17 mm when expressed over the original 100 km².
| Assumed C depth | Catchment estimate |
|---|---|
| 0 mm | 17 mm |
| 60 mm | 29 mm |
| 120 mm | 41 mm |
| No value assumed | Full-catchment mean remains unresolved |
A zero at C is one possible assumed scenario. It is not the meaning of a missing value. Under the illustrative range 0–120 mm for C, the full-catchment mean ranges from 17 to 41 mm. That range is conditional on the chosen bounds.
Renormalising A and B gives weights 50/80 and 30/80. Their weighted mean is 21.25 mm over their combined 80 km². Calling 21.25 mm the whole-catchment estimate adds a new assumption about the unobserved 20 km².
Other approaches might use another gauge, radar evidence or a documented infilling model. The key reporting habit is to label what was measured, what was inferred and how the gap was handled.
Compare two honest ways to handle the gap
One approach keeps the original cells and supplies an explicitly estimated depth for C. Another removes C from the available network and recomputes the nearest-gauge regions using the remaining stations. These are different spatial models. Simply dividing the old A and B weights by 0.80 does not generally reproduce the geometric areas obtained by redrawing the polygons.
For an invented redraw, suppose A takes 60 km² and B 40 km² once C is omitted. The two-gauge estimate becomes 0.60 × 10 + 0.40 × 40 = 22 mm. Renormalising their old areas gave 21.25 mm instead. The difference arises from how the previously unrepresented area was assigned. Neither operation should be described as if C had actually measured that area during the missing period.
The scenario bound has a separate meaning. Saying “if C lies between zero and 120 mm, the mean lies between 17 and 41 mm” is a conditional arithmetic statement. The interval’s credibility depends on why those C bounds are plausible. It is not an automatically generated confidence interval, and it does not cover every possible source of rainfall-estimation error.
CHAPTER 8 OF 12 · Handle incomplete evidence
8. Keep uncertainty and units separate from precision
Suppose the gauge-depth errors are bounded by ±2 mm at A, ±4 mm at B and ±10 mm at C. Treat the areas and weights as fixed for this exercise.
A conservative bound on the absolute weighted-depth error is 0.50 × 2 + 0.30 × 4 + 0.20 × 10 = 4.2 mm. The bound allows the component errors to align in the same direction.
Over 100 km², 4.2 mm corresponds to 420,000 m³. This is an illustrative worst-case bound under the stated limits, not a statistical confidence interval.
It does not include uncertainty from spatial representation, polygon construction or an unsampled storm feature. Reporting 37.0000 mm would not remove those omissions. Extra decimal places can describe arithmetic more finely while leaving the important evidence gap unchanged.
Explore how much one reading can move the mean
With weights fixed, increasing one gauge depth by δ changes the mean by wδ. A five-millimetre revision at A changes the mean by 2.5 mm; the same revision at C changes it by one millimetre. This sensitivity describes the calculation. It does not say A is a more accurate instrument or that C’s location is less scientifically important.
Weight errors matter too. If two square kilometres were mistakenly assigned to A instead of C in the main example, correcting the map transfers weight 0.02 from A to C. The mean increases by 0.02 × (100 − 10) = 1.8 mm. The correction is volume-relevant because the two assigned depths differ. If both depths were equal, this particular area transfer would not change the mean.
These checks make uncertainty concrete. Ask which component is uncertain: the reading, the area, the period, the spatial assumption or more than one. Then calculate the effect of a stated perturbation where possible. Keep the answer labelled as a sensitivity scenario unless a justified probability model supports a statistical interpretation. An exact polygon area is not evidence that rainfall inside it was known exactly.
Change the task to another supplied catchment boundary. The same three gauge totals apply, but the clipped areas are now 20, 30 and 50 km².
The new weights are 0.20, 0.30 and 0.50. The mean is 0.20 × 10 + 0.30 × 40 + 0.50 × 100 = 64 mm.
The simple average remains 50 mm because it never used the boundary. It now lies below, rather than above, the Thiessen estimate. There is no rule that the arithmetic mean must always overestimate rainfall.
The learning transfer is precise: changing the spatial question can change the weights even when every gauge reading stays fixed. Reusing last project’s weights without checking the catchment would answer the wrong geographical question correctly.
Combine subcatchments by area, not by their names
Suppose an upstream subcatchment covers 40 km² with an estimated mean of 25 mm, and a non-overlapping downstream contributing area covers 60 km² with an estimated mean of 45 mm. Under compatible periods and definitions, their combined mean is (40 × 25 + 60 × 45)/100 = 37 mm. Averaging 25 and 45 equally would give 35 mm and silently assign equal areas.
The represented volumes provide the same check: 1,000,000 m³ upstream plus 2,700,000 m³ downstream equals 3,700,000 m³. Make sure “downstream area” here excludes the upstream subcatchment. If it already included that upstream area, adding both volumes would count some land twice. Words describing nested catchments need the same care as polygon boundaries.
This second-level averaging transfers the original principle. Whether the components are gauge cells or supplied subcatchments, a volume-compatible depth uses the relevant area denominator. It also reveals a limit: combining means is valid for this accounting only when their periods, rainfall definitions and spatial coverage are compatible. A daily total and an event total cannot be combined merely because both are expressed in millimetres.
Confirm that the clipped polygons cover the intended catchment without unintended gaps or overlaps, that their areas use a suitable coordinate system and that the weights sum to one.
Confirm that every gauge total covers the same period. A recording gap is different from dry weather; a changed reporting day is different from a changed storm.
Then calculate both ΣwᵢPᵢ and ΣAᵢPᵢ/A. They should agree. Check that the weighted mean lies between the smallest and largest included values when all weights are non-negative.
For independent practice, use areas 25, 25 and 50 km² with depths 20, 60 and 40 mm. The weighted contributions are 5, 15 and 20 mm, giving 40 mm overall. The volume over 100 km² is 4,000,000 m³.
Finally, explain one thing the answer leaves unresolved. For example: the area mean does not reveal whether most rain fell early or late in the period.
Audit a proposed answer
Read this fictional claim: “Gauge C failed, so it was entered as zero. The basin received 17 mm, and the river therefore carried 1,700,000 m³.” There are two independent unsupported steps. Missing observation was converted to a dry reading. Then estimated precipitation volume was converted directly into river outflow without a runoff and storage account. Repairing the first error would not repair the second.
A stronger report states the full-catchment estimate as unresolved unless an explicit missing-data method is applied. It can still give the known contributions and conditional scenarios. If a full estimate is produced, retain the infilling or redraw rule, gauge identities, time window and geometry alongside the result. That record lets a later reader reproduce the calculation and inspect what was assumed.
For delayed practice, explain why the two 37 mm spatial patterns can produce identical represented rainfall volumes yet remain different modelling inputs for a spatial runoff model. Then explain why the arithmetic mean can sit above or below the Thiessen mean. These questions test the relationship between geography and calculation rather than the ability to repeat one weighted sum.
Is the wettest gauge always the most influential?
No. Influence on the weighted mean depends on both its rainfall value and its assigned area. The weights describe area, not confidence or instrument quality.
Can an outside gauge contribute?
Yes, if its nearest-gauge cell intersects the catchment in the chosen construction.
Should weights change every storm?
For a fixed network and boundary, ordinary Thiessen geometry stays the same. A different network, missing station treatment or boundary can require different weights. Changing them should be documented.
Does a correct area mean guarantee a correct runoff peak?
No. Peak flow also depends on timing, spatial distribution, losses, catchment state and routing.
Is renormalising available gauges forbidden?
No. It can be part of an explicit estimation method. Explain the changed support and assumption rather than presenting it as an unchanged complete observation.
Can a gauge’s large weight be interpreted as high confidence?
No. The weight here is a geometric area share. An instrument can represent a large assigned polygon while having an uncertain or unrepresentative reading. A method that weights observations by their uncertainty would be a different calculation, with different assumptions. Do not describe a geometric weight as a measurement-quality score.
Why does the mean stay inside the gauge range?
With non-negative weights summing to one, every contribution is a share of a supplied depth. Their sum cannot exceed the largest supplied depth or fall below the smallest. This is a useful arithmetic check on this weighted mean. It does not mean the actual rainfall everywhere inside the catchment stayed within the measured gauge range; an unsampled local maximum could still exist.
Can a better spatial estimate repair a bad time window?
No. If one gauge covers midnight to midnight and another covers a different reporting day, the spatial calculation combines unlike totals. Matching time support is a separate task from computing polygon areas. A trustworthy estimate needs both. State the actual observation interval and treatment of gaps so later readers can check that the inputs describe the same period.
Methods and assumptions
National Weather Service: Precipitation Measurements compares rainfall-averaging methods. National Weather Service Training Center: mean areal precipitation develops the spatial-estimation context. Links checked on 21 October 2026. All maps described by supplied areas, depths, missing-data scenarios and error bounds here are invented examples.
Understand the geometry and water system
Voronoi Regions: How Nearest-Point Geography Divides Space explains the partitioning rule. The Gold Standard of Hydrology connects rainfall with pathways, storage, measurement and budgets.
Check records before comparing catchments
Double-Mass Curves and Rain-Gauge Consistency examines changes in the observation record. Paired Catchments and Attribution of Change examines what a comparison can establish.
Follow the rainfall into a response
Predicting a Flood Before the River Reaches the Street places rainfall inputs inside a wider modelling chain. Continue through learning that transfers into life and How X Works. This educational benchmark does not certify a particular hydrological model or design.
Follow estimated rainfall toward a runoff response
After distinguishing rainfall from rainfall excess, follow the separate transformation step in The Gold Standard of Hydrology | Unit Hydrographs and the Limits of Shift-and-Add.
Extend the evidence check
An area-weighted depth still needs a compatible time window. Continue with Rainfall Time Steps and Why Equal Storm Totals Can Produce Different Runoff to inspect what a storm total leaves unresolved about timing.