In Ancient Greece each day had 24 equal hours and the Sunrise was at the same hour on each day throughout the year. This is different than how we tell time today, because we vary the time of Sunrise based on the day of the year - the Sunrise in the Summer is at an earlier time than the Sunrise in the Winter. 

The Bronze Disk acted as the "hour hand" of the clock and was connected to the Rotating Axle which rotated once every 24 hours. On the Bronze Disk were markings for each day of the year based on the amount of sunlight and night on that day. As we know, the Sun rises and sets at different times throughout the year and days in the Summer are longer than the days in the Winter in the Northern Hemisphere. Along with the "hour hand" it is though that the Bronze Disk also showed a celestial map of the stars and planets.

The Wire Frame is stationary and its curved lines show the equal 24 hour hour markings throughout the day and night.

The curves on the Bronze Disc are created in three steps - marking the radius of the disc according to days of the year, creating the sunrise and sunset curve according to the time of daylight in Athens, Greece and finally, shifting the curves for each day according to their Zodiac Constellations.

Historians say that the disc marks the time of sunrise and sunset for each day of the year. The day of the year is linked to the radius on the disk. Each day of the year is placed on a separate radius distance on the disk, with the day with the shortest amount of daylight closest to the Center of the disk, and the day with the longest amount of daylight furthest from the disk. However only 180 radius points on the disk are needed because each day has the same exact sunrise and sunset times as one other day of the year. For example January 24th and November 18th have the same sunrise and sunset times. So one point on the Sunrise and Sunset curve represents 2 different days.

The disk will have two curves representing Sunrise on the left and Sunset on the right with the area between the curve at the top of the disk representing daylight and the area between the arcs at the bottom of the disc representing night.

To begin creating the curve Start with the days of the Spring Equinox and Autumnal Equinox which have equal daylight and night times, so the points of the curve for the sunrise curve and sunset curve of these points are placed at the equilibrium of the disk, or at 180 degrees for the sunrise curve and 0 degrees for the sunset curve. 

The shape of the curve is determined by the daylight changes throughout the year in Athens, Greece.

In order to show the Zodiac Constellations throughout the year, the curve for each Zodiac is shifted to its appropriate 30 degree Zodiac area on the disc. So, the curve that was once continuous broken is now broken into 12 yellow curves for sunrise and 12 black curves for sunset.

Principle #1 - Angular Rotation relates to Linear Movement. The Rotating Disc must rotate at the rate of 1 revolution per day. So the linear movement of the chain around the Rod, and the Float rising, must match the speed at which the Rod is rotating.

Principle #2 - The Continuity Principle means the rate of water leaving Container #1 (Q1) is equal to the rate of Water entering Container #2 (Q2).

Principle #3 - The equations of velocity = distance/time (V=D/t), and Q=A x V are used to calculate sizes in the design,

First, we must understand that the Rod connected to the Disk must rotate once every 24 hours. Therefore using the formula v = rω, where v is the Linear Velocity of the Rod, r is the radius of the Rod (0.1 meters), and ω is the Angular Velocity of the Rod. The Angular Velocity of the Rod over the time of 24 hours means ω = 2π/T where T is the period of rotation which is 86400 seconds, and therefore ω = 7.272 × 10^-5 radians/second.

So the Linear Velocity of the Rod is, v = rω = 7.272 × 10^-6 meters/second

The Linear Velocity of the Rod is also the Linear Velocity of the Float which is attached to the Chain.

We can also find the Height that that Float will rise to based on the formula Velocity = Distance/Time, Distance = Velocity x Time. Over the course of the 24 Hours, the Float will rise a Distance = 7.272 × 10^-6 meters/second x 86400 seconds = 0.628 meters. We can use this as the Height of Container #2.

Now we can find the Flow Rates (Q) for the water flow out of Container #1 (Q1) and into Container #2 (Q2). The flow rates will be equal based on the Continuity Equation. The Flow Rate, Q = Area x Velocity of the Fluid.

Let's find the Flow Rate into Container #2 (Q2). We found the Velocity of the Float above and now we can determine the Area of the circular Container #2 (A2). And that can be found using the equation A = π x R^2, where R is the Radius of the circular Container #2. Let's pick a Radius of 0.04 meters for Container #2. Then the Area of Container #2 = 0.005027 meters squared.

Therefore the flow rate Q2 = Area x Velocity of the Fluid = (0.005027 meters squared) x (7.272 × 10^-6 meters/second) = 1.368 × 10^-8 cubic meters/second. We also know that Q2 = Q1 because of the Continuity Equation. Since Q2 = Q1, we could also say that Q2 = Area of Container #1 (A1) x Velocity of the Water flowing out of Container #1 (V1). Since the water is flowing at the same speed from Container #1 into Container #2, the Velocities will be the same as well.

So, Q1 = Q2 = A1 x V1 = A2 x V2. With V1 = V2, then A1 = A2. Therefore the Area of the hole of Container #1 is 0.005027 meters squared.

Finally, we can determine the Height of Container #1. Remember, the Radius of Container #1 is not important, but the Height of Container #1 is important and this must be kept constant in order to ensure constant flow into Container #2. We can use Bernoulli's equation to find the Height of Container #1 (H):

Ptop + ½ ρ vtop^2+ρ g H1 = Pbottom + ½ ρ vbottom^2+ρ g H0

Ptop of the container = Pbottom of the container, so these terms cancel out

vtop is the velocity of the water at the top of Container #1 and it is constant so, vtop = 0

H0 is the height of the water at the bottom of Container #1, so we will make it H0 = 0 m

We are left with 

ρ g H1 = ½ ρ vbottom^2

And we can find H1 = (vbottom^2) / (2 x gravity force) = (7.272 × 10^-6 meters/second) ^2 / (2 x 9.81 m/s^2) = 0.0000000265 m.

Therefore, the Heigh of Container #1 is 0.0265 millimeters.

In order for the Water Clock to work correctly the size of Containers #1 and #2 must be appropriate, and the size of the hole at the bottom of Container #1 must also be correct. There are no specific correct values for these sizes, they are just needed to sized together correctly.

An important principal of physics to understand is that the speed of the Counterweight falling away from the Rotating Axle is equal to the speed of the Float rising. This speed is equal to the circumference of the Rotating Axle divided by the time of 24 hours.

Next, the change in the volume of Container #2 is equal to the change in height of Container #2 multiplied by the change in height of Container #2. This is the rate of water flowing out of Container #1 into Container #2.

Finally, the size of the hole in the bottom of Container #1 must be found......

Once these values are found then the Water Clock should work accurately. And remember - Container #1 must maintain a constant full level and this will probably mean there is spillage of water from the top of Container #1.