# Weight distribution in design of boats

Discussion in 'Stability' started by JohnMarc, Oct 29, 2020.

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### TANSLSenior Member

Some consider that aluminum makes the boats lighter. A boat of 7 m in length, monohull, RPF, usually weighs around 2000 ks, empty. So I think the weight you have calculated is low. I may be wrong, of course.
The boat must achieve a balance between the weights and the buoyancy it experiences. Therefore, in addition to the distribution of the weights, you need to know the position of the Buoyancy Center

Last edited: Oct 29, 2020
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### JohnMarc<--- My ultimate goal

Right okay, yes I have many items still to include in my calculations. I am trying to find the meanings of the various column headers from the spreadsheet, such as Xg Mx Yg My Zg Mz? I understand the content of the columns, just not what the headings abbreviations mean. I will generate as similar a spreadsheet as the one you posted, would that information be enough to calculate the buoyancy centre ?

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### TANSLSenior Member

No, for the CoB you need the body lines plan or a 3D model of the hull(s)
Xg means longitudinal distance from a reference point.
Mx (moment) is the weight times its X

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### BlueBellAhhhhh...

That looks correct if your weight estimates are right.

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No, not silly at all...if you don't know!
You seem to be getting many replies from people that do not understand the basics of stability nor how to quickly estimate it based upon limited data. One can only conclude they've never done such simple calculations before. Which begs the question, why are the commenting on an issue they do not understand??

Ok..so based upon the data you have provided, we can very quickly work out the waterplane inertia, to establish the BM, which is important for stability calc's.
Assuming a length of 6m the beam at waterline of say 0.75m and a shape factor, owing to ends being nor fully squared of say 0.8, we can calculate the waterplane area A as 0.75 x 6 x 0.80 = 3.6m2, as shown below.
Then knowing the inertia I, is a.h^2, we can calculate the waterplane inertia from these simple numbers as shown below, as 4.90m^4:

Then if we assume the full load displacement (until proven otherwise) is roughly 2 tonne or roughly 2.0m^3 volume, the BM is simply I/V = 4.9/2 = 2.45m
With your assumed/give draft of 0.25m... we can say the KB is roughly 0.12m, this gives the KM as 2.45 + 0.12 = 2.57m

So..what is the KG?.. well, usually on most multihulls the KG is about at deck level. But this being a pontoon..it would suggest that the KG would be slightly above this..so let's say 0.25m above.
Thus KG = hull depth + 0.25 = 0.7 + 0.25 = 0.95m

Therefore the GM = 2.57 - 0.95 = 1.62, call it 1.60m. (sorry just noticed I forget to write this down in the hand calc).

If we again assume a simple wall sided hull shape, as it is a pontoon...we can use the wall sided formula to calculate (estimate) the GZ:

GZ = sin(theta)[ GM + 1/2BM.tan^2(theta)]

So we know GM = 1.60 and we know BM = 2.45, we can calculate the GZ at say 2.50 and 5.0 and say 10.0 degrees...= 0.070, 0.140 and 0.284 respectively.

Plot this on a simple graph...as shown.

If you have 6 people of say 75kg..standing right on the deck edge this is known as a heeling moment, and calculated at (6 x 0.75)t x 1.2m =0.54t.m.
Give the displacement fo say 2 tonne, the lever is simply 0.54/2 = 0.27m

So we can now draw this lever arm on the GZ curve...and where the two liens intersect (cross) is approximately the angle of heel.
As shown this si roughly 7-8 degrees.

That's it.

Last edited: Oct 30, 2020
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### TANSLSenior Member

The first thing to do is define the Cog and CoB positions and make them coincide because otherwise all those calculations, which are almost correct (but not totally correct), are useless, for the moment, at all. The procedure is correct, although it is necessary to work with more precision before giving definitive conclusions, but the assumptions that have been made could be totally incorrect. They are based on a pontoon that floats with zero trim and no initial heel, which. at this time, it is impossible to assure. You have to start there and then we will talk about GZs, water planea area inertia and all that stuff.

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### JohnMarc<--- My ultimate goal

I have done an entire 3D drawing for the boat. That is where all my weights come from.

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### JohnMarc<--- My ultimate goal

Thank you Thank you thank you Ad Hoc..... someone who actually "heard" my question, I realise I have lots of work to do regarding finalisation of the weights etc. I am equally aware of the failings in my design, but while finalising the design I thought it would be good to know the various requirements (such as weight distribution) while completing the design so that I may ensure I have some insight into the basics of the requirements rather than redesigning the entire thing when my weaknesses become apparent.
I will take some time out and study what you have sent, once agin many thanks.

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### Mr EfficiencySenior Member

You should post the lines of the boat, say what the installed power will be, the total weight of the boat, I think the risk with uneven loading may develop at speed, and depending on water conditions, planing catamarans are not renowned for being dynamically stable.

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### JohnMarc<--- My ultimate goal

When you say the lines of the boat, may I assume you are referring to a line drawing of the craft showing the dimensions. Will prepare and post shortly

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### HeimfriedSenior Member

A lines plan reveals the shape of the demi hulls, not only the dimensions of the boat.
Just an example here: Hull Design https://hydrobikeuom.wordpress.com/hull-design/

The following link doesn't show a lines plan, but an other way to give an impression of a hull shape by presenting a lot of frames: Berechnung der Schwimmlage von Booten http://www.bootsphysik.de/boot21.php
(button English top right)

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### JohnMarc<--- My ultimate goal

Brilliant, thanks for that Heimfried, much appreciated, will prepare.......

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It was loud clear and very obvious to anyone reading it!

Exactly, it is obvious it is not the finished product as yet...and requires a few more iterations, but it is a very simple and basic question to ask yourself at this prelim stage....and that is called... Design!

Any more Qs.... fire away.

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### JohnMarc<--- My ultimate goal

May I (once again ) ask a silly question, In the lines plan I see the centre dimension of the front section piece has 320 as the centre line..... I am trying to find a reference for that dimension. Can it not be zero at 100 mm increments

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### HeimfriedSenior Member

As it is a katamaran hull it consists of 2 demi hulls. The center of a demi hull is horizontally 320 mmapart from the center of the entire hull. It's a model only, therefore quite small.

(I'm not sure if I understand your question right, if not, explain it a bit more.)

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