1
$\begingroup$

Here a ship that is tilted due an external force . the buoyancy and weight force will create an overturning couple that will rotate the ship in the clockwise direction. Now, the moment equation was defined as Fb*d : where Fb is the buoyant force and d is the distance between the centre of gravity and buoyancy force. I am wondering about which point is this ship rotating? Some refer it to the centre of floatation and some refer it to the centre of gravity . But what is the actual fixed or let's say the pivot point at which the ship is rotating around. How can we know that? It is really mysterious and there is a lack of detailed informations on websites and books. Also if there was a body that is totally immersed (consider it a ball). the same principle would happen. The buyont force will apply a turning moment at a distance from the centre of gravity. but this time the body is fully immersed. Now, will the force act about the centre of gravity also? Because the immersed body has no centre of floatation? I can't understand generally at what point does an immersed or floating body rotate from this buyont force moment.

enter preformatted text here

$\endgroup$
2
  • 1
    $\begingroup$ The instantaneous center of rotation is not really related to this statics problem in any way, so it doesn't matter. But for ship dynamics, the center is complicated to figure out and involves an assessment of the added mass, moment of inertia, applied forces, and dynamic history of the system. $\endgroup$
    – Phil Sweet
    Jul 14, 2022 at 22:03
  • 1
    $\begingroup$ of note, ships aren't built like the right version, most commonly they use flat bottoms so that the center of buoyancy will move to the til more than the center of gravity. Old sailing ships were built like the left version, so are modern submarines. $\endgroup$
    – Tiger Guy
    Jul 15, 2022 at 3:22

4 Answers 4

7
$\begingroup$

As it rotates the centre of rotation shifts, because although the centre of gravity (CG) is fixed relative to the ship itself, the centre of buoyancy (CB) isn't. The centre of buoyancy is a function of the hull shape / submerged volume at any given inclination and draught, which can be quite complex (except for the special case of a perfectly cylindrical hull, which would rotate about the axis of the cylinder).

The two hulls shown in the diagram have different centres of gravity and are therefore two separate cases (not the same ship/load). The ship on the right is in unstable equilibrium since the CG has moved above the metacentre - and the ship will continue to roll until capsize.

$\endgroup$
4
  • 1
    $\begingroup$ Metacentre? It's the metacentric height is what you should be referring to. en.wikipedia.org/wiki/Metacentric_height should explain it all better. $\endgroup$
    – Bib
    Jul 15, 2022 at 8:27
  • $\begingroup$ Metacentric height is (as your wikipedia page states) 'a measurement of the initial static stability of a floating body.' The metacentre is the point of intersection between the vertical line through the CB and the original vertical line through the original CB. The two are related. $\endgroup$
    – PM-14
    Jul 15, 2022 at 9:36
  • $\begingroup$ Amazing answer. So it actually rotate about the centre of gravity? In other words the turning moment is about the CG? $\endgroup$ Jul 15, 2022 at 11:14
  • 1
    $\begingroup$ Thanks. Not quite. When you have a couple (two opposing forces with an offset), the resultant 'turning moment' is the same irrespective of your selected reference point (try it yourself by sketching it out - test moment around a point in between the forces and for a point somewhere remote from both forces). For a non circular-section hull the centre of rotation will move as you rotate - the total transformation is a function of both rotation and translation. $\endgroup$
    – PM-14
    Jul 15, 2022 at 11:32
4
$\begingroup$

As is already mentioned by @PhilM, a constant center of rotation may not be defined for all but trivial ship shapes.

However, if we assume sufficiently small angles (and no waves), then the center of flotation is the point about which rotation takes place (what is considered a sufficiently small angle is of course dependent on the ship in question, but about $\pm10 \,deg$ may serve as a rule of thumb). The center of flotation is equivalent to the center of gravity of the waterplane area.

For longitudinal initial stability, the relationship between the center of gravity, center of buoyancy and center of flotation is touched upon by the SE.Eng. question Ship Longtitudinal Centre of Gravity Position. For transverse initial stability, the same principle applies, although symmetry most often gives a center of flotation in the center axis.

Equivalently, the transverse/longitudinal metacenter is the pivot point of the ship, acting similarly to the pivot point of a pendulum (but recall the small angle assumption).

The following image shows a ship with a small trimming angle. The image is taken from the question linked in the text above. Small trimming angle

To show that roll/trim must happen about the center of flotation, imagine a ship with and without and angle of inclination. Since our problem is static, we know that the buoyancy force must equal the mass force, meaning that the displaced volume must remain the same for the two cases. Depending on the ship shape, the axis of rotation must also change.

The two waterline areas below has the same length and breadth, but not the same center of flotation (center of gravity of the waterplane area). If they were to different ships, they would therefore not rotate about the same axis.

(to find their true axis of rotation, you could make cardboard cutouts of them and balance the cutout on your finger, effectively determining the centers of flotation yourself) enter image description here

$\endgroup$
1
$\begingroup$

There is a good analogy with balance boards. When the radius of the board bottom is greater than the height of your center of mass, all you have to do to keep the balance is to stay still (metacenter is above the center of mass). In the opposite case, you have to constantly shift your weight to keep the balance.

enter image description here

Buoyancy behaves similarly at least around the starting position. With larger rotations, the buoyancy center may go up or down, which would be similar to a balancing board with general surface on the bottom.

$\endgroup$
1
  • $\begingroup$ While bringing up analogies concerning the metacentric height and stability, I've always been fund of the following: A positive metacentric height may be regarded as balancing a ball in a convex parabola, it will roll to a stable position by itself when letting go. A negative metacentric height is then like balancing a ball on a concave parabola, it is possible, but it will most often roll away to one side (unstable). $\endgroup$
    – ToxicOwl
    Jul 15, 2022 at 21:43
-1
$\begingroup$

I think the illustration below can provide you with a better picture of this matter.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.