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!

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.

http://www.stewartcalculus.com/data/..._0706a_stu.pdf