Capsized ship

For each position of the ship floating on the water there is a well defined potential energy of the ship and the displaced water. This energy depends on the orientation of the ship, $E=E(\vec {u})$ , where $\vec{u}$ is a unit vector attached to the ship. The ship is stable if this potential energy increases when the orientation of the ship is changed by a small amount. (This corresponds to a small rotation of $\vec{u}$ around one of the many horizontal axes going through it.) We will show that there is a general symmetry in the problem, and the potential energy of the inverted ship has a very simple and straightforward relationship to the ship in the normal position. In particular, we will show that for $\rho=0.5$ $E(\vec{u})=E(-\vec{u})$ and therefore if $E(\vec{u})$ has a local minimum around $\vec{u}=\vec{u_0}$ , so does $E(-\vec{u})$ around $\vec{u}=-\vec{u_0}$ .

We pick and arbitrary $\vec{u}$ in the vicinity of $\vec{u_0}$ , the equilibrium position. For the calculation of the potential energy we need to add up all of the infinitesimal contributions of $dE= zg\rho dV = =z g \rho A(z) dz$ . Here $g$ is the acceleration of gravity and $A(z)$ is the horizontal cross section area of the ship at elevation $z$ . We will select the water surface for $z=0$ . The potential energy of the ship is $g\rho [\int_{z_{min}}^0 z A(z) dz+\int_0^{z_{max}} z A(z) dz]$ . The potential energy of the displaced water is $-g\rho_w \int_{z_{min}}^0 z A(z) dz$ . The total energy is

E= $g(\rho - \rho_w) \int_{z_{min}}^0 z A(z) dz+g \rho \int_0^{z_{max}} z A(z) dz \text{ . . . . . } (1)$ .

We also need to make sure that we use the correct reference point for the $z$ scale. The position of the waterline of the ship is determined by Archimedes' law:

$g(\rho - \rho_w) \int_{z_{min}}^0 A(z) dz+g \rho \int_0^{z_{max}} A(z) dz=0 \text{ . . . . . } (2)$ .

Notice that the function $A(z)$ is derived from the information we know about the shape of the ship. This function will depend on the orientation of the ship (i.e. $\vec{u}$ ) and it can be rather complicated. However, when we invert the ship, making $\vec{u}'=-\vec{u}$ the $A(z)$ function will be the same, except for a $z'=-z$ replacement. The corresponding equations for the ship turned upside down are:

E'= $g(\rho - \rho_w) \int_{-z_{max}}^0 z A(-z) dz+g \rho \int_0^{-z_{min}} z A(-z) dz$ .

$g(\rho - \rho_w) \int_{-z_{max}}^0 A(-z) dz+g \rho \int_0^{-z_{min}} A(-z) dz$ =0.

Introducing $z'=-z$ and $\rho'=\rho_w-\rho$ , we get

E'= $g\rho' \int_0^{z_{max}} z' A(z') dz'+g (\rho'-\rho_w) \int_{z_{min}}^0 z' A(z') dz' \text{ . . . . . } (3)$ .

$g\rho' \int_0^{z_{max}} A(z') dz'+g (\rho'-\rho_w) \int_{z_{min}}^0 A(z') dz'=0 \text{ . . . . . } (4)$ .

Comparing these two expressions to Eqs. (1) and (2), we recognize that the inversion of the ship and changing the density from $\rho$ to $\rho'=\rho_w-\rho$ yields the same waterline [Eqs. (2) and (4)] and the same potential energy [Eqs. (1) and (3)]. With $\rho = 0.5$ we have $\rho'=\rho_w-\rho=0.5$ as well. Accordingly, in our case the inverted ship is as stable as the upright ship.