Presets
Playback
Vessel
Shape (0 cylinder, 1 cone, 2 uniform-fall bowl), top radius, height and starting fill.
Water flow
Orifice diameter at the bottom, constant inflow from the top, discharge coefficient and how fast the clock runs.
Layers
Model checks
The science
Torricelli's law
Water leaving a small hole a height h below the free surface has speed v = √(2 g h), found by equating the lost potential energy to the kinetic energy (Bernoulli). The flow through the orifice is Qout = Cd a √(2 g h), where Cd (about 0.6 for a sharp hole) accounts for the contraction of the jet.
Level equation
For a vessel of cross-section A(h) and inflow Qin, A(h) dh/dt = Qin − Cd a √(2 g h). For a cylinder with no inflow the solution is h(t) = (√h0 − Cd a √(2g) t / 2A)2, so the fall slows as the head drops. With inflow the level settles at heq = (Qin/Cda)2 / 2g. With a = 0 and constant inflow, a cylinder fills at the constant rate Q/A, as in the float clock of the 3D page.
Making the fall even
The level falls at a constant rate only if A(h) is proportional to √h, i.e. the radius follows r ∝ h1/4. Ancient Egyptian outflow clocks were made as sloping bowls for this reason, and the marks for equal hours are then equally spaced.
Honest simplifications
The model assumes a quasi-steady flow, a constant discharge coefficient, negligible viscosity and surface tension and a vessel wide compared with the orifice. Real clepsydrae also suffer temperature-dependent viscosity, clogging and evaporation, which are ignored here. The animation runs at the chosen multiple of real time.
What to try
- Start with the cylinder: the equal-time marks crowd toward the bottom because the fall slows down.
- Switch to the Uniform-fall bowl: the marks become equally spaced and the fall-rate graph is flat.
- Pick Inflow with a leak: the level rises and flattens at the equilibrium shown in the readout.
- Halve the orifice diameter: the outflow area drops fourfold and the run takes about four times as long.