Paritian

Mathematics

Volume by Water Displacement

The volume of an irregular object from how far the water rises, and its density if you also weigh it.

Results

Volume of the object 1099.557429 cm³
Volume of the object 1.099557 L
Volume of the object 0.001099557
How much it rose 3.5000 cm
Cross-section of the container 314.159265 cm²
Density of the object 2.273642 g/cm³
In water it would Sink
Mass of water pushed aside 1.099557 kg

What this tool does

There is no formula for the volume of a stone. There is, however, a measurement: drop it in water and see how much the level rises. The volume of the object equals the volume of water pushed aside, which is the container's cross-section times the rise. Weigh the object too and you get its density, and from that whether it would float — a density above 1 g/cm³ sinks in water.

Formula

V = π (d ÷ 2)² × (level after − level before) · density = mass ÷ V

Variables

SymbolMeaningUnit
dInside diameter of the containercm
h1Water level beforecm
h2Water level aftercm
mMass of the objectkg
VVolume of the objectcm³
LIVolume of the objectL
M3Volume of the object
DHHow much it rosecm
CSCross-section of the containercm²
DEDensity of the objectg/cm³
FLIn water it would
WDMass of water pushed asidekg

Worked example

  • Inside diameter of the container20 cm
  • Water level before15 cm
  • Water level after18.5 cm
  • Mass of the object2.5 kg
  • Volume of the object1099.557429 cm³
  • Volume of the object1.099557 L
  • Volume of the object0.001099557 m³
  • How much it rose3.5000 cm
  • Cross-section of the container314.159265 cm²
  • Density of the object2.273642 g/cm³
  • In water it wouldSink
  • Mass of water pushed aside1.099557 kg

Limitations

  • Mixing units is the most common source of error. Convert every input to the units shown next to each field before calculating.
  • The calculation runs at full precision and only the display is rounded. If you copy an intermediate value and retype it, small differences can appear.