How to find the depth of immersion of this boat model?

So I am trying to find the depth of immersion of this particular boat model, which has curvilinear edges (subparabolic edges) on each side. The figure of the pontoon is attached with this post.

So now when I place this pontoon in water, the height of the water that rises, called the depth of immersion is to be found. I already know how to calculate the displacement volume, but I have no idea how to find the depth of immersion. Below are the steps taken to find the displacement volume

1. Volume of body

I assumed the whole body to be a rectangle and calculated the area of the rectangle first: B*h

And then, since the curves are subparabolic, it is defined by the equation:

y = (h/b^2)*x^2

Integrating the above equation, gives the area under the curve to be:

bh/3

So subtracting : 2*(bh/3) from (B*h) will give the cross sectional area which

C.S.A = (B*h)-(2*(bh/3))

Multiplying this C.S.A with the length of the pontoon (say L) will give the volume

V = (C.S.A)*L = [(B*h)-(2*(bh/3))] * L

So now that volume of the body is known, the following steps are taken to find the displacement volume

2. Density of body

Density = mass of body / Volume

Note: The mass was found by weighing the pontoon on a weighing machine

3. Specific weight of body

Specific weight of body = density of body * g

Note: g = 9.81 m/s^2

4. Weight of body

Weight of body = Specific weight of body * Vbody

5. Weight of water displaced

Weight of displaced water = Specific weight of water * Displaced volume

Note: Specific weight of water = 9810 N/m^3

6. Archimedes principle

According to Archimedes Principle the weight of the body acting downwards is equal to the weight of displaced acting upwards, so

Wbody = 9810 * Vdisp

7. Displaced Volume

And so the displaced volume of water is given by

Vdisp = Wbody/9180

Now another equation for the displacement volume is:

Vdisp = [(B'*h')-(2*(b'h'/3))] * L

h' is the depth of immersion

and B' is the top width at this displaced volume

The problem is that there are two unknowns in the above equation and I just can't figure out how to find the depth of immersion here. If anyone has any idea please please let me know.

 

Welcome to the forum.
What are you really looking for?
1. the draft in which the weight of water displaced by the hull is equal to the weight of the vessel, or
2. the maximum depth to which the device can be submerged without collapsing structure.

 

Yes I am looking for the draft, which is the vertical distance between the waterline and the keel, so it would be the first one that you mentioned

 

Have you ever heard of the calculation of hydrostatic values for a hull?. That's what you need to calculate: the volume displaced for each draft. When the weight of the displaced volume is equal to the weight of the boat, you've solved your problem.

 

No I have in fact not heard of it. Could you please help me understand how to calculate that? Or let me know of a good source where I can learn about it?

 

This is the basics to start making naval architecture calculations. I think you should learn so many things that it is impossible to teach them from this forum. Contact a naval architect, or na student, close to you, who can give you some lessons.
Good luck.

 

I understand that you cannot teach me through the forum and I am not asking you to. But can you at least advice me on a good book or website?

 

I am confused, initially you said "hydrostatic values helmet" ?

 

Sorry, there was a mistake due to my horrible English. The correct word is "hull", that's why I corrected it.

 

I would first assess your knowledge of physics, mathematics, kinematics, among other disciplines, to know where to start. It's not that I do not want to teach you is that I am not able to.
There is a very good book, which to me seems very good, "Principles of Naval Architecture" from SNAME. Start there if you want.