# Trim calculation problem

Discussion in 'Stability' started by Ltan, Aug 26, 2020.

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1. Joined: Aug 2020
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### LtanNew Member

Hi, I'm studying the stability of a boat that I've designed.

But I have problems with the stability, the boat has a lenght of 99m, the center of gravity of the weights and tanks distribution is in +50,561 m from the stern, and the center of buoyancy at + 50,622 m from stern. Is it normal that my draft at FP is 7,87m and my draft at AP is at 4,903m?
My KB is at 3,394m and KG at 5,488m for this loadcase.

I'm using Maxsurf Stability. If anyone can help me and need the file, i can send it by email, I dont post it here because if part of my thesis.

Thank you very much, I appreciate your help.

Last edited: Aug 26, 2020
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### TANSLSenior Member

Welcome to the forum.
There is something very strange there because, with the CoB position ahead of the CoG, the draft in FP should be less than in AP.

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### LtanNew Member

Thank you. Yes, I dont understand why, and the program is giving me the right displacement for that loadcase, so I dont think that there is a problem with the hull.

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### TANSLSenior Member

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### LtanNew Member

Yes, it is.

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### TANSLSenior Member

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Hello Ltan,

If your boat is in this condition:

This emans it is not in equilibrium. Since the LCB and the LCG must be coincident for the vessel to float freely in equilibrium.

If the LCG and LCB are not coincident (in-line) with each other, this means it is not in equilibrium which means you will end up, most likely with something like:

A condition of trim that does not seem to make sense.
This is a "snap-shoot" of the condition, it is a "transitional" condition.

Thus, ask yourself this question:... when the LCB moves aft, to be aligned with the LCG...what will happen?

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### DCockeySenior Member

I am not a MaxSurf user so I don't know what it reports.

Are the reported LCG and LCB above measured parallel to design waterplane, or are the reported LCG and LCB measured relative to the equilibrium waterplane after the vessel trims?

Is the reported LCB the location of the center of the volume submerged before the vessel trims to equilibrium, or is the reported LCB the location of the center of the volume submerged after the vessel trims to equilibrium?

If the LCG and LCB measured parallel to design waterplane and the reported LCB is the location of the center of the volume submerged after the vessel trims to equilibrium and the vessel is trimmed bow down by 1.67 degrees relative to the design waterplane then the reported values make sense.

But the numerical values of LCG and LCB will only be the same in equilibrium if measured parallel to the equilibrium waterplane. If the measurement is made using a coordinate system inclined to the equilibrium waterplane then the reported distances will be different when the LCG is directly above the LCB. Added: In this situation the difference in numerical values in the coordinated system inclined relative to the equilibrium water plane for LCG relative to the equilibrium LCB will be sine of trim angle * (KCG - KCB) Important - Note that the LCB is the center of the submerged volume in equilibrium, not the submerged volume as designed.

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For what purpose and why would you do that?

Datums are precisely that, datums. And as such, the 'fixed' datums for coordinate defencing on vessels are:
Baseline
Midships
AP - stern/transom
FP - Bow/stem

If, as you're suggesting, then data would need to reference the fixed datum and any reorientation angle.... to that datum.
Which of course is not standard practice at all..since every time weight is added, one needs to constantly redefine the angles for each trim and have a set of data for each "possible angle" about the new equilibrium....ergo, doesn't make sense!

Thus data is always from the horizontal baseline, the trim is about the baseline, one doesn't trim the baseline to fit the horizontal waterline.

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### DCockeySenior Member

I agree datums and a coordinate system fixed to the vessel and which are independent of heel, trim and sinkage makes sense.

A concequence of using such datums and coordinate system is when the vessel is in static equilibrium at a non-zero trim the LCB will have a different distance from the fore/aft datum then the LCG (unless the LCG and LCB are at the same height which is not usually the case). The difference in the distance from the datum is due to the inclination of the datum and coordinate system. The difference is given by the formula above, the produce of the sine of the trim angle and the height difference between the CG and CB. Since the trim angle and sine of the trim angle are small, the difference in LCG and LCG will be small compared to the height difference betweent he CG and CB, and very small crelative to the length of the vessel. For the 99m long vessel described in the first post the difference in LCB and LCG is 0.061m or 0.06% of the length of the vessel. I assume that these longitudinal differences have been traditionally neglected in hydrostatic calculations of equilibrium trim angle because the effect of including them would probably be smaller than other uncertainties or assumptions in the calculations.

Note that exactly the same happens with heel, except transverse rather than longitudinal. Consider a vessel with the usually symmetric side to side hull with the CG higher than the CB as is usual for surface vessels. If the vessel is in equilibrium at zero heel the CG will be on the midships plane and the CB will also be on the midships plane. Move the CG to one side away from the midships plane and allow the vessel to come to the new equilibrium heel angle. The new CB will now be a greater distance from the midships plane than the CG (CG higher than CB).

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### DCockeySenior Member

Is it normal that my draft at FP is 7,87m and my draft at AP is at 4,903m? It may be completely normal depending on the shape of hull and distribution of the volume in the hull. How close is the submerged portion of the hull to being symmetric fore/aft?

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Yes, and no.

When doing an inclining experiment, for example, these values need to be correct. Thus one uses the simplified version of your sine*angle as:
LCG = LCB ± (KG-KB) (trim/LBP)

Since the hydrostatic values, are with the preset datums of baseline/midships etc. All horizontal and vertical planes of reference.

Yet when the Master looks over the side s/he doesn't think...hmm.. what angle are we floating at?... They will simply read the draft marks aft and fwd..and then refer to the set of hydrostatics in the stability book...all using the same preset datums. Thus the hydrostatics will have output tables showing the drafts ranging from zero trim to 0.25m - 2.0m... depending upon the size of vessel, as greater or lesser than these values in both bow and stern trim values. The actual absolute correct value of LCB/LCG in this context, is negligible.

An angle of lol.

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### DCockeySenior Member

Not an angle of lol because the CG was moved off the centerplane and the vessel heeled as a result.

An angle of lol occurs if vessel has a negative metacenter at zero heel and comes to equilibrium at a non-zero heel angle with the CG on the midships plane. Angle of loll - Wikipedia https://en.wikipedia.org/wiki/Angle_of_loll

In the situation I described the vessel is initially in equilibrium at zero heel with the CG on the midships plane. The CG is then moved off the centerplane (possibly by moving weights on the deck) and the vessel comes to an new equilibrium heel angle.

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Quite right..well spotted .... I wasn't thinking (& reading too quickly) . Too many jobs at once....I should focus on my work not the replies to postings!!

Yes an angle of Loll is with an initial negative GM.

In which case, if wasn't in too much of a rush to reply, you're talking about list.

Heel = temporary transverse angle of movement from the vertical.
List = permanent....

Last edited: Aug 28, 2020

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### DCockeySenior Member

Does a vessel heel or list during an inclining test? PNA 1967 says it heels.

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