# Clepsydra or water clock

• Jun 11th 2012, 01:10 PM
jphillips
Clepsydra or water clock
I came across this last night and thought to seek some input.

A 12 hour water clock has a height of 4 feet and a radius (at the top) or 1 foot. The shape is a parabola rotated about the y-axis. What should be the function of the curve and the radius of the hole so the water level will fall at a constant rate?

I'm interested in the original problem, but more in the relationship between the water level and the rate of water exiting the clock.

Here's my dilemma: As the water level falls in the container, the pressure on the water exiting the hole will lessen. So, I assume there would be an inverse relationship between the water level and the rate of water leaving the parabaloid. I then assume that there will be an infinite number of disks of water of height dh. the disks would be A=pi*r^2. But what would govern rate of water exiting the tank?

I did do some research on water clocks and found a LOT of really interesting devices many ancient countries used just by manipulating water in this way!
• Jun 12th 2012, 05:59 AM
mfb
Re: Clepsydra or water clock
With some reasonable approximations, the kinetic energy of water leaving the the container is proportional to the pressure difference inside<->outside, which is proportional to the water level above the hole. The velocity of the water is then proportional to the square root of this value: v2 = 2 h g with the height difference h and the gravitational acceleration g=9.81m/s2.
The flow through the hole is then given by A*v with the area A.
• Jun 12th 2012, 06:15 AM
skeeter
Re: Clepsydra or water clock