SIMPLE HYDROSTATICS

In this section, we consider the hydrostatic properties of two simple bodies, a cube and a tube. These bodies are simple enough so that exact calculations can be made, and MOSES uses two different algorithms for the two cases. Thus, they provide an excellent check of the hydrostatics for any body.

The cube was considered in three positions: trimmed, heeled, and both trimmed and heeled. The side of the cube has a length of $l=$ 50 ft, and a draft $T$, which is measured perpendicular to the waterplane from the deepest submerged point of the body. The equations for the center of buoyancy for the trimmed position and for the heeled position were derived from the center of mass equations of a wedge.

The equations for the center of buoyancy and displacement of the cube in a trimmed position are as follows:

$$ \begin{aligned} X_{cb} &= \frac{2T}{3}\sin\phi \\ Y_{cb} &= \frac{l}{2} \\ Z_{cb} &= \frac{T}{3}\cos\phi \\ \Delta &= \rho l \frac{T^2}{\sin 2\phi} \end{aligned} \tag*{(25)} $$

For the cube in a heeled position, the equations become:

$$ \begin{aligned} X_{cb} &= \frac{l}{2} \\ Y_{cb} &= \frac{2T}{3}\sin\theta \\ Z_{cb} &= \frac{T}{3}\cos\theta \\ \Delta &= \rho l \frac{T^2}{\sin 2\theta} \end{aligned} \tag*{(26)} $$

The equations for the center of buoyancy of a trimmed and heeled cube are more complicated, since two rotations are taking place. It is simple to see that the volume of a cube for a draft $T \le l \sin(\theta) \cos(\phi)$, where the cube has been rotated about the $y$-axis by an angle $\theta$ and then rotated about the $x$-axis by an angle $\phi$, will resemble a tetrahedron. The following equations are used to determine the center of buoyancy and the displacement of the trimmed and heeled cube:

$$ \begin{aligned} X_{cb} &= l - \frac{T}{4\sin\phi} \\ Y_{cb} &= \frac{T}{4\cos\phi\cos\theta} \\ Z_{cb} &= l - \frac{T}{4\cos\phi\sin\theta} \\ \Delta &= \frac{\rho T^3} {(1000)(6)\sin\phi \cos^2\phi \sin\theta \cos\theta} \end{aligned} \tag*{(27)} $$

The coordinate systems are shown in Figures 16, 17, and 18. The global coordinate system with its $x\textrm{-}y$ plane on the waterplane is shown as the coordinate system without primes, and the body coordinate system belonging to the cube is shown as the coordinate system with primes.

Figure 16: Side View of Trimmed Cube

Figure 17: Front View of Heeled Cube

Figure 18: Isometric View of Trimmed and Heeled Cube

The previous equations were used to compute the center of buoyancy and the displacement of the cube for three different positions. The results are presented in the table below, where they are compared to the results produced by MOSES.

Comparison of Hydrostatics of a Cube

Next, we consider the hydrostatics of a tube, which will be calculated for three different combinations of draft, trim, and roll. The tube is 50 ft long with a radius of 5 ft. Here $\theta$ is a pitch angle measured from the global $z$-axis to the body $z’$-axis. For these tests, draft is defined as the vertical distance from the body coordinate system to the mean water level. The geometry is shown in Figure 19, and the body coordinate system is denoted as the $x’\textrm{-}z’$ axis system.

Figure 19: Coordinate System for Buoyancy of a Tube

The buoyancy and its center are determined by integration of the following equations:

$$ \begin{aligned} v &= \int_S \textrm{d}v \\ \Delta &= \rho g v \\ X_{cb} &= \frac{1}{v} \int_S x \, \textrm{d}v \\ Y_{cb} &= \frac{1}{v} \int_S y \, \textrm{d}v \\ Z_{cb} &= \frac{1}{v} \int_S z \, \textrm{d}v \end{aligned} \tag*{(28)} $$

where $S$ is the submerged portion of the tube.

The comparison of the results for three selected cases of draft, trim, and roll is presented in the table below.

Comparison of Hydrostatics of a Tube