# Stability Analysis - Trim problem

Discussion in 'Stability' started by Diego San, Jan 21, 2015.

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### Diego SanJunior Member

Hello everyone,

I would like to ask you about a problem I am having with a new product stability analysis.

With the help of Orca3D I have done an hidrostatic & stability analysis; every value seems right except the trim which is positive and the boat is trimmed by the bow just by 0.376° even with a 45 gal. ballast tank.

From my understanding, a proper trim angle should be between 0 and 5° negative (according to the values from Orca - trimmed by the stern), Am I right? Could you recommend some literature about this topic?

I have to explain more or less how will behave the boat once on water, so if you have any readings or so, I will thank you a lot.

Thank you

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### TANSLSenior Member

It is good that the trim, if any, is negative (sink Stern) but if positive, rarely, does not matter. Such a small trim as you get is perfect. Do not worry at all.(imo)

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### NavalSArtichokeSenior Member

Maybe it's minor, maybe not. We know so little about the OP's vessel, who can say with certainty what this trim causes?

For example, if the trim fully loaded (with ballast) is down by the bow, what will happen when the boat is almost out of fuel? Do the propellers or rudders start to come out of the water? In any event, the handling of the boat with bow trim may also be less than desirable.

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### TANSLSenior Member

Whatever the boat OP, it is best that the boat has trim zero. A trimming 0.376º is practically zero trim, be postive or negative.
Let's be practical, this trim is negligible, let's think that, maybe, the ship sails on a sea with waves.

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### DCockeySenior Member

What determines trim = 0 degrees? Is it the design position of the boat? What loading condition?

With the boat trimmed at 0 degrees where are the center of buoyancy and the center of gravity?

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### dougfrolichSenior Member

cheers

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### TANSLSenior Member

Trim = 0: the float is parallel to the baseline chosen. The designer chooses the baseline that it is more convenient. In actual loading conditions is impossible to have a trim = 0, it would be a fluke.

Cob and Cog are to be calculated the same way as any other load conditions. Both will be on a line perpendicular to the baseline.
In all this there are not "dynamic" forces. Just consider the weight and thrust, and the boat without heel.

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### Diego SanJunior Member

I have estimate the weight and COG assuming full load capacity (ballast, fuel and passengers).

To give you further info, I have placed the fuel tanks just below a COG estimated for the boat without those tanks (I have two 200gal-fuel-tanks) in order to avoid a way too different behavior between the full fuel condition and the low fuel condition. I have not calculate the trim for the latter.

As TANSL answers the base line was chosen, by me, parallel to the boat deck.

Last edited: Jan 23, 2015
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### BarrySenior Member

Diego
Perhaps I am missing something here. You are trying to determine the stability of the boat when it is underway? Is it a planning boat or displacement.
The article referenced by dougfrolich has quite a few basic errors in it.
quote " As we know from basic engineering mechanics, for any structure to be in static equilibrium, the net sum of all the forces on it must be zero" "and the moment is zero"

This is not correct. Considering a 2 dimensional static situation, though it is valid in 3D as well, the forces acting on a body do not have to be zero as any quantity of forces acting on a body can be replaced by one force and one moment.

If you are trying to get a trim angle when planning, and I do not see enough info in the opening comment. There are optimum angles of trim for various constant deadrise hulls to maximize lift and minimize drag.
( and warped as well)
The optimum angle for a 12 degree is about 4degrees and you will have to search the net for a table for other deadrises.

A constant deadrise hull, running at zero degrees of trim will not plane. You are trying to optimize lift with trim with respect to the drag caused by the horizontal component of the hydrodynamic forces on the hull. While drag caused by boundary layer exists, the hydrodynamic force drag opposing the direction of travel is greater for reasonable planing speeds than viscous drag

A bit more information on the shape of the hull, and design speed might help

Diego San likes this.
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### TANSLSenior Member

Barry, non-polemical, and I will not do it, I think you should revise your concepts about physics is concerned. That done, not before, you could continue with the planning theory of fast boats.

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### DCockeySenior Member

Equilibrium (acceleration = zero) requires:
Sum of forces = zero
Sum of moments = zero

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### TANSLSenior Member

Okay, DCockey, that is to have clear ideas and speak clearly. Although it is obvious that if no motion acceleration can not be <> 0.

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### BarrySenior Member

I think that you have not understood the concept
Take a one foot plank, horizontally oriented, put a 10 pound vertically down load on one end and a 10 pound vertically up load on the other. The sum of the forces are zero.
But the moment is 10 foot pounds. In order to stabilize this loading, you need to apply a moment of 10 foot pounds to the plank to stabilize it.

Tansl, you stated that a zero trim is best. Not for a planning boat.
There are many tables available on the net that show optimum trim angles for various constant deadrise hulls. The higher the deadrise, the higher the optimum trim angles
These are actual tank tested derived numbers.

For optimum in this case, I would use a definition of the least amount of thrust required in a horizontal direction to move the hull at a specific speed.

Tansl, for your comment, and perhaps I misunderstand your wording. "although it is obvious that if no motion acceleration can not be zero" With dynamic stability, say an airplane in flight at a constant speed, to me means motion, but there is not acceleration

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### TANSLSenior Member

Again, Barry, review your concepts. When an object moves is because there is a resultant force, nonzero, acting on the object. And whims of science, if there is a force, there is an acceleration <> 0.
As explained very well by DCockey, equilibrium requires also that the sum of moments is zero.
I guess you're right about everything related to planning angle. I did not understand that when the OP spoke of hydrostatic and stability analysis was referring to the boat moving. Generally when these studies are made, it is considered that the ship's forward speed is zero. I do not think I'm wrong on this, or at least I always do these studies in that way.

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### Diego SanJunior Member

TANSL, you are right, I just didn't make myself clear. Certainly I'm talking about static trim, I haven't done a dynamic analysis yet.

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