The Gold Standard of Soil Science | Shrink–Swell and the Moving Volume Behind a Moisture Reading

Use three quantities: dry soil mass Ms, water mass Mw and the current bulk volume V of the soil clod. Bulk volume includes the clod’s solids and internal pores. Here it excludes empty space outside the clod after it pulls away from a container wall.

Gravimetric water content is w = Mw/Ms. Volumetric water content is θ = Vw/V, where Vw is water volume. Dry bulk density is ρb = Ms/V. These quantities answer different questions even though they describe the same specimen.

The link is θ = wρb/ρw, with ρw the water density. This identity uses a bulk density referring to the same specimen, volume boundary and state as θ. It does not authorise reusing a density measured before the specimen changed volume.

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Some soils change volume appreciably as their water state changes. USDA NRCS describes Vertisols as soils with expanding clay minerals that shrink on drying and swell on wetting. That is a specific soil behaviour, not a rule that every soil shrinks by the same amount.

The NRCS Soil Survey Field and Laboratory Methods Manual, section 3.5.4 discusses measures of shrink–swell and bulk-density measurements at defined water states. The measurement state is part of the reported property.

Our examples isolate the denominator issue. They do not predict a real soil’s shrinkage curve from its clay percentage. Real volume change depends on material, structure, constraints and wetting or drying history. Use the arithmetic to recognise what must be measured, rather than to invent a missing measurement.

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Take an idealised clod containing 1.000 kg of dry soil solids. Assume their volume stays at 0.400 L. Assume water density is 1.000 kg/L and there is no loss of solid material or dissolved matter.

Initially, the clod contains 0.400 kg of water and occupies 0.800 L. Its 0.400 L of pore space is entirely water-filled. Later, water mass is 0.200 kg and clod volume is 0.600 L. Its remaining 0.200 L of pore space is again entirely water-filled.

This is an idealised saturated shrinkage segment: a 0.200 L decrease in water volume accompanies a 0.200 L decrease in bulk volume. We choose these states to make the volume accounting transparent. They are not a universal relation between drying and shrinkage.

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The water mass halves from 0.400 to 0.200 kg. Because dry mass is unchanged, gravimetric water content also halves, from 40% to 20% on a dry-mass basis.

Volumetric water content falls from 0.400/0.800 = 0.500 to 0.200/0.600 = 0.333. That is a fall of about 16.7 percentage points, or one third relative to the initial volumetric fraction. It is not a 50% fall because the volume denominator also shrank.

Check the conversion at each state. Initially, 0.400 × 1.250 / 1.000 = 0.500. Later, 0.200 × (1.000/0.600) / 1.000 = 0.333. Each calculation agrees with dividing water volume by current clod volume.

If the original density 1.250 kg/L were reused later, the calculation would give θ = 0.250. That value answers a different question: 0.200 L of water per original 0.800 L. It is not the water fraction of the later 0.600 L clod. Name a reference-volume quantity explicitly if that is what you intend to report.

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The initial pore volume is 0.800 − 0.400 = 0.400 L. The later pore volume is 0.600 − 0.400 = 0.200 L. Both states have zero air volume in this idealised example, so their degree of saturation is 100%.

A volumetric water fraction of 0.333 can therefore describe a saturated specimen when porosity is also 0.333. “Only 33% water by bulk volume” does not mean “33% of the pores are filled.” Those denominators differ.

Now add a third invented state with 0.100 L water and total clod volume 0.580 L. Pore volume is 0.180 L, leaving 0.080 L of air. Degree of saturation is 0.100/0.180 ≈ 55.6%, while θ = 0.100/0.580 ≈ 17.2%.

This third state no longer follows the earlier equal water-loss and shrinkage segment. That is intentional: a pair of earlier states cannot establish the entire drying curve. Check that solids, water and air volumes add to the total before attaching a physical interpretation.

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Suppose the same clod sits inside a rigid 1.000 L container. The container is merely a geometric reference; it supplies no additional soil or water. Initially, 0.200 L lies outside the clod. Later, 0.400 L lies outside it.

Both rows are arithmetically correct. They describe different volumes. The fixed-container water storage halves, matching the water loss. The clod-based fraction falls by one third.

A measurement report should explain whether external gaps and cracks are included in the reference volume. Multiplying a clod fraction by the whole container volume would incorrectly assign the clod’s water fraction to the space outside it.

This distinction becomes especially useful when comparing a laboratory specimen with a field layer. The specimen boundary and the field accounting boundary may not include the same voids.

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Consider a separate idealised layer over 1 m² of ground. Its initial depth is 0.200 m and θ is 0.500. Water volume per unit area is θ times depth: 0.500 × 0.200 = 0.100 m, or 100 mm of water.

Assume the layer shrinks only vertically, its area stays fixed, no solid mass moves across its boundary and its later depth is 0.150 m. If θ later equals one third, water depth is (1/3) × 0.150 = 0.050 m, or 50 mm.

Using the original 0.200 m depth with the later fraction would give 66.7 mm. That mixes two states. The correct result requires the fraction and the volume geometry from the same state.

Actual soil can develop cracks, move laterally and exchange water with neighbouring layers. The simple example does not model those processes. Its transfer lesson is narrower: a volumetric fraction becomes a storage only after the relevant volume or depth is specified consistently.

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A mass-based moisture determination and a bulk-volume determination must refer to compatible material. A remoulded sample, an intact clod and a core containing large cracks can have different volume meanings.

Record the sample identity, water state at volume measurement, dry-mass basis, treatment of stones and external voids, and whether the specimen was disturbed. A density measured on one specimen cannot silently become the density of another simply because both carry the same soil name.

Instrument readings also require their own interpretation. A sensor’s displayed volumetric estimate is not automatically a direct measurement of the changing clod boundary used in our paper model. Calibration, contact and sampled volume need to be established for the instrument and material.

For a practical evidence check, ask whether the report lets another reader reconstruct Ms, Mw and V for each state. If one is missing, identify the resulting uncertainty rather than filling the gap with the initial value.

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In our calculation, dry bulk density rises because a clod loses volume while retaining dry mass. That alone does not establish that external mechanical compaction occurred. The mechanism needs evidence beyond the density number.

Conversely, equal final densities do not prove equal histories or pore arrangements. A mechanically loaded specimen and a drying specimen could reach the same bulk density while differing in water state and structure.

Nor does drying followed by wetting guarantee an exact return to every earlier state. The NRCS methods manual distinguishes its extensibility concept from irreversible shrinkage in some materials. Treat a reversible paper path as an assumption, not a claim that all soil structure repairs itself when water returns.

The existing soil compaction and moisture–density guide develops a different question: density under a specified compaction procedure. Use it to distinguish a laboratory compaction reference from the moisture-dependent volume accounting developed here.

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Knowing how much water a specimen contains does not determine how quickly water will enter, drain or move through it. A changing crack network and a changing soil matrix introduce additional questions about pathways and scale.

NRCS’s Vertisol description notes that open cracks can provide bypass flow in some of these soils. That observation is a reason to investigate pathways, not a licence to infer a particular infiltration rate from our clod volumes.

Likewise, a water fraction alone does not establish plant availability or an irrigation requirement. Those questions need additional soil, crop and water-state information. The guide to infiltration, storage and drainage keeps those quantities distinct.

For any transfer, write the decision first. If the question is water mass lost from a specimen, use a consistent mass balance. If it is storage across a field area, define the spatial boundary. If it is movement, add suitable flow evidence instead of stretching a storage calculation into a rate prediction.

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Check one: convert with a new current density

A specimen has dry mass 0.600 kg, water mass 0.120 kg and current bulk volume 0.450 L. Then w = 0.200 kg/kg and ρb = 1.333 kg/L. Taking ρw = 1.000 kg/L, θ = 0.267 L/L. Directly, 0.120/0.450 gives the same result.

Check two: detect a mixed-state conversion

Suppose that specimen previously occupied 0.500 L. Reusing its previous dry bulk density of 1.200 kg/L would give θ = 0.240. This is water volume divided by the former bulk volume. To report the present clod fraction, use 0.450 L or the corresponding current density.

Check three: distinguish a fixed area from a fixed depth

A fictional layer over fixed area has θ = 0.300 and depth 0.120 m, giving 36 mm water storage. Later θ = 0.250 and depth 0.100 m, giving 25 mm. The loss is 11 mm. The calculation 0.050 × 0.120 m = 6 mm accounts only for the fraction change at the old depth and misses the changing geometry.

Explain each denominator aloud before calculating. If you cannot identify the physical volume it represents, stop at that uncertainty rather than presenting a precise storage number.

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Can dry bulk density change without adding dry soil?

Yes. Ms/V changes when V changes. Whether the change came from shrinkage, loading, disturbance or another cause requires additional evidence.

Is a smaller volumetric fraction proof of proportionally less water?

Only if the relevant volume is held fixed. With a changing volume, compare θV, not θ alone, to recover water volume.

Does the example apply to every clay or field?

No. Its states are chosen for transparent accounting. A real shrinkage relationship must be measured or justified for the material, water states and constraints concerned.

Which route develops the next question?

For foundations, use Soil Science and gravimetric moisture to root-zone water depth. For specimen comparisons, use composite samples and hidden spatial variation and coarse fragments in carbon-stock denominators. Their questions differ, but each requires an explicit reference quantity.

For applications, continue into geotechnical engineering or irrigation engineering. Return to The Gold Standard of Education: learning that transfers into life or the How X Works subject map for the wider learning route.

Primary background references are USDA NRCS’s Vertisols overview and Soil Survey Field and Laboratory Methods Manual, Soil Survey Investigations Report 51, version 2, especially section 3.5.4. All clod, container and layer calculations here are original idealised examples.

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