A catchment received an average of twenty-five millimetres of rain. Its land units have an average runoff fraction of one half. Does that mean twelve and a half millimetres of locally generated runoff? Only under an additional condition. The wettest rain might have fallen on the units with the largest fractions, or on those with the smallest. Those arrangements can produce different event volumes while preserving both reported averages.
The short answer is to multiply the paired local quantities before adding them. Under the deliberately simple model used here, generated volume is the sum of each unit’s area times its rainfall depth times its supplied event runoff fraction. Averaging rainfall and fractions separately can discard the spatial pairing that the calculation needs.
This article develops that calculation with fictional, fully specified numbers. It separates locally generated runoff from routed outlet flow, explains the missing term, and shows how to aggregate without changing the answer. Start with the Watersheds subject guide for boundaries, outlets, nested areas and the wider water balance. The question here is narrower: what information must survive a spatial average before an event-volume calculation can be trusted?
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Start with the question closest to yours, or read in order. Worked chapters remain open; the grouped index helps you return to a check.
ROUTE 1 · CHAPTERS 1–5
Build local volumes
Define the event and partition, convert units and work the paired table.
ROUTE 2 · CHAPTERS 6–10
Explain the missing term
Swap rainfall, derive weighted covariance and test agreement conditions.
ROUTE 3 · CHAPTERS 11–16
Choose conserving weights
Derive effective fractions and preserve volumes through aggregation.
ROUTE 4 · CHAPTERS 17–21
Keep model boundaries
Separate marginal summaries, event assumptions, timing and outlet accounting.
ROUTE 5 · CHAPTERS 22–28
Verify and transfer
Test uncertainty, a new event, inverse information and rounding effects.
Chapter contents
Build local volumes · Chapters 1–5
1. Define the event and the quantity being estimated
2. Draw non-overlapping units inside one boundary
3. Convert depth times area into volume
Explain the missing term · Chapters 6–10
6. Test the product of the two separate averages
7. Swap the rain while preserving both averages
8. Express the missing term with area weights
Choose conserving weights · Chapters 11–16
11. Derive a rainfall-weighted effective fraction
12. Check effective-fraction bounds and dry units
13. Work an unequal-area example
14. Aggregate two cells without losing their answer
Keep model boundaries · Chapters 17–21
17. Keep the runoff fractions attached to their assumptions
18. Separate event volume from peak discharge
19. Keep outlet accounting downstream of local generation
Verify and transfer · Chapters 22–28
22. Transfer to a new three-unit event
23. Audit a published-looking summary
24. Repair, stabilise and extend the learning
25. Reconstruct a missing local fraction only when the record supports it
26. Check how rounding moves through the calculation
CHAPTER 1 OF 28 · Build local volumes
1. Define the event and the quantity being estimated
Imagine two mapped land units observed over the same event. Each has a supplied rainfall depth and a supplied fraction of that event’s rainfall that becomes the particular local runoff quantity we are counting. Treat these fractions as stipulated teaching inputs. They are not universal values for a land-cover label, and this article does not calibrate a catchment model.
Our target is locally generated runoff volume across the units. It is not the peak discharge at an outlet. It is also not automatically the total water measured at that outlet during an arbitrarily selected time window. Routing, storage change, baseflow, exchanges and the choice of window can matter when moving from generation to an outlet record.
Naming the target prevents a successful area calculation from being used to answer a different hydrological question. The mathematical relationships below are exact within the stated local linear model. Their usefulness for an actual watershed depends on evidence for the inputs, the spatial representation and the process assumptions.
CHAPTER 2 OF 28 · Build local volumes
2. Draw non-overlapping units inside one boundary
Let the units have areas Aᵢ, rainfall depths Pᵢ and event runoff fractions Cᵢ. Every point in the modelled area belongs to one unit. The units must not overlap, and together they must cover the area represented by the result. Keep an area outside the study boundary outside the sum.
This sounds simple until a map contains both tributary catchments and the larger catchment containing them. Adding the parent area to all its child areas counts some land twice. A hierarchy is useful for organising information, but a volume sum needs a non-overlapping partition at the chosen level.
A practical paper check is to colour every contributing patch once. If a patch receives two colours because it appears in two rows, repair the accounting before doing arithmetic. If a patch has no row, decide whether the result deliberately excludes it or whether information is missing. A neat mean cannot compensate for an incorrect boundary.
One millimetre over one square kilometre represents one thousand cubic metres. A millimetre is one thousandth of a metre; a square kilometre is one million square metres. Their product is one thousand cubic metres. Write this conversion before using rainfall numbers that look small beside large catchment areas.
For unit i, rainfall input volume is AᵢPᵢ with compatible units. Under our model, generated runoff volume is AᵢCᵢPᵢ. Since Cᵢ is dimensionless and lies between zero and one, that unit’s generated volume cannot exceed its rainfall input volume. This is a model check for the named rainfall-derived quantity, not a general rule for every possible outlet water balance.
If area is supplied in square kilometres and rainfall in millimetres, multiply the numerical product by one thousand to obtain cubic metres. Keep that factor outside or inside every row consistently. Mixing hectares, square kilometres and square metres is a change of units, not a change of physical process.
The total generated volume is V = ΣAᵢCᵢPᵢ. Each product belongs to one place during the same stated event. Multiplication preserves the relationship between that place’s rainfall and its response fraction before addition combines places.
The order matters conceptually even though arithmetic can be rearranged within a product. A list of rainfall values and a separate list of fractions do not identify which values belong together. If the labels disappear, the physical pairings disappear with them.
A useful representation has one row per unit and columns for area, rainfall, fraction, rainfall volume and generated volume. Add the last two columns separately. Their totals answer different questions: how much rain entered the represented area, and how much local runoff the supplied model generates. Dividing the second total by the first later gives an effective fraction with a clear meaning.
Take two units, each one square kilometre. Unit A receives ten millimetres and has supplied fraction 0.2. Unit B receives forty millimetres and has fraction 0.8. All values belong to the same fictional event. There is no uncertainty in this first arithmetic exercise.
Unit A receives ten thousand cubic metres of rain and generates two thousand cubic metres of runoff. Unit B receives forty thousand cubic metres and generates thirty-two thousand. Rainfall input totals fifty thousand cubic metres; generated runoff totals thirty-four thousand.
| Unit | Area in square kilometres | Rainfall in millimetres | Supplied fraction | Generated volume in cubic metres |
|---|---|---|---|---|
| A | 1 | 10 | 0.2 | 2000 |
| B | 1 | 40 | 0.8 | 32000 |
| Total | 2 | Area mean 25 | Area mean 0.5 | 34000 |
Spread across the full two square kilometres, the generated volume corresponds to seventeen millimetres. That depth is an area-normalised summary of the computed volume. It does not say that every point generated seventeen millimetres locally: the two local depths are two and thirty-two millimetres.
CHAPTER 6 OF 28 · Explain the missing term
6. Test the product of the two separate averages
The area-average rainfall is twenty-five millimetres. The area-average fraction is one half. Multiplying those means gives twelve and a half millimetres, which corresponds to twenty-five thousand cubic metres across two square kilometres.
That is nine thousand cubic metres below the row-by-row answer. No conversion mistake caused the difference. The shortcut discarded the fact that the larger rainfall coincided with the larger response fraction. It applied the average fraction to an average rainfall as though their spatial pairing contributed nothing further.
This is why the title says the averaging can change the volume. It does not mean every pair of averages must produce an error, or that averaging itself is forbidden. It means the product of two means requires a condition that should be checked. The next chapters show that condition and how its sign affects the result.
CHAPTER 7 OF 28 · Explain the missing term
7. Swap the rain while preserving both averages
Keep both areas and both fractions unchanged. Now let Unit A receive forty millimetres and Unit B receive ten. The rainfall mean remains twenty-five millimetres, and the fraction mean remains one half. The separate-averages shortcut therefore still returns twenty-five thousand cubic metres.
The local calculation changes. Unit A generates eight thousand cubic metres; Unit B also generates eight thousand. Total generated volume is now sixteen thousand, or eight millimetres over the whole catchment. The same shortcut now overestimates by nine thousand cubic metres.
The two catchments have identical areas, identical rainfall totals and identical lists of fractions. Their different pairings produce eighteen thousand cubic metres of difference in generated runoff. This counterexample identifies exactly what the two summary averages failed to retain. It does not require a more complicated runoff model to demonstrate the information loss.
CHAPTER 8 OF 28 · Explain the missing term
8. Express the missing term with area weights
Let total area be A and let wᵢ = Aᵢ/A. The weights are nonnegative and sum to one. Define P̄ = ΣwᵢPᵢ and C̄ = ΣwᵢCᵢ. Actual generated depth under the model is R = ΣwᵢCᵢPᵢ.
The difference between R and C̄P̄ is Σwᵢ(Cᵢ − C̄)(Pᵢ − P̄). Expanding the brackets proves the identity: the two middle terms each sum to C̄P̄, while the final constant term adds one C̄P̄ back. The remaining expression is the mean of the products minus the product of the means.
This weighted covariance is a description of spatial pairing within our finite set of units. It does not require treating the two units as random draws or conducting a significance test. Its units here are millimetres, because fraction is dimensionless. Multiply it by total area with compatible units to obtain the volume difference.
CHAPTER 9 OF 28 · Explain the missing term
9. Calculate the covariance in both arrangements
In the first arrangement, the fraction deviations are minus 0.3 and plus 0.3. Rainfall deviations are minus fifteen and plus fifteen millimetres. Both paired products are plus 4.5 millimetres. Their equal-weight average is therefore plus 4.5 millimetres.
Add that to the shortcut’s twelve and a half millimetres to recover seventeen. Across two square kilometres, 4.5 millimetres represents nine thousand cubic metres. The identity agrees with the direct local calculation, which is the best first check of the algebra.
After swapping rainfall, each paired deviation product is minus 4.5 millimetres. The shortcut depth plus this negative term becomes eight millimetres. A positive term means the separate-averages shortcut is too low in this model; a negative term means it is too high. Neither sign is a universal description of actual catchments. It depends on the event’s mapped pairing and the supported response model.
CHAPTER 10 OF 28 · Explain the missing term
10. Recognise when the shortcut happens to agree
If every unit has the same supplied fraction, it factors out of the sum. Multiplying that common fraction by area-average rainfall then gives the correct generated depth. If rainfall is uniform across units, its common depth similarly factors out.
More generally, the shortcut agrees whenever the weighted covariance is zero. That can occur even when neither variable is uniform. Zero covariance in a particular partition is an algebraic condition; do not turn it into a broad claim that rainfall and land response are physically unrelated.
For a learner check, use four equal-area units with fraction deviations minus 0.3, minus 0.1, plus 0.1 and plus 0.3, and rainfall deviations plus five, minus fifteen, plus fifteen and minus five millimetres. The four products sum to zero. Adding suitable positive means produces nonuniform valid inputs whose shortcut agrees. Agreement in one constructed case does not justify deleting the pairing information from the next event.
CHAPTER 11 OF 28 · Choose conserving weights
11. Derive a rainfall-weighted effective fraction
Suppose the goal is to express generated volume as one fraction of total rainfall input. When total rainfall input is positive, define Cₑ = ΣAᵢCᵢPᵢ / ΣAᵢPᵢ. Then V = Cₑ times total rainfall input by construction.
This effective fraction weights each local fraction by its share of rainfall volume, rather than by its share of area alone. In the first example, Unit B contributes four fifths of rainfall input. The effective fraction is 0.2 × one fifth plus 0.8 × four fifths, which is 0.68.
After the rain swap, the weights reverse and the effective fraction becomes 0.32. The area-average fraction stays at 0.5 in both cases. The distinction is not a choice between two competing names for the same number. The weights answer different questions. Keep the event and its rainfall distribution attached to an effective fraction instead of treating it as a permanent property of the entire watershed.
CHAPTER 12 OF 28 · Choose conserving weights
12. Check effective-fraction bounds and dry units
With nonnegative rainfall and fractions between zero and one, the rainfall-volume weights are nonnegative and sum to one. Therefore the effective fraction lies between the smallest and largest local fractions receiving positive rainfall. It cannot exceed their maximum merely because a wetter unit receives a larger weight.
A dry unit has zero rainfall-volume weight for this event. It still contributes to the total area and therefore affects area-average depths, but it contributes no rainfall input or generated rainfall-derived runoff in our local model. That difference makes area weights and rainfall weights especially easy to confuse.
If every unit is dry, both numerator and denominator of the effective-fraction ratio are zero. The ratio is undefined. Generated volume is zero under the model, but an event with no rainfall provides no numerical effective fraction through this ratio. Writing an arbitrary zero in its place would hide a missing definition rather than complete the calculation.
Now use a one-square-kilometre unit with rainfall twenty millimetres and fraction 0.1, and a three-square-kilometre unit with rainfall sixty millimetres and fraction 0.5. The area weights are one quarter and three quarters, not one half each.
Rainfall input is twenty thousand plus one hundred and eighty thousand cubic metres, giving two hundred thousand. Generated volume is two thousand plus ninety thousand, giving ninety-two thousand. Across four square kilometres, rainfall mean is fifty millimetres and generated depth is twenty-three.
The area-average fraction is 0.4. Multiplying it by rainfall mean gives twenty millimetres, or eighty thousand cubic metres. The shortcut misses twelve thousand. The weighted covariance is three millimetres, while the rainfall-weighted effective fraction is 92/200 = 0.46. These numbers can all be correct simultaneously because they refer to different averages and targets. A simple unweighted mean of the two rows would not represent this catchment’s area.
CHAPTER 14 OF 28 · Choose conserving weights
14. Aggregate two cells without losing their answer
Imagine merging the two unequal-area units into one reporting unit. Its area is four square kilometres and its rainfall depth is fifty millimetres. Which fraction should accompany it if the merged row must preserve the same generated event volume?
Use the rainfall-weighted effective fraction 0.46. Multiplying four square kilometres by fifty millimetres by 0.46 returns ninety-two thousand cubic metres. Using the area-average fraction 0.4 instead returns eighty thousand and changes the model’s answer during aggregation.
This is a preservation rule for the stated event-volume calculation. It does not prove that one merged fraction will work for a later storm with a different rainfall pattern. Nor does it preserve a hydrograph, a peak or local spatial impacts. A coarser representation can preserve one total while losing information needed for other questions. State which property the aggregation is designed to conserve.
CHAPTER 15 OF 28 · Choose conserving weights
15. Preserve totals through a reporting hierarchy
For several non-overlapping groups of cells, calculate each group’s rainfall input and generated volume first. The whole-area totals are the sums of those group totals. If a group’s rainfall input is positive, it can also report its own effective fraction.
To combine group effective fractions, weight them by group rainfall input volumes. Weighting by group area alone generally loses the same pairing information again, now one level higher in the hierarchy. An area-based mean of group rainfall depths remains appropriate for whole-area rainfall, but it does not automatically provide the weights for averaging response fractions.
This gives a practical reconciliation check: calculate the whole result once from original cells and once from the proposed grouped records. The volumes should agree to the stated rounding precision. If they do not, inspect overlapping areas, omitted cells, wrong weights and premature rounding. Do not explain a bookkeeping discrepancy as a new hydrological process until the arithmetic representation has been checked.
CHAPTER 16 OF 28 · Choose conserving weights
16. Ask what separate marginal lists can establish
Suppose an exercise supplies two equal areas, fractions 0.2 and 0.8, and rainfall depths ten and forty millimetres, but removes their location labels. There are two distinct pairings. We have calculated generated volumes sixteen thousand and thirty-four thousand cubic metres.
Without the pairing, the data identify those two possibilities rather than a unique volume. If mixtures or finer spatial arrangements consistent with the summaries are allowed, intermediate values may also be compatible. The admissible possibilities depend on exactly what the supplied summaries preserve.
For two ordered fractions C₁ ≤ C₂ and ordered rainfalls P₁ ≤ P₂, the aligned product sum exceeds the crossed sum by (C₂ − C₁)(P₂ − P₁). Both differences are nonnegative, which explains the ordering without a memorised rule. This is an information check for a specified finite exercise. It is not a substitute for collecting a spatially suitable rainfall record or validating local runoff responses.
CHAPTER 17 OF 28 · Keep model boundaries
17. Keep the runoff fractions attached to their assumptions
In these examples, each Cᵢ is a supplied event fraction that makes local generated depth equal to CᵢPᵢ. Actual runoff generation can depend on antecedent conditions, storm characteristics, infiltration, surface connectivity and other processes. A single fixed fraction may be a limited representation.
The USGS explanation of surface runoff discusses how precipitation and catchment conditions influence runoff. It supports the need to examine conditions; it does not supply or validate our fictional fractions. Likewise, the US Army Corps of Engineers description of gridded precipitation provides context for retaining rainfall’s spatial and temporal distribution.
Do not interpret 0.8 as an official value for an urban land class or 0.2 as a measured value for vegetation. The local product rule is conditional on the supplied model. A more realistic nonlinear response may require different local calculations before aggregation, while the general habit of preserving the relevant spatial information remains useful.
CHAPTER 18 OF 28 · Keep model boundaries
18. Separate event volume from peak discharge
Volume adds water over an event. Peak discharge describes a rate at a particular time. Two areas can generate equal total volumes but deliver their water on different schedules, producing different combined peaks. Our calculation contains no travel times or time-varying hydrographs and therefore cannot determine that peak.
Do not insert the event rainfall depth from our table into a peak-flow formula as though it were an intensity with the required duration and units. The presence of a letter C in two formulas does not make their inputs or assumptions interchangeable.
The watershed root links to a worked branch on tributary timing and hydrographs. That is the appropriate next question after a generation-volume calculation when timing matters. The transfer lesson is to identify which operation the data support: spatial addition of event volumes, time alignment of discharge records, or a separately justified peak model. Each operation needs the information relevant to its target.
CHAPTER 19 OF 28 · Keep model boundaries
19. Keep outlet accounting downstream of local generation
Suppose locally generated runoff totals thirty-four thousand cubic metres, but a gauge records another total over the selected event window. That disagreement alone does not identify an error in the spatial multiplication. The gauge may measure a different combination of water sources and storage changes over a different effective interval.
A consistent accounting model would state its boundary, time window and contributing terms. Water retained temporarily in channels or depressions can change when it crosses the outlet. Baseflow or other inputs can affect the record. Losses and exchanges need appropriate evidence rather than being invented to close a desired balance.
This is why we have called our target generated volume throughout. The adjective protects the meaning of the result. Before comparing it with an observed outlet total, build the bridge between those quantities and examine whether the required terms were measured or modelled. A more detailed decimal in the generated total cannot replace that bridge.
CHAPTER 20 OF 28 · Keep model boundaries
20. Keep time averaging as a separate decision
Our units all refer to one defined event. If the response fractions or rainfall pattern change within that event, a time-resolved model may be needed. It would have its own paired inputs for each place and interval, followed by the relevant summation and accounting.
Do not assume that an event fraction is the arithmetic mean of fractions from equal time intervals. Even at one location, intervals with different rainfall amounts contribute different rainfall volumes. The appropriate effective ratio follows from the generated and input totals, provided the local model and target remain consistent.
This observation extends the weighting logic without claiming that runoff responds instantly or linearly at every time step. Routing and state-dependent generation can make a simple interval product inadequate. Start by naming what the interval records mean, then decide which model can connect them. A calculation should become more detailed because the question needs that information, not merely because smaller time steps are available.
CHAPTER 21 OF 28 · Keep model boundaries
21. Distinguish model error from input uncertainty
Three different limitations can affect an answer. Rainfall and area measurements can be uncertain. Supplied fractions can be uncertain or poorly transferable. The local linear model itself can be an inadequate representation. These limitations need not move the result in the same direction.
The covariance identity describes the aggregation discrepancy within the chosen inputs and model. It is not a complete error bar for the true watershed. Correcting the separate-averages shortcut does not validate every other assumption in the calculation.
For an evidence check, repeat the local sum under explicitly supported input scenarios and retain the pairing in every scenario. If the fraction and rainfall uncertainties are related, do not quietly treat all combinations as equally plausible. Report what the scenarios represent. A deterministic range from stated admissible inputs is different from a probability interval based on a justified statistical model.
Try three equal-area units, each half a square kilometre. Their rainfall depths are ten, twenty and thirty millimetres; their supplied fractions are 0.6, 0.3 and 0.2. Calculate locally generated volumes before reading the result.
Each local generated depth is six millimetres. Each unit therefore generates three thousand cubic metres, for a total of nine thousand. Rainfall input totals thirty thousand cubic metres, so the effective event fraction is 0.3. Across the full one and a half square kilometres, generated depth is six millimetres.
The area-average fraction is 11/30, approximately 0.3667, while mean rainfall is twenty millimetres. Their product is 22/3 millimetres, corresponding to eleven thousand cubic metres. The shortcut is two thousand too high; covariance is minus 4/3 millimetres. Now reorder the fractions while keeping the rainfall list fixed. Explain which quantity changes and why the two separate means cannot reveal that change.
A short report says, “Average rainfall rose, average runoff fraction stayed unchanged, so generated runoff must have risen proportionally.” Under our model, that conclusion also needs information about spatial pairing, represented areas and how the fractions were obtained. The two means alone do not establish proportional change.
Ask for the mapped or otherwise paired input table, the event definition and the target quantity. Recalculate local products and reconcile them with any grouped totals. Check whether the report has moved from generated water to outlet volume or peak flow without adding the necessary evidence.
The objective is not to reject every summary. A useful summary states what it preserves and what supporting detail remains available. If the local records cannot be recovered, retain the resulting uncertainty rather than reconstructing a convenient pairing. An honest conclusion may be that rainfall input is known while generated volume is not uniquely determined by the remaining summaries.
CHAPTER 24 OF 28 · Verify and transfer
24. Repair, stabilise and extend the learning
To repair the first connection, use two physical paper squares with different rainfall and fraction cards. Calculate each square’s rainfall volume and generated volume. Swap only the rainfall cards and explain the change before introducing covariance notation.
To stabilise, convert the row totals into catchment-average depths and then derive the effective fraction. Ask the learner to say which weights appear in each average. A correct number without a named weight can conceal the same misunderstanding that produced the shortcut.
To extend, merge units and require the grouped calculation to preserve the original volume. Then remove the pairing labels or make one unit dry. The learner demonstrates transfer by recognising an undefined ratio or an unidentified volume, not merely by completing another multiplication. Keep the concrete map beside the symbolic identity until the symbols explain a relationship the learner can already describe.
CHAPTER 25 OF 28 · Verify and transfer
25. Reconstruct a missing local fraction only when the record supports it
Suppose the first equal-area example comes with the total generated volume of thirty-four thousand cubic metres and all inputs except Unit B’s fraction. Unit A’s contribution is known to be two thousand. The remaining thirty-two thousand must belong to B within this complete two-unit model. Dividing by B’s forty thousand cubic metres of rainfall input gives fraction 0.8.
This recovery works because the other contribution, rainfall input and total are all specified for the same target and event. If the total instead came from an outlet gauge with unaccounted storage or other water contributions, the same subtraction would not automatically recover a local generation fraction. Matching units is necessary but does not establish matching meanings.
Now remove both fractions while retaining total generated volume thirty-four thousand. The equation becomes 10Cₐ + 40Cᵦ = 34 when volumes are expressed in thousands of cubic metres. Many pairs can satisfy it. With fractions constrained between zero and one, Cₐ can range from zero to one while Cᵦ ranges from 0.85 down to 0.6 along the corresponding line. The total identifies one weighted combination, not both local responses separately.
Run a proposed pair forward through the local products to check it. Then explain why reproducing one total does not validate a unique spatial model. This is a useful distinction between fitting an aggregate observation and identifying the underlying local quantities. More decimal places in the total do not supply a second independent constraint.
CHAPTER 26 OF 28 · Verify and transfer
26. Check how rounding moves through the calculation
A mapped table may publish rainfall to whole millimetres and fractions to one decimal place. Its resulting volume is then based on rounded inputs. Preserve that distinction when comparing a recalculation with a more detailed original model.
For a single cell, changing rainfall by ΔP while keeping its supplied fraction fixed changes generated depth by CΔP. Changing the fraction by ΔC while keeping rainfall fixed changes it by PΔC. If both change, there is also the product ΔCΔP. These identities show why an error allowance for each input cannot simply be pasted onto the final total without considering the calculation.
For a fictional one-square-kilometre cell with P = forty millimetres and C = 0.8, changing rainfall to forty-one while leaving C fixed raises generated volume by eight hundred cubic metres. Changing C from 0.8 to 0.81 while holding rainfall at forty raises it by four hundred. Changing both gives an increase of one thousand two hundred and ten, including the ten-cubic-metre product term.
These are sensitivity exercises, not asserted measurement errors. If actual admissible ranges are supplied, calculate with those ranges under their stated relationships. Do not infer a probability distribution merely from a rounding convention. The practical habit is to retain enough input precision for reconciliation, round the final result appropriately, and state whether a mismatch comes from rounding or from a different model.
Is multiplying average rainfall by average fraction always wrong? It agrees when the weighted covariance is zero under the supplied model. Does a positive covariance mean rainfall causes a larger local fraction? No. Here it describes the spatial pairing of supplied values, not a causal estimate.
Is the effective fraction an area mean? It is weighted by rainfall input volume. Can it be reused for another storm? That requires support for the new event and model; the calculation alone does not establish transferability. Does a dry event have effective fraction zero? The ratio is undefined when total rainfall input is zero, although generated rainfall-derived volume is zero in our model.
Does the calculation predict a flood peak or certify drainage capacity? It estimates the named local event volume under stated assumptions. Timing, routing and design questions need their own suitable evidence and models. Preserving that boundary is part of understanding the result.
CHAPTER 28 OF 28 · Verify and transfer
28. Continue through the catchment and mathematics routes
For the timing question, read Runoff Hydrographs, Equal Volumes and Different Peaks. For a different boundary-accounting task, use Reconciling Material Loads Between Nested Gauges. Its water-quality load calculations should retain their concentration and time information; our rainfall pairing example does not replace them.
For the mathematical relationship, How Covariance Works supplies the wider explanation. Return here to attach that relationship to mapped rainfall and supplied local fractions. Before leaving, write a one-sentence interpretation of each of the original totals: fifty thousand cubic metres of rainfall input, thirty-four thousand of modelled local runoff, and twenty-five thousand from the separate-averages shortcut. If those sentences describe the same target and different calculations clearly, the numerical difference has become an explanation someone else can inspect.
Return to Watersheds for the physical boundary, nested areas and the distinction between volume and timing. Continue to Urban Hydrology for urban storage, flow and water-balance questions, and rainwater harvesting, runoff coefficients and tank sizing for a roof-and-storage application whose target differs from our spatial pairing exercise.
The USGS and US Army Corps sources linked above were checked on 24 October 2026. They provide process and rainfall-representation context. Our areas, rainfall depths, fractions and covariance examples are original fictional teaching cases, not measurements or official calibration recommendations.
For the broader learning route, use The Gold Standard of Education: learning transfers into life and How X Works. Keep one question at the centre: which relationships survive the summary? An average becomes useful when its reader knows both the total it preserves and the information it leaves behind.