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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Identify the quantities
Chapters 1–3. Follow moisture basis, dry bulk density and the conversion.
Build the depth account
Chapters 4–6. Calculate storage, combine layers and separate available-water quantities.
Check applications and errors
Chapters 7–8. Distinguish net storage from application and trace wrong denominators.
Interpret the evidence
Chapters 9–10. Keep sampling, physical limits and sensor meaning visible.
Practise and continue
Chapters 11–12. Check the worked answers and choose a soil-water route.
Full chapter index · Practice and worked checks · Soil Science subject root
A soil report says “20% moisture”. Does that mean twenty litres of water in every hundred litres of soil? It might mean something else entirely. A percentage based on dry soil mass needs another quantity before it can become a water volume or a depth across a garden bed.
This guide follows an invented laboratory record from sample masses to volumetric water content, then into a layered root-zone water account. We distinguish total water stored, water above a supplied lower reference and the amount needed to reach a supplied target. Each calculation keeps its denominator and spatial boundary visible.
Use The Gold Standard of Soil Science for profiles, texture, structure and sampling. The worked values below are paper examples, not measurements from a real plot or instructions for drying soil. Their purpose is to explain what a soil-water number represents before using it in a decision.
Chapter index
Identify the quantities · Chapters 1–3
1. Ask what the moisture percentage divides by
2. Use dry bulk density to connect mass with volume
3. Derive the conversion rather than memorising a shortcut
Build the depth account · Chapters 4–6
4. Turn volumetric content into equivalent water depth
5. Weight layers by thickness
6. Separate total storage from an available-water account
Check applications and errors · Chapters 7–8
7. A net storage deficit is not the gross irrigation amount
8. See how a wrong denominator travels through the calculation
Interpret the evidence · Chapters 9–10
9. Keep the sample basis and physical limits visible
10. A sensor percentage needs its own interpretation
Practise and continue · Chapters 11–12
11. Practice checks with worked answers
12. Questions readers ask, sources and next reading
CHAPTER 1 OF 12 · Identify the quantities
1. Ask what the moisture percentage divides by
Gravimetric water content on a dry-mass basis divides water mass by dry soil mass. Volumetric water content divides water volume by the total bulk soil volume represented. These are different ratios.
For a supplied record, a sample has wet mass 120 grams and dry soil mass 100 grams, with the container excluded from both. Assume the twenty-gram difference represents water removed under the stated laboratory method.
Dry-basis gravimetric content is 20/100 = 0.20 g/g, or 20%. A wet-basis fraction would instead be 20/120 ≈ 0.1667, or 16.67%. Neither fraction alone tells us the water volume per bulk soil volume.
The units and denominator belong beside the percentage. “Twenty percent” without its basis invites a reader to compare unlike quantities. The problem is not solved by adding a decimal place.
The NRCS Soil Health Educators Guide separates bulk density, moisture and other soil indicators. Keeping those properties distinct helps explain why a second measurement is needed here.
CHAPTER 2 OF 12 · Identify the quantities
2. Use dry bulk density to connect mass with volume
Dry bulk density is dry soil mass divided by the represented bulk volume, including its pore space. It differs from the density of the solid particles alone.
Suppose the same sample represents an intact bulk volume of 80 cubic centimetres. Its dry bulk density is 100/80 = 1.25 g/cm³. This volume record lets us connect the water mass to a fraction of the original soil volume.
Using water density approximately 1 g/cm³ for this teaching calculation, twenty grams of water occupies about twenty cubic centimetres. Dividing by the 80-cubic-centimetre bulk volume gives volumetric content 20/80 = 0.25 cm³/cm³, or 25% by volume.
| Supplied quantity | Value |
|---|---|
| Wet sample mass, excluding container | 120 g |
| Dry soil mass, excluding container | 100 g |
| Represented bulk soil volume | 80 cm³ |
| Water mass under the model | 20 g |
| Dry bulk density | 1.25 g/cm³ |
| Volumetric water content | 0.25 cm³/cm³ |
Notice the result: 20% by dry mass corresponds to 25% by volume in this specific example. It is not a universal conversion factor. A different dry bulk density would change the relationship.
CHAPTER 3 OF 12 · Identify the quantities
3. Derive the conversion rather than memorising a shortcut
Let w = water mass/dry soil mass, ρb = dry soil mass/bulk soil volume, and ρw = water mass/water volume. Multiplying w by ρb and dividing by ρw cancels the masses, leaving water volume/bulk soil volume.
Thus θ = wρb/ρw, where θ denotes volumetric water content. Use compatible units for the two densities. If both are in g/cm³, their ratio is dimensionless.
For the supplied record, θ = 0.20 × 1.25/1.00 = 0.25. Use the fraction 0.20, rather than inserting the percentage number twenty and calling the resulting 25 a volume fraction.
The same calculation works in SI units: ρb = 1,250 kg/m³ and ρw ≈ 1,000 kg/m³. Then θ = 0.20(1,250)/1,000 = 0.25 m³/m³.
The NRCS surface-irrigation handbook chapter connects dry-mass moisture, density and volumetric content. Oregon State University’s soil moisture monitoring guide supplies further context for using volumetric content in a root-zone account.
The derivation is a useful error check. If the remaining units are grams per square centimetre or a bare mass, the calculation has not reached the intended water-volume fraction.
CHAPTER 4 OF 12 · Build the depth account
4. Turn volumetric content into equivalent water depth
Imagine a uniform soil layer 0.40 metres deep with θ = 0.25 throughout its represented area and depth. For one square metre of ground, bulk soil volume is 0.40 cubic metres. Water volume is 0.25 × 0.40 = 0.10 cubic metres.
Spread that water volume over the same square metre and its equivalent depth is 0.10 metres, or 100 millimetres. More generally, stored water depth S = θZ when θ is uniform and Z is layer thickness in matching length units.
One millimetre over one square metre equals one litre. Therefore this model contains 100 litres of water per square metre within the chosen 0.40-metre layer.
It does not mean a visible 100-millimetre pool sits on the surface. Equivalent water depth is an accounting representation of water distributed through soil pores.
It also does not mean plants can extract all one hundred litres. We have calculated total stored water under the volume-content model. Availability involves how water is held, rooting and the conditions of the soil and plants.
State the depth boundary every time. “One hundred millimetres stored in the top forty centimetres” is much more informative than “one hundred millimetres in the soil”. Extending the result to a deeper profile would require evidence for that extra depth.
Real profiles need not have uniform water content. Use a second fictional record: an upper 0.20-metre layer has θ = 0.20, and a lower 0.40-metre layer has θ = 0.30.
The upper layer stores 0.20 × 200 = forty millimetres. The lower stores 0.30 × 400 = 120 millimetres. Total storage over the 0.60-metre profile is 160 millimetres.
| Layer | Thickness | Volumetric content | Equivalent water depth |
|---|---|---|---|
| Upper | 200 mm | 0.20 | 40 mm |
| Lower | 400 mm | 0.30 | 120 mm |
| Whole represented profile | 600 mm | Thickness-weighted mean ≈ 0.2667 | 160 mm |
A simple average of the two contents is 0.25. Multiplying that by 600 millimetres gives 150 millimetres, which is wrong for this supplied profile because the wetter layer is twice as thick.
The appropriate profile mean is (0.20×200 + 0.30×400)/600 ≈ 0.2667. Keeping the layer calculations visible is usually clearer than compressing them into a single average too early.
A point sensor within a layer also needs a representativeness argument before its value describes that whole layer. The table supplies layer-average contents by assumption; an actual field measurement does not automatically come with that guarantee.
CHAPTER 6 OF 12 · Build the depth account
6. Separate total storage from an available-water account
Return to the uniform 400-millimetre layer with current θ = 0.25. Suppose the exercise supplies field-capacity content θFC = 0.30 and a lower reference θWP = 0.12, both volumetric and appropriate to the same represented layer.
The conventional available-water capacity between those supplied references is (0.30−0.12)×400 = 72 millimetres. Current water above the lower reference is (0.25−0.12)×400 = 52 millimetres.
The deficit to the field-capacity reference is (0.30−0.25)×400 = twenty millimetres. The account closes: 52 + 20 = 72 millimetres.
| Quantity in the supplied model | Water depth |
|---|---|
| Total currently stored | 100 mm |
| Storage below the supplied lower reference | 48 mm |
| Current storage above that reference | 52 mm |
| Deficit to supplied field capacity | 20 mm |
| Capacity between the two references | 72 mm |
These are distinct quantities. Saying “the soil contains one hundred millimetres” does not establish that one hundred millimetres is readily available to the plants.
Field capacity and permanent-wilting reference values are contextual descriptors, not identical universal numbers for every soil and plant. Actual root extraction is not uniform through depth. The handbook itself notes limits to a simple root-zone capacity calculation. Use the exercise to understand the account, rather than turn its invented thresholds into a watering instruction.
CHAPTER 7 OF 12 · Check applications and errors
7. A net storage deficit is not the gross irrigation amount
The twenty-millimetre deficit above describes a desired net storage increase within the modelled layer. Water supplied at the surface may take other paths or fail to distribute uniformly through the intended area.
In a separate simplified planning calculation, assume exactly 80% of supplied water becomes the desired additional root-zone storage, with no other changes. Gross application would be 20/0.80 = 25 millimetres. Over ten square metres, that is 250 litres.
This is a stipulated accounting factor, not a recommended efficiency or an irrigation schedule. A real application needs the crop, soil, weather, system, distribution and relevant management advice. The factor should not be borrowed from an unrelated installation simply because the arithmetic is convenient.
Distinguish amount from rate. A 25-millimetre total application does not specify how quickly it arrives. Surface ponding or runoff can depend on that rate and the changing intake conditions.
The previous infiltration comparison guide explains entry rates and their boundaries. The irrigation engineering guide connects supplied water with root-zone service. The three accounts—entry, storage and delivery—work together without becoming the same measurement.
CHAPTER 8 OF 12 · Check applications and errors
8. See how a wrong denominator travels through the calculation
Take the original 120-gram wet and 100-gram dry sample. If someone uses the wet-basis fraction 20/120 ≈ 0.1667 in the dry-basis conversion θ = wρb/ρw, they obtain about 0.2083 instead of 0.25.
Across a 400-millimetre layer, the incorrect value implies 83.33 millimetres stored, rather than one hundred. The arithmetic is internally tidy; the starting ratio is unsuitable for that formula.
Another error inserts wet bulk density where the formula requires dry bulk density. Here wet bulk density is 120/80 = 1.50 g/cm³, while dry bulk density is 1.25. Substituting the wrong density would produce another inconsistent result.
Do not try to repair both errors by changing an unexplained conversion constant until the expected answer appears. Return to the quantity definitions and cancel the units.
The soil compaction and moisture–density guide develops wet and dry density in an engineering context. That is a useful comparison, while its compaction objectives are different from an agricultural root-zone water question.
The general lesson is that an early denominator mistake can survive several correct later multiplications. Trace the calculation back to the measured masses and represented volume before diagnosing the final number alone.
CHAPTER 9 OF 12 · Interpret the evidence
9. Keep the sample basis and physical limits visible
A conversion needs matched definitions. A fine-earth moisture result, a bulk density that treats coarse fragments differently and a total profile volume cannot be combined casually. Record which material fraction and volume each value describes.
Some soils change volume as water content changes. The represented bulk volume and its measurement condition therefore matter. A loose jar of disturbed material should not silently replace the original intact field volume in an otherwise unchanged calculation.
Volumetric water content also has a physical bound: water occupies part of the represented bulk volume. A result above one is impossible for an ordinary water-volume fraction and usually reveals a unit or percentage error. A stronger local check compares the value with appropriate pore-space information.
An apparently plausible result below one can still be wrong. A density borrowed from another site, a missing container mass or an unrepresentative sample can produce a reasonable-looking fraction with a poor basis.
These checks do not require every learner to conduct a professional soil investigation. They teach what questions to ask of a supplied record. Name the material, the volume, the method and the depth before making a claim about a plot.
CHAPTER 10 OF 12 · Interpret the evidence
10. A sensor percentage needs its own interpretation
A device displaying “25%” may report an estimated volumetric water content, a relative scale or another defined output. Read its documentation and calibration basis before treating it as θ = 0.25 in a storage equation.
A tensiometer instead concerns water tension, often expressed in pressure units. It asks how tightly water is held, rather than directly counting a water-volume fraction. Converting tension to content requires an appropriate soil-water relationship, not a universal percent-per-kilopascal multiplier.
Oregon State University’s monitoring guide distinguishes content from tension and explains why sensor placement and calibration matter. We use those distinctions here; no accuracy specification for an actual device is implied.
Now change the setting from a small sample to a surface observation from a larger sensing product. Its sampled depth and area may differ substantially from the crop’s full rooting zone. A precise surface estimate does not independently establish the deeper profile’s storage.
The transfer question is the same as before: what volume of soil does this value represent? The method of obtaining the value has changed, so the representativeness and calibration questions need to change with it.
A useful record preserves the original instrument output and its meaning beside any converted water depth. That allows the next reader to distinguish observation, calibration-based estimate and profile-level inference.
Convert a dry-mass fraction Let w = 0.15 g/g and dry bulk density 1.40 g/cm³. With water density approximated as one g/cm³, θ = 0.15×1.40 = 0.21. Across a uniform 300-millimetre layer, storage is 63 millimetres.
Convert storage into volume Thirty millimetres over eight square metres equals thirty litres per square metre multiplied by eight, or 240 litres. The depth and area must describe the same represented region.
Use two layers A 100-millimetre layer at θ = 0.10 stores ten millimetres. A 300-millimetre layer at θ = 0.20 stores sixty. Total is seventy millimetres; the profile mean is 70/400 = 0.175.
Find the target deficit Current θ = 0.22 and a supplied target θ = 0.28 across 500 millimetres give a thirty-millimetre net storage difference. That is not automatically the gross irrigation requirement or an instruction to apply it immediately.
Repair the record “Moisture 20%, depth forty centimetres” is insufficient to calculate water depth if the percentage basis is unknown. Ask whether it is gravimetric or volumetric and obtain compatible density information where needed.
Distinguish an objective A high dry density can enter a mathematically correct conversion without proving that compaction is desirable for roots. A measurement relationship is not itself a management recommendation.
CHAPTER 12 OF 12 · Practise and continue
12. Questions readers ask, sources and next reading
Can gravimetric content be converted without bulk density? Not into a bulk-volume fraction using this relationship. You need compatible information connecting dry soil mass to represented bulk volume.
Is all stored water available to plants? No. Water content and the conditions of extraction matter. A root-zone available-water account needs appropriate reference properties and rooting information.
Can I average two sensor percentages? First check that they describe the same quantity, valid calibration and the represented layers or areas. Unequal layer thicknesses require suitable weighting when calculating profile storage.
Sources and scope The NRCS surface-irrigation handbook supports the mass-to-volume and root-zone capacity framework. Oregon State University’s soil moisture monitoring guide explains content, tension and monitoring context. All sample masses, layers, thresholds and application factors here are original fictional teaching inputs.
Follow soil, water and density Return to The Gold Standard of Soil Science, compare infiltration, storage and drainage, and continue to irrigation engineering or soil compaction mathematics for their distinct applications.
Connect the learning Learning That Transfers into Life explains the value of preserving a quantity’s meaning when its context changes. The How X Works hub connects the wider library.
Before turning a moisture percentage into litres or millimetres, write its denominator. Then name the volume and depth it represents. Those details make the conversion useful beyond the calculator.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.