Pavement evaluation
Drainage evidence in a pavement investigation
Derive a reservoir water balance, plot storage and discharge, and test outlet sensitivity without confusing the model with field seepage.
Editorial preview · Technical review pending
Drainage observations belong beside pavement observations.
The rain stops; the calculation does not
Two outlets can experience the same storm and tell very different stories six hours later.
The official catalogue includes TRH15, Subsurface Drainage for Roads. It is a source to consult for the relevant investigation and design context. (Department of Transport, n.d.)
Balance the water before blaming the outlet
Governing relationship: dS/dt = I − kS; Qout = kS.
S: stored volume (m³); I: inflow (m³/h); k = 0.5 h⁻¹.
Initially S = 0. Inflow is 2 m³/h for two hours, then zero.
The film first plots inflow (m³/h), then follows the calculation and plots storage (m³). At 2 h, 4 m³ entered, 2.528 remains and 1.472 left. The mass balance closes; storage then decays as inflow stops.
Lumped water-balance model
Rainfall, storage and delayed drainage
The reservoir is a mass balance, not a calibrated pavement moisture model.
Model, calculations and transcriptEquations, worked steps, source notes and a timed reading guide
Understand the model
Stored water changes by inflow minus outflow. Once inflow stops, the remaining store drains gradually rather than disappearing.
Work through the inputs and equations below, then compare the calculated output with the film. All numerical inputs in this study are synthetic.
Inputs and symbol key
- S; t
- Stored water volume (m³); time (h).
- I; Qout
- Inflow and outflow rates (m³/h).
- k
- Linear discharge coefficient (h⁻¹), not hydraulic conductivity.
Values used in this film
- S: stored volume (m³); I: inflow (m³/h); k = 0.5 h⁻¹.
- Initially S = 0. Inflow is 2 m³/h for two hours, then zero.
The mathematical relationships
Equation 1
Source and scopeThe derivative is the storage change per hour. The chosen model makes discharge proportional to storage through k.
Equation 2
Source and scopeSolve the constant-inflow equation for the first two hours, then start the exponential recession from S(2). The two solutions meet continuously.
Work through the example
Use the relationships above with consistent units. These substitutions are our worked example, not values quoted from the source.
Equilibrium during constant inflow
I/k = 2/0.5 = 4
m³
Storage at rain end
4 × (1 − exp(−1)) = 2.528482
m³
Outflow volume by two hours
2 × 2 − 2.528482 = 1.471518
m³
Recession half-life
ln(2)/0.5 = 1.386294
hours
Read the output
At 2 h, 4 m³ entered, 2.528 remains and 1.472 left. The mass balance closes; storage then decays as inflow stops.
Check your understanding
Check S(0) = 0, continuity at rainfall cessation and nonnegative storage. Integrate kS over time and close the water balance. Test the k → 0 limit during rainfall, where storage approaches It.
Assumptions and limits
The reservoir is a mass balance, not a calibrated pavement moisture model.
Return to the full references and method notes to check the cited edition and its applicability.
Timestamped film transcript 25 short entries · select a time to revisit the film
Rainfall, storage and delayed drainage. I = 2 m³/h for 2 h; k = 0.5 h⁻¹; initially empty.
MATHEMATICS · define the relationship and the meaning of each symbol before calculating.
S: stored volume (m³); I: inflow (m³/h); k = 0.5 h⁻¹. Initially S = 0. Inflow is 2 m³/h for two hours, then zero.
INPUT · inflow schedule · read the axis units and the numerical values.
I = 2 m³/h for 2 h; k = 0.5 h⁻¹; initially empty.
At 2 h: storage = 2.528 m³; cumulative inflow = 4 m³. The balance leaves 1.472 m³ discharged.
OUTPUT · stored water · read the axis units and the numerical values.
At 2 h, 4 m³ entered, 2.528 remains and 1.472 left. The mass balance closes; storage then decays as inflow stops.
CALCULATION · substitute the defined inputs; keep units through each intermediate step.
Equilibrium during constant inflow: I/k = 2/0.5 = 4 m³
Storage at rain end: 4 × (1 − exp(−1)) = 2.528482 m³
Outflow volume by two hours: 2 × 2 − 2.528482 = 1.471518 m³
Recession half-life: ln(2)/0.5 = 1.386294 hours
SENSITIVITY · change one declared input · read the axis units and the numerical values.
k = 0.25 /h: 3.14775 m³ stored at 2 h. Two-hour inflow of 2 m³/h; initial storage zero.
k = 0.5 /h: 2.52848 m³ stored at 2 h. Two-hour inflow of 2 m³/h; initial storage zero.
k = 1 /h: 1.72933 m³ stored at 2 h. Two-hour inflow of 2 m³/h; initial storage zero.
Check: Check S(0) = 0, continuity at rainfall cessation and nonnegative storage. Integrate kS over time and close the water balance. Test the k → 0 limit during rainfall, where storage approaches It.
ONE ROW · Two outlets can experience the same storm and tell very different stories six hours later.
PARAMETER MAP · 14701 evaluated cells. The colour limits stay fixed across every row.
Every row receives 2 m³/h for two hours, starts empty and obeys dS/dt = I − kS. Only k changes. The ridge occurs when inflow stops; recession retains the storage already
accumulated. Compare the long tail at small k with the rapid emptying at large k. The integral of outlet flow plus remaining storage equals the 4 m³ input after the storm.
VALUE LANDSCAPE · Height and colour encode Stored water (m³). This is a graph over the two input axes, not automatically a physical pavement shape.
Lumped linear storage, not a spatial seepage solution. The coefficient is assumed; it is not inferred hydraulic conductivity, pavement saturation or a distress predictor.
The reservoir is a mass balance, not a calibrated pavement moisture model.
Download the timed captions (VTT)
The transcript records the on-screen wording. Formula provenance and interpretation are explained in the source note above.
Build a causal argument from located observations
In a synthetic inspection, the log contains one outlet observation, three shoulder observations, two surface-water observations and two ditch observations. There are eight records, but not necessarily eight independent pieces of evidence. Several may describe the same place or the same event.
Case review
The same storm, three different recessions
Would a slow outlet still retain water after rainfall stops?
The discriminating check
Compare dated rainfall, outlet and pavement observations at the same location; distinguish observation from causal explanation.
Evidence supports the comparison
If observations support retained water, design a targeted investigation into geometry and hydraulic conditions.
Evidence is missing or incompatible
If the only evidence is a blocked outlet beside distress, retain alternative explanations and investigate them.
What you can conclude. The storage calculation closes a water balance. It does not prove the cause of a particular pavement defect.
Original worked case. Read the source editions and method notes.
An obstructed outlet beside distressed pavement is worth recording and investigating. The coincidence does not by itself prove that the obstruction caused the distress. A causal explanation needs a plausible mechanism supported by the relevant site evidence and the applicable investigation method.
| Statement | Type of claim |
|---|---|
| The outlet is visibly obstructed in a dated photograph | Observation |
| Water may be retained because of the obstruction | Hypothesis |
| Retained water caused the pavement distress | Causal interpretation requiring support |
| A particular drainage detail will resolve it | Design proposition requiring evaluation |
The official catalogue lists TRH15 for subsurface drainage. That establishes an appropriate source to consult, not a site-specific conclusion or a substitute for reading the relevant method. (Department of Transport, n.d.)
Record weather context, inspection time, accessible features and uncertainty. If a later visit shows a different condition, retain both records. The change may be important to understanding the process; it should not be treated as a reason to overwrite the first visit.
The reservoir experiment that follows tests storage and discharge conservation. It does not calculate hydraulic gradients or spatial seepage through pavement layers.
Calculate storage through and after rainfall
A lumped reservoir makes the conservation argument visible without claiming to solve unsaturated pavement flow. Let S be stored water in m³, I the inflow in m³/h and k a linear drainage coefficient in h⁻¹. With dS/dt = I − kS, the outlet discharge is kS. For I = 2, k = 0.5 and an initially empty store, S(t) = 4(1 − exp(−0.5t)) during the first two hours. Once inflow stops, S(t) = S(2) exp(−0.5(t − 2)).
At two hours, 4 m³ has entered, 2.528 m³ remains and 1.472 m³ has left. The rendered curve follows these solutions. Reducing k would retain more water for longer in this particular reservoir model. Translating that coefficient into a real drainage design needs geometry, hydraulic properties and observations; the animation does not estimate pavement moisture, pore pressure or material damage.
Calculation laboratory
What changes when the outlet coefficient is halved?
| Calculation step | Substitution or result | Units and interpretation |
|---|---|---|
| Equilibrium during constant inflow | I/k = 2/0.5 = 4 | m³ |
| Storage at rain end | 4 × (1 − exp(−1)) = 2.528482 | m³ |
| Outflow volume by two hours | 2 × 2 − 2.528482 = 1.471518 | m³ |
| Recession half-life | ln(2)/0.5 = 1.386294 | hours |
Multiply the governing equation by exp(kt) to obtain d[S exp(kt)]/dt = I exp(kt). Integrating with S(0) = 0 yields the rising storage curve. After inflow stops, the homogeneous equation dS/dt = −kS gives exponential recession from the existing storage, not a reset to zero. The two branches meet continuously at two hours.
Halving k both raises storage at the end of rainfall and doubles the recession half-life. It does not double the two-hour stored volume, because outflow and storage evolve together during the event. This is why a parameter sweep is more informative than animating a generic blue water level. At any selected time the inflow volume must equal accumulated outflow plus the change in storage.
The coefficient k is a lumped scenario parameter, not hydraulic conductivity. Converting it into a field drainage detail would require geometry and hydraulic evidence absent here. This unlimited reservoir also has no overflow threshold, infiltration limit or spatial pore-pressure field. If a finite storage capacity were introduced, the model would need an additional overflow flux to conserve water. The example isolates conservation and recession; the TRH15 catalogue reference identifies drainage guidance to consult and does not authenticate this original illustrative model. (Department of Transport, n.d.)
Controlled sensitivity experiment
| Controlled case | Calculated output | What changes |
|---|---|---|
| k = 0.25 /h | 3.147755 m³ stored at 2 h | Two-hour inflow of 2 m³/h; initial storage zero. |
| k = 0.5 /h | 2.528482 m³ stored at 2 h | Two-hour inflow of 2 m³/h; initial storage zero. |
| k = 1 /h | 1.729329 m³ stored at 2 h | Two-hour inflow of 2 m³/h; initial storage zero. |
What changes when the outlet coefficient is halved?
Compare the cases on a fixed scale. Select a case to inspect its input change and calculated output.
Reproduce and challenge the result
Check S(0) = 0, continuity at rainfall cessation and nonnegative storage. Integrate kS over time and close the water balance. Test the k → 0 limit during rainfall, where storage approaches It.
These are original synthetic calculations. The sensitivity cases vary the stated input while retaining the other assumptions; they are not field observations or treatment thresholds. The film’s calculation and sensitivity chapters use the same numerical cases as this table.
Watch the recession fan out
Every row receives 2 m³/h for two hours, starts empty and obeys dS/dt = I − kS. Only k changes. The ridge occurs when inflow stops; recession retains the storage already accumulated. Compare the long tail at small k with the rapid emptying at large k. The integral of outlet flow plus remaining storage equals the 4 m³ input after the storm.
Reproduce the experiment. S=2(1−exp(−kt))/k for t≤2; S=S(2)exp(−k(t−2)) afterwards. The downloadable table contains 14,701 evaluated cells, with the two input coordinates and output value retained on every row. These are parameter combinations, not that many independent field observations. The equation extends the worked example above; it is original teaching arithmetic, not a new standard or calibrated prediction.
The rain stops; the calculation does not
Two outlets can experience the same storm and tell very different stories six hours later.
Download every calculated cell (CSV) · Open full-size explorer · Equation source and tests (ZIP)
Lumped linear storage, not a spatial seepage solution. The coefficient is assumed; it is not inferred hydraulic conductivity, pavement saturation or a distress predictor.
References and method notes
- Department of Transport (n.d.). Roads: Engineering Standard Guidelines and Manuals. Official catalogue; accessed 5 September 2026. Original source (Accessed: 5 September 2026).TMH22 V8b Draft; TRH12; TRH15; TRH24 catalogue entries
Use the source edition and project requirements that apply to your work. Educational examples are not site-specific design advice.