Body rotation

Hi,

what would be considered as the centre of rotation of a boat or any other floating body?

I'm asking mainly for heel situations, when assuming the mass and CoG stays the same, where the underwater shape changes and CoB moves which results in a change of trim.

When rotating around for example CoG it might happen that the displacement changes. Around which point the body has to be rotated so that the displacement stays the same?

 

In transverse stability calculations, we use the value of the metacentric height: I/V. In this formula, "V" is the displacement (volume under water plane) of the floating object, and "I" is the value of the second moment of the water plane area, calculated around a longitudinal axis of rotation that passes through the centroid of said flotation. It is therefore assumed that the rotation occurs around the aforementioned centroid.
No matter how the ship rotates, its displacement cannot change if its weight has not changed. This is a concept you should analyze and understand.

 

The displacement can't change, it is the weight of the boat.

 

Thank you both for your replies. I know that displacement cannot change. I also know BM = I/V.

I'm programming a Rhino plugin for large angle stability, when simply rotating the hull for, let's say 10°, about CoG the hull is not in equilibrium anymore, LCG and LCB missmatch and also the displacement does not match. This is true since it's a floating body and it's center of rotation is not the same as CoG (in opposite to a non-floating body which normally rotates about CoG).

An improvement was using the centroid of the waterplane area and heeling the hull numerically, but still, a new equilibrium has to be find as LCG and LCB do not macht. Finding the trim leads to a new situation where the displacement has to be found.

All in all, it works but I think it can be simplified by finding that centre of rotation (if it exists).

 

The results are correct. The difference in alignment is what creates the righting moment.

 

So large angle stability is calculated without taking care for trim? I know that I overthink things often, but I doubt that trim is not taken into account when calculating transverse stability.

 

The righting moment is a result of transverse missalignment, I think the longitudinal position of CoB and CoG still should match.

 

No, that's totally wrong. This "difference" creates the heeling moment.

 

To my knowledge it doesn't exist.
I did a website on this some years ago (bootsphysik.de ; bootsphysik.de/boot25m.php) and used numerical iteration to find the new equilibrium of a hull after changing CoG or mass of the boat. I was reading a lot of specific literature and found later in a university script (Hamburg-Harburg) a staterment saying there is no other mathematical method than interation.
[After a new version of php was installed to the server, this website is not properly working anymore, button English is located top right.]

 

Thanks, @Heimfried. So the euqilibrium cannot be found in a simple manner, I will continue my method and can only simplify my program by simplifying the code and make it more compact.

 

There is no other solution than to develop an iteration process, which is made even more complicated, if possible, by the fact that when the heel changes, the boat will automatically change its trim. In any case, by accepting that the boat moves as if it were rotating around axes passing through the center of the waterline, good results can be achieved.

 

Once you are dealing with large angles, you really need a 6 DoF model. Iteration usually isn't too onerous if you are plotting a RM and trim curve in one or two degree increments. You do need to be very careful about how you define roll, pitch, and RM though. The CB and CG have to be coplanar, and that plane has to be perpendicular to your RM vector. This can be set to the original transverse plane and held in earth reference, or it can be defined in body frame - the point being that RM and the CB location have to be treated consistently.

 

Ok, if you really want to answer this, it is time to brush off your vector algebra and matrix calculus....

Physically, most vessels are effectively a solid rigid body (i.e. we are not talking about pool noodles or a half inflated rubber dinghy)...and they always rotate and translate about their Center of Mass...end of subject! These motions (i.e. degrees of freedom) are surge (x fwd), sway (y stbd), heave (z down) roll (phi, clockwise x), pitch (theta, cw y), and yaw (psi, cw z) Note: that depending on where you were taught and whose text/paper you are reading these symbols may change...and I won't even get into substitution variables or dot notation.

However, the vessel, being a solid rigid body, and lighter than its displacement floating statically in a still fluid, it will generally seek one or more positions where it is at equilibrium given the external buoyancy forces. However any change from this position can result in a force, and this force will cause a corresponding change on the body. Therefore we end up with a 6x6 matrix where a change in position of the body about one degree of freedom will result in forces that want to restore the body to equilibrium. This gives us 6 direct couplings (x|x,y|y, etc) and 30 cross couplings which are inverse pairs (i.e. x|phi = 1/phi|x) where a change in one degree of freedom cause a corresponding change in another degree of freedom which may cause a change to a third degree of freedom and so on. We can also consider the response over time; i.e we can take the first derivative with respect to time and get a matrix for drag forces, the second gives an "added mass" matrix, etc.

Returning to the static, zeroth derivative of time, matrix and considering a static floating body of arbitrary shape the force matrix look like this:
x y z phi theta psi
x 0 0 0 0 0 0 Where:
y 0 0 0 0 0 c1 is a function of waterplane area |z
z c1 c2 c3 0 c2 is a function of transverse waterplane area |z
phi c4 c5 0 c3 is a function of longitudinal waterplane area |z
theta c6 0 c4 is a function of transverse waterplane inertia |z
psi 0 c5 is a function of least moment of waterplane inertia |z
c6 is a function of longitudinal waterplane inertia |z
So looking at the simplistic case of roll only where x, y, theta, and psi are constrained, the vessel rolls around its CG but the CG moves in z proportional to 1/c2.

Edit X-post with philsweet.
Edit Edit: Matrix not showing up correctly, added below.
Edit Edit Edit: on re-reading cleaned up some typos, no context change

 

Ohh BTW....
Regarding the previous two post...notice I hadn't even started discussing a non-planer water surface....