# Trim calculation using Moment equilibrium

Discussion in 'Stability' started by Sam M, Jan 6, 2021.

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### Sam MJunior Member

Hello,

How can you calculate Trim for v=0 using moment equilibrium if I simplify the geometry of the ship as shown in the figure (Capture.PNG https://drive.google.com/file/d/1D4hYd1JNBVNRWl-CFxfBM1_lmUkraCJJ/view?usp=sharing)?

I did search on the internet. the methods which I have found for the trim calculation are
1. Trim and Stability for Ships, Boats, Yachts and Barges – Part 2 http://hawaii-marine.com/templates/Trim_Article.htm
3. finding the trim using draft at the perpendicular

the clue which I have is

Solving 3 equations to get 3 variables (LCF, Trim, T)
(1)LCF = function(hull geometry(T, Trim))
(2) Buoyancy force = function(ϱ, g, hull geometry(T, Trim))
(3)Moment Equilibrium (LCB, VCB; LCF, F_G" " , F_B" " )

Kind regards

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### DCockeySenior Member

Are you a student? If so what are you studying, and what level/year are you?

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### TANSLSenior Member

For very small trims, you can use the moment values to trim 1 cm (Mt1cm), which are obtained by calculating the hydrostatics of the boat.
In that case : trim (cm) = (FG * gf - FB * bf) / Mt1cm
But when the trim increases, let's say when it is greater than 1º (it depends on the shapes of the boat), things get complicated. As the hull acquires a certain trim, the CoG turns jointly with the hull while the position of the CoB varies depending on the submerged volume. This volume should be the same as in the starting position. Equilibrium will occur when the CoG and CoB are in the same vertical.
You should, through a trial and error process:
- trim the boat and position it at a draft that produces the same initial displacement.
- calculate the longitudinal position of the CoG and CoB in this position.
- increase or decrease the trim until the two points are aligned.

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### DCockeySenior Member

This is incorrect because there is an external force, F in the drawing. With an external vertical force when the vessel is in equilibrium CoG and CoB will not be vertical alignment unless the vertical force is position so that it passes through CoG or CoB. The correct moment equibrium is shown on the drawing.

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### TANSLSenior Member

@DCockey, there may be "an external vertical force" or not, that does not matter. The resultant of all the forces that represent the weight or weights existing in the ship, is located at a point that I have called "G". What's wrong with it?. Logically, if any additional weight has been added, the first thing to do is calculate the position of the new CoG (maybe I should have explained this previously). What's wrong with it? Your explanations will be received with all my scientific curiosity activated, and I hope they help me to get out of my mistake.

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### DCockeySenior Member

In the drawing that SamM asked about the force F is not located at the point G in the drawing. G in the drawing is distance LCG from the AP which is consistent. F in the drawing is distance LCF from the AP.

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### TANSLSenior Member

Tell me, please, what is wrong with my picture and what I say. I don't see anything wrong but I will be grateful to anyone who helps me correct my mistakes.

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### DCockeySenior Member

I already told you. SamM asked about solving for trim in the situation shown in a drawing. That drawing shows F located at a different location than G and B. Nothing in your drawing or response shows that F is at a different location the G or B.

A combined G and LCG could be calculated based on the G, LCG, F and LCF in the original drawing. However your initial response and drawing did not say anything about calculating a combined G and LCG.

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### TANSLSenior Member

You are right, which is why, in my later post (#5), I say that perhaps I should have explained that point G represents the CoG of the total ship. Once explained this (for the second time), I insist, tell me what is wrong in my posts. You can ask my post for a clarification but that does not imply that it is incorrect.
Apart from all these disquisitions, which lead nowhere, try to see if my explanations can help the OP to solve the problem that he has raised. If they are not correct, and may cause an unacceptable error, please tell me why and explain to the OP the method you propose. Once you know if the OP is a student and the grade or level he has, I am sure that you will give the most appropriate answer to his knowledge and professional qualification.

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### KayakmarathonJunior Member

My experience is limited to canoes and kayaks, where paddler weight far exceeds boat weight. For center of mass of the paddler, I use a single point mass located at the belly button. To find the center of buoyancy of the kayak, I find each cross sectional area, with sections spaced 1 cm apart to make calculations easier. Pick the bow as a reference point for the moment arm of each section. Multiply the distance from the bow by the cross section area to get the sample moment. Add all the moments. Divide by total (boat + paddler) displaced volume.

Displaced volume by simpson's rule is A0/2 + A1+A2+...An-2 +An-1 + An/2

Center of Buoyancy = (A0*L0 + A1*L1 ... + An-1*Ln-1 + An*LWL) / Displaced Volume

LWL = length waterline

For touring kayaks, locate the seat such that the paddler's belly button is 2 to 4 CM afore the center of buoyancy to account for dynamic lift. In olympic kayaks, the paddler's belly button is ~15 cm afore the center of buoyancy to minimize drag at racing speeds.

My designs have no overhang, so the length of the waterline and length overall are the same.

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