Hello ACR,

That the ship turns around a pivot point is a tautology, as such meaningless. In our scenario, the **static** pivot point distance is fixed by the ship's measurements, mass, mass moment of inertia, position and force of the thruster. When Vetus explains how to calculate the thrust needed, it says that the thrust must balance the wind force when the ship is bridled at the stern, leaning at a quay, whatever.

There is not the slightest link with the static pivot point, the center of instantaneous rotation, a virtual point resulting from inertia and external forces acting on the ship. Didn't you read too hastily, did you?

Let us take the ship as in your reference, scaled arbitrarily, with a 10 mt displacement M (dty 1). Her mass moment of inertia, i.e. Iyaw=M x (4² + 10.44²)/12 or 104161.33 if it was a blunt parallelepiped, probably too much, as it gives sqr(Iyaw/M)=3.23 m as radius of gyration, but let us take it for granted.

Computing the distance moved by the CoM as a lateral movement and as a rotational one, then eliminating time, angle of rotation and thruster force shows that the mass moment of inertia is the product of mass times distance CoM to thruster times CoM to static pivot point.

CoM to static pivot point is =10.416133/4.56, 2.28 m behind the CoM.

Now, a 60 kgf thruster...

The initial lateral acceleration is for instance 600N/10000kg or 6 cm/s² and the corresponding rotational acceleration is 4.56*600/104161.33 rad/s² or 90°/min.s (90 degree/min each second, 4 s to reach 360°/min, if at all posible, while disregarding the yaw water resistance)

The higher Iyaw for a given displacement, the more the pivot point lies to the stern. The lighter the ship, the closer to the CoM.

You could do the same for the 110 m ship, with comparable hypotheses, just take 60% of the moment of inertia as computed above, for a better approximation.

Also, as a side note, the pivot point is nowhere to be found in the ships parameters, it is the zero drift angle point when the ship turns, nothing more, nothing less, and is only fixed relative to the ships frame of reference when in steady turn.

The editor is not available to the public, of course, you certainly knew it and can guess also why it is so, but it is possible to find by googling many good textbooks about ship hydrodynamics, not forgetting a revision of classical mechanics and Newton laws.

Regards,

Luc