From Rainfall Rate to Rainfall Depth
Rainfall intensity describes how quickly rain is accumulating. An intensity of \(5\) millimetres per hour means that, at that rate, rainfall depth increases by about \(5\) millimetres during one hour. If the intensity stays constant, finding the accumulated depth is a multiplication. If intensity changes over time, the rate must be accounted for throughout the interval.
The accumulated depth is measured as a vertical layer of water. It is not the volume of water over an entire field or region. The model below treats intensity as representative of the location being studied; converting depth to volume would additionally require an area.
For example, if \(i(t)\) is measured in millimetres per hour and \(t\) in hours, then \(i(t)\,dt\) has units of millimetres. The integral adds these small depth contributions over time. The key practical requirement is that the time variable and the intensity's time unit agree.
Why Integration Gives Accumulated Depth
First consider an interval divided into short subintervals. On a sufficiently short subinterval, the intensity is approximately constant, so the depth accumulated there is approximately intensity multiplied by the subinterval's duration. Adding these contributions approximates the total depth. Refining the partition gives the integral.
Partition \([a,b]\) into subintervals with widths \(\Delta t_k\).
Choose a time \(t_k^*\) in each subinterval. Its rainfall-depth contribution is approximately \(i(t_k^*)\Delta t_k\).
The sum of the contributions approaches the definite integral as the largest subinterval width approaches zero.
In symbols, the approximating sum is \(\sum_k i(t_k^*)\Delta t_k\). By the definition of the Riemann integral, its limit is \(\int_a^b i(t)\,dt\), when \(i\) is integrable. The units also agree at every step: intensity has units of depth/time, duration has units of time, and their product has units of depth.
If \(i\) is continuous, the Fundamental Theorem of Calculus gives a useful interpretation. Define accumulated depth from a fixed starting time \(a\) by \(P(t)=\int_a^t i(s)\,ds\). Then \(P'(t)=i(t)\). For a piecewise-continuous intensity, this derivative equality holds at points where \(i\) is continuous. At a discontinuity, the assigned value of \(i\) at that single point does not affect the integral or accumulated depth.
Worked Example: A Smoothly Changing Intensity
Worked Example: A Smoothly Changing Intensity
Suppose rainfall intensity over a four-hour period is \(i(t)=2+\frac{1}{2}t\) millimetres per hour, where \(t\) is measured in hours from the start. Find the rainfall depth over the period.
Because \(t\) is in hours and \(i(t)\) is in millimetres per hour, integrate from \(t=0\) to \(t=4\):
An antiderivative is \(2t+\frac{1}{4}t^2\). Evaluate it at both endpoints:
Check the calculation by differentiating the antiderivative: \(\frac{d}{dt}(2t+\frac{1}{4}t^2)=2+\frac{1}{2}t=i(t)\). The endpoints also make sense: the intensity rises from \(2\) to \(4\) millimetres per hour, and the computed accumulated depth is positive. The result is a depth, not a rate.
Intervals with Different Intensities
Rainfall intensity may be given by separate rules on successive time intervals. In that case, integrate each rule over the interval where it applies and add the resulting depths. The endpoint values of adjacent pieces do not create an extra contribution: a single instant has duration zero and adds no depth.
Worked Example: Three Constant-Intensity Periods
Worked Example: Three Constant-Intensity Periods
Suppose intensity is \(3\) millimetres per hour for the first \(2\) hours, \(6\) millimetres per hour for the next \(1.5\) hours, and \(2\) millimetres per hour for the final \(1.5\) hours. Find the total depth over the full \(5\) hours.
Compute each interval's depth as its intensity multiplied by its duration. The intervals and contributions are:
Add the interval depths:
Check: the durations add to \(2+1.5+1.5=5\) hours, matching the stated period. Each product has units \((\text{mm/h})(\text{h})=\text{mm}\), so adding them gives a total depth of \(18\) millimetres.
Changing Time Units Correctly
A common source of error is integrating an intensity stated per hour with a time variable measured in minutes. The differential \(dt\) must use the same time unit as the denominator in the intensity units. Either convert the time variable to hours before integrating or include the conversion factor when integrating with respect to minutes.
Worked Example: Intensity Given per Hour, Time Given in Minutes
Worked Example: Intensity Given per Hour, Time Given in Minutes
During a \(30\)-minute shower, let \(t\) denote elapsed time in minutes and suppose the intensity is \(i(t)=4+\frac{t}{30}\) millimetres per hour. Find the accumulated depth.
Since \(t\) is in minutes while the intensity is in millimetres per hour, convert the time differential: one minute is \(\frac{1}{60}\) hour, so \(dt\) minutes corresponds to \(\frac{dt}{60}\) hours. Therefore:
Factor out \(\frac{1}{60}\) and integrate with respect to the numerical time value in minutes:
Check the rate: at the start it is \(4\) millimetres per hour, and at \(30\) minutes it is \(4+\frac{30}{30}=5\) millimetres per hour. The computed depth is positive. Also, differentiating the antiderivative inside the brackets, \(4t+\frac{t^2}{60}\), gives \(4+\frac{t}{30}\), as required; the factor \(\frac{1}{60}\) converts the minute-based differential to hours.
A Common Unit Mistake
In the previous example, omitting the conversion factor and calculating \(\int_0^{30}(4+\frac{t}{30})\,dt=135\) would be wrong. The numerical result \(135\) treats the time increments as hours, even though the interval is \(30\) minutes. The units of this unconverted calculation would be \((\text{mm/h})(\text{min})\), not millimetres.
The correction is to convert minutes to hours, which divides the integral by \(60\): \(135/60=2.25\) millimetres. A quick unit check often catches this kind of error before the final answer is reported.
Accumulation Across a Whole Interval
The definite integral provides a total over a chosen interval, regardless of whether intensity is constant, smoothly changing, or piecewise-defined. If accumulated depth is already known at time \(a\), then the accumulated depth at a later time \(b\) is the starting depth plus the depth added between those times:
This formula follows from additivity of integrals: the accumulation from the original starting time to \(b\) splits into the accumulation up to \(a\) and the accumulation from \(a\) to \(b\). It is useful whenever a rainfall record begins after the accumulation period has already started, or when a total must be updated from a known earlier value.
For a physically meaningful rainfall model, intensity is nonnegative. A zero intensity contributes no additional depth during that period. A model that assigns negative intensity is not representing ordinary rainfall intensity, even though its integral can still be evaluated as a mathematical quantity.
Check Your Understanding
Use units in every calculation, and state whether each answer is a depth or a rate.
- A constant intensity of \(4\) millimetres per hour lasts \(2.5\) hours. What depth accumulates?
- If \(i(t)=1+t\) millimetres per hour for \(0\leq t\leq 3\), with \(t\) in hours, write and evaluate the integral for the accumulated depth.
- An intensity of \(6\) millimetres per hour lasts \(20\) minutes. What conversion is needed before multiplying rate by duration, and what depth results?
- Two intensity rules apply on adjacent intervals. How should their contributions be combined to obtain the total depth?
- Why can changing the stated intensity at just one instant leave the accumulated depth unchanged?
