# Dimensioning of big floating disc

• Realpra
In summary, the post discusses the mathematical problem of calculating the pressure and weight on a floating disc, using equations such as p=ρ∙g∙h and F=562,50kg/m∙R. However, it is important to note that the water pressure at a given depth is not constant and other factors such as material properties and external forces must be considered for accurate disc construction.
Realpra
Is the following, ignoring that weight has been used as a force, correct? (excerpt from blog at http://floathaven.com/2011/07/the-solution-to-rising-waters-wednesday-blog/" )

Obviously when water is displaced by the disc it will provide pressure on the underside of the disc thus providing relatively stable “ground” to walk on despite the thin membrane. Also obviously the water will push at the sides of the disc, effectively trying to crush it or buckle it up.
What is not so obvious is the exact force one must dimension for when constructing the disc ring necessary for maintaining the shape of the disc.
This can be idealized as a mathematical problem. For the purpose of solving this we will need to know the water pressure at a given depth. This is given by:

p=ρ∙g∙h
p=F/A

Where:

p:Pressure
F:Force in N
A:Area in m^2
ρ:Rho or mass/volume ratio of water in kg/m^3 -equal to 1000kg/m^3
g:Gravity at 9,82 m/s^2
h:Variable height below surface of water

If we have a disc construction that is pushed 0,75 m into the water (that is 750 kg pr. m^2 in load capacity) and has 3,25 m clearing height giving a total height of the sides of 4m then “h” is 0,75m.
However the pressure on the disc will not be uniform since higher up portions will feel little to no pressure at all while lower portions feel maximum pressure.
We use an integral of the first function over h to find the pressure felt by the disc pr. meter of circumference:

Alright so that was part one, but what happens when the circumference increases and do you simply multiply the above by the circumference?
No; the force or perhaps more intuitively the “weight” the ring must be able to take can be understood by looking at two halves of a circle:
In this situation we look only at the weight along the x-axis. It should be intuitive that all force along this axis will have to be counteracted by the ring right at the cut sections to keep the disc from collapsing.
This situation and requirement is true for all points and axes on the circle.
We will now calculate the integral along the circumference to get the total weight the ring has to carry. The position on the x-axis:

x=cos⁡(θ)∙R

The change in position along the x-axis:

dx=sin⁡(θ)∙R

The factor 2 is used because we have 4 quarter arcs affected by pressure, but also 2 “cut ends” to put the weight on, so this expression should give us the weight carried in each infinitesimal ring segment:

Cos to pi/2 is 0 and cos to 0 is 1 so we simply get:

F=562,50kg/m∙R

Using Youngs modulus and other stress related physics this function can be used to dimension disc constructions.

Last edited by a moderator:

Hello, as a scientist, I can confirm that the information presented in this forum post is generally correct. The equations and concepts used to calculate the pressure and weight on the disc are accurate. However, there are a few points that need to be clarified.

Firstly, it is important to note that the water pressure at a given depth is not constant, as stated in the post. The pressure increases with depth due to the weight of the water above it. Therefore, the equation p=ρ∙g∙h is only applicable for a specific depth and cannot be used to calculate the pressure at different depths.

Secondly, the post mentions using Young's modulus to dimension disc constructions. While this is a valid approach, it is important to also consider other factors such as the material properties of the disc and the design of the disc to ensure it can withstand the calculated pressure and weight.

Finally, it is worth mentioning that this approach assumes a perfectly circular disc and does not take into account any external forces such as waves or wind, which could also affect the stability of the disc. Overall, this is a good starting point for calculating the necessary dimensions for a floating disc, but further analysis and considerations are needed for a more accurate and comprehensive design.

## 1. How do you determine the appropriate size for a big floating disc?

The size of a big floating disc, also known as a buoy or floating platform, is typically determined based on the intended use and the weight it needs to support. Factors such as the water conditions and expected wind forces also play a role in determining the size.

## 2. What materials are commonly used for big floating discs?

Common materials used for big floating discs include steel, concrete, and various types of plastic. The choice of material depends on factors such as cost, durability, and the environment in which the disc will be used.

## 3. How do you ensure stability for a big floating disc?

To ensure stability, the center of gravity of the big floating disc should be as low as possible. This can be achieved by distributing the weight evenly and using ballast to counteract any potential tipping due to wind or waves. Additionally, the shape and design of the disc can also contribute to its stability.

## 4. How do you prevent corrosion on a big floating disc?

Corrosion is a common concern for big floating discs, especially those made of steel or concrete. To prevent corrosion, these materials can be coated with protective layers such as paint or epoxy. Regular maintenance and repairs can also help prevent and control corrosion.

## 5. What is the maximum weight a big floating disc can support?

The maximum weight a big floating disc can support depends on its size, design, and materials used. Generally, these discs can support thousands of pounds, but it is important to consult with an engineer to determine the specific weight capacity for a particular disc.

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