How to Calculate Buoyancy for RC Boat Displacement and Hull Volume?

Hello, I would like to know how to calculate the buoyancy of a boat


Thanks in advance

www.google.pl -> Archimedes' Principle -> Volume of submerged part * specific gravity of water (or other liquid in which the body is immersed).

You can simply assume that one cubic decimeter = 1 liter = 1 kg.

I don't really understand all this
I have the dimensions of the container 70x30x15 in centimeters.

I'm sorry, but this is the 4th grade of elementary school. Solid volume * specific gravity.
What don't you understand here? The boat is a geometric solid. Its total volume of the hull can be counted or measured experimentally (pour water inside as much as it will enter and pour this water into a cylinder with a scale).

You are interested in the part of the solid that will be submerged to a certain depth. How deep should the hull be submerged?

What sticks out above the water doesn't matter.

You calculate the volume, convert to cubic decimeters and you get the result in kg because 1 dcm3 of water = 1 kg.
You subtract the weight of the boat from this and you get the result of how much equipment and cargo you can add to maintain the assumed draft.

There's no magic here. A bit of multiplication and division.

In your example, if the container is a rectangular prism then 70x30x15 = 31500cm3 1dcm3 = 1000cm3 or 31500/1000 = 31.5dcm3 * 1kg (specific gravity of water 1dcm3) = 31.5kg buoyancy.

up to 31kg load - floats
32kg and more load - sinking.

For example, if the container is to be half submerged, the maximum buoyancy will be 15.75 kg (half of the total buoyancy).

With more complex hull shapes and cross-sections, volume calculations will be more troublesome. But it's doable.

You know, the 4th grade of elementary school was a long time ago and then I didn't have to deal with it at all, so I may not understand something.

But now I know and thanks in advance for your help :)

After calculating, I got 10 kg of buoyancy minus 5 kg in total, so I will be able to take out 5 kg of bait the amount is fine for me

Fw=ro*g*V
Fw - buoyancy force
r - density of water!
g - gravitational acceleration g = approximately 10 m/s2
V - The volume of the boat!

Gravitational acceleration was not needed to calculate the buoyancy of this boat.

abythadon wrote:

I will be able to carry 5 kg of bait

Minus the weight of all tank accessories, engine battery, etc.

Madrik wrote:

pour water inside as much as it comes in and pour this water into the measuring cup with a scale).

It's a bit wrong. The outer volume of a solid should be measured, measuring the center will falsify the result even with the thickness of the hull walls
The measuring principle with displaced water is the same
It is best to use a vessel in which it is easy to count how much water is in it
You measure the height of the water column in the vessel, put the boat inside,
pressing it down by hand or ballasting it to the line of maximum draft and again
you measure the height of the water column
From the difference in height measurements, you can calculate the amount of displaced water 1L=1kg