Why a Sprinkler Turns Geometry into a Water Plan
A lawn sprinkler makes mathematics visible. A rotating jet traces an arc, the wetted ground resembles a circle or circular sector, neighbouring patterns overlap, and a flow measured in litres per minute becomes a depth measured in millimetres. The system looks simple from a distance, but using it thoughtfully involves geometry, rates, unit conversion, sampling, variability and optimisation.
This is an educational model, not a landscape installation guide. Real distribution depends on pressure, nozzle, height, wind, droplet size, slope, soil, vegetation and maintenance. The United States Department of Agriculture Natural Resources Conservation Service publishes an official Sprinkler System conservation practice standard and technical material such as its National Engineering Handbook chapter on sprinkler irrigation. Those sources show that professional design considers far more than drawing circles.
For students, that complexity is a gift. It creates a clean central model and meaningful reasons to test where the model stops. The purpose is not to memorise one “perfect” spacing. It is to understand what radius, angle, area, flow and measured distribution can and cannot tell us.
A Reading Route
- Start with circle and sector area for the basic model.
- Convert water volume into depth to connect rate and units.
- Study overlap before assuming every shared patch is waste.
- Use catch containers to replace appearance with data.
- Follow the student workflow for a careful investigation.
The Geometry of One Sprinkler
The simplest model treats a sprinkler as a point at the centre of a circle. If water reaches radius (r), the full-circle area is
(A=\pi r^2).
For (r=6) m,
(A=\pi(6)^2=36\pi\approx113.10\text{ m}^2).
The squared radius matters. Doubling reach from 3 m to 6 m does not double nominal area; it multiplies area by four. This is the same scale-factor principle that appears in maps, lenses and enlarged diagrams.
Circular Sectors
At a boundary or corner, a sprinkler may be represented as a sector rather than a full circle. If the sector angle is ( heta) degrees, its area is
(A=\frac{\theta}{360^\circ}\pi r^2).
For a 90-degree sector with radius 6 m,
(A=\frac{90}{360}\pi(6)^2=9\pi\approx28.27\text{ m}^2).
The angle is one quarter of a full turn, so the area is one quarter of the full circle. This gives a quick reasonableness check.
Arc Length and Boundary Length
The curved edge of a sector has length
(s=\frac{\theta}{360^\circ}2\pi r).
For the same quarter-circle, (s=3\pi\approx9.42) m. If a problem asks for the full boundary of the sector, add the two radii: (9.42+6+6=21.42) m approximately. Area and perimeter answer different questions; using the correct formula begins with identifying the requested quantity.
Annular Sectors
Some patterns have a simplified inner region that is not treated as uniformly watered. If the outer radius is (R), inner radius is (r), and angle is ( heta), the annular-sector area is
(A=\frac{\theta}{360^\circ}\pi(R^2-r^2)).
For (R=6) m, (r=1) m and ( heta=90^\circ),
(A=\frac14\pi(36-1)=8.75\pi\approx27.49\text{ m}^2).
Notice that subtracting the radii first and squaring, ((R-r)^2), would be wrong. The model subtracts the areas of two sectors, so it uses (R^2-r^2).
A Circle Is a Boundary, Not a Distribution Guarantee
The circle model tells us where water might reach under declared assumptions. It does not say that every point receives the same depth. A real pattern may deposit more near the centre, more in a ring, less at the edge or asymmetrically in wind.
This distinction is central to mathematical modelling:
- coverage boundary asks whether a point is reached;
- application depth asks how much water arrives;
- uniformity compares depths at different points;
- efficiency needs additional decisions about useful water, loss and purpose.
A coloured diagram can look beautifully even while measured distribution varies. Mathematics encourages us to define the metric rather than trust a visual impression.
Radial Profiles
A simplified radial profile records depth (d(r)) as a function of distance from the centre. One fictional profile might fall linearly from 12 mm at the centre to 0 mm at a 6 m edge. Another might peak at 4 m. Both share the same nominal radius, yet their distributions differ.
To estimate total volume from a radial profile, advanced students can divide the circle into thin rings. A ring at radius (r) with thickness (Delta r) has approximate area (2\pi r\Delta r). Multiplying by local depth and summing connects geometry with numerical integration. The model must convert depth to metres before the product represents cubic metres.
Did You Know?
One millimetre of water spread uniformly over one square metre is one litre. This compact relationship follows because
(1\text{ mm}\times1\text{ m}^2=0.001\text{ m}^3=1\text{ L}).
It is one of the most useful unit identities in rainfall and irrigation calculations.
How Litres Become Millimetres of Water
Depth is volume divided by area:
(d=V/A).
If (V) is in cubic metres and (A) in square metres, (d) is in metres. Multiply by 1,000 to convert metres to millimetres. Alternatively, using the identity above, litres divided by square metres gives millimetres directly.
Worked Example: One Thousand Litres
Suppose 1,000 L is distributed uniformly over 100 square metres in an ideal model.
(d=1000\text{ L}/100\text{ m}^2=10\text{ L/m}^2=10\text{ mm}).
If this happens over 30 minutes, the average idealised application rate is
(10\text{ mm}/0.5\text{ h}=20\text{ mm/h}).
The rate does not mean every point actually received 10 mm. That conclusion needs distribution measurements.
Worked Example: Flow and Time
A fictional sprinkler discharges 12 L/min for 30 minutes. The total volume is
(12\times30=360\text{ L}).
If its effective area is 60 square metres and distribution is treated as uniform,
(360/60=6\text{ mm}).
The corresponding average rate is 12 mm/h. A good solution labels the result “idealised average over 60 square metres,” not simply “the lawn gets 6 mm everywhere.”
Dimensional Check
Write units as part of the algebra:
(\frac{\text{L}}{\text{m}^2}=\frac{0.001\text{ m}^3}{\text{m}^2}=0.001\text{ m}=1\text{ mm}).
If the final units do not reduce to length when calculating depth, something is wrong. Unit analysis is an error detector, not a decorative final label.
Flow Rate, Area and Time Form a Three-Way Relationship
Let discharge be (Q) litres per minute, run time be (t) minutes and effective area be (A) square metres. Under a uniform idealisation,
(d=Qt/A) millimetres.
This equation supports several useful comparisons:
- double (Q) while holding (t) and (A) fixed, and depth doubles;
- double (t), and depth doubles;
- double (A) with the same volume, and depth halves;
- multiply both flow and area by the same factor, and average depth is unchanged.
Solving Backwards
If a fictional target average depth is 8 mm over 75 square metres and flow is 15 L/min, required idealised volume is
(8\times75=600\text{ L}).
Time is
(600/15=40\text{ min}).
This is an algebra exercise, not a watering recommendation. A real schedule must consider current weather, soil, plants, restrictions, equipment and professional guidance.
Comparing Sectors Carefully
If one device applies the same total flow to a 90-degree sector instead of a 360-degree circle with the same radius, the idealised area is one quarter as large. If all else remained equal, average application rate would be four times as high. In real equipment, nozzle and control behaviour may change with arc settings, so the ratio is a geometric thought experiment unless verified.
Why Overlap Is Not Automatically Waste
Students often see two overlapping circles and conclude that the shared area is being “watered twice” while the rest is watered once. That may be true in a binary model where each head contributes a constant depth inside a hard boundary. Real patterns usually fade or vary across distance. Overlap can help combine weaker edges into a more even total.
The total depth at a point is modelled by adding contributions:
(D(x,y)=d_1(x,y)+d_2(x,y)+\cdots+d_n(x,y)).
This is superposition in a simplified model. A point reached by two weak edge contributions may receive about the same total as a point near one centre. Whether overlap is helpful depends on measured or documented patterns, spacing and operating conditions.
Union Area Versus Sum of Areas
For two sets (A) and (B),
(|A\cup B|=|A|+|B|-|A\cap B|).
Adding both circle areas double-counts the intersection. Subtract it once to obtain total unique covered area. This inclusion-exclusion principle appears throughout probability, set theory and data analysis.
Two Equal Circles
For advanced students, two circles of radius (r) whose centres are distance (d) apart, where (0\le d\le2r), have intersection area
(A_{int}=2r^2\cos^{-1}(d/2r)-\frac d2\sqrt{4r^2-d^2}).
Use radians for the inverse-cosine term in this formula. Check limiting cases: when (d=0), intersection area is (pi r^2); when (d=2r), it is zero. A grid-count estimate is often more appropriate for younger students and provides a useful comparison with the formula.
Overlap Still Has Costs
More overlap is not always better. It can increase total applied volume, create local excess, require more equipment or fail to correct a poor distribution pattern. The mathematical goal is not to maximise intersection area. It is to choose a metric—perhaps uniform measured depth across a target region—and evaluate trade-offs.
Spacing Patterns Change the Gaps
Place equal-radius circles on a plan. With square spacing, centres form rows and columns. With triangular spacing, neighbouring centres form equilateral triangles. The geometry of the largest uncovered gap differs.
Square Grid
If adjacent centres are distance (s) apart on a square, the centre of each grid cell is (s/\sqrt2) from the four surrounding heads. A binary reach model avoids a gap at that point when
(r\ge s/\sqrt2), or (s\le r\sqrt2).
This condition only says the cell centre lies within nominal reach. It does not guarantee uniform depth.
Equilateral-Triangle Grid
For centres at the vertices of an equilateral triangle of side (s), the triangle centre is (s/\sqrt3) from each vertex. A binary reach model covers that point when
(r\ge s/\sqrt3), or (s\le r\sqrt3).
The comparison is geometrically interesting, but it should not be turned into a universal installation rule. Boundaries, wind, operating pressure, available equipment and site constraints change the real problem.
Boundary Clipping
A circle near the edge of a rectangular lawn extends outside it. Full-circle area overstates area inside the target. Students can estimate useful area by:
- decomposing the intersection into rectangles, sectors and segments;
- using coordinate geometry;
- counting cells on graph paper;
- sampling random points and estimating the fraction inside both shapes.
Different methods create an excellent discussion about approximation and resolution.
From a Pretty Circle to Measured Distribution
Equal containers placed on a regular grid can create a simple catch test under safe, supervised conditions. The containers should have equal openings and be placed consistently. After a fixed test, measure collected depth or volume and record the grid position.
This small experiment does not certify a system. It demonstrates sampling, descriptive statistics and spatial variation.
Mean, Range and Coefficient of Variation
Suppose nine fictional measured depths in millimetres are
(8,9,10,7,11,9,8,10,9).
The mean is 9 mm. The range is (11-7=4) mm. If the population standard deviation is about 1.15 mm, the coefficient of variation is
(CV=(1.15/9)\times100\%\approx12.8\%).
CV expresses spread relative to the mean. Lower CV indicates less relative variation in that dataset, but the value depends on sampling layout, duration and conditions. Do not compare two tests unless methods are comparable.
A Heat Map
Plot each depth at its grid location and shade values using a declared scale. A heat map reveals spatial patterns that one average hides. A mean of 9 mm could come from values clustered near 9 or from a mix of very low and very high values.
Interpolation Needs Caution
Values between containers are unmeasured. Nearest-neighbour shading, bilinear interpolation and smooth contouring make different assumptions. A smooth-looking map is not additional evidence. Always show or preserve the original measurement points.
Wind, Slope and Infiltration Add New Models
A first wind model might shift an ideal distribution by a vector. If every point is translated 0.8 m east and 0.3 m north, the centroid moves by the same vector. This teaches coordinates and vectors, but it does not reproduce droplet aerodynamics.
Vector Magnitude and Direction
The shift magnitude is
(sqrt{0.8^2+0.3^2}\approx0.854\text{ m}).
Its direction north of east is
( an^{-1}(0.3/0.8)\approx20.6^\circ).
The calculation is correct for the declared translation model. The model remains a simplification.
Application Rate and Infiltration
Imagine a fictional soil model with an infiltration limit of 10 mm/h while idealised application rate is 15 mm/h. The difference is 5 mm/h. That does not prove actual runoff of exactly 5 mm/h: storage, slope, changing infiltration and vegetation matter. It flags a possible mismatch for further assessment.
This is a valuable lesson in inequalities. When input rate exceeds a limiting capacity, accumulation or loss may occur. The same structure appears in queues, drainage, network traffic and production systems.
Time Variation
Flow and weather may change during a test. Divide time into intervals and calculate volume in each:
(V\approx\sum Q_i\Delta t_i).
The sum is a discrete approximation to an integral. It is often more honest than multiplying one average flow by total time when the rate varied substantially.
Designing a Model Around a Real Boundary
An irregular target can be represented on a coordinate grid or as a polygon. The modelling sequence matters:
- define the target boundary;
- place candidate centres;
- assign radii and sector angles;
- form the union of predicted coverage;
- clip that union to the target;
- calculate covered target area, uncovered target area and outside-target area;
- add a measured depth model if uniformity is the goal.
Three Different Percentages
Let (T) be target area, (C\cap T) covered target, and (C\setminus T) coverage outside target.
- target coverage percentage: (|C\cap T|/|T|\times100\%);
- uncovered percentage: (|T\setminus C|/|T|\times100\%);
- outside share of predicted coverage: (|C\setminus T|/|C|\times100\%).
These metrics answer different questions. A plan can have high target coverage and still send much of its nominal pattern outside the boundary.
Optimisation Is More Than “Use Fewer Heads”
A fictional objective might combine uncovered area, outside area, variation and equipment count:
(J=aU+bO+cV+dN).
The weights (a,b,c,d) express priorities. Changing them can change the preferred layout. Mathematics does not secretly decide those values; people must declare them. This is an important ethical and practical insight about optimisation.
A Student Workflow for a Reproducible Model
Step One: State the Question
Choose one measurable question, such as “How does centre spacing affect uncovered area in a binary circular model?” Avoid vague aims like “find the best sprinkler system.”
Step Two: Define Geometry and Units
Draw a scale plan. Record coordinate origin, boundary dimensions, centre coordinates, radii and angles. Use metres for length, square metres for area, litres for volume and minutes or hours for time.
Step Three: Calculate an Ideal Baseline
Compute full-circle, sector or annular-sector areas. Form unions without double-counting. Convert total volume to idealised average depth.
Step Four: Add Synthetic or Safely Measured Data
For a classroom-only study, create a declared synthetic catch grid. For an ordinary supervised observation, follow relevant instructions and restrictions, keep electricity and slippery surfaces out of the activity, and do not adjust pressurised equipment.
Step Five: Test Sensitivity
Change one input at a time: radius, angle, spacing, flow or run time. Report how the chosen metric changes. Sensitivity analysis shows which assumptions dominate.
Step Six: Write Limitations
Name wind, pressure, nozzle pattern, terrain, soil and measurement resolution as appropriate. Do not use a neat calculation to imply professional validation.
Comparing Competing Plans Without Hiding the Trade-Offs
Suppose two fictional layouts serve the same 96-square-metre target. Plan A predicts 94 square metres of target coverage and 18 square metres outside the boundary. Plan B predicts 90 square metres of target coverage and 6 square metres outside. Which is better?
There is no answer until the objective is declared. Plan A covers a larger fraction of the target:
(94/96\times100\%\approx97.9\%).
Plan B covers
(90/96\times100\%=93.75\%).
But Plan A also predicts three times as much outside-target area. A decision-maker might value the extra four square metres of target coverage, or might give more weight to the boundary. Mathematics makes the conflict explicit; it does not conceal the value judgement.
A Score Needs Transparent Weights
One fictional score could be
(S=2U+O),
where (U) is uncovered target area and (O) is outside-target area, both in square metres. Plan A has (U=2), so (S_A=2(2)+18=22). Plan B has (U=6), so (S_B=2(6)+6=18). Under this declared scoring rule, the lower score favours B.
If uncovered area is weighted five times instead, Plan A scores (5(2)+18=28), while Plan B scores (5(6)+6=36); the preference reverses. This is an accessible demonstration of sensitivity to weights. A polished score is not objective unless its priorities are justified.
Pareto Thinking
A plan is dominated if another plan is no worse on every chosen metric and better on at least one. If one layout has less uncovered area, less outside area and lower total volume than another, the second is dominated under those metrics. When one layout improves coverage but worsens boundary loss, neither dominates. They sit on a trade-off frontier.
This way of thinking is valuable beyond lawn geometry. It appears in engineering design, transport, finance and public policy whenever goals conflict.
Normalising Different Units
Do not add square metres, litres and dollars directly. Convert each metric to a comparable dimensionless score, such as a fraction of a reference value, before using a weighted sum. Otherwise the numerical scale and unit choice can control the result accidentally.
Estimation Methods for Irregular Coverage
Not every clipped sector has a convenient formula. Numerical estimation lets students handle an irregular lawn, path or planting bed without pretending the boundary is rectangular.
Grid Counting
Overlay a square grid. Count fully covered cells and estimate partially covered cells. If each cell represents 0.25 square metres, 320 full-cell equivalents represent 80 square metres. Repeat with a finer grid. The difference indicates sensitivity to resolution.
Monte Carlo Sampling
Generate random points uniformly inside a bounding rectangle. If (m) of (n) points lie in both target and coverage, estimated intersection area is
(A_{box}(m/n)).
For a 120-square-metre box, if 760 of 1,000 points are inside both, the estimate is 91.2 square metres. A new random sample will differ. Larger samples usually reduce random sampling variation, but they do not correct a wrong boundary model.
Coordinate Clipping
Advanced students can represent the target as a polygon and clip it against a dense binary coverage mask. Keep coordinate units, orientation and boundary rules consistent. Decide whether a point exactly on a boundary counts inside; the choice has little area effect in theory but can affect a finite grid.
Validate One Method with Another
For a case where analytic sector area is known, compare grid and Monte Carlo estimates with the formula. Once the code or spreadsheet behaves sensibly on the known case, apply it to the irregular case. Validation does not prove perfection, but it is stronger than trusting an untested result.
Report Random Variation Honestly
A Monte Carlo result changes from run to run. If 760 of 1,000 points fall inside, the estimated fraction is 0.76. A rough standard error for a proportion is
(\sqrt{p(1-p)/n}).
Using (p=0.76) and (n=1000) gives about 0.0135, or 1.35 percentage points. This calculation relies on independent uniform sampling. Repeating the simulation and showing the spread is more informative than publishing one value to many decimal places.
Separate Numerical Error from Model Error
A fine grid can reduce discretisation error while leaving the underlying circular-reach assumption unchanged. Numerical accuracy asks whether the method solves the chosen model well. Model validity asks whether the chosen model represents the real question well. A result can be numerically precise and physically poor. Students should discuss both.
Keep a final table with one row for each estimate, including method, resolution, result and assumptions. If an analytic value, fine-grid value and simulation value are close, say how close in both square metres and percentage terms. Agreement supports the computation; it does not erase shared assumptions about radius, boundaries or distribution.
Clear records make later correction possible.
Common Misconceptions and Better Questions
“Everything Inside the Circle Gets the Same Amount”
The circle is a reach model. Ask: what measured or documented distribution profile supports uniform depth?
“Overlap Always Wastes Water”
Overlap can combine weaker edge contributions. Ask: what is the total measured depth in the shared region?
“One Millimetre Means Very Little Water”
Across one square metre, it is one litre. Across 100 square metres, it is 100 litres. Ask: over what area?
“A Larger Radius Is Always Better”
Area grows with (r^2), but boundary overspray, pressure and distribution also matter. Ask: better according to which metric and constraints?
“The Mean Proves Uniformity”
Averages can hide spatial extremes. Ask: what are the range, variation and map of values?
“A Formula Is a Watering Recommendation”
The calculation describes a declared model. Ask: what site evidence, current guidance and professional judgement are needed for a real decision?
Where the Model Stops
The classroom geometry usually assumes flat ground, a sharp reach boundary, steady flow, fixed angle, known radius and calm conditions. Real systems can depart from every assumption.
- wind can distort direction and droplet travel;
- pressure can vary among heads and over time;
- nozzles create non-uniform radial profiles;
- elevation affects pressure and runoff;
- soil intake can change during application;
- plants intercept water;
- evaporation and drift create losses;
- clogged or damaged parts change output;
- local rules may restrict timing, equipment or use.
The appropriate conclusion is not “the formula failed.” It is “the formula answered a narrower question.” Strong mathematical thinking keeps that question visible.
Guidance for Students, Parents and Teachers
For Students
Draw the geometry before using a formula. Keep a unit column in every table. When estimating overlap, compare a coarse and fine grid so you can see numerical resolution affect the answer.
For Parents
Ask your child why litres per square metre equal millimetres. Let them prove it with unit conversion. Use synthetic numbers if a physical activity would be unsafe, wasteful or contrary to restrictions.
For Teachers
This topic connects circles, sectors, rates, coordinates, statistics and modelling. Separate a pure geometry lesson from a data lesson, then ask students to reconcile them. Require a limitations paragraph as part of the mathematics, not as an optional disclaimer.
| Stage | Main question | Useful mathematics |
|---|---|---|
| Geometry | Where might water reach? | Circles, sectors, unions |
| Rate | How much average depth? | Flow, time, unit conversion |
| Measurement | How variable is distribution? | Sampling, mean, spread, maps |
| Decision | Which layout meets declared goals? | Constraints, sensitivity, optimisation |
Frequently Asked Questions
What is the area of a 90-degree sprinkler with radius 6 m?
In the ideal sector model, it is (90/360\times\pi\times6^2=9\pi\), about 28.27 square metres.
How do I convert litres over an area into millimetres?
Divide litres by square metres. One litre per square metre equals one millimetre of depth.
Does twice the run time mean twice the water depth?
In a steady-flow, fixed-area ideal model, yes. Real conditions and distribution can vary, so measure when the distinction matters.
Why use overlapping sprinkler patterns?
One reason is that individual patterns may be weaker near their edges. Overlap can improve total distribution, but the result should be evaluated with appropriate measurements or professional design information.
Can I use circle area to choose a real system?
Circle area is only a starting model. Real selection and installation should follow current manufacturer information, local requirements and qualified guidance.
What is a good school investigation?
Use a scale drawing or synthetic catch grid, compare two spacing patterns, calculate target coverage and variation, and document every assumption. Avoid modifying or servicing actual pressurised equipment.
Next Reading and a Final Challenge
Rainfall uses the same volume-area-depth relationship. Continue with Why Mathematics? | Rain Gauges, Catch Area and Rainfall Depth. For agriculture and time accumulation, read Why Mathematics? | Growing Degree Days, Heat Units and Crop Timing. For another flow-and-capacity application, see Why Mathematics? | Swimming Pools, Turnover Time and Filtration Flow. More pathways are collected in the Mathematics Learning Hub.
For a final challenge, create a 12 m by 8 m fictional rectangular target with a 2 m by 3 m exclusion zone. Compare two declared sets of sector centres and radii. Calculate nominal target coverage, outside-target area and ideal average depth from a fixed total volume. Then add a synthetic catch grid and decide whether the layout that covers more area is also the one with less variation.
That tension is the point. Mathematics turns a spinning spray into a set of answerable questions, but it also teaches us not to confuse one answer—such as area—with the whole decision.
A Practical Mathematics Studio
These investigations use drawings, synthetic data or ordinary supervised operation within a manufacturer's instructions. They are learning activities, not installation, compliance, maintenance or repair instructions. Record assumptions and make every result reproducible.
Investigation 1: Map one full circle
Choose a radius, calculate pi r squared, and draw the corresponding circle to scale. Measure whether the plotted radius and computed area are consistent.
Investigation 2: Build a sector
Use A = theta/360 times pi r squared for several angles. Check that 90, 180 and 360 degrees produce one quarter, one half and one whole circle.
Investigation 3: Model an annular sector
Add a non-watered inner radius and calculate theta/360 times pi times the difference of squared radii. Label the geometry before substituting numbers.
Investigation 4: Measure catch cans
Place equal containers on a synthetic grid, assign collected depths, and calculate mean, range and coefficient of variation. State that one small trial does not certify uniformity.
Investigation 5: Convert flow to depth
Divide a declared water volume by watered area, convert cubic metres to litres and metres of depth to millimetres, and reconcile the units.
Investigation 6: Compare run times
For equal flow and different sector angles, predict how application depth changes if the same volume is distributed over a smaller area. Keep real sprinkler controls outside the classroom model.
Investigation 7: Draw head-to-head spacing
Place equal-radius circles on a square grid and mark overlaps and potential dry gaps. Vary spacing while keeping radius fixed.
Investigation 8: Try triangular spacing
Arrange centres on equilateral triangles, calculate centre distances and compare overlap patterns with square spacing. Do not infer field superiority from geometry alone.
Investigation 9: Estimate overlap numerically
For two equal circles, use a trusted calculator or numerical grid to estimate intersection area. Compare with cell-count estimation and discuss resolution error.
Investigation 10: Add a boundary
Clip circular coverage to a rectangular lawn. Estimate watered area inside and outside the boundary and explain why full-circle area overstates useful coverage near edges.
Investigation 11: Model wind displacement
Shift a synthetic catch distribution by a declared vector. Compare centroid before and after without claiming the simple translation captures droplet physics.
Investigation 12: Find precipitation rate
Use total discharge divided by irrigated area and convert to millimetres per hour. Check that doubling flow doubles idealised rate while doubling area halves it.
Investigation 13: Separate infiltration
Compare application rate with a fictional soil infiltration limit. Mark potential runoff time, while stating that real soil assessment needs local professional guidance.
Investigation 14: Plan zones
Partition an irregular plan into regions with different radius or exposure. Explain why one average run time can hide important differences.
Investigation 15: Run sensitivity checks
Change radius, angle, spacing and flow one at a time. Rank which assumption has the greatest effect on your chosen metric.
Investigation 16: Publish an audit trail
Include plan scale, coordinates, radii, sectors, flow assumptions, catch data, calculations, exclusions and a clear statement that it is an educational model.