Bulk response can be inverted to recover the compression volume change.

Bulk response can be inverted to recover the compression volume change. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Bulk pressure can be inverted to recover compression volume

This book uses positive compression magnitudes. Pressure change 2 Pa over bulk modulus 12 Pa sets the volume-change fraction; with starting volume 6 cubic meters the recovered volume change is 1 cubic meter.

ΔV=ΔPKV=2 Pa12 Pa×6 m3=1 m3\Delta V=\frac{\Delta P}{K}V=\frac{2\ \text{Pa}}{12\ \text{Pa}}\times6\ \text{m}^{3}=1\ \text{m}^{3}
Bulk inversion rowPressure, modulus, volume, and volume change are checked magnitudes.pressurevolume changevolume=6 m^3volumeChange=1 m^3pressureChange=2 Pamodulus=12 Pa

Bulk pressure can be inverted to recover compression volume

This book uses positive compression magnitudes. Pressure change 3 Pa over bulk modulus 18 Pa sets the volume-change fraction; with starting volume 6 cubic meters the recovered volume change is 1 cubic meter.

ΔV=ΔPKV=3 Pa18 Pa×6 m3=1 m3\Delta V=\frac{\Delta P}{K}V=\frac{3\ \text{Pa}}{18\ \text{Pa}}\times6\ \text{m}^{3}=1\ \text{m}^{3}
Bulk inversion rowPressure, modulus, volume, and volume change are checked magnitudes.pressurevolume changevolume=6 m^3volumeChange=1 m^3pressureChange=3 Pamodulus=18 Pa

Bulk pressure can be inverted to recover compression volume

This book uses positive compression magnitudes. Pressure change 6 Pa over bulk modulus 16 Pa sets the volume-change fraction; with starting volume 8 cubic meters the recovered volume change is 3 cubic meters.

ΔV=ΔPKV=6 Pa16 Pa×8 m3=3 m3\Delta V=\frac{\Delta P}{K}V=\frac{6\ \text{Pa}}{16\ \text{Pa}}\times8\ \text{m}^{3}=3\ \text{m}^{3}
Bulk inversion rowPressure, modulus, volume, and volume change are checked magnitudes.pressurevolume changevolume=8 m^3volumeChange=3 m^3pressureChange=6 Pamodulus=16 Pa