A river’s response to the second hour of rain can overlap its response to the first. A unit hydrograph makes that overlap calculable: scale a response for each rainfall-excess pulse, shift it to the right time and add the contributions.
We will construct the calculation without hiding it inside software. An invented catchment and short response table let us check every number, compare storms with equal excess depth and identify which assumptions a real application would need.
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
If convolution looks abstract, start with one pulse and two pulses. If you can already calculate it, use the volume check and changed-response example to test the model.
Full chapter index · Practice and checking · Hydrology subject guide
Chapter contents
Define the response · Chapters 1–2
Build the calculation · Chapters 3–4
Check volume and timing · Chapters 5–6
Challenge the fixed model · Chapters 7–8
Practise and inspect · Chapters 9–10
Questions, sources and connected reading · Chapters 11–12
Total rainfall is not automatically direct runoff. A rainfall–runoff model may first account for interception, infiltration and storage through its chosen loss representation. The resulting excess depth then becomes the input to a unit-hydrograph transform.
For this article, the excess depths are supplied. We do not infer them from a rain gauge by assuming every millimetre reaches the outlet.
A unit hydrograph describes a direct-runoff response to a specified unit depth of excess applied over a specified duration and area under its model assumptions. The unit might be one millimetre or one centimetre; the duration might be one hour or another interval.
State all three. “The unit hydrograph” without unit depth, duration and catchment leaves the calculation incomplete. The US Army Corps of Engineers unit-hydrograph reference gives the formal framework.
Name each stage of the calculation
Use four labelled stages: rainfall estimation, loss or excess estimation, direct-runoff transformation, and any subsequent routing. Each stage changes the meaning of the numbers. A depth read from a gauge is not already an outlet flow, and a direct-runoff hydrograph is not automatically the whole river discharge. The unit hydrograph belongs to the transform stage under its stated assumptions.
For a tiny fictional loss example, a one-hour rainfall depth of 5 mm might be represented by a supplied loss of 3 mm and excess of 2 mm. Our first pulse would use the 2 mm. Feeding in all 5 mm would scale the response to two and a half times the intended excess input. That error would occur before any time shifting or summation.
The lesson is to attach a name and unit to every input. If the excess depths were estimated elsewhere, record that source and its assumptions. You do not need to solve every upstream modelling problem to perform a classroom transform, but you do need to know which supplied quantity the transform expects. Correct convolution cannot compensate for choosing the wrong physical input.
Our fictional catchment covers 21.6 km². Its supplied response to 1 mm of excess applied uniformly during one hour is the following discrete sequence.
| Response interval after pulse begins | U: m³/s per mm |
|---|---|
| Hour 1 | 1 |
| Hour 2 | 3 |
| Hour 3 | 2 |
| Later hours | 0 |
We use mean flows within hourly bins. Multiplying a bin value by 3,600 seconds gives its volume. This explicit convention avoids confusing point values on a drawn curve with interval averages.
Check the unit response before using it. Its total volume is (1 + 3 + 2) × 3,600 = 21,600 m³ per millimetre of input. One millimetre over 21.6 km² also gives 21,600 m³. The response therefore has the required volume.
The short rectangular-bin response is a teaching construction. Real hydrographs are smoother, and software may use other ordinate or integration conventions. Keep the convention consistent rather than mixing a bin sum with an unrelated trapezoidal calculation.
Distinguish an interval mean from a point on a curve
Our first output value applies to the interval from zero to one hour after the pulse begins. The second applies from one to two hours, and the third from two to three. Treating them as horizontal bin means makes their volumes unambiguous. Their drawn locations at bin centres would be a display choice, rather than new instantaneous measurements.
If values 1, 3 and 2 were instead instantaneous readings at three timestamps, the area beneath the hydrograph would require a convention between the readings and suitable information at its beginning and end. A rectangular-bin sum and a trapezoidal integration can then give different results because they represent different assumptions about the time series. Neither should be applied solely because it is the formula most recently learned.
The main response was chosen so its bin volumes match one millimetre over 21.6 km². If a proposed response instead had bins [1, 3, 1], its sum would represent only 18,000 m³ under the same convention. That would fail our required 21,600 m³ check. Before using it, investigate whether a tail is missing, the depth unit differs, or the values have another definition.
Place 2 mm of excess in the first one-hour input interval. Under the linear model, multiply every value of the unit response by two.
The first pulse contributes mean direct-runoff flows of 2, 6 and 4 m³/s in the first three output hours. After that, its contribution is zero in this constructed response.
Scaling changes the flow magnitudes, not the response’s time pattern. The response still lasts three bins and has its largest bin in the second hour.
Its represented runoff volume is (2 + 6 + 4) × 3,600 = 43,200 m³. The input-volume check is 2 mm × 21.6 km² × 1,000 = 43,200 m³.
If your scaled curve has twice the depth but its integral remains the same, something in the units or scaling has gone wrong. If its time pattern changes, you have introduced an assumption beyond this fixed linear response.
Change the depth unit without changing the event
One centimetre equals ten millimetres. A response expressed per centimetre of excess therefore has values [10, 30, 20] m³/s per cm for this same catchment and one-hour duration. The 2 mm pulse is 0.2 cm, so multiplying gives [2, 6, 4] m³/s again. The physical event has not changed; only the numerical representation of its depth has.
Mixing the per-centimetre response with the number 2 interpreted as millimetres would produce [20, 60, 40], ten times too large. The shape would still look plausible and every addition could be correct. A volume calculation tied to the catchment and input depth would reveal the inconsistency immediately.
This makes dimensional checking more useful than a final unit label pasted beside an answer. Carry the units through the multiplication: cm × (m³/s per cm) leaves m³/s. When the units do not cancel, stop before summing the contributions. Repairing this foundation is easier than diagnosing a suspicious flood curve after the whole model has been run.
CHAPTER 4 OF 12 · Build the calculation
4. Shift the second pulse and add the overlap
Now add 1 mm of excess in the second input hour. Its response has the same shape as the original unit response, beginning one hour later.
| Output interval | First pulse contribution | Second pulse contribution | Total direct runoff |
|---|---|---|---|
| Hour 1 | 2 × 1 = 2 | 0 | 2 m³/s |
| Hour 2 | 2 × 3 = 6 | 1 × 1 = 1 | 7 m³/s |
| Hour 3 | 2 × 2 = 4 | 1 × 3 = 3 | 7 m³/s |
| Hour 4 | 0 | 1 × 2 = 2 | 2 m³/s |
During the second output hour, the first pulse is in its second response bin while the second pulse is in its first. Their contributions are 6 and 1 m³/s, giving 7.
During the third output hour, their contributions are 4 and 3, again giving 7. The fourth hour contains the final 2 m³/s contribution from the second pulse.
The resulting sequence is [2, 7, 7, 2]. The computation remembers earlier input because its responses have not yet finished. It does not need a claim that today’s water came exclusively from today’s rain.
Read the overlap as two ages of response
In output Hour 2, the first pulse is one bin older than the second. That is why we use its U₂ value alongside the second pulse’s U₁ value. The total is not made by adding both pulses at the same position in the response table. Their different start times are the reason the subscript shift matters.
Try describing Hour 4 without symbols. The first pulse has finished its three-bin response; the second is in its last bin. Only the second contributes. If a spreadsheet still includes a first-pulse contribution in that hour, inspect whether its response tail was extended accidentally. If the second contributes nothing, inspect whether it was shifted two bins instead of one.
The model has a useful sense of memory: earlier excess continues contributing while its response persists. That does not identify the age of individual water molecules leaving the catchment. A hydrograph transform relates an input representation to an output response. Interpreting the physical origin and age of water requires other evidence. Keep a mathematical response history distinct from a tracer-based account of where the observed water came from.
CHAPTER 5 OF 12 · Check volume and timing
5. Write convolution after understanding the sum
Let Pⱼ be the excess depth in input bin j, and Uₖ the unit response in the kth bin after that pulse begins. Contributions outside the supplied response are zero.
The discrete expression is Qₙ = Σⱼ PⱼUₙ₋ⱼ₊₁. Here P is in millimetres, U in m³/s per millimetre, and Q in m³/s. The units show why the depth factor disappears from the final flow.
For n = 3 in our example, Q₃ = P₁U₃ + P₂U₂ = 2 × 2 + 1 × 3 = 7. The symbols describe the same overlap that the table made visible.
The formula is compact because it repeats one rule. Learn the rule before memorising the notation: identify how long ago each input occurred, select that stage of its response and add the surviving contributions.
A unit response with a longer tail would keep earlier pulses in the sum for longer. Cutting the tail prematurely could lose volume even if the first few flow values looked correct.
Build a small calculation that can be inspected
In a spreadsheet, one transparent arrangement puts an output hour in each row and one rainfall pulse in each column. Enter zero before a pulse begins, then its scaled response down the following rows. Sum across the row to obtain Q. The four-row table in this guide is exactly that arrangement. It makes a shift error easier to spot than a single long formula copied without inspection.
There is also a general mass check. When all responses are included and every pulse uses the same response, summing Q across time gives (ΣP)(ΣU). Multiplying by the bin duration gives the total volume. In our example, ΣP = 3 mm and ΣU = 6 m³/s per mm. Their product is 18 m³/s across the bins, which becomes 64,800 m³ after the time factor.
That identity follows from rearranging a finite sum of contributions. It is an independent algebraic check on the numerical table, not proof that the catchment follows the model. If the total matches but the peak occurs in the wrong hour, investigate the shifts: reordering contributions can preserve their sum while changing when the water is represented as arriving.
CHAPTER 6 OF 12 · Check volume and timing
6. Compare equal excess depths with different timing
Our two-pulse storm has 3 mm of excess in total. Its direct-runoff volume is (2 + 7 + 7 + 2) × 3,600 = 64,800 m³, which matches 3 × 21.6 × 1,000.
Put all 3 mm into the first one-hour interval instead. The response becomes [3, 9, 6]. Total volume is unchanged at 64,800 m³, but the maximum hourly mean rises from 7 to 9 m³/s and the response ends a bin earlier.
Now separate the original pulses by an empty hour: inputs [2, 0, 1]. The output becomes [2, 6, 5, 3, 2]. The volume remains 64,800 m³ while the maximum bin is 6 m³/s.
Equal total excess does not imply equal flow timing or peak. Even in a fixed linear model, the input sequence matters. An annual or event total cannot supply information about the sequence that produced it.
Separate a maximum bin from an instantaneous peak
Our largest values are hourly means. A real flow within an hour could rise above or fall below its mean. Calling the 7 m³/s result an established instantaneous peak would add information not contained in our bin data. The safe wording is “maximum hourly mean in this discrete teaching response.” The time resolution belongs in the result.
The reordered input [1, 2] provides another contrast. Its response [1, 5, 8, 4] has the same total volume as [2, 1], but a larger maximum hourly mean. Linearity allows addition; it does not make the response insensitive to input order. The larger later pulse meets a different combination of remaining contributions.
The empty-hour example makes a second point. In [2, 0, 1], the zero input is a specified zero excess during that hour, not a missing rainfall record. A missing input would leave part of the response unresolved. Replacing missing input with zero could create an artificial quiet interval and reduce later predicted flow. Preserve the distinction between known zero and unavailable information before constructing the convolution.
CHAPTER 7 OF 12 · Challenge the fixed model
7. Keep baseflow and subsequent routing separate
The output above is direct runoff under the supplied transform. It is not necessarily the river’s entire discharge.
For a simple arithmetic illustration, suppose a separately specified baseflow is exactly 1 m³/s during all four hours of the two-pulse response. Adding it produces total flows [3, 8, 8, 3] m³/s. Over those four hours the baseflow contributes 14,400 m³.
The 64,800 m³ direct-runoff volume check still applies to the direct component. Comparing the rainfall-excess volume with the 79,200 m³ total would mix two components.
Channel routing downstream is another modelling task. A unit response defined at one outlet does not automatically describe a point farther along the river. Keep the location attached to every hydrograph.
The existing hydrograph-separation guide examines how component estimates themselves can be uncertain. In real work, baseflow is not made certain simply by drawing it beneath the response curve.
Keep component volumes on compatible windows
The four-hour total in the baseflow example is 79,200 m³: 64,800 from the direct component plus 14,400 from the specified background flow. If a report used only the first three hours, the direct component would total (2 + 7 + 7) × 3,600 = 57,600 m³. The missing fourth hour contains the remaining 7,200 m³.
That truncated window would not support the conclusion that the transform lost 7,200 m³. The response simply had not finished inside the chosen accounting period. Check the beginning and end of every volume comparison, especially when several storms overlap or a long tail extends beyond an event label in the rainfall record.
An outlet also has a spatial identity. If a downstream hydrograph is compared with this response, intervening tributaries, storage and routing may matter. A numerical match at one outlet does not establish applicability at another. Time window, location and component definition are three independent checks; using the same units does not make records compatible if any of those three differ.
CHAPTER 8 OF 12 · Challenge the fixed model
8. Test linearity and a fixed response separately
Linearity permits proportional scaling and addition. Time invariance means the same input pulse has the same response shape when it occurs at a different time within the model’s applicable conditions.
Those are useful approximations, not laws that every catchment obeys. Changing flow conditions, connected storage or inundation pathways can change the response. The US Army Corps of Engineers Clark-model discussion describes why variable responses may be needed across events.
Keep a second distinction clear. Soil wetness can change how total rainfall becomes excess. If you feed raw rain into a transform intended for excess, an error may arise before the unit response is applied.
If excess has already been estimated, ask separately whether its transport response remains appropriate. “The catchment was wetter” is not a complete explanation until you identify which part of the modelling chain changed.
Change duration by constructing the new input
Suppose the new unit is still 1 mm of excess, but spread uniformly over two hours. Within our one-hour discrete representation, that is [0.5, 0.5] mm. Convolving with [1, 3, 2] gives hourly mean flows [0.5, 2, 2.5, 1] m³/s. Their sum is 6, preserving the one-millimetre volume of 21,600 m³.
The response differs from the one-hour unit because the input is different. Relabelling [1, 3, 2] as a two-hour response would miss this change. If we then choose two-hour output bins, the compatible averages for this constructed event are (0.5 + 2)/2 = 1.25 and (2.5 + 1)/2 = 1.75 m³/s. Their volumes sum to 3 × 7,200 = 21,600 m³.
This exercise changes both the input duration and, in the second step, the output reporting interval. Keeping those operations separate prevents a common confusion. It is an illustration inside our fixed discrete model, not a universal duration-conversion prescription for an arbitrary observed hydrograph. Other applications need their own compatible representation and evidence.
Apply a 4 mm one-hour excess pulse. The fixed response predicts [4, 12, 8] m³/s, with volume 24 × 3,600 = 86,400 m³.
Suppose an independently supplied teaching event instead has interval mean direct-runoff flows [4, 16, 4]. It has exactly the same volume, 86,400 m³, but a sharper peak than the fixed response predicts.
Volume agreement alone therefore cannot validate the response shape. The mismatch could motivate checks of excess estimation, timing, component separation, measurement and the fixed-response assumption. It does not identify its own cause.
This is a changed-context test: the original two-pulse calculation was correct under its assumptions, yet a new observation can question those assumptions. A useful model is one whose errors can be examined, not one whose arithmetic prevents it from being challenged.
Recover a response, then test it elsewhere
In a perfectly consistent inverse teaching exercise, suppose the excess [2, 1] and output [2, 7, 7, 2] are known, while a three-bin unit response is unknown. The first equation gives 2U₁ = 2, so U₁ = 1. The second gives 2U₂ + U₁ = 7, so U₂ = 3. The third gives 2U₃ + U₂ = 7, so U₃ = 2. The fourth equation checks U₃ = 2.
We have recovered the response used to construct the dataset. That is a calibration exercise, not independent validation. Testing the same values used to determine the response cannot show how well it generalises to another event. The supplied [4, 16, 4] comparison provides a different test because its shape was not used to recover those parameters.
Real inverse problems also inherit errors in the supplied input and output. A small mismatch does not automatically identify the transform as the sole cause. Clock alignment, excess estimation and component separation can all require investigation. Preserve the residuals instead of adjusting the response until every record appears perfect and then describing the fit as proof of universal applicability.
Confirm the catchment and outlet, excess-depth unit, input duration, output interval and integration convention. Check whether the supplied response represents direct runoff or total discharge.
Verify the unit response’s volume. Then calculate one output bin by hand, check the full volume and inspect the time shifts. A one-bin shift error can preserve the sum while moving the peak.
For independent practice, apply a 1 mm pulse in the first hour and a 2 mm pulse in the second. The output should be [1, 5, 8, 4] m³/s. Its sum is 18 m³/s across the bins, giving 64,800 m³ after multiplying by 3,600 seconds.
Do not call the sum alone a volume; the time factor is essential. The volume matches the earlier storm because the total excess is again 3 mm. The peak differs because the larger input occurred later.
Diagnose two errors that a single check can miss
First imagine the entire correct response is shifted one hour late. Its volume remains 64,800 m³, but it misrepresents timing. The volume check passes, while the time-shift check fails. Now imagine every value is multiplied by ten through a centimetre–millimetre mismatch. The timing can remain correct while the volume check fails. Independent checks are useful because different mistakes preserve different features.
For a final exercise, input [1, 0, 1] mm into the same response. The output is [1, 3, 3, 3, 2] m³/s: at Hour 3, the first pulse contributes 2 and the second 1. Its sum is 12, and its volume is 43,200 m³. Explain both the overlap and the total without borrowing an earlier storm’s answer.
A finished explanation includes the named quantity, supplied assumptions, one traceable output calculation, a volume check and a statement of what would challenge the model. That combination makes the method usable when the pulse order, duration, catchment area or reporting convention changes. The aim is control over a relationship, rather than recognition of one familiar hydrograph shape.
Does one unit mean one cubic metre per second?
No. It refers to the specified excess depth. The flow response must integrate to that depth multiplied by the catchment area.
Can I use a one-hour response for three-hour pulses unchanged?
Not by simply relabelling it. Duration conversion or a compatible model is needed because the input representation has changed.
Is this an instantaneous unit hydrograph?
No. The teaching response is for a finite one-hour pulse. An instantaneous response is a separate theoretical construction.
Does convolution prove a catchment is linear?
No. It calculates the consequences of a linear response assumption. Independent data are needed to evaluate its usefulness.
Why not use a detailed model immediately?
A transparent small example helps you inspect units, timing and assumptions. A more detailed model still needs those checks and appropriate evidence for its application.
Can a volume check identify a timing error?
Not always. Shifting the entire response leaves its integral unchanged. Comparing start times and selected individual bins is necessary alongside the volume check. Two hydrographs can carry the same represented amount of water while describing different arrival patterns, with different implications for a time-dependent question.
What if a supplied unit response has a long tail?
Include the tail over a suitable accounting window, or state how it was approximated and what volume remains outside that window. Truncation can change both the total and the overlap with a later event. A short table is convenient for learning, but its convenience does not justify discarding a physically relevant tail from an actual model.
Is an excellent fitted response enough?
It establishes a fit to the data and choices used to obtain it. Generalisation needs comparisons outside that fitting exercise and within the intended conditions of use. Check timing, magnitudes, volume and residual patterns, while retaining uncertainty in rainfall excess and measured discharge. A model can remain useful despite imperfect fit when its purpose, limits and errors are understood.
Formal concepts and model limits
USACE HEC-HMS: Unit Hydrograph Basic Concepts supports the definition, scaling and convolution framework. USACE HEC-HMS: Clark Unit Hydrograph Model explains translation, storage and variable-response considerations. Links checked on 21 October 2026. The catchment, interval means and test events here are original constructions, not a calibrated operational model.
Connect the whole modelling chain
The Gold Standard of Hydrology establishes the water-system foundation. Predicting a Flood Before the River Reaches the Street connects rainfall, runoff, routing and forecasts.
Examine what a hydrograph preserves
Hydrograph Separation and the Limits of Baseflow Estimates examines component interpretation. Flow-Duration Curves and the Drought Sequences They Hide shows what is lost when chronological order disappears.
Continue from evidence to learning
Use Paired Catchments and Attribution of Change for a different comparison question. Return to learning that transfers into life or How X Works. This educational benchmark does not establish an actual flood forecast or approve an engineering design.
Inspect the spatial rainfall input
Before estimating rainfall excess, examine the area represented by each gauge in The Gold Standard of Hydrology | Thiessen Rainfall Weights and the Area a Gauge Represents.
Extend the evidence check
Extend the timing checks with Rainfall Time Steps and Why Equal Storm Totals Can Produce Different Runoff. Examine a response tail through Recession Constants, Time Steps and the Storage Behind a Falling Flow, then evaluate a model score against its actual benchmark in Nash–Sutcliffe Efficiency and the Benchmark Behind a Model Score.