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/" [Broken])(adsbygoogle = window.adsbygoogle || []).push({});

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.

[PLAIN]http://floathaven.com/forum/download/file.php?mode=view&id=14&sid=e975dc489021e1e5702a17e44225d7ef [Broken]

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:

[PLAIN]http://floathaven.com/forum/download/file.php?mode=view&id=11&sid=e975dc489021e1e5702a17e44225d7ef [Broken]

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:

[PLAIN]http://floathaven.com/forum/download/file.php?mode=view&id=13&sid=e975dc489021e1e5702a17e44225d7ef [Broken]

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:

[PLAIN]http://floathaven.com/forum/download/file.php?mode=view&id=12&sid=e975dc489021e1e5702a17e44225d7ef [Broken]

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.1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Dimensioning of big floating disc

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