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Soil bulk density is dry soil mass divided by the total sampled volume, including the spaces between particles:Here $M_s$ is the dry-solid mass. The denominator includes pore space, which distinguishes soil bulk density from soil-particle density. Reporting the dry-solid mass also distinguishes this quantity from the mass of wet soil per sampled volume.^[ref]

A constructed core measurement starts with a cylinder containing $100\mathrm{c}{\mathrm{m}}^3$ of soil. If its wet mass is $150\mathrm{g}$ and its dry-solid mass is $130\mathrm{g}$, then soil bulk density is $130/100=1.30\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$. Using the wet mass would give $1.50\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, a different quantity. The relevant sampled volume must be retained even if the removed material changes shape during drying.

The basic measurement connects a known-volume core with drying and weighing. The USDA guide describes this cylindrical-core approach and identifies a drying device and a weighing scale as needed equipment.^[ref] Sample disturbance, omitted coarse fragments or inconsistent volume conventions can change what the result represents.

Soil bulk density supplies a useful conversion between amounts per mass and amounts per volume. For example, suppose a measured substance contributes $10\mathrm{mg}$ per kilogram of dry soil. With the stipulated soil bulk density $1300\mathrm{kg}{\mathrm{m}}^{-3}$, its amount per bulk volume isThus identical mass-based measurements need not imply identical amounts in an equal soil volume. Packing and particle composition both affect soil bulk density; it is not a universal numerical constant.^[ref]

Soil-particle density compares dry particle mass with the volume assigned to the particles themselves:Spaces between particles are excluded from $V_s$. By contrast, the denominator of soil bulk density includes those spaces. A measurement must specify how particle volume is determined, especially when the particles themselves contain pores.^[ref]

One measurement principle uses a pycnometer: determine the dry sample mass, then infer its volume from the displaced liquid mass and the liquid's density. Trapped gas, access to internal pores and the chosen volume convention matter when interpreting the displacement. The standard distinguishes fine soil from gravel and stones, and its stated particle-volume convention includes internal particle pores while excluding spaces between particles.^[ref]

A constructed mixture shows why particle composition matters. Assume additive volumes, no internal pore complication, and two kinds of particles with densities $2.60$ and $1.30\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$. Equal masses of $13\mathrm{g}$ occupy $5$ and $10\mathrm{c}{\mathrm{m}}^3$, respectively. The combined soil-particle density israther than the arithmetic mean $1.95\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$. For mass fractions $w_i$ and constituent densities ${\rho }_i$, the same additive-volume model gives ${\rho }_s^{-1}={\sum }_i{w}_i/{\rho }_i$.

For compatible volume conventions, a particle-and-pore model relates the two densities to porosity:This identity requires the same dry mass in both density definitions. An assumed soil-particle density therefore makes the derived porosity an estimate under that assumption, rather than an independent measurement of pore volume.^[ref]

Water-filled pore space is the fraction of a soil sample's pore volume occupied by water. If $V_w$ is water volume and $V_p$ is total pore volume, the fraction is $V_w/V_p$; multiplying by one hundred expresses it as a percentage. The denominator is the pore volume, whereas volumetric soil water content uses the entire sample volume.^[ref]

Let total sample volume be $V$, porosity be $\epsilon =V_p/V$, and volumetric soil water content be $\theta =V_w/V$. ThenFor matching dry-solid mass $M_s$, soil bulk density is ${\rho }_b=M_s/V$ and soil-particle density is ${\rho }_s=M_s/V_s$. Under an additive solid-and-pore volume model,These identities require compatible volume definitions; shrinkage or swelling can change the measured sample geometry.

Consider two hypothetical samples with $\theta =0.30$ and ${\rho }_s=2.65\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$. For ${\rho }_b=1.325\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, porosity is $0.50$ and water-filled pore space is $0.60$. For ${\rho }_b=1.59\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, the corresponding values are $0.40$ and $0.75$. Equal water per total volume fills different fractions of the available pores. This compares two stipulated samples, rather than assuming a compaction process preserves every other quantity.

The 1984 study related water-filled porosity to carbon dioxide production, oxygen uptake and nitrous oxide production. Its reported maximum aerobic activity near sixty percent applies to the studied conditions, rather than a universal threshold.^[ref] Pore occupancy describes geometry; pore connectivity and biological demand also matter when interpreting gas exchange.