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Some slides associated to the book Handbook of Marine Craft Hydrodynamics and Motion Control (2021 edition) are provided by the author of the book here.

At page 12 of the slides for Chapter 4, the moment due to the restoration forces (gravity, buoyancy) of a floating vessel is computed, expressed in the body frame of the vessel. According to page 14, such moment is computed in the Center of Flotation; however, the lever arm computed at page 12 is the following:

$$r_{\text{GM}}^b = \begin{bmatrix} -\text{GM}_L \sin\theta \\ \text{GM}_T \sin\phi \\ 0 \end{bmatrix}$$

where $GM_L$ is the longitudinal metacentric height, $GM_T$ is the transverse metacentric height, $\theta$ is the pitch angle of the vessel, $\phi$ is its roll angle, and the superscript $b$ indicates that the arm is represented in the vessel body frame.

The geometrical meaning of the second component of the arm is shown in a figure on page 10, as follows:

enter image description here

where $CB$ stands for Center of Buoyancy, and $CG$ is the Center of Gravity.

Then the moment is computed as follows: $$\mathbf{m}_r^b=\mathbf{r}_{GM}^b\times \mathbf{f}_b^b$$

where $\it{f_b^b}$ is the buoyancy force expressed in body frame.

By looking at this, it seems to me that the moment was computed about the Center of Gravity, not about the Center of Flotation, as claimed later.

Is this right, or is there any implicit assumption here that makes the assertion on the slides correct (for example the vessel is boxed-shaped)?

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  • $\begingroup$ the moment would be the same about either, it's just the distance between the verticals of the 2. And moments apply equally anywhere on a body. This righting moment only works on flat-bottom ships. $\endgroup$
    – Tiger Guy
    Apr 2 at 21:50
  • $\begingroup$ @TigerGuy "This righting moment only works on flat-bottom ships." eh?? For significant heel angles, say greater than 2 degrees, The waterplane begins to change significantly, and you need to use to Z = Mt(phi) curve to evaluate RM. RM is only affected by actual waterplane, displacement, and CG. Bottom shape matters not at all. $\endgroup$
    – Phil Sweet
    Apr 4 at 15:35
  • $\begingroup$ @PhilSweet, how about round? $\endgroup$
    – Tiger Guy
    Apr 5 at 4:36
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    $\begingroup$ For a round bottom, the waterplane never changes, so the RM doesn't change when the boat heels. Stable boats will remain stable. Unstable boats will stay unstable. $\endgroup$
    – Phil Sweet
    Apr 5 at 9:35
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The gravitational force and the buoyancy force create what is known as a moment couple (or sometimes just couple).

Essentially they are two equal (because buoyancy and gravity should be equal in a boat), parallel and opposite facing forces, when have a zero translational effect and produce only rotation. See the following example

enter image description here

For the above example it doesn't matter which point in space you calculate the moment of the two forces. will always get $F\cdot d$.

So as pointed by TigerGuy, it doesn't matter which point you take as reference for the moment.

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  • $\begingroup$ The reason for this is that we are taking a case where all forces are statically balanced. In a dynamic system, this doesn't work. It also gets a bit weird when pitch and roll are coupled. In body reference system, there is no coupling, but it is more common to use the navigation or seakeeping reference system, and those have coupling terms. $\endgroup$
    – Phil Sweet
    Apr 4 at 15:48

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